🧠:Okay, let's tackle this geometry problem. It's about proving two things: first, that triangle PQR is equilateral, and second, that the centroid of triangle PQR lies on segment AB. Hmm, let's start by understanding the setup.We have a segment AB with a random point M on it (not A or B). On one side of AB, there's a point C forming an equilateral triangle ABC. On the other side of AB, points D and E are chosen so that triangles AMD and MBE are also equilateral. Then, P, Q, R are the centroids (intersection of medians) of triangles ABC, AMD, and MBE respectively. We need to show that triangle PQR is equilateral and its centroid is on AB.First, let me sketch this in my mind. AB is the base segment. C is on one side, D and E on the other. Each of ABC, AMD, MBE is equilateral. The centroids P, Q, R are each a third of the way from the vertices along the medians. So, maybe coordinates can help here? Using coordinate geometry might be a systematic way to approach this.Let me assign coordinates. Let's set AB on the x-axis for simplicity. Let A be at (0,0) and B at (b,0) where b > 0. Point M is a random point on AB, so let's say M is at (m,0), where 0 < m < b.Now, triangle ABC is equilateral on one side of AB. Let's choose the upper half-plane for C. The coordinates of C can be determined. For an equilateral triangle on AB, the height is (√3/2)*AB. Since AB is from (0,0) to (b,0), the length AB is b. So the height is (√3/2)*b. Thus, C would be at (b/2, (√3/2)*b). Wait, but if AB is from (0,0) to (b,0), then the midpoint of AB is at (b/2, 0), so moving up by (√3/2)*b gives C as (b/2, (√3/2)*b).Similarly, triangles AMD and MBE are equilateral but on the other side of AB (lower half-plane). Let me confirm: the problem states "on the other side" for D and E. So, since C is on one side, D and E are on the opposite side, which I just considered as lower half-plane.So, triangle AMD is equilateral. A is at (0,0), M is at (m,0). So, to construct an equilateral triangle AMD on the lower side. The point D would be such that AMD is equilateral. Similarly, triangle MBE: M is at (m,0), B is at (b,0), so E is the third vertex of the equilateral triangle on the lower side.Let me compute coordinates for D and E.Starting with triangle AMD. Points A(0,0), M(m,0). The length AM is m. So, the third vertex D should be such that AMD is equilateral. To find D, since it's on the lower half-plane, the coordinates can be found by rotating point M around A by -60 degrees (clockwise) because it's on the other side. Rotation matrix for -60 degrees is:[ cos(-60) -sin(-60) ][ sin(-60) cos(-60) ]Which is:[ 0.5 √3/2 ][ -√3/2 0.5 ]But since we are rotating point M(m,0) around A(0,0) by -60 degrees, the new coordinates D would be:x = m * 0.5 - 0 * (√3/2) = 0.5my = m * (-√3/2) + 0 * 0.5 = - (√3/2)mSo D is at (0.5m, - (√3/2)m)Similarly, for triangle MBE. Points M(m,0) and B(b,0). The length MB is (b - m). To find E, which is the third vertex of the equilateral triangle on the lower half-plane. So, we can rotate point B around M by -60 degrees clockwise. Wait, or maybe rotate point M around B? Hmm, need to be careful.Wait, to construct triangle MBE with base MB and E on the lower half-plane. So starting from M(m,0) to B(b,0). To get E, we can rotate vector MB by -60 degrees around M. Alternatively, since we need E such that MBE is equilateral.The rotation method: taking vector from M to B, which is (b - m, 0), rotating it by -60 degrees to get the vector from M to E.Rotation matrix for -60 degrees:[ cos(-60) -sin(-60) ] = [ 0.5 √3/2 ][ sin(-60) cos(-60) ] [ -√3/2 0.5 ]So, vector MB is (b - m, 0). Rotating this vector by -60 degrees gives:x' = (b - m)*0.5 - 0*√3/2 = 0.5(b - m)y' = (b - m)*(-√3/2) + 0*0.5 = - (√3/2)(b - m)Therefore, the coordinates of E are M plus this vector:E_x = m + 0.5(b - m) = 0.5(b + m)E_y = 0 - (√3/2)(b - m) = - (√3/2)(b - m)So E is at (0.5(b + m), - (√3/2)(b - m))Alright, so now we have coordinates for C, D, E.C: (b/2, (√3/2)b)D: (0.5m, - (√3/2)m)E: (0.5(b + m), - (√3/2)(b - m))Now, centroids P, Q, R of triangles ABC, AMD, MBE.The centroid of a triangle is the average of its three vertices.So, centroid P of ABC:A(0,0), B(b,0), C(b/2, (√3/2)b)So,P_x = (0 + b + b/2)/3 = (3b/2)/3 = b/2P_y = (0 + 0 + (√3/2)b)/3 = (√3/6)bSo P is at (b/2, (√3/6)b)Centroid Q of AMD:A(0,0), M(m,0), D(0.5m, - (√3/2)m)Q_x = (0 + m + 0.5m)/3 = (1.5m)/3 = 0.5mQ_y = (0 + 0 + (-√3/2)m)/3 = (-√3/6)mSo Q is at (0.5m, - (√3/6)m)Centroid R of MBE:M(m,0), B(b,0), E(0.5(b + m), - (√3/2)(b - m))R_x = (m + b + 0.5(b + m))/3Let me compute that:First, numerator: m + b + 0.5b + 0.5m = (m + 0.5m) + (b + 0.5b) = 1.5m + 1.5b = 1.5(m + b)So R_x = 1.5(m + b)/3 = 0.5(m + b)Similarly, R_y:0 + 0 + (-√3/2)(b - m) all over 3So R_y = (-√3/6)(b - m)Therefore, R is at (0.5(m + b), - (√3/6)(b - m))So now we have coordinates for P, Q, R.P: (b/2, (√3/6)b)Q: (0.5m, - (√3/6)m)R: (0.5(m + b), - (√3/6)(b - m))Now, need to prove that triangle PQR is equilateral.To prove triangle PQR is equilateral, we can show that all sides are equal in length. Let's compute the distances PQ, QR, RP.First, let's compute PQ.Coordinates of P: (b/2, (√3/6)b)Coordinates of Q: (0.5m, - (√3/6)m)Difference in x: 0.5m - b/2 = 0.5(m - b)Difference in y: - (√3/6)m - (√3/6)b = - (√3/6)(m + b)So distance PQ squared:[0.5(m - b)]² + [ - (√3/6)(m + b) ]²= (0.25)(m - b)^2 + ( (3/36) )(m + b)^2= (0.25)(m - b)^2 + (1/12)(m + b)^2Similarly, compute QR.Coordinates of Q: (0.5m, - (√3/6)m)Coordinates of R: (0.5(m + b), - (√3/6)(b - m))Difference in x: 0.5(m + b) - 0.5m = 0.5bDifference in y: - (√3/6)(b - m) - ( - (√3/6)m ) = - (√3/6)(b - m) + (√3/6)m = - (√3/6)b + (√3/6)m + (√3/6)m = - (√3/6)b + (√3/3)mWait, let's compute it step by step.QR's y difference:R_y - Q_y = [ - (√3/6)(b - m) ] - [ - (√3/6)m ]= - (√3/6)(b - m) + (√3/6)m= - (√3/6)b + (√3/6)m + (√3/6)m= - (√3/6)b + (√3/3)mSo QR's distance squared:(0.5b)^2 + [ - (√3/6)b + (√3/3)m ]^2= (0.25b²) + [ (√3/3)m - (√3/6)b ]²Let me factor out √3/6:= (0.25b²) + [ (√3/6)(2m - b) ]²= 0.25b² + (3/36)(2m - b)^2= 0.25b² + (1/12)(2m - b)^2Now, let's compute PR.Coordinates of P: (b/2, (√3/6)b)Coordinates of R: (0.5(m + b), - (√3/6)(b - m))Difference in x: 0.5(m + b) - b/2 = 0.5m + 0.5b - 0.5b = 0.5mDifference in y: - (√3/6)(b - m) - (√3/6)b = - (√3/6)(b - m + b) = - (√3/6)(2b - m)Wait, let me check:PR's y difference:R_y - P_y = [ - (√3/6)(b - m) ] - [ (√3/6)b ]= - (√3/6)(b - m) - (√3/6)b= - (√3/6)(b - m + b )= - (√3/6)(2b - m )Therefore, distance PR squared:(0.5m)^2 + [ - (√3/6)(2b - m) ]^2= 0.25m² + (3/36)(2b - m)^2= 0.25m² + (1/12)(2b - m)^2Now, to check if PQ, QR, PR are equal, we need to see if their squared distances are equal.First, let's compute PQ², QR², PR².PQ² = 0.25(m - b)^2 + (1/12)(m + b)^2QR² = 0.25b² + (1/12)(2m - b)^2PR² = 0.25m² + (1/12)(2b - m)^2Hmm, this seems complicated. Maybe there's a pattern or symmetry here. Let's try to expand these expressions.First, expand PQ²:0.25(m - b)^2 + (1/12)(m + b)^2= 0.25(m² - 2mb + b²) + (1/12)(m² + 2mb + b²)= (0.25m² - 0.5mb + 0.25b²) + ( (1/12)m² + (1/6)mb + (1/12)b² )Combine like terms:m²: 0.25 + 1/12 = (3/12 + 1/12) = 4/12 = 1/3mb: -0.5 + 1/6 = (-3/6 + 1/6) = -2/6 = -1/3b²: 0.25 + 1/12 = (3/12 + 1/12) = 4/12 = 1/3Thus, PQ² = (1/3)m² - (1/3)mb + (1/3)b² = (1/3)(m² - mb + b²)Similarly, let's expand QR²:0.25b² + (1/12)(2m - b)^2= 0.25b² + (1/12)(4m² -4mb + b²)= 0.25b² + (4/12 m² - 4/12 mb + 1/12 b²)= 0.25b² + (1/3 m² - 1/3 mb + 1/12 b²)Convert 0.25b² to 3/12 b²:= 3/12 b² + 4/12 m² - 4/12 mb + 1/12 b²Combine terms:m²: 4/12 = 1/3mb: -4/12 = -1/3b²: 3/12 + 1/12 = 4/12 = 1/3So QR² = (1/3)m² - (1/3)mb + (1/3)b² = same as PQ²Now, PR²:0.25m² + (1/12)(2b - m)^2= 0.25m² + (1/12)(4b² -4mb + m²)= 0.25m² + (4/12 b² -4/12 mb + 1/12 m²)Convert 0.25m² to 3/12 m²:= 3/12 m² + 4/12 b² -4/12 mb + 1/12 m²Combine terms:m²: 3/12 +1/12 =4/12=1/3mb: -4/12= -1/3b²:4/12=1/3So PR² = (1/3)m² - (1/3)mb + (1/3)b²Hence, all three squared distances are equal. Therefore, PQ = QR = PR, meaning triangle PQR is equilateral. That proves part (a).Now part (b): the centroid of triangle PQR lies on segment AB.The centroid of triangle PQR is the average of points P, Q, R.Let's compute the centroid's coordinates.Centroid (G) coordinates:G_x = (P_x + Q_x + R_x)/3G_y = (P_y + Q_y + R_y)/3Compute G_x:P_x = b/2, Q_x = 0.5m, R_x = 0.5(m + b)Sum: b/2 + 0.5m + 0.5(m + b) = b/2 + 0.5m + 0.5m + 0.5b = (b/2 + 0.5b) + (0.5m + 0.5m) = b + mDivide by 3: (b + m)/3G_x = (b + m)/3G_y:P_y = (√3/6)bQ_y = - (√3/6)mR_y = - (√3/6)(b - m)Sum: (√3/6)b - (√3/6)m - (√3/6)(b - m)Factor out √3/6:√3/6 [ b - m - (b - m) ] = √3/6 [ b - m - b + m ] = √3/6 [ 0 ] = 0Therefore, G_y = 0So the centroid G is at ( (b + m)/3 , 0 ), which is on the x-axis, hence on segment AB since AB is from (0,0) to (b,0). The x-coordinate (b + m)/3 must lie between 0 and b because m is between 0 and b.Indeed, since 0 < m < b, then 0 < (b + m)/3 < (b + b)/3 = (2b)/3 < b. Wait, but (2b)/3 is less than b. Wait, actually, (b + m)/3 is between (b + 0)/3 = b/3 and (b + b)/3 = 2b/3. Both b/3 and 2b/3 are between 0 and b, so yes, the centroid lies on AB.Therefore, part (b) is proven.So, by using coordinate geometry and calculating the centroids' coordinates, we found that triangle PQR has all sides equal and its centroid is on AB. The key steps were assigning coordinates, computing centroids, and then verifying distances and centroid location through algebra.Final Answera) boxed{triangle PQR text{ is equilateral}} b) boxed{text{The centroid of } triangle PQR text{ lies on segment } AB}

✅:To solve the problem, we start by assigning coordinates to the points. Let ( A ) be at ( (0, 0) ) and ( B ) at ( (b, 0) ). Point ( M ) is on segment ( AB ) at ( (m, 0) ). The coordinates of points ( C ), ( D ), and ( E ) are determined as follows:- Point ( C ) is the third vertex of the equilateral triangle ( ABC ) on the upper half-plane. The coordinates of ( C ) are ( left( frac{b}{2}, frac{sqrt{3}}{2}b right) ).- Point ( D ) is the third vertex of the equilateral triangle ( AMD ) on the lower half-plane. Using a rotation matrix, the coordinates of ( D ) are ( left( frac{m}{2}, -frac{sqrt{3}}{2}m right) ).- Point ( E ) is the third vertex of the equilateral triangle ( MBE ) on the lower half-plane. Using a rotation matrix, the coordinates of ( E ) are ( left( frac{b + m}{2}, -frac{sqrt{3}}{2}(b - m) right) ).Next, we find the centroids ( P ), ( Q ), and ( R ) of triangles ( ABC ), ( AMD ), and ( MBE ) respectively:- Centroid ( P ) of ( ABC ) is the average of the coordinates of ( A ), ( B ), and ( C ): [ P = left( frac{b}{2}, frac{sqrt{3}}{6}b right) ]- Centroid ( Q ) of ( AMD ) is the average of the coordinates of ( A ), ( M ), and ( D ): [ Q = left( frac{m}{2}, -frac{sqrt{3}}{6}m right) ]- Centroid ( R ) of ( MBE ) is the average of the coordinates of ( M ), ( B ), and ( E ): [ R = left( frac{b + m}{2}, -frac{sqrt{3}}{6}(b - m) right) ]To prove that triangle ( PQR ) is equilateral, we calculate the distances between ( P ), ( Q ), and ( R ). Using the distance formula, we find that the squared distances ( PQ^2 ), ( QR^2 ), and ( PR^2 ) are all equal to ( frac{1}{3}(m^2 - mb + b^2) ). Hence, all sides are equal, proving that ( triangle PQR ) is equilateral.For part (b), we find the centroid of ( triangle PQR ). The centroid's coordinates are the average of ( P ), ( Q ), and ( R ):[left( frac{b + m}{3}, 0 right)]This point lies on the x-axis (segment ( AB )) since the y-coordinate is 0 and the x-coordinate ( frac{b + m}{3} ) is between 0 and ( b ). Final Answera) boxed{triangle PQR text{ is equilateral}} b) boxed{text{The centroid of } triangle PQR text{ lies on segment } AB}

🔑:Let's analyze and solve each part step-by-step.# Part (a): Proving that the triangle PQR is equilateral1. Let point T be the intersection of lines AQ and BR.2. Since P is the intersection of the medians of triangle ABC, and similarly for Q and R with triangle AMD and triangle MBE respectively, they can be described as follows: - Median AP intersects with BC at P - Median AQ intersects with MD at Q - Median BR intersects with ME at R3. The quadrilateral APBT forms a rhombus with angles 60^circ and 120^circ at vertices A and T. Triangles ART and BRT are equilateral. - Since triangle ABC is equilateral, AP = PB = AB / sqrt{3}. - Similarly, for triangles AMD and MBE, we get: [ AQ = frac{AM}{sqrt{3}}, quad BR = frac{BM}{sqrt{3}} ] 4. Given that AM + BM = AB, we can denote: [ AP = AQ + BR = AT = BT ]5. Since QT = BR and PT = PB, triangles QTP and RBP are congruent due to: - Their angles QTP and RBP are 60^circ each. - Their corresponding sides. 6. Additionally, the angle between the corresponding sides PT and PB is also 60^circ, making triangle QTP a 60^circ rotation about point P clockwise to triangle RBP. 7. Thus, the sides QP and RP of triangle PQR are equal and the angle QPR between them is 60^circ. 8. Therefore, triangle PQR must be equilateral. Conclusion: [ boxed{PQR text{ is equilateral}} ]# Part (b): Proving that the intersection of the medians of triangle PQR lies on segment AB1. Observe that angles PQR and PTR are 60^circ each, thus point Q lies on the circumcircle of triangle PTR.2. Since triangle PQR and triangle PTR are congruent with their circumscribed circles coinciding, the same circumcenter is valid for both triangles.3. Denote the centroid or intersection of the medians of triangle PQR as point S. - Point S is also the center of the circumcircle of triangle PQR which coincides with the circumcircle of triangle PTR. 4. Thus, point S must lie on the perpendicular bisector of segment AB to the chord RT of this circumcircle.5. Therefore, point S lies on segment AB, as required. Conclusion: [ boxed{text{The centroid of } triangle PQR text{ lies on segment } AB} ]These detailed solutions conclude parts (a) and (b).

🧠:Alright, let's tackle this probability problem. Hmm, the question is asking: How many trials need to be conducted so that the probability of the event occurring at least 75 times is 0.9? The probability of success in each trial is 0.8, and the trials are independent. Okay, so we need to find the smallest number of trials n such that P(X ≥ 75) = 0.9, where X is the number of successes in n trials.First, since each trial is independent and has the same probability of success, this is a binomial distribution problem. The binomial distribution has parameters n (number of trials) and p (probability of success), which in this case is 0.8. So, X ~ Bin(n, 0.8). We need to find the smallest n such that P(X ≥ 75) ≥ 0.9.But working with the binomial distribution directly can be tricky, especially for large n. Maybe we can approximate it with a normal distribution? The conditions for normal approximation are usually np ≥ 10 and n(1-p) ≥ 10. Let's check later if these hold, but assuming n is large enough, which it probably will be since we're dealing with 75 successes, the approximation should be valid.The normal approximation for a binomial distribution has mean μ = np and variance σ² = np(1-p), so σ = sqrt(np(1-p)). We want P(X ≥ 75) ≥ 0.9. To use the normal approximation, we should apply the continuity correction. So P(X ≥ 75) is approximately equal to P(Y ≥ 74.5), where Y is a normal variable with μ = 0.8n and σ = sqrt(n*0.8*0.2) = sqrt(0.16n).We need to find n such that P(Y ≥ 74.5) = 0.9. Since we want the probability that Y is greater than or equal to 74.5 to be 0.9, this means that 74.5 should be the critical value corresponding to the 10th percentile of the distribution (because 1 - 0.9 = 0.1). Wait, no. If we're looking at P(Y ≥ 74.5) = 0.9, then 74.5 is at the 10th percentile. Because 90% of the distribution is to the right of it. Hmm, that seems counterintuitive. Let me think.Wait, no. If we want P(Y ≥ 74.5) = 0.9, that means 90% of the probability mass is above 74.5, so 74.5 is the value such that only 10% of the time Y is less than 74.5. Therefore, 74.5 corresponds to the 10th percentile of the distribution. So, using the standard normal distribution, we can write:74.5 = μ - z * σwhere z is the z-score corresponding to the 10th percentile. Looking at standard normal tables, the z-score for which P(Z ≤ z) = 0.1 is approximately -1.28. Because the 10th percentile is on the left side of the mean. So, z = -1.28.Therefore, substituting into the equation:74.5 = 0.8n - 1.28 * sqrt(0.16n)Let me write this equation out:0.8n - 1.28 * sqrt(0.16n) = 74.5This is an equation in terms of n. Let's simplify sqrt(0.16n). sqrt(0.16n) is sqrt(0.16) * sqrt(n) = 0.4 * sqrt(n). Therefore, the equation becomes:0.8n - 1.28 * 0.4 * sqrt(n) = 74.5Calculate 1.28 * 0.4: 1.28 * 0.4 = 0.512So:0.8n - 0.512 sqrt(n) - 74.5 = 0Let's let x = sqrt(n). Then, n = x². Substituting into the equation:0.8x² - 0.512x - 74.5 = 0This is a quadratic equation in terms of x. Let's write it as:0.8x² - 0.512x - 74.5 = 0Multiply all terms by 1000 to eliminate decimals:800x² - 512x - 74500 = 0Hmm, maybe that's messy. Alternatively, solve the quadratic equation as is. The quadratic equation is:0.8x² - 0.512x - 74.5 = 0Using the quadratic formula:x = [0.512 ± sqrt(0.512² + 4*0.8*74.5)] / (2*0.8)First, compute discriminant D:D = (0.512)^2 + 4 * 0.8 * 74.5Calculate 0.512^2: 0.512 * 0.512 = 0.262144Then, 4 * 0.8 = 3.2; 3.2 * 74.5 = 238.4So D = 0.262144 + 238.4 = 238.662144Then sqrt(D) ≈ sqrt(238.662144) ≈ 15.45So x ≈ [0.512 ± 15.45]/1.6We are looking for positive x, so take the positive root:x ≈ (0.512 + 15.45)/1.6 ≈ (15.962)/1.6 ≈ 9.976So x ≈ 9.976, which is sqrt(n). Therefore, n ≈ (9.976)^2 ≈ 99.52Since n must be an integer, we round up to 100. But wait, this seems low. If n is 100, then the expected number of successes is 0.8*100=80. So 75 is below the mean. But we want the probability that X is at least 75 to be 0.9. Since 75 is below the mean (80), the probability of being above 75 should be more than 0.5. Wait, but the answer here is n≈100. But wait, maybe our approximation is off?Wait, but when we used the normal approximation, we set up the equation as 74.5 = 0.8n - 1.28*sqrt(0.16n). Solving that gave us n≈99.52, so n=100. But let's check if this is correct. Let's compute for n=100:Mean μ = 80, σ = sqrt(100*0.8*0.2) = sqrt(16) = 4So, using continuity correction, P(X ≥75) ≈ P(Y ≥74.5) = P(Z ≥ (74.5 - 80)/4) = P(Z ≥ -1.375) ≈ 1 - Φ(-1.375) = Φ(1.375)Looking up Φ(1.375) in standard normal tables: 1.375 is 1.37 + 0.005. The value for 1.37 is 0.9147, and 1.38 is 0.9162. So, interpolating, 1.375 is halfway between 1.37 and 1.38, so approximately (0.9147 + 0.9162)/2 ≈ 0.91545. Therefore, the probability is approximately 0.9154, which is about 0.915, which is higher than 0.9. So n=100 gives a probability of ~0.915, which is more than 0.9. So maybe n=99? Let's check n=99.For n=99, μ=0.8*99=79.2, σ=sqrt(99*0.8*0.2)=sqrt(15.84)=≈3.98Continuity correction: P(X ≥75) ≈ P(Y ≥74.5) = P(Z ≥ (74.5 -79.2)/3.98) = P(Z ≥ (-4.7)/3.98) ≈ P(Z ≥ -1.181)Φ(-1.181) is the probability below -1.181, which is 1 - Φ(1.181). Looking up Φ(1.18) is 0.8810, so Φ(-1.18) = 0.1190. Therefore, P(Z ≥ -1.18) = 1 - 0.1190 = 0.8810, which is approximately 0.881, less than 0.9. Therefore, n=99 gives probability ~0.881, which is less than 0.9, and n=100 gives ~0.915, which is more than 0.9. So according to the normal approximation, n=100 is the answer.But wait, the problem says "to expect the event to occur at least 75 times with a probability of 0.9". So n=100.But let me check if the normal approximation is valid here. The rule of thumb is np ≥10 and n(1-p)≥10. For n=100, np=80 and n(1-p)=20, both greater than 10. So the approximation is valid. Therefore, the answer is 100.But wait, another approach is to use the inverse of the binomial distribution. However, since n is large, exact computation might be difficult without a calculator. Alternatively, maybe using the Poisson approximation? But p=0.8 is not small, so Poisson isn't suitable. So normal approximation is the way to go.Alternatively, maybe solving using the Central Limit Theorem. Let's see. The Central Limit Theorem tells us that the binomial distribution approximates a normal distribution for large n. So our approach is correct.But let me double-check. If we use n=100, the z-score is (74.5 -80)/4 = -1.375, and the probability is 0.915, which is more than 0.9. So n=100 is sufficient. If we try n=95.Wait, let's check n=95. μ=76, σ=sqrt(95*0.8*0.2)=sqrt(15.2)=≈3.898P(X ≥75)≈P(Y ≥74.5)=P(Z ≥(74.5-76)/3.898)=P(Z≥-0.385)=1 - Φ(-0.385)=Φ(0.385)≈0.65, which is much less than 0.9. So n=95 is too low. Similarly, n=100 gives 0.915, which is sufficient.But maybe there's a more precise method. The exact answer would require summing the binomial probabilities from k=75 to n for various n until the sum is ≥0.9. But that's computationally intensive without a calculator. Alternatively, maybe using the Wilson-Hilferty approximation or another method? But probably the normal approximation is acceptable here.Alternatively, maybe the question is asking for the expected number of trials such that the expected number of successes is 75. But no, the wording is "expect the event to occur at least 75 times with a probability of 0.9". So it's about probability, not expectation. The expectation would be np, so if they wanted the expected number of successes to be 75, then n=75/0.8=93.75, which rounds to 94. But that's different from the question's requirement.Therefore, going back, using the normal approximation, n=100 is the answer. Let me check once more.Alternatively, maybe set up the equation correctly. The requirement is P(X ≥75) ≥0.9. Using the normal approximation with continuity correction:P(X ≥75) ≈ P(Y ≥74.5) = 0.9Which translates to:P(Y ≤74.5) =0.1Therefore, the z-score corresponding to 74.5 is:z = (74.5 -0.8n)/sqrt(0.16n) = -1.2816 (since the 10th percentile is z≈-1.28)Therefore:(74.5 -0.8n)/sqrt(0.16n) = -1.2816Multiply both sides by sqrt(0.16n):74.5 -0.8n = -1.2816 * sqrt(0.16n)Multiply both sides by -1:0.8n -74.5 =1.2816 * sqrt(0.16n)Let me write this as:0.8n -1.2816*sqrt(0.16n) -74.5 =0Which is the same equation as before. Let me write sqrt(0.16n) as 0.4sqrt(n):0.8n -1.2816*0.4*sqrt(n) -74.5=0Calculate 1.2816*0.4=0.51264So:0.8n -0.51264sqrt(n) -74.5=0Let x = sqrt(n), so n =x²:0.8x² -0.51264x -74.5=0Multiply through by, say, 1000 to eliminate decimals:800x² -512.64x -74500=0Use quadratic formula:x = [512.64 ± sqrt(512.64² +4*800*74500)]/(2*800)Calculate discriminant D:D= (512.64)^2 +4*800*74500First, 512.64²: approx (500)^2 + 2*500*12.64 +12.64² =250000 +12640 +159.7696≈250000+12640=262640+159.77≈262799.77Then, 4*800*74500=3200*74500=3200*74,500. Let's compute 3200*70,000=224,000,000; 3200*4,500=14,400,000. Total=224,000,000 +14,400,000=238,400,000Therefore, D≈262,799.77 +238,400,000≈238,662,799.77sqrt(D)≈sqrt(238,662,800)≈15,443.5 (since 15,443^2=238,486, 249; 15,444^2=238,517, 136) but actually, sqrt(238,662,799.77)≈15,443.5Therefore, x≈ [512.64 +15,443.5]/1600 ≈ (15,956.14)/1600≈9.97258Thus, x≈9.9726, so n≈x²≈99.45, so n≈99.45. Since n must be integer, round up to 100. So, same result. Therefore, the answer is 100 trials.But let me verify with n=100 as before. The exact probability using binomial is difficult, but we can check if the approximation was accurate enough. Given that the normal approximation suggested ~91.5%, which is above 0.9, but what's the exact probability?But calculating the exact binomial probability for n=100, p=0.8, k=75 to 100 would require summing C(100, k)*(0.8)^k*(0.2)^{100−k} from k=75 to 100. This is tedious by hand, but maybe we can use a calculator or a statistical table. Alternatively, use another approximation.Alternatively, the exact probability can be approximated with software, but since we don't have that here, we rely on the normal approximation which is standard for such problems.Therefore, the answer is 100 trials.But let me check for n=95 again, just to be thorough. For n=95, μ=76, σ≈sqrt(95*0.8*0.2)=sqrt(15.2)=3.898z=(74.5 -76)/3.898≈-1.5/3.898≈-0.385. P(Z≥-0.385)=0.65, as before. Not enough.n=99: μ=79.2, σ≈3.98z=(74.5-79.2)/3.98≈-4.7/3.98≈-1.18. P(Z≥-1.18)=0.881, still less than 0.9.n=100: μ=80, σ=4z=(74.5-80)/4≈-1.375. P(Z≥-1.375)=0.915, which is above 0.9.Therefore, the minimal n is 100.Another way to look at it: Since 75 is 5 less than the mean of 80 when n=100. The standard deviation is 4, so 75 is 1.25σ below the mean. The probability of being above -1.25σ is Φ(1.25)=0.8944, but with continuity correction, it's 74.5 which is 74.5-80=-5.5, which is -5.5/4≈-1.375, giving Φ(1.375)≈0.915. So this aligns.Therefore, the answer is 100 trials.

✅:To determine the number of trials ( n ) required such that the probability of an event (with success probability ( p = 0.8 )) occurring at least 75 times is 0.9, we use the normal approximation to the binomial distribution. Here's the step-by-step solution:1. Model the Problem: - Let ( X sim text{Binomial}(n, 0.8) ). - We need ( P(X geq 75) geq 0.9 ).2. Normal Approximation: - For large ( n ), approximate ( X ) by ( Y sim mathcal{N}(mu, sigma^2) ), where: [ mu = 0.8n, quad sigma = sqrt{0.16n}. ] - Apply continuity correction: ( P(X geq 75) approx P(Y geq 74.5) ).3. Z-Score Equation: - The 10th percentile of the normal distribution corresponds to ( z = -1.28 ). - Set up the equation for the critical value: [ 74.5 = mu - zsigma implies 74.5 = 0.8n - 1.28sqrt{0.16n}. ]4. Solve for ( n ): - Substitute ( sqrt{0.16n} = 0.4sqrt{n} ): [ 0.8n - 0.512sqrt{n} - 74.5 = 0. ] - Let ( x = sqrt{n} ), leading to a quadratic equation: [ 0.8x^2 - 0.512x - 74.5 = 0. ] - Solve using the quadratic formula: [ x approx 9.976 implies n approx x^2 approx 99.52. ] - Round up to the nearest integer: ( n = 100 ).5. Verification: - For ( n = 100 ), ( mu = 80 ), ( sigma = 4 ). - Compute ( P(Y geq 74.5) ): [ Z = frac{74.5 - 80}{4} = -1.375 implies P(Z geq -1.375) approx 0.915 geq 0.9. ]Final Answer: (boxed{100}) trials are needed.

🔑:1. Given Data: The problem provides the following information: - Probability of occurrence of the event in each independent trial: ( p = 0.8 ). - Probability of not occurring in each trial: ( q = 1 - p = 0.2 ). - Required number of successful outcomes: ( k_1 = 75 ). - Number of trials: ( n ). - Desired probability: ( P_n = 0.9 ).2. Application of the Central Limit Theorem (Laplace's Theorem): To determine the number of trials ( n ), we will use the integral theorem of Laplace. The theorem states: [ P_n(k_1; n) = Phileft( x'' right) - Phileft( x' right) ] where: [ x' = frac{k_1 - np}{sqrt{npq}}, quad x'' = frac{k_2 - np}{sqrt{npq}} ]3. Substitute Given Data: Since ( k_2 = n ) and ( k_1 = 75 ), we get: [ 0.9 = Phileft( frac{n - 0.8n}{sqrt{n cdot 0.8 cdot 0.2}} right) - Phileft( frac{75 - 0.8n}{sqrt{n cdot 0.8 cdot 0.2}} right) ]4. Simplify the Expressions: Simplify the inner expressions involving ( n ): [ 0.9 = Phileft( frac{0.2n}{sqrt{0.16n}} right) - Phileft( frac{75 - 0.8n}{0.4sqrt{n}} right) ]5. Further Simplification: This can be simplified to: [ 0.9 = Phileft( frac{sqrt{n}}{2} right) - Phileft( frac{75 - 0.8n}{0.4sqrt{n}} right) ]6. Assumption Based on Distribution Behavior: Notice that since ( n > 75 ), we can assume: [ frac{sqrt{n}}{2} > sqrt{frac{75}{2}} approx 4.33 ] Given that the Laplace function (Phi(x)) is increasing and (Phi(4) approx 0.5), we can approximate: [ Phileft( frac{sqrt{n}}{2} right) approx 0.5 ]7. Expected Probability Calculation: Thus, we rewrite our initial equality: [ 0.9 = 0.5 - Phileft( frac{75 - 0.8n}{0.4sqrt{n}} right) ]8. Solve for ( Phi left( frac{75 - 0.8n}{0.4 sqrt{n}} right) ): [ Phileft( frac{75 - 0.8n}{0.4sqrt{n}} right) = -0.4 ] From the standard normal distribution table, (Phi(1.28) approx 0.4), so: [ frac{75 - 0.8n}{0.4sqrt{n}} = -1.28 ]9. Solve for ( n ): Solve the equation: [ 75 - 0.8n = -1.28 cdot 0.4 sqrt{n} ] [ 75 - 0.8n = -0.512 sqrt{n} ] Convert it into a quadratic equation in terms of ( sqrt{n} ): [ 0.8n - 75 = 0.512 sqrt{n} ] Let ( x = sqrt{n} ): [ 0.8x^2 - 75 = 0.512x ] [ 0.8x^2 - 0.512x - 75 = 0 ] Solving the quadratic equation for ( x ): [ Delta = (0.512)^2 + 4 cdot 0.8 cdot 75 ] [ Delta = 0.262144 + 240 ] [ Delta = 240.262144 ] [ x = frac{0.512 pm sqrt{240.262144}}{2 cdot 0.8} ] [ x = frac{0.512 pm 15.50}{1.6} ] We take the positive root: [ x approx frac{16.012}{1.6} approx 10 ] Hence ( sqrt{n} approx 10 ) leading to: [ n = 10^2 = 100 ] Conclusion: [ boxed{100} ]

🧠:Okay, let's tackle this problem step by step. So, the question is about the sequence formed by the sum of the digits of multiples of nine. We need to figure out a few things: first, where does 81 appear for the first time, and what number follows it? Then, we need to determine whether 27 appears four times in a row before 36 appears three times in a row. Lastly, we need to comment on the alternation of numbers in the sequence.Starting with the first part: when does 81 first appear in the digit sums of multiples of nine? Let me recall that the digit sum of a multiple of 9 is always 9, right? Because 9 is a divisor of the number, so the sum of digits should be a multiple of 9. Wait, but in the example given, after 99, the sum is 18. So, the digit sum can be 9, 18, 27, etc. So, 81 is a much higher digit sum. Wait, 81 is actually the maximum possible digit sum for a number with, say, 9 digits all being 9. So, to get a digit sum of 81, the number must be a multiple of 9 and consist of 81 / 9 = 9 nines. So, the number would be 999,999,999. Let me check: 999,999,999 divided by 9 is 111,111,111. So, that multiple of 9 is at position 111,111,111. But wait, the sequence of multiples of 9 is 9, 18, 27, 36, ..., so the nth term is 9n. Therefore, 999,999,999 is 9n, so n = 111,111,111. Therefore, the digit sum here is 9*9 = 81, so the 111,111,111th term in the sequence of digit sums is 81. Then, the next multiple of 9 after 999,999,999 is 1,000,000,008, whose digit sum is 1 + 0 + ... + 0 + 8 = 9. So, the number following 81 in the digit sum sequence is 9. Therefore, the first occurrence of 81 is at position 111,111,111, followed by 9.But wait, let me verify if there is a smaller number with a digit sum of 81. Because 81 is 9*9, the number must have at least 9 digits, all 9s. Any smaller number would have fewer digits, so the maximum digit sum would be less. For example, a number with 8 digits, all 9s, would have a digit sum of 72. So, indeed, the first time the digit sum 81 occurs is at 999,999,999, which is 9*111,111,111. So, position 111,111,111. Then, the next number is 1,000,000,008, digit sum 9. So that's the first part.Now, the second question: which appears first in the sequence, the number 27 four times in a row or the number 36 three times in a row? Let's think. So, we need to look for runs of 27s and 36s in the digit sum sequence. To do this, we have to find four consecutive multiples of 9 where each has a digit sum of 27, and similarly three consecutive multiples of 9 with digit sum 36. Which occurs first?First, let's recall that the digit sum of a multiple of 9 is 9, 18, 27, 36, etc. So, 27 is a digit sum for numbers where, for example, the number has three 9s (sum 27) or combinations that add up to 27. Similarly, 36 would be four 9s or combinations.To get four consecutive 27s, we need four consecutive multiples of 9, each with digit sum 27. Similarly, three consecutive 36s.So, perhaps we can look for numbers around the magnitude where digit sums start being 27 and 36. Let's consider when digit sums start exceeding 18. For example, 999 is a multiple of 9 (999 = 9*111), digit sum 27. Then, 1008 is 9*112, digit sum 1+0+0+8=9. Wait, so perhaps the numbers with digit sums 27 are numbers like 999, 1998, 2997, etc., but these may not be consecutive multiples of 9.Wait, let's think. Let's take consecutive multiples of 9 and check their digit sums.Starting from 9:9: 918: 927: 936: 945: 954: 963: 972: 981: 990: 999: 18108: 9117: 9126: 9135: 9144: 9153: 9162: 9171: 9180: 9189: 18198: 18207: 9216: 9225: 9234: 9243: 9252: 9261: 9270: 9279: 18288: 18297: 18306: 9... So, up to here, digit sums are mostly 9, with some 18s. Let's go higher. Maybe when numbers have three digits or more, the digit sums can be 27. For example, 999: digit sum 27. Then, 999 + 9 = 1008: digit sum 9. So, 999 is 9*111, sum 27. Next multiple is 1008, sum 9. Then, 1017: 1+0+1+7=9. Wait, so 999 is followed by numbers that drop back to digit sum 9. Therefore, getting four consecutive 27s would require four consecutive multiples of 9 each with digit sum 27. Similarly, 36 would be even higher digit sums.Alternatively, maybe there's a range of numbers where digit sums stay high. For example, numbers like 999999, which is 9*111111, digit sum 54. But adding 9 each time, the digit sum can decrease or increase. For instance, 999999 +9 = 1000008, digit sum 1+8=9. So again, it drops. Hmm, so maybe consecutive high digit sums are rare.Alternatively, perhaps numbers like 8999999991 or something, where adding 9 cycles through carries. Wait, maybe a number like 999999999 (digit sum 81) followed by 1000000008 (digit sum 9). So, similar pattern.But how do we get a run of 27s? Let's think. Suppose we have a number like 999, which is 9*111, digit sum 27. Then, 999 +9=1008, digit sum 9. So, that's not consecutive. What if we have a number that when you add 9, the digit sum remains 27? For example, suppose the number is 9999, which is 9*1111, digit sum 36. Adding 9 gives 10008, digit sum 9. Not helpful.Wait, maybe another example: 9891. Let's check: 9891 divided by 9 is 1099, so it's a multiple of 9. Digit sum: 9+8+9+1=27. Then, 9891 +9=9900, which is 9*1100. Digit sum: 9+9+0+0=18. So, that's lower. Hmm.Alternatively, 9981: 9+9+8+1=27. Next multiple is 9990: 9+9+9+0=27. Then, next is 9999: 9+9+9+9=36. Then, 10008: 1+0+0+8=9. So, here we have two consecutive 27s (9981, 9990), then 36, then 9. So, two in a row. To get four in a row, we need four multiples of 9 in a row where each has digit sum 27.Similarly, for three 36s in a row.So, perhaps we need to find four numbers in the sequence of multiples of 9 where each has digit sum 27. Let's think about how the digit sum changes when we add 9. Adding 9 can either increase the digit sum by 9 (if there's no carryover), but usually, adding 9 causes a carryover, which reduces the digit sum by 9*k, where k is the number of trailing 9s. For example, 999 +9=1008: digit sum goes from 27 to 9, a decrease of 18. Similarly, 9999 +9=10008: digit sum from 36 to 9, decrease of 27.But sometimes, adding 9 might not cause as much decrease. For example, 189 +9=198: digit sum remains 18 (1+8+9=18, 1+9+8=18). Similarly, 198 +9=207: digit sum 9. So, here adding 9 to 189 gives 198, same digit sum.Similarly, 279 +9=288: digit sum 18 for both. 288 +9=297: digit sum 18. 297 +9=306: digit sum 9. So here, three consecutive numbers with digit sum 18. So, that's a run.Similarly, maybe for digit sum 27. Let's see. Suppose we have a number like 8991: 8+9+9+1=27. Add 9: 9000, digit sum 9. Not helpful. What about 9891: 9+8+9+1=27. Add 9: 9900, digit sum 18. Not helpful.Wait, maybe a different example. Let's take 6993: 6+9+9+3=27. Add 9: 7002, digit sum 9. Hmm. What about 7992: 7+9+9+2=27. Add 9: 8001, digit sum 9. Still not helpful.Wait, what if we have a number like 1989: 1+9+8+9=27. Add 9: 1998, digit sum 1+9+9+8=27. Then, add 9: 2007, digit sum 9. So here, we have two consecutive 27s: 1989 and 1998. Then, 2007 is 9. So, that's two in a row. If we can find a way to have more.Similarly, 2997: 2+9+9+7=27. Add 9: 3006, digit sum 9. Hmm.Wait, perhaps looking at numbers with more digits. Let's consider 9999: digit sum 36. Not helpful. Wait, maybe 8999999991 (9 digits). Let's compute: 8+9*8 +1=8+72+1=81. Wait, too high.Wait, let's think of 5-digit numbers. 9999 is four digits, digit sum 36. Let's see a 5-digit number: 99990: digit sum 9+9+9+9+0=36. Not helpful. Maybe 98919: 9+8+9+1+9=36. Not 27.Alternatively, maybe 3-digit numbers: 999:27. But adding 9 gives 1008:9. So, no.Wait, perhaps 4-digit numbers. Let's take 1989: as above, 27. Then 1998:27. Then 2007:9. So, two in a row.What if we look at 5994:5+9+9+4=27. Add 9:6003, digit sum 6+0+0+3=9. Nope.Wait, 6993:6+9+9+3=27. Add 9:7002:7+0+0+2=9.Alternatively, 1899:1+8+9+9=27. Add 9:1908:1+9+0+8=18. Hmm.Wait, maybe we need to find numbers where adding 9 doesn't cause the digit sum to drop. For example, numbers that have trailing 9s. For instance, 999: adding 9 gives 1008. But if a number ends with a 9, adding 9 would increment the last digit to 10, causing a carryover. So, if we have a number like 8999999991 (as before), adding 9 would turn the last 1 to 10, making it 89999999910, but actually, the number is 8999999991 +9=89999999910? Wait, no: 8999999991 is a 10-digit number. Adding 9 would make it 8999999991 +9=89999999910? Wait, no. Wait, 8999999991 is 8,999,999,991. Adding 9 gives 8,999,999,991 +9=9,000,000,000. Which has digit sum 9. So, digit sum goes from 81 to 9.Alternatively, maybe numbers that have multiple trailing 9s. For example, 1999999998: digit sum 1+9*8 +8=1+72+8=81. Adding 9:1999999998 +9=1999999998+9=19999999917? Wait, no. Wait, 1999999998 +9=1999999998+10-1=1999999998+10=19999999908 -1=19999999907? Wait, no. Wait, 1999999998 +9=1999999998 +10 -1=2000000008 -1=2000000007. Digit sum 2+7=9. So again, drops to 9.Hmm. So perhaps the only way to have consecutive digit sums of 27 is when adding 9 doesn't cause a carryover that reduces the digit sum. For example, if a number ends with a digit less than 9, then adding 9 just increments that digit, so digit sum increases by 9. Wait, but if you can increment a digit without carryover, then digit sum increases by 9. But multiples of 9 have digit sums that are multiples of 9. So, if you have a number with digit sum 27, and you add 9 without causing a carryover, then the next number's digit sum would be 27 +9=36. But that's not 27. So, in that case, you can't have two 27s in a row.Alternatively, if adding 9 causes some digits to carry over, but the total digit sum remains the same. For example, in the case of 1989 and 1998. Let's check:1989: 1+9+8+9=27Add 9:1998:1+9+9+8=27So, digit sum remains 27. How does that happen?When you add 9 to 1989:1989+   9=1998So, the last two digits are 89 +9=98, so no carryover beyond the tens place. So, the thousands digit remains 1, the hundreds digit 9 becomes 9, the tens digit 8 becomes 9, and the units digit 9 becomes 8. Wait, no: 1989 is 1 9 8 9. Adding 9 to the units digit: 9 +9=18, so write 8, carryover 1. Then, tens digit: 8 +1=9. No carryover. Hundreds digit: 9. Thousands digit:1. So, 1998. So, digit sum:1+9+9+8=27. So, in this case, adding 9 causes a carryover in the units digit, but the tens digit doesn't carry over. Therefore, the digit sum decreases by 9 (from 9 to 8 in the units digit) but increases by 9 in the tens digit (from 8 to 9). So, net change 0. Therefore, the digit sum remains the same.Therefore, it's possible to have consecutive numbers with the same digit sum if adding 9 causes a carryover that redistributes the digits but keeps the sum the same. So, in this example, we have two consecutive 27s. Similarly, maybe there are longer sequences.Let's try to find three consecutive 27s. Let's take 1998 and add 9:1998 +9=2007. Digit sum:2+0+0+7=9. So, that breaks the sequence. So, only two in a row.What if we have a different starting number? Let's try 2997:2+9+9+7=27. Add 9:3006:3+0+0+6=9. No good.How about 1899:1+8+9+9=27. Add 9:1908:1+9+0+8=18. No.Another example: 2799:2+7+9+9=27. Add 9:2808:2+8+0+8=18. Hmm.Wait, let's look for numbers where adding 9 causes a similar redistribution. Let's take 3699:3+6+9+9=27. Add 9:3708:3+7+0+8=18. No.Alternatively, 4599:4+5+9+9=27. Add 9:4608:4+6+0+8=18. No.Hmm. Let's try with more digits. Let's consider a 5-digit number: 19899:1+9+8+9+9=36. Not 27. 18999:1+8+9+9+9=36. 17991:1+7+9+9+1=27. Add 9:17991+9=18000:1+8+0+0+0=9. No.Wait, let's think of 6-digit numbers. 199899:1+9+9+8+9+9=45. No. Maybe 189999:1+8+9+9+9+9=45. Not helpful. 198891:1+9+8+8+9+1=36. Hmm.Alternatively, perhaps numbers where multiple digits are involved in the carryover. For example, 9999: digit sum 36. Add 9:10008:9. Not helpful. 8991:8+9+9+1=27. Add 9:9000:9. Not helpful.Wait, maybe 3-digit numbers with two 9s. 999:27. Add 9:1008:9. No.Alternatively, 4-digit numbers: 1989 and 1998 as before. Two in a row. Let's check another pair. Let's say 2898:2+8+9+8=27. Add 9:2907:2+9+0+7=18. No.Wait, 3591:3+5+9+1=18. No.Wait, 3699:3+6+9+9=27. Add 9:3708:18.Hmm. Maybe another example: 6993:6+9+9+3=27. Add 9:7002:9.No. It's challenging to find even three 27s in a row. So far, we have two. Let's see if there's a way to get three.Wait, let's take 1989:27. Then 1998:27. Then 2007:9. No. What if we have another number after 1998? But adding 9 gives 2007:9. So, only two.Alternatively, maybe starting with a different number. Let's say 59892:5+9+8+9+2=33. Not 27.Wait, this approach might not be effective. Maybe there's a mathematical way to determine when consecutive multiples of 9 have the same digit sum. Since adding 9 can either increase the digit sum by 9 (if no carryover) or decrease it by 9k -9, where k is the number of trailing 9s. Wait, when you add 9 to a number, each trailing 9 becomes 0 with a carryover. So, the digit sum decreases by 9 for each 9 that is turned to 0, and increases by 1 in the next digit. For example, adding 9 to ...x999... becomes ...(x+1)000... So, digit sum decreases by 9*(number of trailing 9s) and increases by 1 if x is not 9. So, the total change is -9k +1, where k is the number of trailing 9s. Therefore, for the digit sum to remain the same, -9k +1 +9m =0, where m is the change due to other digits. Wait, maybe not. Alternatively, the total change in digit sum is (sum after) - (sum before) = [sum of new digits] - [sum of old digits]. When you add 9, you increment the last digit by 9. If that causes a carryover, you subtract 10 from that digit and add 1 to the next digit. If the next digit is 9, it overflows, etc. So, each carryover reduces a digit by 10 and increments the next digit by 1. Therefore, the total change in digit sum for each carryover is -10 +1 = -9. So, if you have k trailing 9s, adding 9 will turn them into 0s and increment the (k+1)th digit by 1. Therefore, the total digit sum change is -9k +1. Therefore, for the digit sum to remain the same, we need:Sum_after = Sum_before + (-9k +1) = Sum_beforeTherefore, -9k +1 =0 => 9k=1. But 9k=1 is impossible for integer k. Therefore, it's impossible for the digit sum to remain exactly the same when adding 9 unless there's some other effect. Wait, but in the earlier example of 1989 to 1998, the digit sum remained 27. Let's see:1989:1+9+8+9=271998:1+9+9+8=27So, the digit sum didn't change. According to the previous formula, when adding 9 to 1989, how many trailing 9s are there? 1989 ends with 89, so no trailing 9s. Wait, then according to the formula, the change should be -9*0 +1=+1. But the digit sum remained the same. Contradiction. Therefore, our previous reasoning must be flawed.Wait, let's analyze 1989 +9=1998.Original digits:1 9 8 9Adding 9 to the last digit:9 +9=18, so write 8, carryover 1.Now, the next digit is 8 +1=9, no carryover.So, digits become:1 9 9 8.Digit sum:1+9+9+8=27. So, the digit sum changed as follows:Original digits:1+9+8+9=27New digits:1+9+9+8=27So, the units digit decreased by 1 (from 9 to 8), the tens digit increased by 1 (from 8 to 9), and the other digits stayed the same. So, net change: (-1) + (+1) =0. Therefore, total digit sum remains the same.Therefore, in this case, adding 9 caused a single carryover from the units to the tens digit, which resulted in a decrease of 1 in the units digit and an increase of 1 in the tens digit, keeping the sum the same. Therefore, this is a special case where there is a carryover, but the digit sum remains the same. Therefore, our earlier formula needs adjustment.So, the formula for digit sum change when adding 9 depends on the number of trailing digits that are 9. If there are k trailing 9s, then adding 9 turns those k 9s into 0s and increments the (k+1)th digit by 1. Therefore, the digit sum change is -9k +1. However, in the case of partial trailing digits that are not all 9s, the change can be different.For example, adding 9 to a number ending with 89:1989: ends with 89. Adding 9 turns 89 into 98. So, units digit 9 ->8 (decrease by1), tens digit 8->9 (increase by1). Therefore, total digit sum change: -1 +1=0.Similarly, adding 9 to a number ending with 899: say 19899. Adding 9 would turn 899 into 900. So, units digit 9->0 (-9), tens digit 9->0 (-9), hundreds digit 8->9 (+1). Total change: -9 -9 +1= -17. Then, digit sum changes by -17.Wait, but in reality, adding 9 to 19899:1 9 8 9 9+         9=1 9 8 9 9+9=1 9 8 9 18. So, carryover:1 9 8 (9+1)=10, so 0 with carryover.So, 1 9 (8+1)=9 0 8. Wait, no, let's do it step by step.19899 +9:Start from the right:Units digit:9 +9=18. Write 8, carryover 1.Tens digit:9 +1=10. Write 0, carryover 1.Hundreds digit:8 +1=9. No carryover.So, the number becomes 19908.Digit sum:1+9+9+0+8=27. Original was 1+9+8+9+9=36. So, digit sum decreased by 9. Wait, but according to our previous partial analysis, the change should be -9*2 (for two trailing 9s) +1= -18 +1= -17. But actual change is -9. So, there's inconsistency.Wait, the digit sum of 19899 is 36, and after adding 9 it's 27. So, change of -9. Which matches 36-27=9. But according to trailing 9s, there are two trailing 9s. So, formula would predict -9*2 +1= -17, but actual change is -9. Therefore, the formula is not applicable here because not all trailing digits after the first non-9 digit are 9s. So, the formula only applies when all k trailing digits are 9s. In this case, the number ends with 899, not 999, so only one trailing 9. Wait, but 19899 ends with 99, right? Wait, 19899 is 1 9 8 9 9. Yes, the last two digits are 9s. So, trailing 9s=2. Then adding 9 turns them into 0s and increments the next digit. So, the hundreds digit is 8, becomes 9. Therefore, digit sum change: -9*2 +1= -17. But actual digit sum decreased by 9. Contradiction. Therefore, the formula is incorrect.Wait, what's happening here. The original number is 19899, digit sum 1+9+8+9+9=36. After adding 9, it becomes 19908, digit sum 1+9+9+0+8=27. So, the change is -9. Let's see where the digits changed:- The last two digits (tens and units) were 99, became 08. So, sum decreased by 9+9=18, increased by 0+8=8. Net change: -10.- The hundreds digit was 8, became 9. Increased by 1.- The other digits remained the same.Total change: -10 +1= -9. Which matches. So, the formula needs to consider not only the number of trailing 9s but also the digits affected by the carryover. Therefore, the correct way to compute the digit sum change when adding 9 is:For each trailing 9 turned into 0, subtract 9, and add 1 for the carryover. However, if the carryover propagates beyond the trailing 9s, it might affect other digits.In the case of 19899:- Trailing two 9s become 0s: subtract 18.- The next digit (hundreds place) is 8, becomes 9: add 1.- Total change: -18 +1= -17. But actual digit sum change is -9. Therefore, discrepancy. Hence, the formula is not straightforward.Therefore, it's better to analyze specific cases rather than rely on a general formula.Given that, let's return to the original problem: whether 27 appears four times in a row before 36 appears three times in a row.Given the difficulty in finding even three 27s in a row, maybe 36 three times in a row is more likely to appear first? Or maybe neither appears, but according to the problem, we need to determine which occurs first.Alternatively, perhaps we need to look for numbers where adding 9 causes the digit sum to stay the same. For example, in the case of 1989 and 1998, two 27s in a row. Similarly, maybe there's a way to have three or four such numbers.Alternatively, consider larger numbers. For example, a number like 1999999998: digit sum 1+9*8+8=1+72+8=81. Adding 9 gives 1999999998+9=1999999998+10-1=2000000007. Digit sum 2+7=9. Not helpful.Alternatively, let's think of numbers where multiple digits can be incremented without reducing the digit sum. For example, a number like 111111111111111111111111 (many 1s). But adding 9 would turn some 1s into 0s and carryover.Wait, perhaps a different approach. Let's think about the fact that the digit sum of 9n is 9, 18, 27, etc. So, the digit sum is 9 times the number of digits when the number is all 9s, but for other numbers, it's less.To get a digit sum of 27, the number must have a digit sum of 27, which could be three 9s (999), or combinations like 9,9,9 or 9,8,10 (but digits can't exceed 9). So, the minimum number of digits needed for digit sum 27 is 3 (9+9+9). Similarly, for 36, it's four 9s (9999).Now, to get four consecutive 27s, we need four consecutive multiples of 9 with digit sum 27. Similarly, three consecutive 36s.Looking at the examples we had earlier, 1989 and 1998 are two consecutive multiples of 9 with digit sum 27. Then, 2007 has digit sum 9. So, only two in a row. Are there other instances where two or three consecutive 27s occur?Let's try to find another pair. Let's take 2898:2+8+9+8=27. Add 9:2907:2+9+0+7=18. No.How about 3699:3+6+9+9=27. Add 9:3708:3+7+0+8=18.Another example: 4599:4+5+9+9=27. Add 9:4608:4+6+0+8=18.Hmm. What about 5499:5+4+9+9=27. Add 9:5508:5+5+0+8=18.Still no luck. Let's check 6399:6+3+9+9=27. Add 9:6408:6+4+0+8=18.7299:7+2+9+9=27. Add 9:7308:7+3+0+8=18.8199:8+1+9+9=27. Add 9:8208:8+2+0+8=18.9099:9+0+9+9=27. Add 9:9108:9+1+0+8=18.So, all these examples only give two consecutive 27s if we start with a number ending with 99, but adding 9 turns the last two digits to 08 and increments the third digit, leading to a digit sum of 18. Except for the case of 1989 and 1998.Wait, let's check another pair. Let's take 2997:2+9+9+7=27. Add 9:3006:3+0+0+6=9. No.Wait, maybe a 5-digit number. Let's try 18999:1+8+9+9+9=36. No. 19899:1+9+8+9+9=36. 17991:1+7+9+9+1=27. Add 9:18000:1+8+0+0+0=9. No.Another 5-digit number: 19989:1+9+9+8+9=36. No. 18981:1+8+9+8+1=27. Add 9:18990:1+8+9+9+0=27. Then add 9:19008:1+9+0+0+8=18. So, here we have two consecutive 27s:18981 and 18990. Then 19008:18. So, another example of two in a row.But still, not three. To get four in a row, we need four consecutive numbers where adding 9 each time doesn't decrease the digit sum below 27. This seems difficult.Alternatively, maybe considering numbers with more digits where adding 9 causes a carryover that maintains the digit sum. For example, a number like 1999998:1+9+9+9+9+9+8=54. Add 9:1999998+9=2000007:2+7=9. No.Wait, let's consider a number like 9999: digit sum 36. Add 9:10008:9. Not helpful.Alternatively, 999999999999 (12 nines): digit sum 108. Add 9:1000000000008: digit sum 1+8=9.No. So, these large numbers with many 9s just drop to 9 when adding 9.Alternatively, let's think of a number like 111111111111111111111111111 (27 ones): digit sum 27. But this is not a multiple of 9. To be a multiple of 9, the digit sum must be a multiple of 9, which 27 is. But 111...111 (27 ones) is 111...111. Let's check if it's divisible by 9. Since the digit sum is 27, which is divisible by 9, yes, it is divisible by 9. So, 111...111 (27 ones) is a multiple of 9. Then, adding 9 to it would give 111...111 +9=111...1110 +9=111...1119. But wait, adding 9 to the last digit:111...111 (27 digits) +9=111...1120. Because the last digit 1 becomes 10, so carryover. So, the last digit 0, carryover 1 to the previous digit, which was 1, becomes 2. So, the number becomes 111...1120. Digit sum:27 -1 +2 -0=28? Wait, let's compute digit sum:Original:27 ones: sum 27.New number:26 ones, followed by 2 and 0. So, sum=26*1 +2 +0=28. Not a multiple of 9. Therefore, this is not a multiple of 9. Wait, but 111...111 (27 ones) is a multiple of 9, but adding 9 gives a number that's not a multiple of 9? Wait, that can't be. Wait, multiples of 9 plus 9 should also be a multiple of 9. So, 111...111 (27 ones) is 9n. Then, 9n +9=9(n+1), so it must be a multiple of 9. But according to the digit sum, 28 is not a multiple of 9. Contradiction.Therefore, there must be a mistake in the digit sum calculation. Let's check:If the original number is 111...111 (27 ones), adding 9 would turn the last digit 1 into 10, so write 0, carryover 1. The next digit is 1, becomes 1+1=2. So, the number becomes 111...11120 (25 ones, then 2, then 0). So, digit sum is 25*1 +2 +0=27. Ah, yes! Because we have 25 ones, one 2, and one 0. 25+2=27. Therefore, digit sum remains 27. Therefore, the new number is also a multiple of 9 with digit sum 27. Therefore, adding 9 to 111...111 (27 ones) gives 111...1120, which also has digit sum 27. Therefore, these two numbers are consecutive multiples of 9 with digit sum 27. Then, adding 9 again:111...1120 +9=111...1129. Let's compute digit sum:111...1129:25 ones, then 2, then 9. Digit sum=25*1 +2 +9=36. So, digit sum increases to 36. Then, adding 9 gives 111...1138:25 ones, 3, 8. Digit sum=25+3+8=36. Wait, 36 is a multiple of 9. Then, adding 9 again:111...1138 +9=111...1147:25 ones, 4,7. Digit sum=25+4+7=36. Wait, this seems that once you reach a digit sum of 36, adding 9 keeps the digit sum the same? Wait, no, let's check:Wait, 111...1129 is 25 ones, 2,9. Digit sum=25+2+9=36. Then, adding 9:The last digit is 9 +9=18. Write 8, carryover 1.Next digit is 2 +1=3.So, the number becomes 111...1138. Digit sum:25 ones +3 +8=36.Adding 9 again:1138 +9=1147. Digit sum:25 ones +4 +7=36. Similarly, each time you add 9, the last digit increases by 9, causing a carryover that increments the next digit by1. So, the digit sum remains 36 until you reach a digit that is not 9. Wait, but in this case, after 111...1129, adding 9 repeatedly keeps the digit sum at 36. So, these are consecutive multiples of 9 with digit sum 36. Therefore, if you start with 111...111 (27 ones), then add 9, you get 111...1120 (digit sum 27), then add 9 to get 111...1129 (36), and then adding 9 repeatedly gives numbers with digit sum 36. Therefore, here we have two consecutive 27s followed by multiple 36s. But not four 27s.But wait, the first two numbers:111...111 (27 ones) and 111...1120 both have digit sum 27. Then the next number,111...1129, has 36. So, only two 27s in a row.But what if we create a number that allows adding 9 multiple times while keeping the digit sum 27? For example, a number like 111111111111111111111111111 (27 ones). Adding 9 gives 111...1120 (digit sum 27). Then, adding 9 gives 111...1129 (digit sum 36). So, only two in a row.Alternatively, if we can find a number where adding 9 multiple times causes the digit sum to remain 27. For example, a number like 111...111 followed by some other digits. Let's try a different number.Consider 111111111111111111111111111 (27 ones): digit sum 27. Add 9:111...1120: digit sum 27. Add 9:111...1129: digit sum 36. So, only two.Alternatively, take a number like 222222222222222222222222222 (27 twos). Digit sum 54. Not helpful.Alternatively, 333...333 (9 threes): digit sum 27. Add 9:333...333 +9=333...3342. Wait, 333...333 (9 threes) is 333,333,333. Adding 9 gives 333,333,342. Digit sum:3*8 +4 +2=24 +6=30. Not 27. So, no.Alternatively, 444444444444444444444444444 (27 fours): digit sum 108. Add 9: ...4453. Digit sum decreases.Not helpful.Alternatively, let's consider a number with digit sum 27 and trailing zeros. For example, 999000:9+9+9=27. Add 9:999009:9+9+9+0+0+9=36. No. Or 999000: add 9=999009: digit sum 36.Alternatively, a number like 198900000:1+9+8+9=27, rest zeros. Add 9:198900009:1+9+8+9+0+0+0+0+9=36. No.Hmm. It seems challenging to find four consecutive 27s. However, the example with 111...111 (27 ones) and 111...1120 gives two in a row. If we can find a different setup where adding 9 multiple times keeps the digit sum at 27, but I can't think of any.On the other hand, for digit sum 36, which requires four 9s or equivalent. For example, 9999: digit sum 36. Adding 9:10008: digit sum 9. Not helpful. But maybe numbers like 999999999999 (12 nines): digit sum 108. Not helpful. Alternatively, a number like 111111111111111111111111111111 (36 ones): digit sum 36. Adding 9 would give 111...1110: digit sum 36 -1 +1=36. Wait, similar to the previous example.Let's check:Original number: 36 ones, digit sum 36. Add 9: last digit becomes 10, carryover. So, the last digit is 0, next digit is 1+1=2. So, the number is 35 ones, followed by 2 and 0. Digit sum:35*1 +2 +0=37. Not 36. Therefore, digit sum increases by 1. Not helpful.Wait, but 36 ones is divisible by 9 (digit sum 36). Adding 9 gives 36 ones +9. But digit sum increases by 9? No, it depends on the carryover. Wait, adding 9 to 111...111 (36 ones):Last digit becomes 1+9=10. Write 0, carryover 1. The next digit is 1+1=2. So, the number becomes 111...1120 (34 ones, 2, 0). Digit sum:34*1 +2 +0=36. Therefore, digit sum remains 36. Then, adding 9 again:34 ones, 2, 0 +9=34 ones, 2,9. Digit sum:34 +2 +9=45. Then, adding 9 again:34 ones, 3,8. Digit sum:34 +3 +8=45. So, here, the digit sum increases to 45 and remains there. Therefore, in this case, we have two consecutive 36s: the original 36 ones and the next number 34 ones, 2,0. Then, the third number is 34 ones,2,9: digit sum 45. Therefore, two consecutive 36s. But not three.However, let's check:Number:111...111 (36 ones) digit sum 36.Add 9:111...1120 (34 ones,2,0) digit sum 36.Add 9:111...1129 (34 ones,2,9) digit sum 34+2+9=45.So, only two 36s in a row.Similarly, if we take a number like 9999: digit sum 36. Add 9:10008: digit sum 9. Not helpful.Alternatively, 999999999: digit sum 81. Add 9:1000000008: digit sum 9. No.Alternatively, a number like 8999999999999999999991: very long number with many 9s. But adding 9 would turn it into 9000000000000000000000: digit sum 9.Therefore, it seems difficult to get three consecutive 36s.But wait, let's consider the number 111111111111111111111111111111 (36 ones). Adding 9 gives 111...1120 (34 ones,2,0) with digit sum 36. Adding 9 again gives 111...1129 (34 ones,2,9) digit sum 45. So, only two 36s.Alternatively, what if we take a number like 222222222222222222222222222222 (36 twos). Digit sum 72. Add 9:222...2229: digit sum 72 -2 +9=79. Not helpful.Alternatively, 333...333 (12 threes): digit sum 36. Add 9:333...3342: digit sum 36 -3 +4 +2=39. Not helpful.Hmm. This is getting complicated. Let's think differently. Maybe the answer is that 27 four times in a row never appears, and neither does 36 three times. But the problem states that we need to determine which appears first. Therefore, one of them must appear.Alternatively, maybe in larger numbers, there are sequences where multiple 27s or 36s occur. For example, consider a number that is structured such that adding 9 multiple times causes minimal disruption to the digit sum.Let's try to construct such a number. Suppose we have a number with several trailing 9s and a preceding digit that can absorb the carryover without changing the digit sum. For example, a number like 1999999998: digit sum 1+9*8 +8=81. Adding 9 gives 2000000007: digit sum 9. Not helpful.Alternatively, consider a number like 198198198198. Digit sum:1+9+8+1+9+8+1+9+8+1+9+8= (1+9+8)*4=18*4=72. Add 9:198198198198 +9=198198198207. Digit sum:72 -8 +7=71. Not helpful.Alternatively, a number with repeating patterns that allow carryovers to propagate without affecting the digit sum much. For example, 111...111 with certain spacing.Alternatively, let's think about the fact that to get four consecutive 27s, we need four multiples of 9 such that each has digit sum 27. Given that adding 9 to a number with digit sum 27 can either keep the digit sum the same (as in 1989→1998) or decrease it (as in 999→1008). Therefore, to get four in a row, we need three instances where adding 9 doesn't decrease the digit sum.But in the case of 1989 and 1998, we have two in a row. To get a third, we would need 1998 +9=2007, which has digit sum 9. So, no. Therefore, even if we find a different pair, the third number breaks the sequence.Similarly, for 36s: to get three in a row, we need three multiples of 9 with digit sum 36. Let's see if that's possible.Take 9999: digit sum 36. Add 9:10008: digit sum 9. No.Take 8999999991: digit sum 8+9*8 +1=81. Add 9:9000000000: digit sum 9. No.Wait, what about 111111111111111111111111111111 (36 ones): digit sum 36. Add 9:111...1120: digit sum 36. Add 9:111...1129: digit sum 45. So, two 36s in a row. Then, adding 9 again:111...1138: digit sum 45. So, no third 36.Alternatively, take a number like 9999999999999999 (16 nines): digit sum 144. Add 9:10000000000000008: digit sum 1+8=9. No.Alternatively, a number like 444444444444444444444444444 (27 fours): digit sum 108. Add 9:444444444444444444444444453: digit sum 108 -4 +5 +3=112. Not helpful.Alternatively, consider numbers where digit sum is maintained by balancing carryovers. For example, a number like 199999999999999999999999998: very long number with many 9s and a 1 and 8 at the end. But adding 9 would turn it into 200000000000000000000000007: digit sum 9.Alternatively, this is getting too convoluted. Maybe there's a mathematical principle I'm missing.Wait, the key might be that for a number to have a digit sum of 27, it must be divisible by 9 (which it is, since it's a multiple of 9), and the digit sum is 27. Similarly for 36.Now, the numbers we are considering are 9n, where n is a positive integer. The digit sum of 9n is S(9n).We need to find the smallest n such that S(9n)=27, S(9(n+1))=27, S(9(n+2))=27, S(9(n+3))=27 for four in a row.Similarly, for three in a row with S(9n)=36.From our earlier examples, the smallest n where S(9n)=27 is n=111 (9*111=999). The next n=112 gives 1008 (digit sum 9). So, no.But we also found n=221:9*221=1989 (digit sum 27). Then n=222:1998 (digit sum 27). Then n=223:2007 (digit sum 9). So, two in a row.Similarly, n=333:2997 (digit sum 27). Next n=334:3006 (digit sum 9). No.So, the earliest occurrence of two 27s in a row is at n=221 and 222. To get four in a row, we need three more. But where?Similarly, let's check higher numbers. For example, n= 1881:9*1881=16929. Digit sum 1+6+9+2+9=27. Next, n=1882:16938:1+6+9+3+8=27. Next, n=1883:16947:1+6+9+4+7=27. Next, n=1884:16956:1+6+9+5+6=27. So, here, four consecutive numbers:16929,16938,16947,16956—all with digit sum 27. Therefore, four in a row.Wait, this seems promising. Let's verify:9*1881=16929: digit sum 1+6+9+2+9=27.9*1882=16938:1+6+9+3+8=27.9*1883=16947:1+6+9+4+7=27.9*1884=16956:1+6+9+5+6=27.9*1885=16965:1+6+9+6+5=27.So, actually, starting from n=1881 to n=1885, five consecutive numbers with digit sum 27. Therefore, the first occurrence of four consecutive 27s is at positions 1881,1882,1883,1884.Similarly, for three 36s in a row, we need three consecutive multiples of 9 with digit sum 36. Let's see.The smallest number with digit sum 36 is 9999 (9*1111). Let's check the next few multiples:9999: digit sum 36.10008: digit sum 9.10017: digit sum 9.... So, no. Next, maybe a larger number.For example, 999999999999 (12 nines): digit sum 108. Adding 9 gives 1000000000008: digit sum 9.Not helpful.Alternatively, consider numbers like 111111111111111111111111111111 (36 ones): digit sum 36. Adding 9 gives 111...1120: digit sum 36. Adding 9 gives 111...1129: digit sum 45. So, two 36s in a row. Then, adding 9 again gives 111...1138: digit sum 45. So, only two.Alternatively, let's look for numbers where adding 9 maintains the digit sum at 36. Let's try:Take 9999: digit sum 36. Add 9:10008:9. No.Take 8999999999991: digit sum 8+9*12 +1=8+108+1=117. No.Alternatively, consider 444444444444 (12 fours): digit sum 48. Add 9:444444444453: digit sum 48 -4 +5 +3=52. Not helpful.Alternatively, 333333333333 (12 threes): digit sum 36. Add 9:333333333342: digit sum 33 +4 +2=39. No.Wait, 333333333333: digit sum 36. Add 9:333333333342: digit sum 39. No.Hmm. What if we take a number like 1999999999998: digit sum 1+9*12 +8=1+108+8=117. No.Alternatively, 111111111111111111111111111111111111 (36 ones): digit sum 36. Adding 9:111...1120 (34 ones, 2, 0): digit sum 34+2+0=36. Adding 9 again:111...1129 (34 ones,2,9): digit sum 34+2+9=45. So, two in a row. Then, adding 9 again:111...1138: digit sum 34+3+8=45. So, only two 36s.But perhaps there's a different number where three 36s occur. Let's think.Suppose we have a number like 9999999999999999 (16 nines): digit sum 144. Add 9:10000000000000008: digit sum 9. No.Alternatively, let's consider a number like 1899999999999999999. Let's compute its digit sum:1+8+9*16=1+8+144=153. Which is divisible by 9, but digit sum is 153. Not helpful.Alternatively, think of a number structured such that adding 9 three times doesn't decrease the digit sum below 36. For example, numbers with multiple trailing 9s and a preceding digit that can absorb the carryover without reducing the digit sum.But I'm not sure. It seems that getting three 36s in a row is as difficult as getting four 27s. However, in our previous example, we found that starting at n=1881, there are five consecutive multiples of 9 with digit sum 27. Therefore, four in a row occurs at n=1881-1884.So, the first occurrence of four 27s in a row is at position 1881. What about three 36s in a row? Have we found any?Alternatively, maybe three 36s occur later. To check, let's find numbers where three consecutive multiples of 9 have digit sum 36.For example, consider 9999 (n=1111): digit sum 36. Next multiple:10008:9. No.Another example: 111111111111111111111111111111 (36 ones): digit sum 36. Next multiple:111...1120: digit sum 36. Next:111...1129:45. So, two in a row.Another example: let's take n=4444:9*4444=39996: digit sum 3+9+9+9+6=36. Next multiple:40005:4+0+0+0+5=9. No.Another number: 999999999999 (12 nines): digit sum 108. Next multiple:1000000000008:9. No.Wait, maybe a different structure. For example, 1999999999999999998: digit sum 1+9*17 +8=1+153+8=162. No.Alternatively, 5999999994:5+9*8 +4=5+72+4=81. No.Hmm. It seems challenging to find three 36s in a row. Therefore, based on the earlier example where four 27s occur starting at n=1881, which is earlier than any possible three 36s, we can conclude that 27 four times in a row appears first.Finally, the last question: what else can be said about the alternation of numbers in the sequence? Well, the sequence of digit sums of multiples of 9 is always a multiple of 9. The digit sums alternate between 9, 18, 27, etc., but typically, most numbers have digit sum 9, and higher digit sums occur less frequently. The alternation is not regular; there are long stretches of 9s, with occasional higher values like 18, 27, etc. However, when a higher digit sum occurs, it's often followed by a drop back to 9, unless there's a carryover structure that allows maintaining the digit sum, as seen in the consecutive 27s example. Therefore, the alternation is irregular, with most terms being 9, and higher sums appearing sporadically, sometimes in consecutive terms due to specific number structures.

✅:Step-by-Step Explanation and Answers:1. Position of the First 81 and the Following Number: - First Occurrence of 81: The digit sum of 81 requires a number with nine 9's, which is (999,999,999). This is the (111,111,111)th multiple of 9 (since (999,999,999 = 9 times 111,111,111)). - Number Following 81: The next multiple of 9 is (1,000,000,008), whose digit sum is (1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 8 = 9). Answer: - 81 first appears at position 111,111,111, followed by 9.2. 27 Four Times vs. 36 Three Times: - Consecutive 27s: The earliest occurrence of four consecutive 27s starts at the (1881)th multiple of 9. These multiples are (9 times 1881 = 16929), (9 times 1882 = 16938), (9 times 1883 = 16947), and (9 times 1884 = 16956), all with digit sum 27. - Consecutive 36s: Finding three consecutive 36s is more challenging. The earliest examples of two consecutive 36s occur in larger numbers (e.g., (111...111) with 36 ones followed by (111...1120)), but three in a row are not observed before the four 27s. Answer: - 27 appears four times in a row first.3. Alternation in the Sequence: - The sequence predominantly consists of 9s. Higher digit sums (18, 27, 36, etc.) occur less frequently and typically revert to 9 due to carryover effects. However, specific number structures (e.g., (16929, 16938, 16947, 16956)) can maintain higher sums consecutively by balancing carryovers. Answer: - The sequence alternates irregularly, dominated by 9s. Higher sums appear sporadically and can occasionally repeat consecutively due to carryover patterns but generally reset to 9 afterward.Final Answers: - 81 first appears at position 111,111,111, followed by 9. - 27 four times in a row appears before 36 three times in a row. - The sequence mainly alternates with 9, occasionally interrupted by higher sums that can repeat under specific structural conditions.

🔑:1. Sum of Digits of Numbers Divisible by 9: We have the sequence of numbers divisible by 9 given by: [ 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, ldots ] When observing their sums of digits, we initially get: [ 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 18, ldots ]2. First Occurrence of 81 in Sequence: For a sum of digits to be 81, the number itself must be at least 9 digits long, since: [ 9 cdot 9 = 81 ] The smallest 9-digit number whose digits sum to 81 is (999,999,999). Therefore, in the given sequence, 81 will appear at the position of: [ boxed{111,111,111} ] The next sum after (81) would be the sum of digits of (1,000,000,008), which is 9. Hence, the next number in the sequence will be: [ 9 ]3. Appearance of 27 Four Times Consecutively vs. 36 Three Times Consecutively: - For 27 to appear consecutively, we need to find four numbers whose digits add to 27. Possible candidates include: [ 3969 (3+9+6+9=27), 3978 (3+9+7+8=27), 3987 (3+9+8+7=27), 3996 (3+9+9+6=27) ] These correspond to sums of 27 in the given sequence<*>, which would appear as: [ 27, 27, 27, 27 text{ consecutively (before any occurrence of 36)} ] - For 36 to appear three times consecutively, we need: [ 9999 (9+9+9+9=36) ] This would appear in the sequence as three consecutive occurrences of 36: [ 36, 36, 36 ] Hence, 27 will appear four times consecutively before 36 appears three times in the sequence. Thus: [ boxed{4 text{ times } 27} ] _Conclusion:_ - First occurrence of 81: (boxed{111,111,111}) - Next number after 81 in the sequence: (boxed{9}) - Sequence pattern analysis: (27) appears four times before (36) appears thrice. The alternating pattern of the numbers follows the non-increasing sequence in the given sequence, due to properties of numbers divisible by 9 divided by 9.

🧠:Okay, so I have this problem where I need to prove 25 different relationships in triangle ABC. The first one is a + c = 2b, and then there are several others involving sines, cosines, cotangents, areas, radii, etc. Hmm. Let me start with the first one and see if I can find a pattern or a key relationship that might help with the others. Maybe all these are connected through some initial condition or given property of the triangle. Wait, the problem just says "In triangle ABC, prove the following relationships," so it's possible that all these relationships are true under some specific condition, maybe that a + c = 2b is given? But the first one is part of what needs to be proven. Wait, no, maybe all these are to be proven in sequence, each building on the previous? That might be, but 25 is a lot. Alternatively, maybe triangle ABC has some special properties, like being a specific type of triangle such as isosceles or having angles in arithmetic progression, or sides in a particular ratio. Let me check if the first few equations can hint at a particular type of triangle.Looking at 1: a + c = 2b. So the sum of sides a and c is twice side b. If this is a general triangle, this would imply some relationship between the sides. If I can assume that triangle ABC has sides in the ratio 2:1:1 or something? Wait, but a + c = 2b. For example, if b is the middle side, then a + c = 2b. Maybe the triangle is such that angles A and C are related in a way that their sines add up to twice the sine of B, as in equation 2. So maybe using the Law of Sines here. Let's try that.Law of Sines states that a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius. If a + c = 2b, then substituting a = 2R sin A, c = 2R sin C, and b = 2R sin B, we get 2R sin A + 2R sin C = 2*(2R sin B). Dividing both sides by 2R, sin A + sin C = 2 sin B. Hey, that's exactly equation 2! So equation 1 and equation 2 are equivalent via the Law of Sines. Therefore, if equation 1 holds, equation 2 must hold and vice versa. So maybe the triangle is such that a + c = 2b, which implies sin A + sin C = 2 sin B, and then the rest of the equations follow from that. So perhaps all these equations are under the condition that a + c = 2b, which is given in equation 1. Therefore, we can take a + c = 2b as given and then prove the other equations. Alternatively, maybe they are all interconnected such that proving one helps prove the others. But since the problem says "prove the following relationships," maybe we need to prove each one, possibly using previous ones.Alternatively, maybe the triangle has angles in arithmetic progression, such that A + C = 2B, but since in a triangle A + B + C = π, then 2B + B = π, so 3B = π, so B = π/3. Wait, that might be possible. If B is 60 degrees, then the angles A and C would satisfy A + C = 2π/3. Alternatively, if sides a, b, c are in an arithmetic sequence with b as the middle term, so a + c = 2b. So perhaps the triangle has sides in arithmetic progression. Let me check if that's a valid assumption.If the sides are in arithmetic progression, then a = b - d, c = b + d for some common difference d. Then angles would adjust accordingly. But perhaps we can use the Law of Sines and Law of Cosines to relate the angles. Let's try.If a + c = 2b, then using Law of Sines: sin A + sin C = 2 sin B. Let me see if that holds. Let's suppose that angles A, B, C are such that A + C = 2B. Wait, but in a triangle, A + B + C = π. So if A + C = 2B, then 2B + B = π => 3B = π => B = π/3. So angles would be B = 60 degrees, and A + C = 120 degrees. Then, maybe that's the key here. So the triangle has angle B = 60 degrees, and angles A and C sum to 120 degrees. Then, using this, we can derive all the other equations. Let me check if that's the case.For example, equation 3: cos((A - C)/2) = 2 cos((A + C)/2). If A + C = 120 degrees, then (A + C)/2 = 60 degrees, so cos(60°) = 0.5. Then, 2 cos((A + C)/2) = 2 * 0.5 = 1. So equation 3 would become cos((A - C)/2) = 1. But cos(theta) = 1 implies theta = 0, so (A - C)/2 = 0 => A = C. So if A = C, then angles A and C are both 60 degrees, making the triangle equilateral. But in that case, a + c = 2b would mean 2a = 2a, which is trivial, but angle B is also 60 degrees. However, equation 3 seems to suggest that cos((A - C)/2) = 2 * cos(60°) = 1, but unless A = C, which would make it 1. However, if the triangle is not equilateral, but angle B is 60 degrees and A ≠ C, then equation 3 would require cos((A - C)/2) = 1, which is only possible if A = C. Therefore, this seems contradictory unless the triangle is equilateral. But in that case, all the equations would hold trivially. But perhaps there is another approach.Wait, maybe I made a mistake here. Let me re-examine equation 3. It says cos[(A - C)/2] = 2 cos[(A + C)/2]. If A + C = 120°, then cos[(A - C)/2] = 2 cos(60°) = 2*(0.5) = 1. Therefore, cos[(A - C)/2] = 1 implies (A - C)/2 = 0 => A = C. So in this case, the triangle must be isosceles with A = C. So if a + c = 2b, and the triangle is isosceles with A = C, then sides a and c would be equal. But a + c = 2b would mean 2a = 2b => a = b. So if a = b = c, then it's equilateral. Hmm, but if the triangle is equilateral, all angles are 60°, and all relationships would hold. But perhaps the problem is more general. Maybe the triangle isn't necessarily equilateral but satisfies a + c = 2b. However, from the above, if a + c = 2b and the triangle is isosceles (a = c), then it's equilateral. But if the triangle isn't isosceles, then A ≠ C, and equation 3 would require cos[(A - C)/2] = 1, which would force A = C. Therefore, perhaps the only triangle satisfying a + c = 2b is the equilateral triangle. But that's not true because if we take a triangle with sides 3, 3, 3, then a + c = 6 = 2*3, which is 6 = 6. But any triangle with a + c = 2b doesn't have to be equilateral. For example, sides 1, 2, 3? Wait, no, that doesn't satisfy triangle inequality. Let's take sides 2, 3, 4. Then a + c = 2 + 4 = 6, which is not 2*3 = 6. Wait, 2 + 4 = 6, which is equal to 2*3. So in this case, sides 2, 3, 4 would satisfy a + c = 2b. Wait, but in this triangle, angles are not equal, so it's not equilateral. So this triangle would have sides 2, 3, 4. Let's check angle B. Using Law of Cosines: cos B = (a² + c² - b²)/(2ac) = (4 + 16 - 9)/(2*2*4) = (11)/(16). So angle B is arccos(11/16) ≈ 46.567 degrees, not 60 degrees. But according to previous logic, if a + c = 2b, then from equation 1, using Law of Sines, sin A + sin C = 2 sin B. Let's check with sides 2, 3, 4. Let's find angles using Law of Sines.a = 2, opposite angle A; b = 3, opposite angle B; c = 4, opposite angle C.Law of Sines: 2/sin A = 3/sin B = 4/sin C.Let me compute angles:First, compute angle B using Law of Cosines as before: cos B = (4 + 16 - 9)/(2*2*4) = 11/16 ≈ 0.6875. So angle B ≈ 46.567°. Then sin B ≈ sqrt(1 - (11/16)^2) = sqrt(1 - 121/256) = sqrt(135/256) = (3*sqrt(15))/16 ≈ 0.726.Then, 2/sin A = 3/sin B ≈ 3/0.726 ≈ 4.132. So sin A ≈ 2 / 4.132 ≈ 0.484, so angle A ≈ 28.96°. Similarly, 4/sin C = 4.132, so sin C ≈ 4 / 4.132 ≈ 0.968, so angle C ≈ 75.52°. Check A + B + C ≈ 28.96 + 46.567 + 75.52 ≈ 151.05, which is not 180. Wait, that can't be. There's a mistake here. Wait, perhaps I miscalculated. Let me recalculate.Wait, sides are a=2, b=3, c=4. Law of Cosines for angle B:cos B = (a² + c² - b²)/(2ac) = (4 + 16 - 9)/(2*2*4) = 11/16 ≈ 0.6875, which is correct. So angle B is arccos(11/16) ≈ 46.567°, sin B ≈ sqrt(1 - (121/256)) = sqrt(135/256) = (3*sqrt(15))/16 ≈ 0.72618.Law of Sines: 2/sin A = 3/sin B ≈ 3/0.72618 ≈ 4.1318.So sin A = 2 / 4.1318 ≈ 0.484, angle A ≈ arcsin(0.484) ≈ 28.96°, then angle C ≈ 180 - 46.567 - 28.96 ≈ 104.473°.Then, sin C ≈ sin(104.473°) ≈ sin(75.527°) ≈ 0.968. Then, checking sin A + sin C ≈ 0.484 + 0.968 ≈ 1.452. 2 sin B ≈ 2*0.726 ≈ 1.452. So equation 2 holds. So in this triangle, equations 1 and 2 hold. But angle B is not 60°, but equations 3 to 25 might also hold? Let's check equation 3: cos[(A - C)/2] = 2 cos[(A + C)/2].Given angles A ≈ 28.96°, C ≈ 104.473°, so A - C ≈ -75.513°, divided by 2: -37.756°, so cos(-37.756°) ≈ cos(37.756°) ≈ 0.790. On the other hand, A + C ≈ 133.433°, so (A + C)/2 ≈ 66.716°, cos(66.716°) ≈ 0.395. Then 2 * 0.395 ≈ 0.79. So equation 3 holds. Interesting! So even though angle B isn't 60°, equation 3 holds. So my previous assumption that angle B is 60° was incorrect. Therefore, the triangle with sides 2, 3, 4 satisfies a + c = 2b, and equations 1, 2, 3 hold. Therefore, there must be a different relationship here. So perhaps this triangle has some other properties, and the equations can be proven based on a + c = 2b.So, the key is that given a + c = 2b, we can use the Law of Sines and Law of Cosines to derive all other relationships. Let me start working through each equation step by step.1. Prove that (a + c = 2b).Wait, but this is given as the first equation to prove. Wait, the problem says "prove the following relationships," so all 25 are to be proven. However, the first one is (a + c = 2b). Wait, maybe there's a typo or misunderstanding. If all 25 are to be proven, maybe they are interconnected. Alternatively, maybe all these relationships are true given some initial condition not stated here. However, the problem statement is a bit unclear. It says "In triangle ABC, prove the following relationships," which suggests that all these relationships hold in any triangle ABC. But that's not possible because, for example, in a general triangle, a + c = 2b would not hold. Hence, the problem must be missing some initial condition. Alternatively, maybe all these relationships are equivalent or follow from each other, starting from one given. Alternatively, perhaps the problem is in the context where a + c = 2b is given, and the rest are to be proven under that condition. The way the problem is phrased is a bit ambiguous.Wait, looking at the problem again: "In triangle ABC, prove the following relationships: 1. (a + c = 2b) 2. (sin A + sin C = 2 sin B) ..." So it's saying in triangle ABC, all these 25 relationships hold. But that can't be true unless the triangle has specific properties. Therefore, the correct interpretation must be that the triangle ABC satisfies one of these conditions, and the rest follow. However, the problem states "prove the following relationships," which could mean that they are all equivalent or that they are consequences of each other. But given the number of relationships, it's more likely that they are all consequences of a particular property of the triangle, which might be given by the first equation, (a + c = 2b). So perhaps the problem is to assume that in triangle ABC, (a + c = 2b), and then prove the other 24 relationships. Alternatively, maybe all these relationships are equivalent and can be derived from each other.Given the problem's structure, the most plausible scenario is that all these relationships hold in a triangle where (a + c = 2b), and we need to prove each of them under that condition. Therefore, we can take (a + c = 2b) as given and then use that to prove the remaining 24 equations. Alternatively, some of them may be used to prove others. Since the user listed all 25, it's expected to prove each one, possibly using previous results. Given the extensive list, this will require a systematic approach, possibly starting from the first equation and using trigonometric identities, Laws of Sines and Cosines, inradius, circumradius formulas, etc.Let me begin with equation 1: (a + c = 2b). If this is a given condition, then we can use it to derive the other equations. However, the problem states "prove the following relationships," so perhaps all of them are to be proven without assuming any prior condition. But that's impossible unless there's a specific triangle type. Alternatively, maybe all these equations are interconnected such that proving one helps in proving others. However, given the time constraints, it's more efficient to assume that (a + c = 2b) is a given condition and then derive the rest from there. Let's proceed under that assumption.1. (a + c = 2b)Assuming this is given, perhaps as a property of the triangle.2. (sin A + sin C = 2 sin B)Using the Law of Sines: (a = 2R sin A), (b = 2R sin B), (c = 2R sin C). Then (a + c = 2b) becomes (2R (sin A + sin C) = 2 * 2R sin B), dividing both sides by 2R: (sin A + sin C = 2 sin B). Hence, proven.3. (cos frac{A - C}{2} = 2 cos frac{A + C}{2})From equation 2: (sin A + sin C = 2 sin B). Using the sum-to-product identity: (sin A + sin C = 2 sin frac{A + C}{2} cos frac{A - C}{2}). Therefore:(2 sin frac{A + C}{2} cos frac{A - C}{2} = 2 sin B).Divide both sides by 2:(sin frac{A + C}{2} cos frac{A - C}{2} = sin B).But in any triangle, (A + B + C = pi), so (A + C = pi - B). Therefore, (frac{A + C}{2} = frac{pi - B}{2}). So:(sin left(frac{pi - B}{2}right) cos frac{A - C}{2} = sin B).But (sin left(frac{pi - B}{2}right) = sin left(frac{pi}{2} - frac{B}{2}right) = cos frac{B}{2}).So:(cos frac{B}{2} cos frac{A - C}{2} = sin B).Now, express (sin B) as (2 sin frac{B}{2} cos frac{B}{2}):(cos frac{B}{2} cos frac{A - C}{2} = 2 sin frac{B}{2} cos frac{B}{2}).Divide both sides by (cos frac{B}{2}) (assuming (cos frac{B}{2} neq 0)):(cos frac{A - C}{2} = 2 sin frac{B}{2}).But from equation 3, we need to prove (cos frac{A - C}{2} = 2 cos frac{A + C}{2}).Wait, but from the previous step, we have (cos frac{A - C}{2} = 2 sin frac{B}{2}). Let's see. Since (A + C = pi - B), then (frac{A + C}{2} = frac{pi - B}{2}), so (cos frac{A + C}{2} = cos left(frac{pi}{2} - frac{B}{2}right) = sin frac{B}{2}). Therefore, (2 cos frac{A + C}{2} = 2 sin frac{B}{2}). Comparing with the previous equation, (cos frac{A - C}{2} = 2 sin frac{B}{2} = 2 cos frac{A + C}{2}). Hence, equation 3 is proven.4. (sin^2 frac{A}{2} + sin^2 frac{C}{2} = cos B)Use the identity (sin^2 theta = frac{1 - cos 2theta}{2}):(sin^2 frac{A}{2} + sin^2 frac{C}{2} = frac{1 - cos A}{2} + frac{1 - cos C}{2} = 1 - frac{cos A + cos C}{2}).We need to show this equals (cos B). Therefore, need to show (1 - frac{cos A + cos C}{2} = cos B).Rearranged: (2 - (cos A + cos C) = 2 cos B).So need to prove (cos A + cos C = 2 - 2 cos B).Wait, but we can use the identity for sum of cosines in a triangle. Alternatively, use the fact that (A + C = pi - B), so maybe use sum-to-product formulas.(cos A + cos C = 2 cos frac{A + C}{2} cos frac{A - C}{2}).From equation 3, (cos frac{A - C}{2} = 2 cos frac{A + C}{2}). Therefore:(cos A + cos C = 2 cos frac{A + C}{2} * 2 cos frac{A + C}{2} = 4 cos^2 frac{A + C}{2}).But (frac{A + C}{2} = frac{pi - B}{2}), so:(cos A + cos C = 4 cos^2 left(frac{pi - B}{2}right) = 4 sin^2 frac{B}{2}).Therefore, going back to the previous equation:(2 - (cos A + cos C) = 2 - 4 sin^2 frac{B}{2}).But (2 - 4 sin^2 frac{B}{2} = 2(1 - 2 sin^2 frac{B}{2}) = 2 cos B) (using the double-angle identity: (cos B = 1 - 2 sin^2 frac{B}{2})).Therefore, (2 - (cos A + cos C) = 2 cos B), which rearranged gives:(sin^2 frac{A}{2} + sin^2 frac{C}{2} = cos B).Hence, equation 4 is proven.5. (cot frac{A}{2} + cot frac{C}{2} = 2 cot frac{B}{2})First, recall that (cot frac{theta}{2} = frac{1 + cos theta}{sin theta}).Alternatively, use formula for cotangent in terms of semiperimeter and sides.But maybe use identities related to the triangle's inradius and semiperimeter. Alternatively, use the formula (cot frac{A}{2} = frac{s - a}{r}), where (s) is the semiperimeter.Let’s recall that in any triangle:(cot frac{A}{2} = frac{s - a}{r}),(cot frac{C}{2} = frac{s - c}{r}),(cot frac{B}{2} = frac{s - b}{r}).Therefore, (cot frac{A}{2} + cot frac{C}{2} = frac{s - a + s - c}{r} = frac{2s - (a + c)}{r}).But since (a + c = 2b), this becomes (frac{2s - 2b}{r} = frac{2(s - b)}{r} = 2 cot frac{B}{2}), since (cot frac{B}{2} = frac{s - b}{r}). Hence, equation 5 is proven.6. (cot frac{A}{2} cot frac{C}{2} = 3)Using the same formulas as above, (cot frac{A}{2} = frac{s - a}{r}), (cot frac{C}{2} = frac{s - c}{r}). Therefore, their product is (frac{(s - a)(s - c)}{r^2}).But we need to relate this to known quantities. Alternatively, use trigonometric identities.From equation 3: (cos frac{A - C}{2} = 2 cos frac{A + C}{2}). Let's square both sides:(cos^2 frac{A - C}{2} = 4 cos^2 frac{A + C}{2}).Using the identity (cos^2 x = frac{1 + cos 2x}{2}):Left side: (frac{1 + cos(A - C)}{2}).Right side: 4 * (frac{1 + cos(A + C)}{2}) = 2(1 + cos(A + C)).But (A + C = pi - B), so (cos(A + C) = cos(pi - B) = -cos B). Therefore:Left side: (frac{1 + cos(A - C)}{2}).Right side: 2(1 - cos B).Thus:(frac{1 + cos(A - C)}{2} = 2(1 - cos B)).Multiply both sides by 2:(1 + cos(A - C) = 4(1 - cos B)).Therefore:(cos(A - C) = 4(1 - cos B) - 1 = 3 - 4 cos B).This actually is equation 18. So we can use this later.But back to equation 6. Let's consider using the formula for (cot frac{A}{2} cot frac{C}{2}).Alternatively, use the identity:(cot frac{A}{2} cot frac{C}{2} = frac{cos frac{A}{2} cos frac{C}{2}}{sin frac{A}{2} sin frac{C}{2}}).Let me compute this ratio. From equations involving sum and product.First, from equation 5: (cot frac{A}{2} + cot frac{C}{2} = 2 cot frac{B}{2}). Let's denote (x = cot frac{A}{2}) and (y = cot frac{C}{2}). Then, x + y = 2 cot frac{B}{2}), and we need to find xy = 3.But maybe relate to another equation. Let's use equation 4: (sin^2 frac{A}{2} + sin^2 frac{C}{2} = cos B).Express (sin^2 frac{A}{2} = frac{1 - cos A}{2}), similarly for C. Then:(frac{1 - cos A}{2} + frac{1 - cos C}{2} = cos B).Which simplifies to:(1 - frac{cos A + cos C}{2} = cos B).From earlier, we have (cos A + cos C = 4 sin^2 frac{B}{2}). Therefore:(1 - 2 sin^2 frac{B}{2} = cos B), which is true since (cos B = 1 - 2 sin^2 frac{B}{2}). So this checks out.Alternatively, perhaps use formulae involving cotangents. Recall that:(cot frac{A}{2} = frac{cos frac{A}{2}}{sin frac{A}{2}}), and similarly for C.But maybe use the formula for (cot frac{A}{2} cot frac{C}{2}) in terms of sides.Given that (a + c = 2b), and using semiperimeter (s = frac{a + b + c}{2} = frac{2b + b}{2} = frac{3b}{2}).Then, (s - a = frac{3b}{2} - a), (s - c = frac{3b}{2} - c), and (s - b = frac{3b}{2} - b = frac{b}{2}).From equation 5: (cot frac{A}{2} + cot frac{C}{2} = 2 cot frac{B}{2} = 2 * frac{s - b}{r} = 2 * frac{b/2}{r} = frac{b}{r}).But from equation 5: (cot frac{A}{2} + cot frac{C}{2} = frac{s - a}{r} + frac{s - c}{r} = frac{2s - (a + c)}{r}). Since (a + c = 2b), this is (frac{2s - 2b}{r} = frac{2(s - b)}{r} = 2 * frac{b/2}{r} = frac{b}{r}). So that checks out.But for the product (cot frac{A}{2} cot frac{C}{2} = frac{(s - a)(s - c)}{r^2}).Given (s = 3b/2), (s - a = (3b/2 - a)), (s - c = (3b/2 - c)). Since (a + c = 2b), then (c = 2b - a). Therefore, (s - c = 3b/2 - (2b - a) = 3b/2 - 2b + a = -b/2 + a).Similarly, (s - a = 3b/2 - a).Thus, the product ((3b/2 - a)(-b/2 + a)).Let me compute this:((3b/2 - a)(a - b/2) = (3b/2 - a)(a - b/2)).Let me expand this:= 3b/2 * a - 3b/2 * b/2 - a * a + a * b/2= (3ab/2) - (3b²/4) - a² + (ab/2)= (3ab/2 + ab/2) - 3b²/4 - a²= 2ab - 3b²/4 - a²But since a + c = 2b, and c = 2b - a, we can express this in terms of a and b. Let's also use the Law of Cosines to find a relation between a and b.From equation 1: a + c = 2b => c = 2b - a.Using Law of Cosines on angle B:b² = a² + c² - 2ac cos BSubstitute c = 2b - a:b² = a² + (2b - a)^2 - 2a(2b - a) cos BExpand (2b - a)^2 = 4b² - 4ab + a²:b² = a² + 4b² - 4ab + a² - 2a(2b - a) cos BSimplify:b² = 2a² + 4b² - 4ab - 4ab cos B + 2a² cos BBring all terms to one side:0 = 2a² + 4b² - 4ab - 4ab cos B + 2a² cos B - b²Simplify:0 = 2a² + 3b² - 4ab - 4ab cos B + 2a² cos BHmm, this seems complicated. Maybe there's a better approach.Alternatively, use the fact that from equation 3, we have (cos frac{A - C}{2} = 2 cos frac{A + C}{2}). Earlier, we found that this implies (cos(A - C) = 3 - 4 cos B) (equation 18). Maybe use this.But perhaps there's a trigonometric identity that can link the product of the cotangents. Let's consider that:(cot frac{A}{2} cot frac{C}{2} = frac{cos frac{A}{2} cos frac{C}{2}}{sin frac{A}{2} sin frac{C}{2}}).Using product-to-sum identities:(cos frac{A}{2} cos frac{C}{2} = frac{1}{2} [cos frac{A - C}{2} + cos frac{A + C}{2}]).(sin frac{A}{2} sin frac{C}{2} = frac{1}{2} [cos frac{A - C}{2} - cos frac{A + C}{2}]).Therefore, the ratio becomes:(frac{frac{1}{2} [cos frac{A - C}{2} + cos frac{A + C}{2}]}{frac{1}{2} [cos frac{A - C}{2} - cos frac{A + C}{2}]} = frac{cos frac{A - C}{2} + cos frac{A + C}{2}}{cos frac{A - C}{2} - cos frac{A + C}{2}}).From equation 3: (cos frac{A - C}{2} = 2 cos frac{A + C}{2}). Let's denote (X = cos frac{A + C}{2}), so (cos frac{A - C}{2} = 2X).Substituting:Numerator: (2X + X = 3X).Denominator: (2X - X = X).Therefore, the ratio is (3X / X = 3). Hence, (cot frac{A}{2} cot frac{C}{2} = 3). Equation 6 is proven.7. (a cos^2 frac{C}{2} + c cos^2 frac{A}{2} = frac{3b}{2})Start by expressing (cos^2 frac{C}{2} = frac{1 + cos C}{2}) and (cos^2 frac{A}{2} = frac{1 + cos A}{2}). Substitute into the left side:(a cdot frac{1 + cos C}{2} + c cdot frac{1 + cos A}{2} = frac{a + c}{2} + frac{a cos C + c cos A}{2}).From equation 1, (a + c = 2b), so first term is (2b/2 = b). Therefore:= (b + frac{a cos C + c cos A}{2}).Need to show that this equals (3b/2), so:(b + frac{a cos C + c cos A}{2} = frac{3b}{2}).Subtract b from both sides:(frac{a cos C + c cos A}{2} = frac{b}{2}).Multiply both sides by 2:(a cos C + c cos A = b).Now, recall the projection formula: in any triangle, the length of side b is equal to the sum of the projections of the other two sides onto it. Specifically, (b = a cos C + c cos A). This is a standard result in trigonometry. Hence, this equation holds true in any triangle. Therefore, the expression simplifies to (b + frac{b}{2} = frac{3b}{2}), proving equation 7.8. (h_b = 3r)(h_b) is the altitude from B, and (r) is the inradius.The area (Delta) can be expressed as (frac{1}{2} b h_b). Also, the area is (r cdot s), where (s) is the semiperimeter.From equation 1, (a + c = 2b), so the perimeter is (a + b + c = 2b + b = 3b), hence semiperimeter (s = frac{3b}{2}).Therefore, (Delta = r cdot frac{3b}{2}).Also, (Delta = frac{1}{2} b h_b).Equating the two expressions for area:(frac{1}{2} b h_b = frac{3b}{2} r).Divide both sides by (frac{b}{2}):(h_b = 3r). Hence, equation 8 is proven.9. (cos A + cos C = 4 sin^2 frac{B}{2})From earlier in equation 4's proof, we found that (cos A + cos C = 4 sin^2 frac{B}{2}). Hence, equation 9 is already proven.10. The quadratic equation ((sin B - sin C) x^2 + (sin C - sin A) x + sin A - sin B = 0) has equal roots.For a quadratic equation (Ax^2 + Bx + C = 0) to have equal roots, the discriminant must be zero: (B^2 - 4AC = 0).Compute discriminant D:([(sin C - sin A)]^2 - 4(sin B - sin C)(sin A - sin B)).Expand the first term:((sin C - sin A)^2 = sin^2 C - 2 sin A sin C + sin^2 A).Expand the product:(4(sin B - sin C)(sin A - sin B) = 4[sin B sin A - sin^2 B - sin C sin A + sin C sin B]).Thus, discriminant D:(sin^2 C - 2 sin A sin C + sin^2 A - 4[sin A sin B - sin^2 B - sin A sin C + sin B sin C]).Expand the terms inside the brackets:= (sin^2 C - 2 sin A sin C + sin^2 A - 4 sin A sin B + 4 sin^2 B + 4 sin A sin C - 4 sin B sin C).Combine like terms:- (sin^2 A + sin^2 C)- (-2 sin A sin C + 4 sin A sin C = 2 sin A sin C)- (-4 sin A sin B)- (4 sin^2 B)- (-4 sin B sin C)So D becomes:(sin^2 A + sin^2 C + 2 sin A sin C - 4 sin A sin B + 4 sin^2 B - 4 sin B sin C).Now, recall from equation 2: (sin A + sin C = 2 sin B). Let’s denote (S_A = sin A), (S_C = sin C), (S_B = sin B). Then, (S_A + S_C = 2 S_B).We can use this to substitute in the discriminant expression.Let's compute each term:1. (sin^2 A + sin^2 C = (S_A + S_C)^2 - 2 S_A S_C = (2 S_B)^2 - 2 S_A S_C = 4 S_B^2 - 2 S_A S_C).2. (2 sin A sin C = 2 S_A S_C).3. (-4 sin A sin B = -4 S_A S_B).4. (4 sin^2 B = 4 S_B^2).5. (-4 sin B sin C = -4 S_B S_C).Putting all together:D = [4 S_B^2 - 2 S_A S_C] + [2 S_A S_C] - 4 S_A S_B + 4 S_B^2 - 4 S_B S_CSimplify term by term:- 4 S_B^2 - 2 S_A S_C + 2 S_A S_C = 4 S_B^2.- -4 S_A S_B + 4 S_B^2 -4 S_B S_C.So total D = 4 S_B^2 -4 S_A S_B +4 S_B^2 -4 S_B S_C.Combine like terms:4 S_B^2 +4 S_B^2 =8 S_B^2-4 S_A S_B -4 S_B S_C= -4 S_B (S_A + S_C) = -4 S_B (2 S_B) = -8 S_B^2.Thus, D =8 S_B^2 -8 S_B^2=0.Therefore, the discriminant is zero, so the quadratic has equal roots. Equation 10 is proven.11. (sin^2 frac{A}{2} + sin^2 frac{B}{2} + sin^2 frac{C}{2} = cos^2 frac{B}{2})Use the identity (sin^2 theta = frac{1 - cos 2theta}{2}):Left side:(frac{1 - cos A}{2} + frac{1 - cos B}{2} + frac{1 - cos C}{2} = frac{3 - (cos A + cos B + cos C)}{2}).We need to show this equals (cos^2 frac{B}{2} = frac{1 + cos B}{2}).Therefore, require:(frac{3 - (cos A + cos B + cos C)}{2} = frac{1 + cos B}{2}).Multiply both sides by 2:(3 - (cos A + cos B + cos C) = 1 + cos B).Rearranged:(3 - cos A - cos C - 2 cos B = 1).So:(- cos A - cos C - 2 cos B = -2).Multiply by -1:(cos A + cos C + 2 cos B = 2).From equation 9: (cos A + cos C = 4 sin^2 frac{B}{2}). And since (4 sin^2 frac{B}{2} = 2(1 - cos B)).Therefore, substituting:(2(1 - cos B) + 2 cos B = 2 - 2 cos B + 2 cos B = 2). Hence, equality holds. Therefore, equation 11 is proven.12. (cos A + 2 cos B + cos C = 2)From equation 9: (cos A + cos C = 4 sin^2 frac{B}{2}). Also, (4 sin^2 frac{B}{2} = 2(1 - cos B)). Therefore, substitute:Left side: (2(1 - cos B) + 2 cos B = 2 - 2 cos B + 2 cos B = 2). Hence, proven.13. (2 sin frac{A}{2} sin frac{C}{2} = sin frac{B}{2})Use the identity (2 sin frac{A}{2} sin frac{C}{2} = cos frac{A - C}{2} - cos frac{A + C}{2}).From equation 3: (cos frac{A - C}{2} = 2 cos frac{A + C}{2}).Therefore:(2 sin frac{A}{2} sin frac{C}{2} = 2 cos frac{A + C}{2} - cos frac{A + C}{2} = cos frac{A + C}{2}).But (A + C = pi - B), so (frac{A + C}{2} = frac{pi - B}{2}), thus:(cos frac{pi - B}{2} = sin frac{B}{2}).Hence, (2 sin frac{A}{2} sin frac{C}{2} = sin frac{B}{2}). Equation 13 is proven.14. (r = 2R sin^2 frac{B}{2})Recall that the inradius (r = frac{Delta}{s}) and circumradius (R = frac{abc}{4Delta}).From equation 8: (h_b = 3r). Also, (h_b = frac{2Delta}{b}). Therefore:(frac{2Delta}{b} = 3r) => (r = frac{2Delta}{3b}).Also, (R = frac{abc}{4Delta}).We need to express (r) in terms of R and (sin^2 frac{B}{2}).Express (Delta = frac{1}{2}ab sin C), but perhaps use other relations. Alternatively, use the formula (r = 4R sin frac{A}{2} sin frac{B}{2} sin frac{C}{2}). But I need to check if this formula is correct.Yes, in any triangle, (r = 4R sin frac{A}{2} sin frac{B}{2} sin frac{C}{2}).But from equation 6: (cot frac{A}{2} cot frac{C}{2} = 3). Let's see:First, express (r = 4R sin frac{A}{2} sin frac{B}{2} sin frac{C}{2}).We need to relate this to equation 14. Let's compute it:From equation 13: (2 sin frac{A}{2} sin frac{C}{2} = sin frac{B}{2}), so (sin frac{A}{2} sin frac{C}{2} = frac{1}{2} sin frac{B}{2}).Therefore:(r = 4R cdot frac{1}{2} sin frac{B}{2} cdot sin frac{B}{2} = 2R sin^2 frac{B}{2}). Hence, equation 14 is proven.15. (b = 2r cot frac{B}{2})From the formula in a triangle: (cot frac{B}{2} = frac{s - b}{r}). Hence, (s - b = r cot frac{B}{2}).But from equation 5, semiperimeter (s = frac{3b}{2}). Therefore:(s - b = frac{3b}{2} - b = frac{b}{2} = r cot frac{B}{2}).Thus, (b = 2 r cot frac{B}{2}). Hence, equation 15 is proven.16. (s = 3r cot frac{B}{2})From equation 15: (b = 2 r cot frac{B}{2}).From equation 1: (a + c = 2b), so perimeter is (a + b + c = 3b). Therefore, semiperimeter (s = frac{3b}{2}).Substitute (b = 2r cot frac{B}{2}) into (s = frac{3b}{2}):(s = frac{3 * 2r cot frac{B}{2}}{2} = 3r cot frac{B}{2}). Hence, equation 16 is proven.17. (Delta = 3r^2 cot frac{B}{2})Area (Delta = r * s). From equation 16: (s = 3r cot frac{B}{2}).Therefore, (Delta = r * 3r cot frac{B}{2} = 3r^2 cot frac{B}{2}). Hence, equation 17 is proven.18. (cos (A - C) = 3 - 4 cos B)From earlier in equation 3's proof, we derived this. Specifically, using equation 3 and squaring both sides, we arrived at (cos(A - C) = 3 - 4 cos B). Hence, equation 18 is proven.19. (2 cos A cos C = 3 - 5 cos B)Start from equation 18: (cos(A - C) = 3 - 4 cos B).Also, (cos(A - C) = cos A cos C + sin A sin C).From equation 2: (sin A + sin C = 2 sin B). Square both sides:(sin^2 A + 2 sin A sin C + sin^2 C = 4 sin^2 B).From identity (sin^2 A + sin^2 C = 1 - cos^2 A + 1 - cos^2 C = 2 - (cos^2 A + cos^2 C)). Not sure if helpful.Alternatively, use (sin^2 A + sin^2 C = 2 - cos^2 A - cos^2 C). Not sure.Alternatively, rearrange equation 18:(cos A cos C + sin A sin C = 3 - 4 cos B).From equation 21 (which we can use if proven, but let's check later), (sin A sin C = 3 sin^2 frac{B}{2}).But let's proceed step by step.From equation 18:(cos A cos C + sin A sin C = 3 - 4 cos B).From equation 21: (sin A sin C = 3 sin^2 frac{B}{2}).Substitute:(cos A cos C + 3 sin^2 frac{B}{2} = 3 - 4 cos B).But (3 sin^2 frac{B}{2} = 3 * frac{1 - cos B}{2} = frac{3}{2} - frac{3}{2} cos B).Therefore:(cos A cos C + frac{3}{2} - frac{3}{2} cos B = 3 - 4 cos B).Rearrange:(cos A cos C = 3 - 4 cos B - frac{3}{2} + frac{3}{2} cos B = frac{3}{2} - frac{5}{2} cos B).Multiply both sides by 2:(2 cos A cos C = 3 - 5 cos B). Hence, equation 19 is proven.20. (1 + cos A cos C = 5 sin^2 frac{B}{2})From equation 19: (2 cos A cos C = 3 - 5 cos B). Therefore, (cos A cos C = frac{3 - 5 cos B}{2}).Add 1 to both sides:(1 + cos A cos C = 1 + frac{3 - 5 cos B}{2} = frac{2 + 3 - 5 cos B}{2} = frac{5 - 5 cos B}{2} = frac{5(1 - cos B)}{2}).But (1 - cos B = 2 sin^2 frac{B}{2}), so:(1 + cos A cos C = frac{5 * 2 sin^2 frac{B}{2}}{2} = 5 sin^2 frac{B}{2}). Hence, equation 20 is proven.21. (sin A sin C = 3 sin^2 frac{B}{2})From equation 2: (sin A + sin C = 2 sin B).Square both sides:(sin^2 A + 2 sin A sin C + sin^2 C = 4 sin^2 B).From equation 18: (cos(A - C) = 3 - 4 cos B). Also, (cos(A - C) = cos A cos C + sin A sin C).From equation 19: (2 cos A cos C = 3 - 5 cos B). Hence, (cos A cos C = frac{3 - 5 cos B}{2}).Substitute into (cos(A - C)):(frac{3 - 5 cos B}{2} + sin A sin C = 3 - 4 cos B).Multiply all terms by 2 to eliminate denominator:(3 - 5 cos B + 2 sin A sin C = 6 - 8 cos B).Rearrange:(2 sin A sin C = 6 - 8 cos B - 3 + 5 cos B = 3 - 3 cos B).Divide by 2:(sin A sin C = frac{3}{2} - frac{3}{2} cos B = frac{3}{2}(1 - cos B) = 3 sin^2 frac{B}{2}).Hence, equation 21 is proven.22. (Delta = 6 R^2 sin B sin^2 frac{B}{2})From equation 14: (r = 2R sin^2 frac{B}{2}).From equation 17: (Delta = 3r^2 cot frac{B}{2}).Substitute r from equation 14:(Delta = 3(2R sin^2 frac{B}{2})^2 cot frac{B}{2} = 3 * 4 R^2 sin^4 frac{B}{2} * cot frac{B}{2}).Simplify:= 12 R^2 sin^4 frac{B}{2} * frac{cos frac{B}{2}}{sin frac{B}{2}} = 12 R^2 sin^3 frac{B}{2} cos frac{B}{2}).But (12 sin^3 frac{B}{2} cos frac{B}{2} = 6 * 2 sin frac{B}{2} cos frac{B}{2} * sin^2 frac{B}{2} = 6 sin B sin^2 frac{B}{2}) (since (2 sin frac{B}{2} cos frac{B}{2} = sin B)).Hence, (Delta = 6 R^2 sin B sin^2 frac{B}{2}). Equation 22 is proven.23. (left(tan frac{A}{2} + tan frac{C}{2}right) tan frac{B}{2} = frac{2}{3})From equation 5: (cot frac{A}{2} + cot frac{C}{2} = 2 cot frac{B}{2}). Let's take reciprocal of both sides:(frac{1}{cot frac{A}{2} + cot frac{C}{2}} = frac{1}{2 cot frac{B}{2}}).But (tan frac{A}{2} + tan frac{C}{2} = frac{1}{cot frac{A}{2}} + frac{1}{cot frac{C}{2}} = frac{cot frac{C}{2} + cot frac{A}{2}}{cot frac{A}{2} cot frac{C}{2}}).From equation 5: numerator is (2 cot frac{B}{2}), and from equation 6: denominator is 3. Therefore:(tan frac{A}{2} + tan frac{C}{2} = frac{2 cot frac{B}{2}}{3}).Therefore, (left(tan frac{A}{2} + tan frac{C}{2}right) tan frac{B}{2} = frac{2 cot frac{B}{2} tan frac{B}{2}}{3}).Since (cot theta tan theta = 1), this simplifies to (frac{2}{3}). Hence, equation 23 is proven.24. (a sin^2 frac{C}{2} + c sin^2 frac{A}{2} = frac{b}{2})Use the identity (sin^2 frac{theta}{2} = frac{1 - cos theta}{2}):Left side:(a cdot frac{1 - cos C}{2} + c cdot frac{1 - cos A}{2} = frac{a + c}{2} - frac{a cos C + c cos A}{2}).From equation 1: (a + c = 2b), so first term is (b). From the projection formula, (a cos C + c cos A = b). Therefore:= (b - frac{b}{2} = frac{b}{2}). Hence, equation 24 is proven.25. (cos A cot frac{A}{2} + cos C cot frac{C}{2} = 2 cos B cot frac{B}{2})Express each cotangent in terms of sine and cosine:(cot frac{A}{2} = frac{cos frac{A}{2}}{sin frac{A}{2}} = frac{1 + cos A}{sin A}) using the identity (cot frac{theta}{2} = frac{1 + cos theta}{sin theta}).Therefore:(cos A cot frac{A}{2} = cos A cdot frac{1 + cos A}{sin A}).Similarly for the other terms.Compute each term:Left side:(cos A cdot frac{1 + cos A}{sin A} + cos C cdot frac{1 + cos C}{sin C}).= (frac{cos A (1 + cos A)}{sin A} + frac{cos C (1 + cos C)}{sin C}).From the Law of Sines, (frac{a}{sin A} = frac{c}{sin C} = 2R). Let’s denote (k = 2R), so (sin A = frac{a}{k}), (sin C = frac{c}{k}).Therefore:= (frac{cos A (1 + cos A)}{a/k} cdot a + frac{cos C (1 + cos C)}{c/k} cdot c).Wait, perhaps a better approach. Let's multiply numerator and denominator by k:= (k left[ frac{cos A (1 + cos A)}{a} + frac{cos C (1 + cos C)}{c} right]).But since (k = 2R = frac{a}{sin A} = frac{c}{sin C}), so (a = k sin A), (c = k sin C). Substitute:= (k left[ frac{cos A (1 + cos A)}{k sin A} + frac{cos C (1 + cos C)}{k sin C} right]).Simplify k:= (left[ frac{cos A (1 + cos A)}{sin A} + frac{cos C (1 + cos C)}{sin C} right]).= (cos A cot frac{A}{2} + cos C cot frac{C}{2}).Wait, this just brings us back. Maybe another approach.Express (cos A cot frac{A}{2}):= (cos A cdot frac{cos frac{A}{2}}{sin frac{A}{2}}).= (cos A cdot frac{cos frac{A}{2}}{sin frac{A}{2}}).Using the identity (cos A = 1 - 2 sin^2 frac{A}{2}):= ((1 - 2 sin^2 frac{A}{2}) cdot frac{cos frac{A}{2}}{sin frac{A}{2}}).= (frac{cos frac{A}{2}}{sin frac{A}{2}} - 2 sin frac{A}{2} cos frac{A}{2}).= (cot frac{A}{2} - sin A).Similarly, (cos C cot frac{C}{2} = cot frac{C}{2} - sin C).Therefore, the left side becomes:(cot frac{A}{2} - sin A + cot frac{C}{2} - sin C).= (cot frac{A}{2} + cot frac{C}{2} - (sin A + sin C)).From equation 5: (cot frac{A}{2} + cot frac{C}{2} = 2 cot frac{B}{2}).From equation 2: (sin A + sin C = 2 sin B).Therefore:= (2 cot frac{B}{2} - 2 sin B).The right side is (2 cos B cot frac{B}{2}).Therefore, need to show:(2 cot frac{B}{2} - 2 sin B = 2 cos B cot frac{B}{2}).Divide both sides by 2:(cot frac{B}{2} - sin B = cos B cot frac{B}{2}).Rearrange:(cot frac{B}{2} - cos B cot frac{B}{2} = sin B).Factor out (cot frac{B}{2}):(cot frac{B}{2} (1 - cos B) = sin B).Express (cot frac{B}{2} = frac{cos frac{B}{2}}{sin frac{B}{2}}):(frac{cos frac{B}{2}}{sin frac{B}{2}} (1 - cos B) = sin B).Note that (1 - cos B = 2 sin^2 frac{B}{2}):Left side becomes:(frac{cos frac{B}{2}}{sin frac{B}{2}} cdot 2 sin^2 frac{B}{2} = 2 sin frac{B}{2} cos frac{B}{2} = sin B).Which equals the right side. Hence, equation 25 is proven.Final AnswerAll relationships (1) through (25) are proven as a consequence of the given condition in triangle ABC where a + c = 2b. Each boxed result corresponds to the respective equation number.boxed{1}, boxed{2}, boxed{3}, boxed{4}, boxed{5}, boxed{6}, boxed{7}, boxed{8}, boxed{9}, boxed{10}, boxed{11}, boxed{12}, boxed{13}, boxed{14}, boxed{15}, boxed{16}, boxed{17}, boxed{18}, boxed{19}, boxed{20}, boxed{21}, boxed{22}, boxed{23}, boxed{24}, boxed{25}

✅:Given a triangle ( triangle ABC ) with sides ( a, b, c ) opposite to angles ( A, B, C ) respectively, we need to prove 25 relationships. The key condition used is ( a + c = 2b ), from which all other relationships are derived using trigonometric identities, the Law of Sines, the Law of Cosines, and properties of the triangle's inradius ( r ), circumradius ( R ), area ( Delta ), semiperimeter ( s ), and altitude ( h_b ). Key Steps and Proofs:1. ( a + c = 2b ) - Given condition.2. ( sin A + sin C = 2 sin B ) - Using the Law of Sines: ( a = 2R sin A ), ( b = 2R sin B ), ( c = 2R sin C ), we get ( sin A + sin C = 2 sin B ).3. ( cos frac{A - C}{2} = 2 cos frac{A + C}{2} ) - Using sum-to-product identities and the fact ( A + C = pi - B ), we derive ( cos frac{A - C}{2} = 2 cos frac{A + C}{2} ).4. ( sin^2 frac{A}{2} + sin^2 frac{C}{2} = cos B ) - Using the identity ( sin^2 theta = frac{1 - cos 2theta}{2} ) and trigonometric identities, we show ( cos A + cos C = 4 sin^2 frac{B}{2 ).5. ( cot frac{A}{2} + cot frac{C}{2} = 2 cot frac{B}{2} ) - Using the formula for cotangent in terms of semiperimeter and inradius, we derive the result.6. ( cot frac{A}{2} cot frac{C}{2} = 3 ) - Using trigonometric identities and previous results, we show the product equals 3.7. ( a cos^2 frac{C}{2} + c cos^2 frac{A}{2} = frac{3b}{2} ) - Using projection formula and trigonometric identities.8. ( h_b = 3r ) - Relating area expressions using altitude and inradius.9. ( cos A + cos C = 4 sin^2 frac{B}{2} ) - Derived from previous results and trigonometric identities.10. Quadratic equation with equal roots - Calculating discriminant and showing it equals zero.11. ( sin^2 frac{A}{2} + sin^2 frac{B}{2} + sin^2 frac{C}{2} = cos^2 frac{B}{2} ) - Using trigonometric identities and previous results.12. ( cos A + 2 cos B + cos C = 2 ) - Derived from previous results.13. ( 2 sin frac{A}{2} sin frac{C}{2} = sin frac{B}{2} ) - Using product-to-sum identities and previous results.14. ( r = 2R sin^2 frac{B}{2} ) - Relating inradius and circumradius using trigonometric identities.15. ( b = 2r cot frac{B}{2} ) - Using semiperimeter and inradius relationships.16. ( s = 3r cot frac{B}{2} ) - Relating semiperimeter and inradius.17. ( Delta = 3r^2 cot frac{B}{2} ) - Relating area, inradius, and semiperimeter.18. ( cos (A - C) = 3 - 4 cos B ) - Derived from squaring previous results.19. ( 2 cos A cos C = 3 - 5 cos B ) - Using trigonometric identities and previous results.20. ( 1 + cos A cos C = 5 sin^2 frac{B}{2} ) - Derived from previous results.21. ( sin A sin C = 3 sin^2 frac{B}{2} ) - Using trigonometric identities and previous results.22. ( Delta = 6 R^2 sin B sin^2 frac{B}{2} ) - Relating area, circumradius, and trigonometric identities.23. ( left(tan frac{A}{2} + tan frac{C}{2}right) tan frac{B}{2} = frac{2}{3} ) - Using reciprocal relationships and previous results.24. ( a sin^2 frac{C}{2} + c sin^2 frac{A}{2} = frac{b}{2} ) - Using projection formula and trigonometric identities.25. ( cos A cot frac{A}{2} + cos C cot frac{C}{2} = 2 cos B cot frac{B}{2} ) - Using trigonometric identities and previous results. Final AnswerAll relationships (1) through (25) are proven as a consequence of the given condition in ( triangle ABC ) where ( a + c = 2b ). Each boxed result corresponds to the respective equation number.[boxed{1}, boxed{2}, boxed{3}, boxed{4}, boxed{5}, boxed{6}, boxed{7}, boxed{8}, boxed{9}, boxed{10}, boxed{11}, boxed{12}, boxed{13}, boxed{14}, boxed{15}, boxed{16}, boxed{17}, boxed{18}, boxed{19}, boxed{20}, boxed{21}, boxed{22}, boxed{23}, boxed{24}, boxed{25}]

🔑:Given we need to prove the equivalences among a set of conditions (1) to (25).1. Given: (a+c = 2b) - Using the Law of Sines: [ a = 2R sin A, quad b = 2R sin B, quad c = 2R sin C ] - Therefore, (2R sin A + 2R sin C = 2 cdot 2R sin B) - Simplifying, we get: [ sin A + sin C = 2 sin B quad text{(2)} ]2. Prove the transition from (2) to (3): (sin A + sin C = 2 sin B Leftrightarrow cos frac{A-C}{2} = 2 cos frac{A+C}{2}) - Using the sum-to-product identities: [ sin A + sin C = 2 sin frac{A+C}{2} cos frac{A-C}{2} ] - Given (sin A + sin C = 2 sin B): [ 2 sin frac{A+C}{2} cos frac{A-C}{2} = 2 cdot 2 sin frac{A+C}{2} cos frac{A+C}{2} ] - Simplifying ( sin frac{A+C}{2} ne 0 ): [ cos frac{A-C}{2} = 2 cos frac{A+C}{2} quad text{(3)} ]3. Prove the transition from (3) to (4): - Given (cos frac{A-C}{2} = 2 cos frac{A+C}{2}): - Squaring both sides: [ cos^2 frac{A-C}{2} = 4 cos^2 frac{A+C}{2} ] - Using the half-angle identity: [ cos^2 x = frac{1 + cos 2x}{2} ] - Applying for (frac{A-C}{2}) and (frac{A+C}{2}): [ frac{1 + cos(A - C)}{2} = 4 cdot frac{1 + cos(A + C)}{2} ] [ 1 + cos(A - C) = 2(1 + cos(A + C)) ] [ 1 + cos(A - C) = 2 + 2 cos(A + C) ] [ cos(A - C) = 1 + 2 cos(A + C) ] - Verifying against (4): [ sin^2 frac{A}{2} + sin^2 frac{C}{2} = frac{1 - cos A}{2} + frac{1 - cos C}{2} ] [ = 1 - frac{cos A + cos C}{2} ] [ = 1 - 2 cos^2 frac{A+C}{2} quad (text{since} cos x = 2 cos^2 frac{x}{2} - 1) ] [ = cos B quad text{(4)} ]From these steps, the remaining conditions can be proved in a similar transition manner for each pair as stated in the problem.Conclusion:[boxed{text{These step-by-step equivalences complete the proof that each condition correctly implies the subsequent one and vice versa.}}]

🧠:Okay, so I need to figure out how to insert operation symbols and maybe parentheses into the expression 1/4 1/4 1/4 1/4 so that the result is 2. Hmm, let's start by understanding the problem. The original expression is four fractions of 1/4 next to each other, and I can add operations like +, -, *, /, and parentheses between them. The goal is to make the whole thing evaluate to 2. First, let me write down the original expression as given: (1/4)(1/4)(1/4)(1/4). Wait, but without any operations, this is just multiplying all four 1/4s together, which would be (1/4)^4 = 1/256. That's way too small. So clearly, I need to change the operations to get a much larger result, specifically 2. Since the original multiplication gives a very small number, maybe I need to use division somewhere because dividing by a small number can give a larger result. Also, adding or subtracting could help, but considering that all the numbers are fractions, adding them would just give a sum less than 1. Let me think step by step.Let me list out the four 1/4 terms: 1/4, 1/4, 1/4, 1/4. I need to insert operations between each pair. Let's denote the operations between them as op1, op2, op3. So the expression would be ((1/4) op1 (1/4)) op2 (1/4) op3 (1/4). But depending on where parentheses are placed, the order of operations can change. Maybe I need to consider different ways of grouping them.Alternatively, since all the numbers are the same, maybe there's a clever combination. Let me try some operations:First, let's try division. If I divide 1/4 by 1/4, that equals 1. For example, (1/4) / (1/4) = 1. If I can create such divisions, maybe I can get numbers that add up to 2. For example, if I have two divisions each giving 1, and then add them together. But wait, there are four 1/4s. So maybe (1/4)/(1/4) + (1/4)/(1/4) = 1 + 1 = 2. But does that work with the four terms? Let's see:If we write ((1/4) / (1/4)) + ((1/4) / (1/4)), that would use all four 1/4s. Each division is between two 1/4s, and then the two results are added. That would indeed give 1 + 1 = 2. Let me check that. First pair: (1/4)/(1/4) = 1. Second pair: (1/4)/(1/4) = 1. Then 1 + 1 = 2. Yes, that works. So the expression would be (1/4 ÷ 1/4) + (1/4 ÷ 1/4). So inserting division symbols between the first and second, and third and fourth terms, and then a plus between the two results. But how does that look in terms of the original expression without parentheses? The original expression is written as 1/4 1/4 1/4 1/4. So inserting operations:First, between the first and second 1/4: ÷Then, between the second and third 1/4: +Then between the third and fourth 1/4: ÷Wait, but that would be 1/4 ÷ 1/4 + 1/4 ÷ 1/4. Let me parse that. According to order of operations, division and multiplication have higher precedence than addition. So each division would be done first, then the addition. So (1/4 ÷ 1/4) is 1, then (1/4 ÷ 1/4) is 1, so 1 + 1 = 2. So the expression would be:1/4 ÷ 1/4 + 1/4 ÷ 1/4But written without parentheses, is that the case? Let's check. If you have 1/4 ÷ 1/4 + 1/4 ÷ 1/4, according to operator precedence, divisions are done left to right before additions. So first, 1/4 ÷ 1/4 is 1, then the next division is 1/4 ÷ 1/4 is 1, then adding them gives 2. So parentheses aren't strictly necessary here because of operator precedence, but to make it clearer, we might add them. However, the problem allows inserting parentheses if necessary. So maybe the minimal answer is just inserting ÷ and + in the right places.So inserting ÷ between the first and second, then + between the second and third, then ÷ between the third and fourth. So:1/4 ÷ 1/4 + 1/4 ÷ 1/4Which would evaluate to 2. Let's verify:First division: (1/4) / (1/4) = 1Second division: (1/4) / (1/4) = 1Adding them: 1 + 1 = 2. Perfect.Alternatively, is there another way? Let's explore other possibilities in case I missed something.Suppose we try multiplication and division. For example, (1/4 * 1/4) / (1/4 / 1/4). Let's compute that:First, numerator: (1/4 * 1/4) = 1/16Denominator: (1/4 / 1/4) = 1So total: (1/16) / 1 = 1/16. Not 2. Not helpful.Another idea: 1/4 + 1/4 + 1/4 + 1/4 = 1. Still too small.How about combining operations. For example, 1/4 / (1/4 / (1/4 / 1/4)). Let's compute innermost first:Innermost: 1/4 / 1/4 = 1Then next layer: 1/4 / 1 = 1/4Then outer layer: 1/4 / (1/4) = 1. So total is 1. Not 2.Alternatively, maybe using subtraction: 1/4 - 1/4 is 0, which isn't helpful. Multiplication would decrease the value. So subtraction and multiplication probably not useful here. So division and addition seem the most promising.Another approach: think of 2 as a target and how to reach it with four 1/4s. Since 1/4 is 0.25, maybe 0.25 * 0.25 is 0.0625, but 0.25 / 0.25 is 1. So combining divisions and additions.Wait, the earlier solution seems valid. Let me check again:(1/4 ÷ 1/4) + (1/4 ÷ 1/4) = 1 + 1 = 2. That uses all four 1/4s. The operations are two divisions and one addition. Parentheses are used to group the divisions first. But in the original problem, the expression is written as four 1/4s with operations inserted. So inserting ÷, +, ÷ between them would give the expression 1/4 ÷ 1/4 + 1/4 ÷ 1/4. Since division has higher precedence than addition, the divisions are performed first even without parentheses. So the expression is evaluated as (1/4 ÷ 1/4) + (1/4 ÷ 1/4) = 1 + 1 = 2. Therefore, the solution is to insert ÷ between the first and second, + between the second and third, and ÷ between the third and fourth 1/4s.Is there another possible solution? Let's see. Maybe using a combination of operations with different parentheses. For example:( (1/4 + 1/4) / (1/4) ) / 1/4. Let's compute:First, 1/4 + 1/4 = 1/2Then divide by 1/4: (1/2) / (1/4) = 2Then divide by 1/4: 2 / (1/4) = 8. Not 2. So that's too much.Alternatively, (1/4 + 1/4) / (1/4 * 1/4). Let's compute:Numerator: 1/4 + 1/4 = 1/2Denominator: 1/4 * 1/4 = 1/16So total: (1/2) / (1/16) = 8. Again 8, not 2.Another attempt: ((1/4) / (1/4) / (1/4)) * (1/4). Let's compute:First division: (1/4) / (1/4) = 1Then next division: 1 / (1/4) = 4Then multiply by 1/4: 4 * 1/4 = 1. Not 2.Alternatively, ((1/4) / (1/4)) * ((1/4) / (1/4)) = 1 * 1 = 1. Still 1.Wait, maybe using subtraction and division. For example:(1/4 + 1/4 + 1/4) / (1/4). Compute:Numerator: 1/4 + 1/4 + 1/4 = 3/4Denominator: 1/4So (3/4) / (1/4) = 3. Not 2.Alternatively, (1/4 + 1/4) / (1/4 - 1/4). But denominator is zero, undefined.Hmm. Maybe combining division and multiplication. Let's see:(1/4) / (1/4) * (1/4) / (1/4). That would be 1 * 1 = 1.Alternatively, 1/4 / (1/4 / 1/4) * 1/4. Let's compute:Inside the denominator first: 1/4 / 1/4 = 1Then 1/4 / 1 = 1/4Then 1/4 * 1/4 = 1/16. No.Alternatively, ((1/4) * (1/4)) / ((1/4) / (1/4)). So numerator is 1/16, denominator is 1, so 1/16. Not helpful.Another angle: What if we use exponents? But the problem says insert operation symbols, which are typically +, -, *, /, and parentheses. Probably not exponents unless specified. Let me check the original problem: "insert operation symbols and, if necessary, parentheses". Operation symbols usually refer to basic arithmetic operations. So exponents might not be allowed unless mentioned. So maybe sticking to +, -, *, /.Wait, but another thought: Could we write something like 1/(4 - 4) + 1/(4 - 4)? But that would involve changing the 1/4s into different numbers, which isn't allowed. The problem states to insert operation symbols between the existing terms, so we can't modify the 1/4s themselves, just insert operations between them.So the original terms are four separate 1/4s. So each 1/4 is a separate entity. Therefore, we can't combine the denominators or anything like that. So each 1/4 is as it is, and we need to put operations between them.So the first solution still seems valid: (1/4 ÷ 1/4) + (1/4 ÷ 1/4) = 2. Let me confirm once again:1/4 ÷ 1/4 = (1/4) / (1/4) = 1. Then another 1/4 ÷ 1/4 = 1. Adding them gives 2. Yes.So the expression is 1/4 ÷ 1/4 + 1/4 ÷ 1/4. Since we need to insert operations between the four 1/4s, the operations would be ÷, +, ÷. So between the first and second 1/4: ÷, between second and third: +, between third and fourth: ÷. So the expression is structured as:(1/4 ÷ 1/4) + (1/4 ÷ 1/4)But written without parentheses (relying on order of operations), it would still be the same because division precedes addition. So the divisions are done first, then the addition. Hence, the expression evaluates to 2.Therefore, the answer is inserting ÷, +, and ÷ between the four 1/4s. Let me write that properly:The original expression is:1/4 1/4 1/4 1/4Inserting operations:1/4 ÷ 1/4 + 1/4 ÷ 1/4Which is equal to 2.Alternatively, with parentheses for clarity:(1/4 ÷ 1/4) + (1/4 ÷ 1/4) = 2But even without parentheses, due to operator precedence, it would still evaluate correctly.Is there any other possible solution? Let me think again.Suppose we do something like 1/4 / (1/4 + 1/4) / 1/4. Let's compute:First, 1/4 + 1/4 = 1/2Then 1/4 / (1/2) = 1/4 * 2 = 1/2Then 1/2 / 1/4 = 2. Wait, this gives 2, but let's check the expression:The expression would be 1/4 ÷ (1/4 + 1/4) ÷ 1/4. Let's parse this:First, compute the addition in the parentheses: 1/4 + 1/4 = 1/2.Then the expression becomes 1/4 ÷ 1/2 ÷ 1/4. Division is left associative, so:(1/4 ÷ 1/2) ÷ 1/4 = (0.25 ÷ 0.5) ÷ 0.25 = (0.5) ÷ 0.25 = 2.So that's another way. Let's check how many 1/4s are used. The expression is:1/4 ÷ (1/4 + 1/4) ÷ 1/4. So that uses four 1/4s: the first one, then the next two are added in the denominator, and the last one. Wait, no, actually, the original expression is four 1/4s. Let me count:1/4 [1st] ÷ (1/4 [2nd] + 1/4 [3rd]) ÷ 1/4 [4th]. Yes, that uses all four 1/4s. So this is another valid solution. Let's verify:1/4 ÷ (1/4 + 1/4) ÷ 1/4 = 1/4 ÷ (1/2) ÷ 1/4 = (1/4 ÷ 1/2) ÷ 1/4 = (0.5) ÷ 1/4 = 0.5 / 0.25 = 2. Correct.So this is another possible solution. So inserting ÷ between first and second, + between second and third, and ÷ between third and fourth, but with parentheses around the second and third terms. So the expression would be 1/4 ÷ (1/4 + 1/4) ÷ 1/4.But does this use all four 1/4s? Yes:First term: 1/4Operation1: ÷Second term: (1/4 + 1/4) [which uses the second and third 1/4s]Operation2: ÷Third term: 1/4 [fourth term]Wait, but in this case, the second and third 1/4s are grouped together with a +, and then divided by the fourth. So the structure is:1/4 ÷ (1/4 + 1/4) ÷ 1/4. That uses all four 1/4s. So this is a valid solution as well. So there are at least two different solutions.Interesting. So depending on where parentheses are placed, different expressions can yield 2.So first solution: (1/4 ÷ 1/4) + (1/4 ÷ 1/4) = 2Second solution: 1/4 ÷ (1/4 + 1/4) ÷ 1/4 = 2Are both correct? Let me verify the second one again:Step-by-step:1/4 ÷ (1/4 + 1/4) ÷ 1/4First, compute the denominator in the first division: 1/4 + 1/4 = 1/2So the expression becomes 1/4 ÷ 1/2 ÷ 1/4Now, division is left associative, so perform the first division: 1/4 ÷ 1/2 = (1/4) * (2/1) = 1/2Then divide that result by 1/4: 1/2 ÷ 1/4 = (1/2) * (4/1) = 2Yes, correct. So this is another valid expression. Therefore, there are at least two ways to insert operations and parentheses to get 2.But the problem says "insert operation symbols and, if necessary, parentheses". So perhaps both solutions are acceptable.But the first solution doesn't require parentheses because of operator precedence, while the second one does require parentheses to group the addition first. So maybe both are valid, but the problem might expect one of them. However, since the problem allows inserting parentheses if necessary, both are correct.But let's check if there are more solutions. For example:Another possibility: ((1/4 ÷ 1/4) + 1/4) ÷ 1/4. Let's compute:First, 1/4 ÷ 1/4 = 1Then add 1/4: 1 + 1/4 = 5/4Then divide by 1/4: (5/4) ÷ (1/4) = 5. Not 2.Alternatively, (1/4 + 1/4 + 1/4) ÷ 1/4. That would be (3/4) ÷ 1/4 = 3. Not 2.Another attempt: (1/4 * 1/4) + (1/4 ÷ 1/4). Let's compute:First multiplication: 1/16Second division: 1Sum: 1/16 + 1 = 17/16 ≈ 1.0625. Not 2.Hmm. How about ((1/4 ÷ 1/4) ÷ 1/4) - 1/4. Let's compute:First division: 1/4 ÷ 1/4 = 1Then divide by 1/4: 1 ÷ 1/4 = 4Subtract 1/4: 4 - 1/4 = 15/4 = 3.75. Not 2.Alternatively, 1/4 ÷ (1/4 ÷ (1/4 ÷ 1/4)). Let's compute innermost first:1/4 ÷ 1/4 = 1Then next layer: 1/4 ÷ 1 = 1/4Then outer layer: 1/4 ÷ (1/4) = 1. So total 1. Not 2.Another idea: 1/4 / 1/4 / 1/4 / 1/4. That's ((1/4 / 1/4) / 1/4) / 1/4 = (1 / 1/4) / 1/4 = 4 / 1/4 = 16. Too big.Alternatively, 1/4 * 1/4 / 1/4 / 1/4 = (1/16 / 1/4) / 1/4 = (1/4) / 1/4 = 1.Not helpful.Alternatively, combining different operations. For example:1/4 + 1/4 + 1/4 + 1/4 = 1. Not 2.1/4 - 1/4 + 1/4 - 1/4 = 0. Not helpful.So far, the two valid solutions are:1. (1/4 ÷ 1/4) + (1/4 ÷ 1/4) = 22. 1/4 ÷ (1/4 + 1/4) ÷ 1/4 = 2Therefore, both are correct. The problem might accept either one. However, let's check if the problem allows using parentheses. Since it says "if necessary", and in the first solution, parentheses aren't strictly necessary because division precedes addition, but they might be added for clarity. However, written without any parentheses, the first expression would still work due to order of operations, while the second requires parentheses to group the addition first.Therefore, both are valid answers, but perhaps the first one is more straightforward.Alternatively, the problem might have only one intended solution. Let's see.Another approach: Maybe using multiplication and division together.For example: (1/4 ÷ 1/4) * (1/4 ÷ 1/4) = 1 * 1 = 1. Not 2.Alternatively, (1/4 ÷ 1/4) + 1/4 * 1/4 = 1 + 1/16 = 17/16. No.Alternatively, 1/4 ÷ 1/4 + 1/4 + 1/4 = 1 + 0.5 = 1.5. Not 2.Another idea: Maybe combining three operations. For example:(1/4 ÷ (1/4 + 1/4)) * 1/4. Wait, that would be:1/4 ÷ (1/2) = 0.5Then 0.5 * 1/4 = 0.125. Not 2.Alternatively, ((1/4 + 1/4) ÷ 1/4) - 1/4. Compute:(1/2 ÷ 1/4) = 22 - 1/4 = 1.75. Not 2.Hmm. So I think the only two valid solutions are the two I found earlier.But let's verify once again:First solution:1/4 ÷ 1/4 + 1/4 ÷ 1/4As per order of operations, divisions first:1 + 1 = 2. Correct.Second solution:1/4 ÷ (1/4 + 1/4) ÷ 1/4First compute the addition inside parentheses:1/4 ÷ (1/2) ÷ 1/4 = (1/4 ÷ 1/2) ÷ 1/4 = (0.5) ÷ 0.25 = 2. Correct.So both are valid. Therefore, the problem has two solutions. However, the problem might expect one of them. Let me check if there are any more constraints. The original problem says "insert operation symbols and, if necessary, parentheses". So possibly both are acceptable. But let's check if there's a way to use subtraction or multiplication in a non-obvious way.Alternatively, is there a way to use mixed operations with multiplication and division? For example:1/4 ÷ 1/4 × 1/4 ÷ 1/4. That would be 1 × 1 = 1. Not helpful.Alternatively, 1/4 × 1/4 ÷ 1/4 ÷ 1/4. Compute:First, 1/4 × 1/4 = 1/16Then 1/16 ÷ 1/4 = 1/4Then 1/4 ÷ 1/4 = 1. Not 2.Hmm. Another angle: If we can use parentheses to change the order of operations beyond just grouping additions or divisions. But I think we've covered the main possibilities.So in conclusion, there are at least two valid ways to insert operations and parentheses to make the expression equal 2. Both involve division and addition in different configurations. The first solution uses two divisions and an addition, while the second uses a combination of division and addition within parentheses. Both are correct. However, since the problem might expect one answer, perhaps the first one is more straightforward without parentheses.But let me check the original problem statement again: it allows inserting parentheses if necessary, so both solutions are acceptable. However, the problem may expect the answer with minimal parentheses. The first solution can be written without parentheses due to operator precedence, while the second requires parentheses around the addition. Therefore, both are correct, but depending on the problem's expectations, either could be acceptable.Alternatively, maybe the intended solution is the first one because it's more direct: two divisions and an addition. The second one is a bit more complex with nested divisions and an addition. But both are mathematically correct.Given that, I should present both solutions. However, the problem says "the value of the resulting expression is equal to 2", so any correct answer is acceptable. Therefore, the solutions are:1. (1/4 ÷ 1/4) + (1/4 ÷ 1/4) = 22. 1/4 ÷ (1/4 + 1/4) ÷ 1/4 = 2But perhaps the problem expects one of them, maybe the first. Alternatively, the problem might have more solutions. Let me see if I can find another one.Wait, another thought: What if we use subtraction and division. For example:(1/4 ÷ 1/4) - (1/4 ÷ 1/4) = 0. Not helpful.Alternatively, 1/4 ÷ 1/4 ÷ (1/4 ÷ 1/4). That's 1 ÷ 1 = 1. No.Alternatively, 1/4 + 1/4 ÷ 1/4 - 1/4. Compute:First division: 1/4 ÷ 1/4 = 1Then the expression becomes 1/4 + 1 - 1/4 = (1/4 - 1/4) + 1 = 0 + 1 = 1. Not 2.Alternatively, 1/4 + 1/4 + 1/4 ÷ 1/4. Compute division first: 1/4 ÷ 1/4 = 1Then add: 1/4 + 1/4 + 1 = 0.5 + 1 = 1.5. Not 2.Another idea: (1/4 + 1/4) ÷ (1/4 ÷ 1/4). Compute:Numerator: 0.5Denominator: 1Result: 0.5. Not 2.Hmm. It seems challenging to find another solution. Therefore, I think the two solutions I found are the only ones. Thus, I'll present both as possible answers, but check if one is more likely expected.Given the problem's structure, perhaps the first solution is more straightforward, using two divisions and an addition without needing parentheses (due to order of operations). The second solution requires parentheses to group the addition first. Depending on the problem's preference, both are correct, but the first may be the primary solution.Therefore, the final answer is either:[left( frac{1}{4} div frac{1}{4} right) + left( frac{1}{4} div frac{1}{4} right) = 2]or[frac{1}{4} div left( frac{1}{4} + frac{1}{4} right) div frac{1}{4} = 2]Both are correct. To confirm once again:For the first expression:[left( frac{1}{4} div frac{1}{4} right) + left( frac{1}{4} div frac{1}{4} right) = 1 + 1 = 2]For the second expression:[frac{1}{4} div left( frac{1}{4} + frac{1}{4} right) div frac{1}{4} = frac{1}{4} div frac{1}{2} div frac{1}{4} = left( frac{1}{4} div frac{1}{2} right) div frac{1}{4} = frac{1}{2} div frac{1}{4} = 2]Yes, both work. Therefore, the problem has two valid solutions. However, the problem might expect one, so I should check if there are any others. After thorough checking, I can't find any other solutions, so these two are the correct ones.

✅:To solve the problem of inserting operation symbols and parentheses into the expression (frac{1}{4} frac{1}{4} frac{1}{4} frac{1}{4}) to make it equal to 2, we explore different combinations of operations and groupings. Here are the two valid solutions: Solution 1: Using Division and AdditionInsert division ((div)) between the first and second (frac{1}{4}), addition ((+)) between the second and third (frac{1}{4}), and division ((div)) between the third and fourth (frac{1}{4}). This leverages the precedence of division over addition:[left( frac{1}{4} div frac{1}{4} right) + left( frac{1}{4} div frac{1}{4} right)]Calculation:[left( frac{1}{4} div frac{1}{4} right) + left( frac{1}{4} div frac{1}{4} right) = 1 + 1 = 2] Solution 2: Using Parentheses to Group AdditionInsert division ((div)) between the first and second (frac{1}{4}), parentheses around the second and third (frac{1}{4}) with addition ((+)), and division ((div)) between the third and fourth (frac{1}{4}):[frac{1}{4} div left( frac{1}{4} + frac{1}{4} right) div frac{1}{4}]Calculation:1. Compute the grouped addition first:[frac{1}{4} + frac{1}{4} = frac{1}{2}]2. Perform the divisions:[frac{1}{4} div frac{1}{2} div frac{1}{4} = left( frac{1}{4} div frac{1}{2} right) div frac{1}{4} = frac{1}{2} div frac{1}{4} = 2] Final AnswerBoth expressions evaluate to 2. The solutions are:[boxed{left( frac{1}{4} div frac{1}{4} right) + left( frac{1}{4} div frac{1}{4} right)}]or[boxed{frac{1}{4} div left( frac{1}{4} + frac{1}{4} right) div frac{1}{4}}]

🔑:```markdownGiven the expression 1/4, 1/4, 1/4, 1/4, we need to arrange them with appropriate arithmetic operations (and brackets if necessary) so that the resulting value equals 2.To find such an arrangement, consider the following steps:1. Step 1: Identify possible operations We have addition (+), subtraction (-), multiplication (*), and division (/). Given that we are trying to achieve a value of 2, division could be a useful operation.2. Step 2: Experiment with division We start by considering the operation that might simplify our fractions most effectively. Let's try: [ frac{1}{4} div frac{1}{4} ] Simplify this expression: [ frac{frac{1}{4}}{frac{1}{4}} = 1 ] This gives 1, and we need to get 2 eventually. We can use addition next.3. Step 3: Insert addition Let's place an addition operation: [ 1 + 1 = 2 ] We need to form a 1 using remaining terms: Consider again: [ frac{1}{4} div frac{1}{4} = 1 ] Now combining all parts together: [ ( frac{1}{4} div frac{1}{4} ) + ( frac{1}{4} div frac{1}{4} ) = 1 + 1 = 2 ]This process can be confirmed by several valid orderings:- Another valid arrangement:[( frac{1}{4} div ( frac{1}{4} + frac{1}{4} )) div frac{1}{4} = 2 ]Thus, multiple solutions like the above ones satisfy the conditions. Conclusion:(boxed{ frac{1}{4} div frac{1}{4} + frac{1}{4} div frac{1}{4} = 2 }) or (boxed{ frac{1}{4} div ( frac{1}{4} + frac{1}{4} ) div frac{1}{4} = 2 })```

🧠:Okay, let me try to figure out this problem. So the question is asking which of the given numbers divide exactly into 15 plus the square root of 49. The options are 3, 4, 5, 7, and 11. Hmm, let me break this down step by step.First, I need to compute the value of (15 + sqrt{49}). The square root of 49 is straightforward because 7 times 7 is 49. So (sqrt{49} = 7). That simplifies the expression to (15 + 7), which is 22. Wait, so the problem is essentially asking which of these numbers are divisors of 22?But hold on, let me confirm that. The original expression is (15 + sqrt{49}). Since (sqrt{49}) is 7, adding that to 15 gives 22. So yes, the number we're looking at is 22. Now, we need to check which of the options divide 22 exactly, meaning without leaving a remainder.Let me list the options again: (A) 3, (B) 4, (C) 5, (D) 7, (E) 11. Let's check each one.Starting with option A, which is 3. Dividing 22 by 3 gives approximately 7.333... That's not a whole number, so 3 doesn't divide 22 exactly. So A is out.Next, option B is 4. Dividing 22 by 4 is 5.5. Again, that's not an integer, so 4 isn't a divisor of 22. So B is not correct.Option C is 5. Dividing 22 by 5 gives 4.4, which is also not an integer. So C isn't a divisor either.Option D is 7. Dividing 22 by 7 gives approximately 3.1428... Hmm, that's about 3.14, which is pi, but here it's just a decimal. So 7 doesn't divide 22 exactly. Wait, but 7 times 3 is 21, and 22 minus 21 is 1, so there's a remainder of 1. So D is also not a divisor.Finally, option E is 11. Dividing 22 by 11 gives exactly 2. That's a whole number, so 11 divides 22 exactly. Therefore, E is the correct answer.Wait, but let me double-check my calculations to make sure I didn't make a mistake. So 15 + sqrt(49) is 15 + 7, which is indeed 22. Then checking each divisor:- 22 ÷ 3 = 7.333... Not an integer.- 22 ÷ 4 = 5.5. Not an integer.- 22 ÷ 5 = 4.4. Not an integer.- 22 ÷ 7 ≈ 3.1428. Not an integer.- 22 ÷ 11 = 2. Exactly, no remainder.So yes, that seems correct. Therefore, the answer should be E) 11.But just to be thorough, maybe I should consider if the question is trying to trick me in any way. The original expression is (15 + sqrt{49}). Sometimes, square roots can have both positive and negative values. Wait, the principal square root is always non-negative, so sqrt(49) is 7, not -7. So even if there were a negative root, the principal one is positive. So that part is fine. The expression is definitely 22.Another possible point of confusion could be if "divide exactly into" is interpreted differently. But in mathematics, if a number divides exactly into another, it means it's a divisor of that number, so there's no remainder. So 11 divides 22 exactly because 22 divided by 11 is 2 with no remainder. All the other options leave a remainder, so they don't divide exactly.Therefore, I think my reasoning is solid here. The answer is E) 11.

✅:To determine which numbers divide exactly into (15 + sqrt{49}):1. Simplify the expression: [ 15 + sqrt{49} = 15 + 7 = 22 ]2. Check divisibility of 22 by each option: - (A) 3: (22 div 3 approx 7.33) (not an integer). - (B) 4: (22 div 4 = 5.5) (not an integer). - (C) 5: (22 div 5 = 4.4) (not an integer). - (D) 7: (22 div 7 approx 3.14) (not an integer). - (E) 11: (22 div 11 = 2) (exact division, no remainder). Conclusion: The only number that divides 22 exactly is 11.Answer: (boxed{E})

🔑:To determine which of the given numbers divide exactly into ( 15 + sqrt{49} ), we will first simplify the expression.1. Simplify ( 15 + sqrt{49} ): [ 15 + sqrt{49} ] We know that ( sqrt{49} = 7 ), so we can substitute: [ 15 + 7 = 22 ]2. Evaluate each option to see which number divides ( 22 ) exactly: - Option (A) 3: [ frac{22}{3} = 7.3333 ldots quad Rightarrow quad 3 text{ does not divide 22 evenly} ] - Option (B) 4: [ frac{22}{4} = 5.5 quad Rightarrow quad 4 text{ does not divide 22 evenly} ] - Option (C) 5: [ frac{22}{5} = 4.4 quad Rightarrow quad 5 text{ does not divide 22 evenly} ] - Option (D) 7: [ frac{22}{7} approx 3.142857 quad Rightarrow quad 7 text{ does not divide 22 evenly} ] - Option (E) 11: [ frac{22}{11} = 2 quad Rightarrow quad 11 text{ divides 22 exactly} ]3. Conclusion: Among the options provided, the only integer that divides ( 22 ) exactly is ( 11 ).[boxed{E}]