🧠:Okay, let's try to figure out this Arbor Day sapling problem. Hmm, so we have two people, A and B, planting saplings. The problem says that person A takes 1/3 more time to plant them alone than person B. Then, when they work together, B plants 36 more saplings than A by the time they finish. We need to find the total number of saplings. Alright, let me break this down step by step.First, let's parse the time difference. If B takes time t to plant all the saplings alone, then A takes t + (1/3)t = (4/3)t time to do the same job alone. So, A is slower than B. That makes sense. Their rates of planting must be different. Maybe I should consider their rates per unit time.Let’s denote the total number of saplings as N. Since they’re planting the same total number when working alone, the rates can be calculated based on the time they take. Let’s see.If B can plant N saplings in time t, then B's planting rate is N/t saplings per unit time. Similarly, A's rate would be N/(4/3)t = (3N)/(4t) saplings per unit time. So, A's rate is 3/4 of B's rate. That might be useful.Now, when they work together, their combined rate should be the sum of their individual rates. So, combined rate = N/t + 3N/(4t) = (4N + 3N)/(4t) = 7N/(4t) saplings per unit time. Wait, but when they work together, they are planting the same total N saplings, so the time taken together would be N divided by their combined rate. Let me check that.Time taken when working together, let's call it T, would be T = N / (7N/(4t)) = (4t)/7. So, together, they finish in 4t/7 time. Hmm, okay. Now, during this time T, how many saplings does each person plant?Person A's rate is 3N/(4t), so in time T, A plants (3N/(4t)) * (4t/7) = (3N/7). Similarly, person B's rate is N/t, so in time T, B plants (N/t) * (4t/7) = 4N/7. The problem states that B plants 36 more saplings than A. So, 4N/7 - 3N/7 = N/7 = 36. Therefore, N = 36 * 7 = 252. Is that right?Wait, let me verify. If N is 252, then B plants 4/7 *252 = 144 saplings, and A plants 3/7 *252 = 108. The difference is 144 - 108 = 36. That matches. So the total saplings are 252. That seems straightforward, but let me think again to make sure I didn't skip any steps.Wait, another way to approach this is by considering their rates when working together. Let me define variables more carefully.Let’s let the time B takes alone be t. Then, time A takes alone is (4/3)t. So, their rates are:Rate of B = N / tRate of A = N / (4/3 t) = 3N / (4t)Combined rate = N/t + 3N/(4t) = (7N)/(4t)Time taken together: N / (7N/(4t)) ) = 4t/7. So, same as before.In this time, A plants (3N/(4t)) * (4t/7) = 3N/7B plants (N/t) * (4t/7) = 4N/7Difference: 4N/7 - 3N/7 = N/7 = 36 => N=252. Same result.But maybe the problem is trickier? Let me check if the interpretation of "1/3 more time" is correct. The problem says, "It takes person A 1/3 more time to plant them alone than it takes person B." So, if B takes t time, A takes t + (1/3)t = (4/3)t. So, yes, that's correct. So A is 1/3 slower, which means 1/3 more time. So that part is right.Another way people might get confused is if they thought A's rate is 1/3 less than B's. But no, the problem states the time difference, not the rate difference. So, since time and rate are inversely related, if A takes 4/3 the time of B, then A's rate is 3/4 of B's rate. Which we accounted for.So, all steps check out. Therefore, the total number of saplings is 252.Wait, but let me think again. When they work together, do they split the work based on their rates? That is, the amount each plants is proportional to their rates. Since the total work is N, and the time is same for both when working together, the number each plants is (rate of A) * time and (rate of B) * time. Which we calculated as 3N/7 and 4N/7, respectively. So the difference is N/7 = 36, so N=252. Seems correct.Alternatively, maybe using efficiency. If A is 3/4 as efficient as B, then when working together, their contributions would be in the ratio of 3:4. Because efficiency ratio is 3:4 (B being 4, A being 3 if we consider rates). Wait, no. Wait, B's rate is higher, so if A's rate is 3/4 of B's, then the ratio of their rates is 3:4 (A:B). So, when working together, the amount each plants would be in the ratio 3:4. Therefore, the difference is 1 part, which is 36, so total parts are 7, so total saplings 7*36=252. Same answer. So this is another way to see it.Yes, that's even simpler. If their rates are in the ratio 3:4, then the work done by each is in the ratio 3:4. The difference is 1 ratio unit, which is 36, so total ratio units are 7, hence total saplings 252. So that's a quicker way.So either approach gets to the same answer. Therefore, the answer is 252.Final AnswerThe total number of saplings is boxed{252}.

✅:Let's denote the total number of saplings as ( N ). Person B takes time ( t ) to plant all the saplings alone, so person A takes ( frac{4}{3}t ) time (since A takes ( frac{1}{3} ) more time than B). The planting rates for A and B are:- Rate of B: ( frac{N}{t} ) saplings per unit time.- Rate of A: ( frac{N}{frac{4}{3}t} = frac{3N}{4t} ) saplings per unit time.When working together, their combined rate is:[frac{N}{t} + frac{3N}{4t} = frac{4N}{4t} + frac{3N}{4t} = frac{7N}{4t} text{ saplings per unit time}]The time taken to plant all ( N ) saplings together is:[T = frac{N}{frac{7N}{4t}} = frac{4t}{7}]During this time ( T ), the number of saplings planted by each person is:- Saplings planted by A: ( frac{3N}{4t} times frac{4t}{7} = frac{3N}{7} )- Saplings planted by B: ( frac{N}{t} times frac{4t}{7} = frac{4N}{7} )The difference in the number of saplings planted by B and A is given as 36:[frac{4N}{7} - frac{3N}{7} = frac{N}{7} = 36]Solving for ( N ):[N = 36 times 7 = 252]Thus, the total number of saplings is (boxed{252}).

🔑:To find the total number of saplings, we need to determine the time and efficiency differences between person A (甲) and person B (乙). Let's denote the time person B needs to plant one sapling as 1 unit (since this is given as a reference). From the problem statement, we know several key points:- Person A's single planting time is (1 + frac{1}{3}) units of time.- Person B plants 36 more saplings than person A when they work together.First, we’ll calculate the work efficiency for both persons.1. Calculate the work efficiencies: - Person B's efficiency (work done per unit time) is: [ text{Efficiency}_B = frac{1 text{ sapling}}{1 text{ unit time}} = 1 text{ sapling/unit time} ] - Person A's efficiency: [ text{Efficiency}_A = frac{1 text{ sapling}}{1 + frac{1}{3} text{ unit time}} = frac{1 text{ sapling}}{frac{4}{3} text{ unit time}} = frac{3}{4} text{ sapling/unit time} ]2. Expressing the number of saplings planted together: - Since B plants 36 more saplings than A, let's denote the number of saplings planted by A as (n). - Thus, the number of saplings planted by B is (n + 36).3. Total number of saplings planted together in terms of (n): - Total saplings ( T ) when both work together can be expressed as: [ T = n + (n + 36) = 2n + 36 ] - Since efficiencies provide the rates and the time taken to plant the total saplings is the same for both, we can establish an equation for the work done: [ n cdot frac{4}{3} + (n + 36) = 2n + 36 ]4. Solving for ( n ): [ frac{4}{3} n = n + 36 ] [ frac{4}{3} n - n = 36 ] [ frac{1}{3} n = 36 ] [ n = 36 times 3 = 108 ]5. Total saplings ( T ): [ T = 2n + 36 = 2 times 108 + 36 = 216 + 36 = 252 ]Conclusion:[boxed{252}]

🧠:Okay, so I need to prove that in this inscribed quadrilateral ABCD where AB equals BC, and there's a point N on side CD such that angle DNB is 90 degrees, the sum AD + NC equals DN. Hmm, let's start by drawing a diagram to visualize the problem. First, since ABCD is cyclic (inscribed quadrilateral), opposite angles should sum to 180 degrees. Also, AB = BC, so triangle ABC is isosceles with AB = BC. That might mean angles at A and C are equal? Wait, in triangle ABC, if AB = BC, then angles at A and C are equal? Wait, no, wait. In triangle ABC, sides AB and BC are equal, so the base is AC, so the base angles at A and C would be equal. But actually, in triangle ABC, sides AB and BC are equal, so the angles opposite them, which are angles at C and A, respectively, would be equal. Wait, no, hold on. In triangle ABC, side AB is opposite angle C, and side BC is opposite angle A. So if AB = BC, then angle C = angle A. So angles at A and C in triangle ABC are equal. But since ABCD is cyclic, angle A and angle C of the quadrilateral might relate differently. Hmm, maybe I need to consider the properties of cyclic quadrilaterals here.Another thing given is that point N is on CD such that angle DNB is 90 degrees. So DN is perpendicular to BN. Maybe I can use some right triangle properties here. Since angle DNB is 90 degrees, triangle DNB is right-angled at N. Maybe I can apply the Pythagorean theorem here? But I need to relate AD, NC, and DN. Let me see.Since ABCD is cyclic, let me recall that in a cyclic quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. But not sure if that's applicable here. Alternatively, maybe using power of a point. The point N is on CD, and we have a right angle at N. If I consider point N, perhaps the power of point N with respect to the circumcircle of ABCD? The power of a point N with respect to a circle is equal to the square of the tangent from N to the circle, which is also equal to ND * NC (since CD is a chord) but wait, power of a point N with respect to the circle would be ND * NC if CD is a chord. But since angle DNB is 90 degrees, maybe there's a relation here.Alternatively, since angle DNB is 90 degrees, point N lies on the circle with diameter DB. So, point N is on both CD and the circle with diameter DB. Therefore, N is the intersection of CD and the circle with diameter DB. Maybe using circle geometry here. Let me note that.Since DB is the diameter of this circle, any angle subtended by DB on the circumference is a right angle. So if N is on that circle, angle DNB is 90 degrees, which is given. So that's a property we can use. Therefore, N is the intersection point of CD and the circle with diameter DB. So perhaps if I can find coordinates or use coordinate geometry to model this problem?Alternatively, maybe using coordinate geometry. Let me try setting up coordinates. Let me place the cyclic quadrilateral ABCD in a coordinate system. Let me assume that points are placed such that calculations can be simplified. For instance, maybe let point B be at the origin. But since ABCD is cyclic, maybe it's better to center the circle at some point. Wait, maybe let me set up coordinates such that the circumcircle of ABCD is centered at the origin. But that might complicate things. Alternatively, set point B at (0,0), and use the fact that AB = BC. Let's try that.Let me let point B be at the origin (0,0). Since AB = BC, let's set point A at (a, 0) and point C at (0, a) for some a, but that might not necessarily form a cyclic quadrilateral. Hmm, maybe this approach is too simplistic. Alternatively, since AB = BC, triangle ABC is isosceles, so let me consider triangle ABC with AB = BC. Let me suppose AB = BC = 1 for simplicity, and then build coordinates accordingly.Let me place point B at (0,0). Let me place point A at (-1,0), so AB = 1 unit. Then, since BC = AB = 1, point C should be at some point such that the distance from B(0,0) to C is 1. Let's place point C at (0,1). Then, triangle ABC has points A(-1,0), B(0,0), C(0,1). Now, ABCD is a cyclic quadrilateral, so point D must lie somewhere on the circumcircle of ABC. Wait, but the circumcircle of triangle ABC can be found. Let's compute that.The three points A(-1,0), B(0,0), C(0,1). Let me find the circumcircle equation. The general equation of a circle is x² + y² + Dx + Ey + F = 0. Plugging in point A(-1,0): (-1)² + 0² + D*(-1) + E*0 + F = 1 - D + F = 0. So 1 - D + F = 0. Point B(0,0): 0 + 0 + 0 + 0 + F = 0 => F = 0. Then from point A: 1 - D + 0 = 0 => D = 1. Point C(0,1): 0 + 1² + D*0 + E*1 + F = 1 + E + 0 = 0 => E = -1. So the equation of the circle is x² + y² + x - y = 0. Completing the square: x² + x + y² - y = 0. (x + 0.5)² - 0.25 + (y - 0.5)² - 0.25 = 0. So (x + 0.5)^2 + (y - 0.5)^2 = 0.5. So center at (-0.5, 0.5) and radius sqrt(0.5).Now, point D is another point on this circle. Let's parameterize point D. Let me use an angle parameter θ to find coordinates of D. Alternatively, since ABCD is a quadrilateral, the order is important. The points are A, B, C, D in order. So after point C(0,1), point D should be somewhere on the circle such that the quadrilateral is ABCD. Let me see. Let's pick a point D on the circle. Let me choose θ as the angle parameter from the positive x-axis. Wait, but the circle is centered at (-0.5, 0.5). Maybe parameterizing with angle θ from the center. Let me parametrize point D as:x = -0.5 + sqrt(0.5) * cosθy = 0.5 + sqrt(0.5) * sinθSo for different θ, point D moves around the circle. Then, once D is chosen, point N is on CD such that angle DNB is 90 degrees. Hmm, this seems complicated. Maybe there's a better way.Alternatively, since N is on CD and angle DNB is 90 degrees, as I thought earlier, N lies on the circle with diameter DB. Therefore, N is the intersection of CD and the circle with diameter DB. So perhaps if I can find coordinates of D, then find the circle with diameter DB, find its intersection with CD, which is point N, and then compute AD, NC, DN to check if AD + NC = DN.But this might involve a lot of algebra. Let me try to proceed step by step.First, let's fix coordinates as before:A(-1,0), B(0,0), C(0,1), and D is a point on the circle centered at (-0.5, 0.5) with radius sqrt(0.5). Let me pick a specific D to make computation easier. Maybe let's choose D such that CD is horizontal? Wait, CD connects C(0,1) to D. If CD is horizontal, then D would have the same y-coordinate as C, which is 1. Let's check if such a point exists on the circle.Plug y = 1 into the circle equation:(x + 0.5)^2 + (1 - 0.5)^2 = 0.5(x + 0.5)^2 + 0.25 = 0.5(x + 0.5)^2 = 0.25x + 0.5 = ±0.5So x = 0 or x = -1But point C is at (0,1), so x=0 would be point C itself. The other solution is x = -1, so D would be (-1,1). Let's check if this is on the circle:(-1 + 0.5)^2 + (1 - 0.5)^2 = (-0.5)^2 + (0.5)^2 = 0.25 + 0.25 = 0.5. Yes, correct. So point D(-1,1). Then, CD is from (0,1) to (-1,1), which is indeed horizontal.But in this case, quadrilateral ABCD would have points A(-1,0), B(0,0), C(0,1), D(-1,1). Let's see if this is cyclic. All points lie on the circle we found earlier, so yes. Now, let's check if this configuration satisfies the conditions.Point N is on CD such that angle DNB is 90 degrees. CD is from (-1,1) to (0,1). Wait, no, D is (-1,1), C is (0,1). Wait, CD is from C(0,1) to D(-1,1). So CD is the line segment from (0,1) to (-1,1). So any point N on CD would have coordinates (x,1) where x ranges from 0 to -1. But angle DNB is 90 degrees. Let's find such a point N.Point N is on CD: coordinates (t,1) where t is between -1 and 0.We need angle DNB = 90 degrees. So vectors ND and NB should be perpendicular.Point D is (-1,1), N is (t,1), B is (0,0).Vector ND = D - N = (-1 - t, 0)Vector NB = B - N = (-t, -1)Their dot product should be zero:(-1 - t)(-t) + (0)(-1) = t(1 + t) + 0 = t + t² = 0So t² + t = 0 => t(t + 1) = 0 => t = 0 or t = -1But t is between -1 and 0 (on CD from C(0,1) to D(-1,1)), so t = 0 would be point C and t = -1 would be point D. But angle at N would be 90 degrees only at these endpoints? But the problem states that N is on side CD, not including the endpoints? Wait, but in this configuration, the only points on CD where angle DNB is 90 degrees are C and D themselves. But angle at C: point N=C, then angle DNB would be angle DNC, but N=C, so angle DCB. But angle DCB is not necessarily 90 degrees. Similarly, at N=D, angle DNB would be angle DDB, which is undefined since N=D and B are distinct points.Hmm, this suggests that in this specific case, there are no such points N on CD except C and D, which are trivial and don't satisfy the condition. Therefore, this choice of D may not be valid because the problem states there exists a point N on CD (presumably not at C or D) such that angle DNB is 90 degrees. So maybe my choice of D(-1,1) is invalid. Therefore, I need to choose a different D.Alternatively, perhaps the issue is with how I placed the points. Let me try a different approach. Instead of fixing coordinates immediately, let's consider general properties.Given ABCD is cyclic with AB = BC. Let's denote AB = BC = x. Since ABCD is cyclic, the opposite angles sum to 180 degrees. So angle A + angle C = 180°, angle B + angle D = 180°. Also, since AB = BC, triangle ABC is isosceles with AB = BC, so angles at A and C are equal. Wait, in triangle ABC, angle at B is angle ABC, and sides AB = BC, so the base is AC, so angles at A and C are equal. Therefore, angle BAC = angle BCA.But in the quadrilateral ABCD, angle at A is angle DAB, and angle at C is angle BCD. Since ABCD is cyclic, angle DAB + angle BCD = 180°, because they are opposite angles. But angle BAC is part of angle DAB, and angle BCA is part of angle BCD. Hmm, maybe this is getting too convoluted.Alternatively, let's consider using the property of cyclic quadrilaterals that the product of the diagonals is equal to the sum of the products of opposite sides. But not sure. Wait, that's Ptolemy's theorem. For a cyclic quadrilateral, Ptolemy's theorem states that AC * BD = AB * CD + AD * BC. Since AB = BC, let's denote AB = BC = k. Then Ptolemy's theorem becomes AC * BD = k * CD + AD * k. So AC * BD = k (CD + AD). Not sure if this helps directly, but maybe.Alternatively, since we have a right angle at N, perhaps we can use some reflection properties. For example, in problems involving right angles, reflecting a point over a line often helps. Let me consider reflecting point B over the line CD. Wait, but I don't know where CD is. Alternatively, since angle DNB is 90°, perhaps N is the foot of the perpendicular from B to CD. Wait, but if that's the case, then BN would be the altitude from B to CD. But in that case, angle DNB is 90°, so N is the foot of the perpendicular from B to CD. But in that case, is AD + NC = DN?Wait, if N is the foot of the perpendicular from B to CD, then DN is the distance from D to N along CD, and NC is from N to C. So DN + NC = DC, but the problem states AD + NC = DN. That would mean AD = DN - NC = DC - 2 NC. Not sure. Wait, maybe that approach is not correct.Alternatively, maybe since angle DNB is 90°, and ABCD is cyclic, there is some relation between the triangles. Let me consider triangles DNB and maybe others.Wait, another thought. Since ABCD is cyclic, points A, B, C, D lie on a circle. Let me consider the circumcircle. Then, perhaps using power of a point. The power of point N with respect to the circumcircle of ABCD should be equal to NA * ND = NC * NB, but wait, power of a point N with respect to the circle is equal to the product of the lengths from N to the points of intersection with the circle. Since N is on CD, which is a chord, the power of N is NC * ND. Also, since angle DNB is 90°, and if BN is tangent to the circle, then power of N would be BN². But BN is not necessarily tangent. Wait, unless there's some tangent property here.Alternatively, since angle DNB = 90°, and if we can show that BN is tangent to the circumcircle of ABCD at some point, then power of N would be BN² = NC * ND. But I don't know if BN is tangent. Let me check.If BN is tangent to the circumcircle at B, then angle NBD would be equal to angle BAD (tangent-chord angle). But since ABCD is cyclic, angle BAD is equal to angle BCD. Hmm, not sure. Alternatively, maybe BN is tangent to another circle. Wait, earlier thought was that N lies on the circle with diameter DB. So point N is on CD and on the circle with diameter DB, which gives angle DNB = 90°. So maybe there are two circles here: the circumcircle of ABCD and the circle with diameter DB. Their intersection is points B and N. Wait, but N is on CD, so unless CD passes through B, which it doesn't in general. Wait, but in our coordinate example earlier, CD was from (0,1) to (-1,1), which doesn't pass through B(0,0). So the circle with diameter DB would have center at midpoint of DB. If D is (-1,1) and B is (0,0), midpoint is (-0.5, 0.5), which is the same as the center of the circumcircle of ABCD. Wait, in that specific case, the circle with diameter DB is the same as the circumcircle of ABCD? Because the center was at (-0.5, 0.5) and radius sqrt(0.5). Let's check the distance from center (-0.5, 0.5) to D(-1,1): sqrt[(-0.5 +1)^2 + (0.5 -1)^2] = sqrt[(0.5)^2 + (-0.5)^2] = sqrt(0.25 + 0.25) = sqrt(0.5), which is the radius. Similarly for point B(0,0): sqrt[(-0.5 - 0)^2 + (0.5 -0)^2] = same sqrt(0.25 + 0.25) = sqrt(0.5). So in this specific case, the circle with diameter DB is the same as the circumcircle of ABCD. But then point N is on CD and on this circle, which would be points C and D. But in that case, as we saw earlier, only C and D lie on both CD and the circle. Hence, angle DNB would be 90 degrees only at those points, which are the endpoints. But in the problem statement, N is a point on CD, so this suggests that in this configuration, the problem's condition is not satisfied unless N coincides with C or D. Therefore, my initial coordinate choice might be invalid.Therefore, perhaps my coordinate system is not general enough. Let me instead consider a different configuration where point D is not such that CD is horizontal.Let me try again with a different point D. Let's choose point D such that the quadrilateral ABCD is convex and cyclic, with AB = BC. Let me go back to the coordinate system with A(-1,0), B(0,0), C(0,1), and find another point D on the circle.Let me parameterize point D. Let's take an angle θ parameterizing the position of D on the circle. The circle has center (-0.5, 0.5) and radius sqrt(0.5). So coordinates of D can be written as:x = -0.5 + sqrt(0.5) * cosθy = 0.5 + sqrt(0.5) * sinθLet me pick θ = 135 degrees (3π/4 radians). Then cosθ = -√2/2, sinθ = √2/2.Thus,x = -0.5 + sqrt(0.5)*(-√2/2) = -0.5 - (sqrt(0.5)*sqrt(2)/2) = -0.5 - ( (sqrt(1))/2 ) = -0.5 - 0.5 = -1y = 0.5 + sqrt(0.5)*(√2/2) = 0.5 + (sqrt(0.5)*sqrt(2)/2) = 0.5 + (1/2) = 1So point D is (-1,1), which is the same as before. So that's the same point. Let me choose θ = 45 degrees (π/4 radians).cosθ = √2/2, sinθ = √2/2x = -0.5 + sqrt(0.5)*(√2/2) = -0.5 + (sqrt(1)/2) = -0.5 + 0.5 = 0y = 0.5 + sqrt(0.5)*(√2/2) = 0.5 + 0.5 = 1So point D would be (0,1), but that's point C. Not allowed. θ = 90 degrees (π/2):cosθ = 0, sinθ = 1x = -0.5 + 0 = -0.5y = 0.5 + sqrt(0.5)*1 ≈ 0.5 + 0.707 ≈ 1.207But checking if this is a valid point. Wait, the circle has radius sqrt(0.5) ≈ 0.707, so moving from center (-0.5, 0.5) up by 0.707 in y-direction gives y ≈ 1.207, which is above point C(0,1). But then CD would be from C(0,1) to D(-0.5,1.207). Let's see if in this case there is a point N on CD such that angle DNB is 90 degrees.So point D(-0.5,1.207). CD is from C(0,1) to D(-0.5,1.207). Let's parametrize CD. Let parameter t go from 0 to 1, with t=0 at C and t=1 at D.Coordinates of N: x = 0 + t*(-0.5 - 0) = -0.5ty = 1 + t*(1.207 - 1) = 1 + 0.207tWe need angle DNB = 90 degrees. So vectors ND and NB should be perpendicular.Point N is (-0.5t, 1 + 0.207t)Vector ND = D - N = (-0.5 + 0.5t, 1.207 - (1 + 0.207t)) = (-0.5 + 0.5t, 0.207 - 0.207t)Vector NB = B - N = (0 - (-0.5t), 0 - (1 + 0.207t)) = (0.5t, -1 - 0.207t)Dot product must be zero:(-0.5 + 0.5t)(0.5t) + (0.207 - 0.207t)(-1 - 0.207t) = 0Let me compute each term:First term: (-0.5 + 0.5t)(0.5t) = (-0.5)(0.5t) + 0.5t * 0.5t = -0.25t + 0.25t²Second term: (0.207 - 0.207t)(-1 - 0.207t) = 0.207*(-1) + 0.207*(-0.207t) -0.207t*(-1) -0.207t*(-0.207t)= -0.207 - 0.042849t + 0.207t + 0.042849t²So combining:Total equation:(-0.25t + 0.25t²) + (-0.207 - 0.042849t + 0.207t + 0.042849t²) = 0Combine like terms:t²: 0.25t² + 0.042849t² ≈ 0.292849t²t terms: -0.25t -0.042849t + 0.207t ≈ (-0.25 -0.042849 +0.207)t ≈ (-0.292849 + 0.207)t ≈ -0.085849tConstants: -0.207So equation is approximately:0.292849t² -0.085849t -0.207 = 0Multiply through by 1000 to eliminate decimals:292.849t² -85.849t -207 = 0Approximate coefficients:292.85t² -85.85t -207 ≈ 0Using quadratic formula:t = [85.85 ± sqrt(85.85² + 4*292.85*207)] / (2*292.85)Compute discriminant:85.85² ≈ 7370.224*292.85*207 ≈ 4*292.85*207 ≈ 4*60617.95 ≈ 242,471.8So sqrt(7370.22 + 242,471.8) ≈ sqrt(249,842) ≈ 499.84Thus,t ≈ [85.85 ± 499.84]/585.7First solution:t ≈ (85.85 + 499.84)/585.7 ≈ 585.69/585.7 ≈ 1.0Second solution:t ≈ (85.85 - 499.84)/585.7 ≈ (-413.99)/585.7 ≈ -0.707Since t must be between 0 and 1, the only valid solution is t≈1.0, which is point D itself. Again, this suggests that in this configuration, the only point on CD satisfying angle DNB=90° is D itself, which is trivial. Hence, this approach might not be working. Maybe my coordinate system is not suitable or the problem requires a more synthetic approach.Let me try a different strategy. Let's consider the cyclic quadrilateral ABCD with AB=BC. Let's denote AB=BC=k. Let's construct triangle ABC with AB=BC, then point D is somewhere on the circumcircle. Then, point N is on CD such that angle DNB=90°, and we need to show AD + NC = DN.Perhaps using the properties of cyclic quadrilaterals and right triangles. Since angle DNB is 90°, triangle DNB is right-angled at N. Let's consider reflecting point D over BN to get a point such that DN is equal to the reflection. But not sure. Alternatively, since BN is the altitude, maybe some similar triangles.Alternatively, since ABCD is cyclic, then angles ADB and ACB are equal (since they subtend the same arc AB). But AB=BC, so angle ACB = angle BAC. Maybe relating these angles.Alternatively, consider triangle ADN and triangle BNC. If we can show some relationship between them that leads to AD + NC = DN. Maybe using the fact that AB=BC and some triangle congruency or similarity.Alternatively, using coordinate geometry but choosing a better coordinate system. Let me try placing point B at the origin (0,0), point A at (-a,0), and point C at (0,a) to make AB=BC=a√2. Wait, AB would be the distance from (-a,0) to (0,0) which is a. BC is from (0,0) to (0,a), which is a. So AB=BC=a. Then, the circumcircle of ABC can be determined.Points A(-a,0), B(0,0), C(0,a). The circumcircle of triangle ABC. Let's find its equation. The general circle equation through three points. Let's compute.Using determinant method:|x y x² + y² 1|For three points:A(-a,0): (-a, 0, a², 1)B(0,0): (0,0,0,1)C(0,a): (0,a, a²,1)The equation of the circle is given by determinant:|x y x² + y² 1|| -a 0 a² 1 || 0 0 0 1 || 0 a a² 1 |Expanding the determinant:x |0 a² 1| - y |-a a² 1| + (x² + y²)|-a 0 1| -1 |-a 0 a²| |0 0 1| |0 0 1| |0 a 1| |0 a a²|But this seems complicated. Alternatively, using perpendicular bisectors.The perpendicular bisector of AB: AB is from (-a,0) to (0,0). Midpoint is (-a/2, 0). The line AB is horizontal, so perpendicular bisector is vertical line x = -a/2.The perpendicular bisector of BC: BC is from (0,0) to (0,a). Midpoint is (0, a/2). The line BC is vertical, so perpendicular bisector is horizontal line y = a/2.The intersection of x = -a/2 and y = a/2 is the center of the circle: (-a/2, a/2). The radius is the distance from center to A(-a,0):sqrt[ (-a/2 + a)^2 + (a/2 - 0)^2 ] = sqrt[ (a/2)^2 + (a/2)^2 ] = sqrt( a²/4 + a²/4 ) = sqrt(a²/2) = a/√2.Therefore, the equation of the circumcircle is (x + a/2)^2 + (y - a/2)^2 = (a/√2)^2 = a²/2.Now, point D is another point on this circle. Let's parameterize point D. Let's use an angle θ. So coordinates:x = -a/2 + (a/√2) cosθy = a/2 + (a/√2) sinθNow, CD is the line segment from C(0,a) to D(x,y). Point N is on CD such that angle DNB=90°. We need to find N and prove AD + NC = DN.Let me parametrize point N as a point along CD. Let parameter t be from 0 to 1, where t=0 is C and t=1 is D. Then coordinates of N are:N_x = 0 + t(x - 0) = txN_y = a + t(y - a) = a + t(y - a)Since D is on the circle, x and y are given above.We need angle DNB = 90°, so vectors ND and NB are perpendicular.Vector ND = D - N = (x - tx, y - (a + t(y - a))) = (x(1 - t), y - a - t(y - a)) = (x(1 - t), (y - a)(1 - t))Vector NB = B - N = (0 - tx, 0 - (a + t(y - a))) = (-tx, -a - t(y - a))Dot product must be zero:[x(1 - t)]*(-tx) + [(y - a)(1 - t)]*(-a - t(y - a)) = 0Factor out (1 - t):(1 - t)[ -tx^2 + (y - a)(-a - t(y - a)) ] = 0Since N is not equal to D (t ≠1 ), we can divide by (1 - t): -tx^2 + (y - a)(-a - t(y - a)) = 0Expand the second term:(y - a)(-a) - t(y - a)^2 = -a(y - a) - t(y - a)^2Thus: -tx^2 -a(y - a) -t(y - a)^2 = 0Factor out -t: -t(x^2 + (y - a)^2) -a(y - a) = 0Multiply both sides by -1:t(x^2 + (y - a)^2) + a(y - a) = 0Solve for t:t = - [a(y - a)] / [x^2 + (y - a)^2]But since D is on the circle, x and y satisfy (x + a/2)^2 + (y - a/2)^2 = a²/2. Let's expand this:x² + a x + (a²)/4 + y² - a y + (a²)/4 = a²/2So x² + y² + a x - a y + a²/2 = a²/2Therefore, x² + y² + a x - a y = 0Now, let's compute x^2 + (y - a)^2:x² + (y - a)^2 = x² + y² - 2a y + a²From the circle equation: x² + y² = -a x + a ySubstitute:x² + (y - a)^2 = (-a x + a y) - 2a y + a² = -a x - a y + a²Therefore, the expression for t becomes:t = - [a(y - a)] / [ -a x - a y + a² ] = - [a(y - a)] / [ -a(x + y - a) ] = - [ (y - a) ] / [ -(x + y - a) ] = (y - a) / (x + y - a)So t = (y - a)/(x + y - a)Now, we need to compute AD, NC, and DN in terms of t and show that AD + NC = DN.First, let's compute AD. AD is the distance from A(-a,0) to D(x,y):AD = sqrt[(x + a)^2 + (y - 0)^2] = sqrt[(x + a)^2 + y^2]NC is the distance from N(tx, a + t(y - a)) to C(0,a):NC = sqrt[(tx - 0)^2 + (a + t(y - a) - a)^2] = sqrt[(tx)^2 + (t(y - a))^2] = t sqrt[x^2 + (y - a)^2]DN is the distance from D(x,y) to N(tx, a + t(y - a)):DN = sqrt[(x - tx)^2 + (y - (a + t(y - a)))^2] = sqrt[(x(1 - t))^2 + ((y - a)(1 - t))^2] = (1 - t) sqrt[x^2 + (y - a)^2]Therefore, AD + NC = sqrt[(x + a)^2 + y^2] + t sqrt[x^2 + (y - a)^2]We need to show this equals DN = (1 - t) sqrt[x^2 + (y - a)^2]So:sqrt[(x + a)^2 + y^2] + t sqrt[x^2 + (y - a)^2] = (1 - t) sqrt[x^2 + (y - a)^2]Subtract t sqrt[x^2 + (y - a)^2] from both sides:sqrt[(x + a)^2 + y^2] = (1 - 2t) sqrt[x^2 + (y - a)^2]But this seems complicated. Maybe we need to express sqrt[(x + a)^2 + y^2] in terms of the circle equation.From the circle equation:x² + y² = -a x + a yTherefore, (x + a)^2 + y^2 = x² + 2a x + a² + y^2 = (x² + y^2) + 2a x + a² = (-a x + a y) + 2a x + a² = a x + a y + a² = a(x + y + a)So sqrt[(x + a)^2 + y^2] = sqrt[a(x + y + a)]Similarly, sqrt[x^2 + (y - a)^2] = sqrt[x^2 + y^2 - 2a y + a²] = sqrt[(-a x + a y) - 2a y + a²] (using x² + y² = -a x + a y from circle equation)= sqrt[-a x - a y + a²] = sqrt[-a(x + y - a)]But this sqrt is of a negative expression unless x + y - a < 0. Wait, the expression inside the sqrt is -a(x + y - a). Given that sqrt is defined, the argument must be non-negative. So -a(x + y - a) >=0. Since a >0 (length), then -(x + y - a) >=0 => x + y - a <=0 => x + y <=a. Therefore, sqrt[-a(x + y - a)] = sqrt[a(a - x - y)].Therefore, sqrt[x^2 + (y - a)^2] = sqrt[a(a - x - y)]So now, the equation we need to verify is:sqrt[a(x + y + a)] + t sqrt[a(a - x - y)] = (1 - t) sqrt[a(a - x - y)]Divide both sides by sqrt[a]:sqrt(x + y + a) + t sqrt(a - x - y) = (1 - t) sqrt(a - x - y)Let me denote s = sqrt(a - x - y). Then, since x + y <=a, s is real.Thus:sqrt(x + y + a) + t s = (1 - t)sRearrange:sqrt(x + y + a) = (1 - t)s - t s = (1 - 2t)sBut t = (y - a)/(x + y - a). Let me compute (1 - 2t):1 - 2t = 1 - 2(y - a)/(x + y - a) = [ (x + y - a) - 2(y - a) ] / (x + y - a) = [x + y - a - 2y + 2a]/(x + y -a) = [x - y + a]/(x + y -a)Therefore:sqrt(x + y + a) = [ (x - y + a)/(x + y - a) ] * sBut s = sqrt(a - x - y)Thus:sqrt(x + y + a) = [ (x - y + a)/(x + y - a) ] * sqrt(a - x - y)Square both sides:x + y + a = [ (x - y + a)^2 / (x + y - a)^2 ] * (a - x - y )Multiply both sides by (x + y - a)^2:(x + y + a)(x + y - a)^2 = (x - y + a)^2 (a - x - y )Let me let u = x + y. Then the equation becomes:(u + a)(u - a)^2 = (x - y + a)^2 (a - u )But this substitution might not help much. Let's expand both sides.Left side: (u + a)(u - a)^2 = (u + a)(u² - 2a u + a²) = u³ - 2a u² + a² u + a u² - 2a² u + a³ = u³ - a u² - a² u + a³Right side: (x - y + a)^2 (a - u ) First, note that x - y + a. From the circle equation:x² + y² = -a x + a yLet me try to express x - y in terms of u = x + y.Let me compute (x - y)^2 = x² - 2xy + y² = (x² + y²) - 2xy = (-a x + a y) - 2xy = -a x + a y - 2xy.Not sure. Alternatively, maybe express x - y + a:x - y + a = x - y + a.From the circle equation: x² + y² + a x - a y =0Let me see if I can find a relation between x - y and u.Let me consider:From u = x + y.Let v = x - y.Then x = (u + v)/2y = (u - v)/2Substitute into circle equation:x² + y² + a x - a y =0[(u + v)^2 /4 + (u - v)^2 /4] + a*(u + v)/2 - a*(u - v)/2 =0Compute:[(u² + 2uv + v² + u² - 2uv + v²)/4] + [a(u + v) - a(u - v)]/2 =0[ (2u² + 2v²)/4 ] + [ a*2v ]/2 =0(u² + v²)/2 + a v =0Multiply by 2:u² + v² + 2a v =0So v² + 2a v + u² =0This is a quadratic in v:v² + 2a v + u² =0But solving for v:v = [ -2a ± sqrt(4a² -4*1*u²) ] /2 = -a ± sqrt(a² - u²)But since v is real, then a² - u² >=0 => u² <=a² => |u| <=a, which aligns with our previous conclusion that x + y <=a.Therefore, x - y + a = v + a = [ -a ± sqrt(a² - u²) ] + a = ± sqrt(a² - u²)Therefore, (x - y + a)^2 = ( sqrt(a² - u²) )^2 = a² - u²Thus, the right side of the equation is:(x - y + a)^2 (a - u ) = (a² - u²)(a - u) = (a - u)(a + u)(a - u) = (a - u)^2(a + u)Left side: (u + a)(u - a)^2 = (u + a)(a - u)^2Comparing both sides:(u + a)(a - u)^2 = (a - u)^2(a + u)Yes, both sides are equal. Therefore, the equation holds true.Thus, our earlier equation sqrt(x + y + a) = [ (x - y + a)/(x + y - a) ] * sqrt(a - x - y) is indeed valid, since both sides when squared lead to identities. Therefore, the original equality AD + NC = DN holds.Therefore, the proof is complete.Final Answerboxed{AD + NC = DN}

✅:Given an inscribed quadrilateral (ABCD) with (AB = BC). On the side (CD), there is a point (N) such that (angle DNB = 90^circ). We need to prove that (AD + NC = DN).1. Cyclic Quadrilateral Properties: - Since (ABCD) is cyclic, the opposite angles sum to (180^circ). - (AB = BC) implies (triangle ABC) is isosceles with (angle BAC = angle BCA).2. Right Angle at (N): - Point (N) lies on (CD) such that (angle DNB = 90^circ). This implies (N) is on the circle with diameter (DB).3. Coordinate System and Circle Equation: - Place (B) at the origin ((0,0)), (A) at ((-a,0)), and (C) at ((0,a)) to form an isosceles triangle (ABC). - The circumcircle of (triangle ABC) has center ((-a/2, a/2)) and radius (a/sqrt{2}).4. Parameterization of Point (D): - Point (D) is parameterized on the circumcircle as (x = -a/2 + (a/sqrt{2}) cos theta) and (y = a/2 + (a/sqrt{2}) sin theta).5. Point (N) on (CD): - Parametrize (N) along (CD) with parameter (t). Coordinates of (N) are ((tx, a + t(y - a))).6. Vectors and Dot Product: - Vectors (ND) and (NB) are perpendicular, leading to the equation for (t): [ t = frac{(y - a)}{(x + y - a)} ]7. Distances Calculation: - Calculate distances (AD), (NC), and (DN) using the parameterized coordinates and verify the equation (AD + NC = DN).8. Verification: - Using the circle equation and algebraic manipulation, show that the equation holds true, confirming (AD + NC = DN).Thus, the proof is complete, and the final answer is:[boxed{AD + NC = DN}]

🔑:1. Construct a Point: - Let us extend the segment DA beyond point A and mark a point K on this extension such that AK = NC (Refer to the figure provided).2. Triangle Congruence: - Note that triangles triangle AKB and triangle CNB are congruent. - Given that AB = BC by the problem's condition and AK = NC by construction. - Angles angle KAD and angle NCB are inscribed angles in the cyclic quadrilateral ABCD. 3. Show the Congruence: - We have angle KAD = angle NCB since they are inscribed angles subtended by the same arc BD in the cyclic quadrilateral ABCD. - Therefore, by the side-angle-side (SAS) criterion, triangle AKB cong triangle CNB.4. Right Angles: - This congruence shows that BK = BN and angle BKA = angle BNC = 90^circ.5. Quadrilateral Analysis: - Observe that triangles triangle DBN and triangle DBK are right triangles with shared hypotenuse BD. - Since BK = BN, triangles triangle DBN and triangle DBK are congruent by the hypotenuse-leg (HL) criterion.6. Calculate the Distances: - As a result of this congruence, we get DN = DK.7. Sum of Segments: - Recall that DK = DA + AK due to our construction, i.e., [DK = DA + AK] - By construction, AK = NC.8. Final Equality: - Substitute AK with NC in the previous equation to get [DN = DA + NC] - This completes our proof that [AD + NC = DN] Conclusion:[boxed{AD + NC = DN}]

🧠:Okay, let's try to figure out this probability problem. So, there are three boxes: A, B, and C. They have 100, 50, and 80 balls respectively. Each box has some number of black balls. We know that box A has 15 black balls. The problem says that if we pick a box at random and then draw a ball from it, the probability of getting a black ball is 101/600. We need to find the maximum number of black balls that can be in box C.First, let me make sure I understand the problem correctly. We have three boxes with different numbers of balls. Each has some black balls. Box A's black balls are given as 15. The probability of choosing a box uniformly at random (so each box has a 1/3 chance of being selected) and then drawing a black ball is 101/600. We need to find the maximum possible number of black balls in box C, given this information.So, probability is (1/3)(probability from A + probability from B + probability from C) = 101/600.Let me write that as an equation. Let's denote the number of black balls in box B as b and in box C as c. Then:(1/3)[(15/100) + (b/50) + (c/80)] = 101/600.Is that right? Let me check. Since each box is chosen with probability 1/3, then the total probability is the average of the individual probabilities from each box. So yes, that equation should hold.So, simplifying the equation:Multiply both sides by 3:(15/100) + (b/50) + (c/80) = 101/200.Because 101/600 multiplied by 3 is 101/200.Now, let's compute 15/100. That's 0.15. 101 divided by 200 is 0.505. So:0.15 + (b/50) + (c/80) = 0.505.Subtract 0.15:(b/50) + (c/80) = 0.505 - 0.15 = 0.355.So, (b/50) + (c/80) = 0.355.We can write this equation as:(8b + 5c)/400 = 0.355.Multiply both sides by 400:8b + 5c = 0.355 * 400.Calculating 0.355 * 400. Let's see, 0.35 * 400 = 140, and 0.005 * 400 = 2, so total is 142. Therefore:8b + 5c = 142.So, the equation we have is 8b + 5c = 142. Now, we need to find the maximum possible c (number of black balls in box C) given that b and c must be non-negative integers, and also that b cannot exceed the total number of balls in box B, which is 50, and c cannot exceed 80.So, our variables are constrained by:0 ≤ b ≤ 50,0 ≤ c ≤ 80,and 8b + 5c = 142.We need to find the maximum integer c such that there exists an integer b within 0 to 50 that satisfies the equation.So, how to approach this? Let's solve for b in terms of c:8b = 142 - 5c,so,b = (142 - 5c)/8.Since b must be a non-negative integer, (142 - 5c) must be divisible by 8 and result in a non-negative integer.Therefore, 142 - 5c ≡ 0 mod 8.So, 5c ≡ 142 mod 8.But 142 divided by 8 is 17*8 + 6, so 142 ≡ 6 mod 8. Therefore,5c ≡ 6 mod 8.We need to solve for c such that 5c ≡ 6 mod 8.Multiplying both sides by the inverse of 5 modulo 8. Since 5*5=25 ≡ 1 mod 8, so inverse of 5 is 5 mod 8.Thus,c ≡ 6*5 ≡ 30 ≡ 6 mod 8.Therefore, c ≡ 6 mod 8. So, c can be written as 8k + 6, where k is a non-negative integer.Now, c must also satisfy 0 ≤ c ≤ 80, and b = (142 - 5c)/8 must be between 0 and 50.So, let's write c as 8k + 6. Then, substitute into b:b = (142 - 5*(8k + 6))/8 = (142 - 40k - 30)/8 = (112 - 40k)/8 = 14 - 5k.So, b = 14 - 5k.Now, since b must be ≥ 0 and ≤ 50, let's find the possible k.First, b ≥ 0:14 - 5k ≥ 0 ⇒ 5k ≤ 14 ⇒ k ≤ 14/5 ⇒ k ≤ 2.8. Since k is integer, k ≤ 2.Second, b ≤ 50:14 - 5k ≤ 50 ⇒ -5k ≤ 36 ⇒ k ≥ -7.2. Since k is a non-negative integer (since c =8k +6 must be non-negative), so k ≥ 0.Thus, possible k values are 0, 1, 2.Therefore, c = 8k + 6, for k=0,1,2.So:For k=0: c=6, b=14-0=14.Check if b=14 is ≤50: yes.For k=1: c=14, b=14 -5=9.For k=2: c=22, b=14 -10=4.Wait, that gives us c=6,14,22. But we need to find the maximum c. Wait, but 22 is higher than 14 and 6. But is that the maximum possible?Wait, but according to this, the possible c's are 6,14,22. But maybe there's a mistake here.Wait, hold on. If k=0,1,2, then c=6,14,22. But these are c values such that c ≡6 mod8. But maybe there's a larger c that also satisfies c ≡6 mod8 and still gives b=(142 -5c)/8 ≥0.Wait, but according to our earlier analysis, k can only be up to 2, because b=14-5k ≥0 requires k≤2.8, so k=0,1,2. Therefore, the maximum c in this case is 22. But 22 is much less than 80. So, why is the answer 22? That seems low.Wait, perhaps there's a mistake in my reasoning here. Let me check again.We have c ≡6 mod8, so possible c values are 6,14,22,30,..., up to 78 (since c ≤80). But our previous reasoning using k=0,1,2 only gives up to c=22. That seems inconsistent.Wait, perhaps my substitution was wrong.Wait, when I set c=8k +6, but maybe k can be larger? But then, according to the expression for b=14 -5k, if k increases beyond 2, then b becomes negative, which is not allowed.But is there another way to represent c? Wait, perhaps the equation 5c ≡6 mod8 allows for solutions where c is larger, but with b still non-negative.Wait, but according to the equation:b=(142 -5c)/8 must be ≥0 ⇒142 -5c ≥0 ⇒5c ≤142 ⇒c ≤28.4.Therefore, c cannot exceed 28.4. Since c must be an integer, c ≤28.But wait, that contradicts with the earlier conclusion that c can be up to 78.Wait, that seems like a critical point. So, let's check again.From the equation 8b +5c=142, and since b and c are non-negative integers, we can find the maximum possible c by minimizing b. The minimal b is 0. If b=0, then 5c=142 ⇒c=142/5=28.4. But c must be an integer, so c=28. Then 5c=140, so 8b=142 -140=2 ⇒b=2/8=0.25. Not an integer. Therefore, that's not possible.Wait, so if we take c=28, then 5c=140, 142-140=2, so 8b=2 ⇒b=0.25, which is not an integer. So, invalid.So, maybe c=28 is too high. Let's check c=28. So 5*28=140. 142-140=2. 8b=2 ⇒b=0.25. Not integer. So invalid.Next, c=27. 5*27=135. 142-135=7. 8b=7 ⇒b=7/8. Not integer.c=26. 5*26=130. 142-130=12. 8b=12 ⇒b=12/8=1.5. Not integer.c=25. 5*25=125. 142-125=17. 8b=17 ⇒b=17/8=2.125. Not integer.c=24. 5*24=120. 142-120=22. 8b=22 ⇒b=22/8=2.75. Not integer.c=23. 5*23=115. 142-115=27. 8b=27 ⇒b=27/8=3.375. Not integer.c=22. 5*22=110. 142-110=32. 8b=32 ⇒b=4. Which is integer. So, c=22, b=4. That's valid.Similarly, c=21. 5*21=105. 142-105=37. 8b=37 ⇒b=37/8≈4.625. Not integer.So, seems like the next valid c is 22. Then going down:c=22, b=4.c=14, b=9.c=6, b=14.Wait, so according to this, the maximum possible c is 22.But that seems contradictory to the initial thought that maybe c could be higher. But given the equation 8b +5c=142, with b≥0, c≥0, the maximum c would be when b is minimized. However, since 142 divided by 5 is 28.4, the maximum possible integer c would be 28, but as shown above, that doesn't give an integer b. So, the next lower c that gives integer b is 22. Then, 22 is the maximum c. Therefore, the answer is 22.But wait, the problem is asking for the maximum number of black balls that can be in box C. But according to the equation, 22 is the maximum possible. However, box C has 80 balls, so theoretically, c can be up to 80. But given the probability constraint, it's limited to 22. Hmm.But let me check if I made a mistake in the equation setup.Original probability:(1/3)(15/100 + b/50 + c/80) = 101/600.Multiplying both sides by 3:15/100 + b/50 + c/80 = 101/200.Convert 15/100 to 3/20, which is 0.15.101/200 is 0.505.So 0.15 + (b/50) + (c/80) = 0.505.Then, (b/50) + (c/80) = 0.355.Multiply both sides by 400 to eliminate denominators:8b +5c = 142.Yes, that's correct.So 8b +5c =142, with b ≤50, c ≤80, and both non-negative integers.Therefore, solving for c, we need 5c =142 -8b.Therefore, 142 -8b must be divisible by 5 and non-negative.So, 142 -8b ≥0 ⇒8b ≤142 ⇒b ≤17.75. So b ≤17.Therefore, b can be from 0 to 17, but since b also has to be such that (142 -8b) is divisible by 5.Therefore, 142 -8b ≡0 mod5.Compute 142 mod5: 142/5=28*5 +2 ⇒142≡2 mod5.Similarly, 8b mod5: 8≡3 mod5 ⇒8b≡3b mod5.Therefore, 2 -3b ≡0 mod5 ⇒ -3b ≡-2 mod5 ⇒3b ≡2 mod5.Multiply both sides by inverse of 3 mod5. The inverse of 3 mod5 is 2, since 3*2=6≡1 mod5.Thus, b ≡2*2=4 mod5. Therefore, b≡4 mod5. So possible b values are 4,9,14,19,... But since b ≤17, the possible b are 4,9,14.So, b=4,9,14.For each of these, compute c:For b=4:5c=142 -8*4=142 -32=110 ⇒c=22.For b=9:5c=142 -72=70 ⇒c=14.For b=14:5c=142 -112=30 ⇒c=6.So, the possible pairs are (b,c)=(4,22),(9,14),(14,6).Thus, the maximum c is 22. Therefore, the maximum number of black balls in box C is 22.But wait, the problem states that each box has a "certain number of black balls," so they must be non-negative integers. The solutions found are all integers, so 22 is valid. Therefore, the answer is 22. So, the maximum possible number is 22.But let me check once more to ensure that there are no mistakes in the reasoning.We started with the probability equation, converted it into an equation with b and c, solved for the constraints, found that b must be congruent to 4 mod5, leading to possible b values 4,9,14, which give c=22,14,6. Therefore, c=22 is the maximum possible. Since 22 is less than 80, which is the total number of balls in box C, that's feasible. So, 22 is the correct answer.I think that's solid. So, the maximum number of black balls in box C is 22.Final Answerboxed{22}

✅:Given three boxes (A), (B), and (C) with 100, 50, and 80 balls respectively, we know box (A) contains 15 black balls. The probability of drawing a black ball from a randomly chosen box is (frac{101}{600}). We need to determine the maximum number of black balls in box (C).First, we set up the probability equation. Let (b) be the number of black balls in box (B) and (c) be the number in box (C). The probability equation is:[frac{1}{3} left( frac{15}{100} + frac{b}{50} + frac{c}{80} right) = frac{101}{600}]Multiplying both sides by 3:[frac{15}{100} + frac{b}{50} + frac{c}{80} = frac{101}{200}]Converting fractions to decimals:[0.15 + frac{b}{50} + frac{c}{80} = 0.505]Subtracting 0.15 from both sides:[frac{b}{50} + frac{c}{80} = 0.355]Multiplying through by 400 to clear denominators:[8b + 5c = 142]We need integer solutions for (b) and (c) such that (0 leq b leq 50) and (0 leq c leq 80). Solving for (b):[8b = 142 - 5c implies b = frac{142 - 5c}{8}](b) must be a non-negative integer. Therefore, (142 - 5c) must be divisible by 8. Solving the congruence (5c equiv 6 pmod{8}), we find (c equiv 6 pmod{8}). Thus, (c = 8k + 6) for non-negative integers (k).Substituting (c = 8k + 6) into the equation for (b):[b = frac{142 - 5(8k + 6)}{8} = 14 - 5k](b) must be non-negative, so (14 - 5k geq 0 implies k leq 2). Possible values of (k) are 0, 1, 2, leading to:- (k = 0): (c = 6), (b = 14)- (k = 1): (c = 14), (b = 9)- (k = 2): (c = 22), (b = 4)The maximum (c) is 22. Verifying, (8b + 5c = 142) holds for (b = 4) and (c = 22).Thus, the maximum number of black balls in box (C) is (boxed{22}).

🔑:1. Define Variables and Given Information Let the number of black balls in boxes ( B ) and ( C ) be ( y ) and ( x ) respectively. - Number of balls in box ( A ) = 100 - Black balls in box ( A ) = 15 - Number of balls in box ( B ) = 50 - Number of balls in box ( C ) = 80 - Given probability of picking a black ball = ( frac{101}{600} )2. Set up the Probability Equation The total probability of picking a black ball follows from considering each box separately and then summing: [ text{Probability} = frac{1}{3} times frac{15}{100} + frac{1}{3} times frac{y}{50} + frac{1}{3} times frac{x}{80} ]3. Substitute the Given Probability [ frac{1}{3} times frac{15}{100} + frac{1}{3} times frac{y}{50} + frac{1}{3} times frac{x}{80} = frac{101}{600} ]4. Simplify the Equation [ frac{1}{3} times frac{15}{100} = frac{15}{300} = frac{1}{20} ] [ frac{1}{3} times frac{y}{50} = frac{y}{150} ] [ frac{1}{3} times frac{x}{80} = frac{x}{240} ] Therefore: [ frac{1}{20} + frac{y}{150} + frac{x}{240} = frac{101}{600} ]5. Combine into a Common Denominator Since 300 is the Least Common Multiple (LCM) of 20, 150, and 240, convert and solve: [ frac{15}{300} + frac{2y}{300} + frac{5x}{600} = frac{101}{600} ] Converting all to a common denominator 600 and simplifying: [ frac{30}{600} + frac{4y}{600} + frac{5x}{600} = frac{101}{600} ]6. Remove Denominators and Solve [ 30 + 4y + 5x = 101 ] [ 4y + 5x = 71 ]7. Find Possible Values for ( y ) Since ( 4y + 5x = 71 ) is a linear Diophantine equation, express ( x ) in terms of ( y ): [ x = frac{71 - 4y}{5} ] For ( x ) to be an integer, ( 71 - 4y ) must be divisible by 5: [ 71 - 4y equiv 0 pmod{5} ] Simplify for allowable values: [ -4y equiv -1 pmod{5} ] [ 4y equiv 1 pmod{5} ] Using inverse modulo: [ y equiv 4 pmod{5} ] Hence possible integer values are ( y = 4, 9, 14, ... ) and so forth, bounded by ( y leq frac{71}{4} approx 17.75 ).8. Substitute and Check Maximum ( y ) To maximize ( x ), find minimal ( y ): - For ( y = 4 ): [ x = frac{71 - 4(4)}{5} = frac{71 - 16}{5} = frac{55}{5} = 11 ] - For ( y = 9 ): [ x = frac{71 - 4(9)}{5} = frac{71 - 36}{5} = frac{35}{5} = 7 ]9. Determine if larger ( y ) results in valid ( x ). [ y = 14 quad (not valid as it surpasses 17 limit) ]Thus, the maximum number of black balls in box ( C ) is when minimal ( y ) is chosen, verifying:[boxed{22}]

🧠:Okay, so there's this problem about Anya, Borya, and Vasya walking to school. They each have different constant speeds, and if someone catches up to another, they slow down to walk together. The question is whether all three can arrive together at school on both days, given the different departure orders.First, let me parse the problem again. On the first day, Anya leaves first, then Borya, then Vasya. Some pair arrives together. The next day, the departure order is reversed: Vasya first, then Borya, then Anya. The question is: can all three arrive together in the second day?Wait, no. The problem says: "Can all three arrive at school together?" So maybe it's asking if in both days, all three arrive together? Or is it asking if in the second day, all three can arrive together, given that in the first day a pair arrived together?Wait, the original problem states: "One day, Anya left first, followed by Borya, and then Vasya. Some pair of them arrived at school together. The next day, Vasya left first, followed by Borya, and then Anya. Can all three arrive at school together?" So the first day, departure order Anya, Borya, Vasya. Then a pair arrives together. The next day, departure order Vasya, Borya, Anya. The question is whether in this second day, can all three arrive together?Or maybe, the question is whether it's possible that in both days, all three arrive together? Hmm, the wording is a bit ambiguous, but probably the problem is asking if, given the departure orders on the two days, is it possible that on the next day (the second scenario), all three arrive together. Because the first day already had a pair arriving together, and the second day's departure order is different, can all three arrive together in that case?Alternatively, maybe the problem is asking whether in both days, all three can arrive together. But given that the first day's departure order leads to a pair arriving together, can the second day's departure order lead to all three arriving together. Hmm.Wait, the problem says: "Some pair of them arrived at school together. The next day, Vasya left first, followed by Borya, and then Anya. Can all three arrive at school together?" So the first day, when they left in Anya, Borya, Vasya order, a pair arrived together. Then the next day, when they left in Vasya, Borya, Anya order, can all three arrive together. So the answer is either yes or no, with an explanation.So we need to determine whether such a scenario is possible. Let's think about how the catching up works.Let me denote their speeds as v_A (Anya), v_B (Borya), v_V (Vasya). Since they have different speeds, we can assume v_A ≠ v_B ≠ v_V. Also, when a faster person catches up to a slower one, they start moving together at the slower person's speed. So if someone is caught, they merge into a group moving at the slower speed. If another person catches up to that group, they also join.In the first day: Anya leaves first, then Borya, then Vasya. Let's assume departure times are t=0 for Anya, t=t1 for Borya, t=t2 for Vasya. But the problem doesn't specify the departure intervals. Wait, actually, the problem says "they walk to school at constant but different speeds, without looking back or waiting for each other." However, if one catches up to another, they slow down. The departure order is given, but the problem doesn't specify the time intervals between their departures. Hmm. Wait, maybe the problem assumes that each leaves immediately after the previous one? But that's not stated. Wait, the original problem says: "Anya left first, followed by Borya, and then Vasya." Similarly for the next day. So maybe they leave at the same time intervals each day. But the problem doesn't specify the intervals. So perhaps we need to assume that the departure times are sequential with some fixed intervals, but since the problem doesn't specify, maybe the intervals can be arbitrary. However, since the answer might depend on the intervals, perhaps we can set the intervals such that in the first day, a pair arrives together, and in the second day, all three arrive together.Alternatively, maybe the problem is designed so that regardless of the intervals, given the speeds, if a pair arrived on the first day, then on the second day, all three can or cannot arrive together. But this is unclear.Wait, perhaps the key is that the departure order is reversed, so the fastest person now leaves last, which might allow for all three to merge. Let's think step by step.First, let's consider the first day: Anya leaves first, then Borya, then Vasya. A pair arrives together. Let's figure out which pair that could be. To have a pair arrive together, someone must catch up to another. Let's suppose that the speeds are such that the later person is faster, so they can catch up.If Anya is the slowest, then Borya, then Vasya is the fastest. So Vasya would have the highest speed. But in that case, Vasya would catch up to Borya, and then Vasya would slow down to Borya's speed. Then, Borya (and Vasya) might catch up to Anya, and then all three would walk at Anya's speed, arriving together. But the problem states that only a pair arrived together on the first day. So maybe in this scenario, Vasya catches up to Borya, and they arrive together, but don't catch Anya. Alternatively, Borya catches up to Anya, and then Vasya catches up to them both.Alternatively, if Anya is the fastest, then she would never be caught. Then Borya, leaving after Anya but slower, would not catch her. Vasya, leaving after Borya, if he's faster than Borya, would catch Borya, and then they walk together. If Vasya is faster than Borya, then Vasya would catch Borya, forming a pair that arrives together. But since Anya is the fastest, she arrives first, and then the pair arrives later. So in this case, a pair arrives together, which matches the first day's condition. But then on the next day, when departure order is Vasya, Borya, Anya. If Vasya is slowest, Borya is medium, Anya is fastest. Then Vasya leaves first, then Borya, then Anya. If Anya is fastest, she would catch Borya, then catch Vasya. But if she's much faster, she might catch both, forming a group. But would all three arrive together?Wait, let's formalize this. Let's denote the distance to school as D (unknown, but same for all). Let the speeds be v_A > v_B > v_V. So Anya is fastest, Vasya is slowest.First day: departure order Anya, Borya, Vasya. So Anya leaves at t=0, Borya at t=τ, Vasya at t=2τ (assuming equal intervals, but the problem doesn't specify). However, the problem doesn't mention the time intervals between departures, so maybe we can choose them such that the desired catching up happens.But perhaps we can assume that the time between departures is 1 unit for simplicity.But to make it general, let's suppose the time between departures is Δt. So Anya departs at t=0, Borya at t=Δt, Vasya at t=2Δt.In the first scenario, with Anya first. If Anya is the fastest, she will not be caught by anyone. Borya is next; if Vasya is faster than Borya, he would catch Borya. But if Vasya is slower, he won't. Wait, but in our assumption, v_A > v_B > v_V, so Vasya is slowest. Therefore, Vasya cannot catch Borya, and Borya cannot catch Anya. So in this case, all three arrive separately, with Anya first, then Borya, then Vasya. But the problem states that a pair arrived together. Therefore, our assumption that v_A > v_B > v_V must be incorrect in this case.Therefore, for a pair to arrive together on the first day, someone must have been caught. Therefore, the departure order is Anya (first), then Borya, then Vasya. So if Vasya is faster than Borya, who is faster than Anya, then Vasya would catch Borya, and then Borya would catch Anya. Wait, no. Let's think carefully.Suppose the speeds are such that the later departures are faster. So Vasya is fastest, then Borya, then Anya slowest. But departure order is Anya first, then Borya, then Vasya. In this case, Borya is faster than Anya, so Borya will catch Anya. Similarly, Vasya is faster than Borya, so Vasya will catch Borya.Wait, let's suppose speeds v_V > v_B > v_A. So Vasya is fastest, Borya next, Anya slowest.First day: Anya departs first. Then Borya departs later. Since Borya is faster than Anya, he will catch up to Anya. The time it takes for Borya to catch Anya is D / (v_B - v_A), but actually, we need to consider the head start.Wait, let's model this mathematically.Let’s denote:- Distance to school: S (same for all)- Speeds: v_A, v_B, v_V (different constants)- Departure times: Anya at t=0, Borya at t=τ, Vasya at t=2τ (assuming equal intervals τ between departures)But actually, the problem doesn’t specify the intervals. So maybe the intervals are arbitrary. But since the problem doesn't specify, perhaps we can choose the intervals to make the catching up possible. However, the problem states that they leave in that order, but doesn't say after how much time. This is a bit ambiguous.Alternatively, maybe the time between departures is zero, but that doesn't make sense. So perhaps the key is that they leave one after another, with some non-zero time in between, but the exact interval is not specified, so we can choose intervals to suit our needs.But to solve the problem, perhaps we need to consider that regardless of the intervals, given the speeds, can such a scenario happen. Or maybe the intervals are fixed, and we have to determine based on the speeds.This is a bit confusing. Let's assume that the time between departures is 1 unit for simplicity. Then Anya departs at t=0, Borya at t=1, Vasya at t=2.If Vasya is the fastest, he might catch up to Borya and Anya. Similarly, Borya might catch Anya.Alternatively, if Anya is the fastest, then she arrives first, and no one catches her. If Borya is slower than Anya but faster than Vasya, then Vasya might catch Borya if Vasya is faster.Wait, this is getting complicated. Maybe we need to set up equations.Let’s formalize the first day:Departure order: Anya (t=0), Borya (t=τ), Vasya (t=2τ). Assume the time between departures is τ.Speeds: Let's denote them as v_A, v_B, v_V. All different.If someone is caught, they merge and proceed at the slower speed.In the first day, a pair arrives together. So either Borya catches Anya, Vasya catches Borya, or Vasya catches Anya directly. Depending on the speeds.Let’s suppose that Borya is faster than Anya. Then Borya will catch Anya at some time. Let’s compute when.Anya departs at t=0, speed v_A.Borya departs at t=τ, speed v_B > v_A.The time taken for Borya to catch Anya is (distance Anya has covered when Borya starts) / (v_B - v_A) = (v_A * τ) / (v_B - v_A). So the catch-up time after Borya departs is (v_A τ)/(v_B - v_A). Therefore, the total time from t=0 is τ + (v_A τ)/(v_B - v_A) = τ(1 + v_A/(v_B - v_A)) = τ v_B / (v_B - v_A).After catching up, they proceed at Anya's speed v_A? Wait, no. If Borya catches Anya, then according to the problem, they slow down to walk together with the one they caught up to. So the group would proceed at Anya's speed. Then Vasya, who departs at t=2τ, if his speed is higher than v_A, would catch up to the group.Wait, but Vasya's speed v_V compared to v_A? If v_V > v_A, then Vasya would eventually catch the group. But if Vasya is slower than Anya, he won't. So if Borya is faster than Anya, and Vasya is faster than Anya, then Vasya would catch the group (Anya and Borya) moving at v_A. If Vasya is slower than Anya, he won't catch up.But in the first day, only a pair arrives together. So if Borya catches Anya, forming a group moving at v_A, and Vasya is slower than v_A, then Vasya won't catch up. Thus, Anya and Borya arrive together at time T = τ v_B / (v_B - v_A) + (S - distance covered when caught)/v_A. Wait, maybe we need to compute the total time.Alternatively, let's compute the time when Borya catches Anya. Let’s assume distance to school is S.Anya's position at time t is v_A * t.Borya starts at t=τ, so his position is v_B*(t - τ) for t >= τ.They meet when v_A t = v_B (t - τ)Solving for t: v_A t = v_B t - v_B τt(v_B - v_A) = v_B τt = (v_B τ)/(v_B - v_A)At this time, they meet, and then proceed at Anya's speed v_A.The remaining distance to school is S - v_A * t = S - v_A*(v_B τ)/(v_B - v_A)The time to cover the remaining distance is [S - v_A*(v_B τ)/(v_B - v_A)] / v_A = S/v_A - (v_B τ)/(v_B - v_A)Therefore, total arrival time for Anya and Borya is t + [remaining time] = (v_B τ)/(v_B - v_A) + [S/v_A - (v_B τ)/(v_B - v_A)] = S/v_AWait, that's interesting. So regardless of when Borya catches Anya, the total time for Anya and Borya to arrive is S/v_A, which is the same as if Anya walked alone. That makes sense because after catching up, they continue at Anya's speed. So the arrival time is the same as Anya's original arrival time. Wait, that can't be right. Let me check the math.Let’s take S as the total distance.Anya's arrival time without any catching is S / v_A.When Borya catches Anya at time t = (v_B τ)/(v_B - v_A), the distance covered by Anya is v_A * t = v_A * (v_B τ)/(v_B - v_A). The remaining distance is S - v_A t.The time to cover the remaining distance is (S - v_A t)/v_A = S/v_A - t.Thus, total arrival time is t + (S/v_A - t) = S/v_A, which is indeed the same as Anya's original arrival time. So even though Borya catches up, the total arrival time for both is the same as if Anya walked alone. Therefore, if Vasya is slower than Anya, he will arrive later, so the pair (Anya and Borya) arrive together at time S/v_A, and Vasya arrives later. If Vasya is faster than Anya, he would catch up to the group, but since the group is moving at Anya's speed, which is slower than Vasya's original speed, he would merge and the entire group would arrive at S/v_A. But wait, if Vasya is faster than Anya, he will catch the group.So let's see. Vasya departs at t=2τ. His speed is v_V. If v_V > v_A, then he will catch the group (Anya and Borya) who are moving at speed v_A.The position of the group at time t is:Before merging with Borya, Anya was at v_A t. After merging at t = (v_B τ)/(v_B - v_A), they continue at v_A. So the position from then on is v_A * (v_B τ)/(v_B - v_A) + v_A*(t - (v_B τ)/(v_B - v_A)) = v_A t.So the group's position is always v_A t, same as Anya's original path.Therefore, Vasya, starting at t=2τ, has to catch up to the group which is at position v_A t when he starts.The distance between Vasya and the group at t=2τ is v_A * 2τ - 0 (since Vasya starts at position 0). Wait, no. Wait, Vasya starts at the same starting point, right? They all live on the same stair landing. So at t=2τ, Vasya starts from position 0, while the group is at position v_A * 2τ.So the distance Vasya needs to cover is v_A * 2τ.Vasya's speed is v_V. The relative speed between Vasya and the group is v_V - v_A (if v_V > v_A).Time to catch up: (v_A * 2τ) / (v_V - v_A)This time is measured from Vasya's departure at t=2τ. So the catch-up time is t_catch = 2τ + (2τ v_A)/(v_V - v_A)At this time, the group's position is v_A * t_catch = v_A [2τ + (2τ v_A)/(v_V - v_A)]But Vasya catches up when his position equals the group's position:v_V (t_catch - 2τ) = v_A t_catchWhich simplifies to:v_V (t_catch - 2τ) = v_A t_catchBut t_catch = 2τ + (2τ v_A)/(v_V - v_A)Plugging in:v_V [ (2τ + (2τ v_A)/(v_V - v_A)) - 2τ ] = v_A [2τ + (2τ v_A)/(v_V - v_A)]Simplify left side:v_V [ (2τ v_A)/(v_V - v_A) ) ] = v_A [2τ + (2τ v_A)/(v_V - v_A)]Divide both sides by 2τ:v_V [ v_A / (v_V - v_A) ) ] = v_A [1 + v_A / (v_V - v_A) ]Multiply both sides by (v_V - v_A):v_V v_A = v_A [ (v_V - v_A) + v_A ]Simplify RHS:v_V v_A = v_A v_VWhich holds true. Therefore, the catch-up time is correct.After catching up, Vasya joins the group, and they proceed at speed v_A. The remaining distance to school is S - v_A t_catch.The time to cover the remaining distance is (S - v_A t_catch)/v_A = S/v_A - t_catchTotal arrival time is t_catch + (S/v_A - t_catch) = S/v_ATherefore, all three arrive together at S/v_A.But wait, the problem states that on the first day, only a pair arrived together. But according to this, if Vasya is faster than Anya, all three would arrive together. Therefore, in order to have only a pair arriving together on the first day, Vasya must be slower than Anya. So that Vasya cannot catch up to the group (Anya and Borya), who are moving at Anya's speed.Therefore, for the first day to have only a pair arriving together, we must have v_A < v_B (so Borya catches Anya) and v_V < v_A (so Vasya can't catch the group). Then Anya and Borya arrive together at S/v_A, and Vasya arrives later at S/v_V.But wait, if v_V < v_A, then Vasya is slower than Anya, so he would arrive later than Anya's group. But Anya and Borya arrive together at S/v_A, so Vasya arrives at S/v_V, which is later. Therefore, only a pair arrives together. That satisfies the first day's condition.So speeds: v_B > v_A > v_VFirst day: Anya, Borya, Vasya depart. Borya catches Anya, they arrive together at S/v_A. Vasya arrives later.Then, the next day, departure order is Vasya, Borya, Anya.So Vasya departs first, then Borya, then Anya.We need to see if all three can arrive together.Let's analyze the second day.Departure order: Vasya (t=0), Borya (t=τ), Anya (t=2τ). Again assuming the same interval τ.Speeds: v_B > v_A > v_V.But wait, in this case, the fastest is Borya, then Anya, then Vasya. But departure order is Vasya (slowest) first, then Borya (fastest), then Anya.So Borya departs at t=τ, speed v_B. Since he's the fastest, he will catch up to Vasya.Similarly, Anya departs at t=2τ, speed v_A. If Anya is faster than Vasya, she can catch up to Vasya. But Borya is faster than Anya, so Borya would catch both Vasya and Anya.Let’s compute.First, Borya catches Vasya.Vasya departs at t=0, position at time t is v_V t.Borya departs at t=τ, position at time t >= τ is v_B (t - τ).They meet when v_V t = v_B (t - τ)Solving: v_V t = v_B t - v_B τt(v_B - v_V) = v_B τt = (v_B τ)/(v_B - v_V)At this time, they meet and proceed at Vasya's speed v_V.Then, Anya departs at t=2τ. Her speed is v_A, which is less than v_B but greater than v_V.So Anya will try to catch up to the group (Vasya and Borya) moving at speed v_V.The group's position at time t >= (v_B τ)/(v_B - v_V) is:At time t_meet = (v_B τ)/(v_B - v_V), the group has position v_V t_meet.After that, they continue moving at speed v_V, so their position at time t is v_V t_meet + v_V (t - t_meet) = v_V t.Wait, that's the same as if Vasya had been walking alone. So merging with Borya doesn't change the group's position over time; they continue moving as if Vasya was walking alone. Because after merging, they proceed at Vasya's speed. Therefore, the group's position is v_V t for t >= t_meet.Anya departs at t=2τ. Her position at time t >= 2τ is v_A (t - 2τ).She needs to catch up to the group at position v_V t.So setting v_A (t - 2τ) = v_V tSolving for t:v_A t - 2τ v_A = v_V tt(v_A - v_V) = 2τ v_At = (2τ v_A)/(v_A - v_V)But we need to check if this t is greater than t_meet (when Borya caught Vasya).t_meet = (v_B τ)/(v_B - v_V)Compare with t = (2τ v_A)/(v_A - v_V)We need to see if (2τ v_A)/(v_A - v_V) > (v_B τ)/(v_B - v_V)Simplify:2 v_A / (v_A - v_V) > v_B / (v_B - v_V)Cross-multiplying (since denominators are positive because speeds are positive and v_A > v_V, v_B > v_V):2 v_A (v_B - v_V) > v_B (v_A - v_V)Expand both sides:2 v_A v_B - 2 v_A v_V > v_A v_B - v_B v_VSubtract v_A v_B from both sides:v_A v_B - 2 v_A v_V > -v_B v_VAdd 2 v_A v_V to both sides:v_A v_B > 2 v_A v_V - v_B v_VFactor right side:v_A v_B > v_V (2 v_A - v_B)But since v_B > v_A (from the first day's condition where Borya is faster than Anya), then 2 v_A - v_B is positive only if 2 v_A > v_B. But since v_B > v_A, 2 v_A could be greater or less than v_B.Wait, this is getting messy. Maybe we can assign numerical values to test.Let’s choose speeds such that v_B > v_A > v_V. For example:Let’s take v_V = 1 m/s, v_A = 2 m/s, v_B = 3 m/s.Interval τ = 1 second.First day: Anya departs at t=0, Borya at t=1, Vasya at t=2.First day, Borya catches Anya:t = (v_B τ)/(v_B - v_A) = (3*1)/(3-2) = 3 seconds.At t=3, Borya catches Anya. They proceed at 2 m/s. Vasya departs at t=2, speed 1 m/s. He can't catch up. So Anya and Borya arrive together at S = 2*3 + 2*(T - 3). Wait, total arrival time is S/v_A = S/2.But let's compute S. Let's assume the school is 6 meters away. Then arrival time for Anya and Borya is 6/2 = 3 seconds? But Anya walked for 3 seconds at 2 m/s: 6 meters. Yes. So arrival time is 3 seconds. Vasya departed at t=2, needs to walk 6 meters at 1 m/s: takes 6 seconds, arrives at t=2+6=8 seconds. So pair arrives together at 3 seconds, Vasya arrives at 8.Next day: departure order Vasya, Borya, Anya. Vasya at t=0, Borya at t=1, Anya at t=2.First, Borya catches Vasya:t_meet = (v_B τ)/(v_B - v_V) = (3*1)/(3-1)= 1.5 seconds.At t=1.5, Borya catches Vasya. They proceed at 1 m/s.Anya departs at t=2. Her speed is 2 m/s. The group (Vasya and Borya) is at position v_V * t = 1 * 1.5 = 1.5 meters at t=1.5, and after that, they continue at 1 m/s. So at t=2, their position is 1.5 + 1*(2 -1.5) = 1.5 + 0.5 = 2 meters.Anya starts at t=2, position 0. She needs to catch up to the group at position 2 meters, moving at 1 m/s.Relative speed: 2 -1 =1 m/s.Distance to catch up: 2 meters.Time to catch up: 2/1 =2 seconds. So she catches up at t=2 +2=4 seconds.At t=4, the group's position is 2 +1*(4-2)=2 +2=4 meters.But Anya's position is 2*(4-2)=4 meters. Yes, they meet at 4 meters.Then, they merge and proceed at 1 m/s. The remaining distance to school is 6 -4=2 meters. Time to finish: 2/1=2 seconds. Arrival time:4 +2=6 seconds.But the group (Vasya, Borya, Anya) arrives together at 6 seconds.But what about the original group's progress? Wait, school is 6 meters away. So arrival time is 6 seconds. But in the first day, Anya and Borya arrived at 3 seconds. Here, they arrive at 6 seconds. So all three arrive together on the second day.Wait, this seems to satisfy the condition. So in this example, on the first day, Anya and Borya arrive together, Vasya arrives later. On the second day, all three arrive together. Therefore, the answer is yes.But we need to confirm that in this example, the first day's condition holds (a pair arrives together) and the second day all three arrive together.Yes, in the first day, Anya and Borya arrive together at 3 seconds, Vasya arrives at 8. In the second day, all arrive at 6 seconds. So it works.Therefore, the answer is yes, it is possible.But wait, let's verify with another example to ensure it's not a coincidence.Take v_V=1, v_A=3, v_B=4. τ=1. School distance=12 meters.First day: Anya departs t=0. Borya t=1. Vasya t=2.Borya catches Anya at t=(4*1)/(4-3)=4 seconds. They merge and proceed at 3 m/s. They have covered 3*4=12 meters, so arrival time is 4 seconds. Vasya departs at t=2, speed 1 m/s. Arrives at 2 +12/1=14 seconds. So pair arrives at 4 seconds.Next day: Vasya departs t=0, Borya t=1, Anya t=2.Borya catches Vasya at t=(4*1)/(4-1)=1.333... seconds. Position at t=1.333: 1*1.333=1.333 meters. Then proceed at 1 m/s. Anya departs at t=2. The group's position at t=2 is 1.333 +1*(2 -1.333)=1.333 +0.666=2 meters.Anya needs to catch up from 0 to 2 meters. Her speed is3 m/s, group's speed 1 m/s. Relative speed 2 m/s. Time to catch up:2/2=1 second. Catches up at t=2+1=3 seconds. Position:3*(3-2)=3 meters. Group's position:2 +1*(3-2)=3 meters. Merge, proceed at1 m/s. Remaining distance:12-3=9. Time:9/1=9 seconds. Arrival time:3+9=12 seconds.But school is 12 meters, so arrival at12 seconds. However, Vasya and Borya were moving at1 m/s after merge. From t=1.333 to t=3, they moved from1.333 to3 meters. Then from3 to12 meters takes9 seconds. Yes, total time from t=0 is12 seconds. So all three arrive together.In the first day, Anya and Borya arrived at4 seconds, Vasya at14. Second day, all arrive at12 seconds. So yes, possible.Therefore, the answer is yes, all three can arrive together on the second day.

✅:Step-by-Step Explanation and Answer:To determine if all three can arrive together on the second day, we analyze the scenarios based on their departure orders and speeds.Key Assumptions:1. Speeds: Let their speeds be ( v_A ) (Anya), ( v_B ) (Borya), ( v_V ) (Vasya), with ( v_B > v_A > v_V ).2. Departure Intervals: Assume equal intervals ( tau ) between departures for simplicity.3. School Distance: Denote the distance to school as ( S ).---First Day (Departure Order: Anya → Borya → Vasya):1. Borya catches Anya: Since ( v_B > v_A ), Borya catches Anya at time ( t = frac{v_B tau}{v_B - v_A} ). They merge and proceed at ( v_A ).2. Vasya's Progress: Vasya departs later and moves slower (( v_V < v_A )), so he cannot catch up. Thus, Anya and Borya arrive together at ( frac{S}{v_A} ), while Vasya arrives later at ( frac{S}{v_V} ).---Second Day (Departure Order: Vasya → Borya → Anya):1. Borya catches Vasya: Borya (( v_B )) catches Vasya (( v_V )) at ( t = frac{v_B tau}{v_B - v_V} ). They merge and proceed at ( v_V ).2. Anya catches the Merged Group: Anya (( v_A )), though slower than Borya, is faster than the merged group (( v_A > v_V )). She catches them at ( t = frac{2tau v_A}{v_A - v_V} ).3. All Merge and Arrive Together: After merging, all three proceed at ( v_V ). The total time to school is calculated to be ( frac{S}{v_V} ), ensuring all three arrive simultaneously.---Example with Concrete Values:- Let ( v_V = 1 , text{m/s}, v_A = 2 , text{m/s}, v_B = 3 , text{m/s} ), ( tau = 1 , text{s} ), and ( S = 6 , text{m} ). - First Day: Anya and Borya arrive at ( 3 , text{s} ); Vasya arrives at ( 8 , text{s} ). - Second Day: All three merge and arrive at ( 6 , text{s} ).---Conclusion:By strategically choosing speeds where ( v_B > v_A > v_V ), it is possible for all three to arrive together on the second day despite the reversed departure order. The critical factor is the relative speeds and merging dynamics.Answer: Yes, all three can arrive at school together on the second day.

🔑:1. Let us assume that Anya walks slower than Borya, and Vasya walks much slower than both of them.2. On the first day: - Anya leaves first, followed by Borya, and then Vasya. - Borya, walking faster than Anya, will eventually catch up to her. - Once Borya catches up to Anya, the two will walk together at Anya's speed. - Since Vasya is much slower, he will not be able to catch up to Anya and Borya by the time they reach school.3. On the following day: - Vasya leaves first, followed by Borya, and then Anya. - Since Vasya walks much slower, Borya, walking faster, will catch up to him. - After Borya catches up to Vasya, they continue walking together at Vasya's speed. - Anya, who is also faster than Vasya, will eventually catch up to both Borya and Vasya. - Hence, all three of them will arrive at school together. Conclusion(boxed{text{They can all arrive at school together.}})

🧠:Okay, so I need to find the radius of a sphere that passes through four points in a cube. The cube is labeled ABCD A₁B₁C₁D₁, and the edge length is 1. The sphere goes through vertices A and C, and the midpoints E and F of edges B₁C₁ and C₁D₁ respectively. Hmm, let's start by visualizing the cube and locating these points.First, let me sketch the cube mentally. Cube ABCD A₁B₁C₁D₁. So, the base is ABCD, and the top face is A₁B₁C₁D₁. The edges are all length 1. Let me recall the coordinates of the cube vertices. If I place the cube in a 3D coordinate system with vertex A at the origin, then:- A is (0, 0, 0)- B is (1, 0, 0)- C is (1, 1, 0)- D is (0, 1, 0)- A₁ is (0, 0, 1)- B₁ is (1, 0, 1)- C₁ is (1, 1, 1)- D₁ is (0, 1, 1)Okay, so points E and F are midpoints of B₁C₁ and C₁D₁. Let's compute their coordinates.Midpoint F of B₁C₁: B₁ is (1, 0, 1) and C₁ is (1, 1, 1). The midpoint would be average of coordinates:F_x = (1 + 1)/2 = 1F_y = (0 + 1)/2 = 0.5F_z = (1 + 1)/2 = 1So F is (1, 0.5, 1)Similarly, midpoint E of C₁D₁: C₁ is (1, 1, 1), D₁ is (0, 1, 1). Midpoint E:E_x = (1 + 0)/2 = 0.5E_y = (1 + 1)/2 = 1E_z = (1 + 1)/2 = 1Thus, E is (0.5, 1, 1)So the four points the sphere passes through are:A: (0, 0, 0)C: (1, 1, 0)F: (1, 0.5, 1)E: (0.5, 1, 1)Now, to find the radius R of the sphere passing through these four points, I need to determine the equation of the sphere and then compute its radius.A sphere in 3D space is generally given by the equation:(x - h)^2 + (y - k)^2 + (z - l)^2 = R^2where (h, k, l) is the center and R is the radius.Since the sphere passes through these four points, plugging each point into the equation will give us four equations to solve for h, k, l, and R. However, since all four points lie on the sphere, we can set up the equations and solve for the unknowns.Let me write the equations:For point A (0,0,0):(0 - h)^2 + (0 - k)^2 + (0 - l)^2 = R^2 ⇒ h² + k² + l² = R² --- (1)For point C (1,1,0):(1 - h)^2 + (1 - k)^2 + (0 - l)^2 = R² ⇒ (1 - h)² + (1 - k)² + l² = R² --- (2)For point F (1, 0.5, 1):(1 - h)^2 + (0.5 - k)^2 + (1 - l)^2 = R² --- (3)For point E (0.5, 1, 1):(0.5 - h)^2 + (1 - k)^2 + (1 - l)^2 = R² --- (4)So, equations (1) through (4). Let's subtract equation (1) from equation (2) to eliminate R².Equation (2) - Equation (1):(1 - h)^2 + (1 - k)^2 + l² - (h² + k² + l²) = 0Expanding (1 - h)^2 = 1 - 2h + h², (1 - k)^2 = 1 - 2k + k²So:[1 - 2h + h² + 1 - 2k + k² + l²] - [h² + k² + l²] = 0Simplify:1 - 2h + h² + 1 - 2k + k² + l² - h² - k² - l² = 0Simplify more:(1 + 1) + (-2h - 2k) + (h² - h²) + (k² - k²) + (l² - l²) = 0Which becomes:2 - 2h - 2k = 0 ⇒ 2 = 2h + 2k ⇒ h + k = 1 --- (5)Okay, so equation (5): h + k = 1.Now, let's subtract equation (1) from equation (3):Equation (3) - Equation (1):(1 - h)^2 + (0.5 - k)^2 + (1 - l)^2 - (h² + k² + l²) = 0Expanding each term:(1 - h)^2 = 1 - 2h + h²(0.5 - k)^2 = 0.25 - k + k² [Wait, (0.5 - k)^2 = (0.5)^2 - 2*0.5*k + k² = 0.25 - k + k²](1 - l)^2 = 1 - 2l + l²So plugging back:[1 - 2h + h² + 0.25 - k + k² + 1 - 2l + l²] - [h² + k² + l²] = 0Simplify:1 -2h + h² + 0.25 -k + k² +1 -2l + l² - h² - k² - l² = 0Simplify term by term:1 + 0.25 +1 = 2.25-2h -k -2lh² - h² = 0k² - k² = 0l² - l² = 0So total:2.25 -2h -k -2l = 0 ⇒ -2h -k -2l = -2.25 ⇒ 2h + k + 2l = 2.25 --- (6)Similarly, subtract equation (1) from equation (4):Equation (4) - Equation (1):(0.5 - h)^2 + (1 - k)^2 + (1 - l)^2 - (h² + k² + l²) = 0Expanding:(0.5 - h)^2 = 0.25 - h + h²(1 - k)^2 = 1 - 2k + k²(1 - l)^2 = 1 - 2l + l²So:[0.25 - h + h² + 1 - 2k + k² + 1 - 2l + l²] - [h² + k² + l²] = 0Simplify:0.25 - h + h² +1 -2k + k² +1 -2l + l² - h² - k² - l² = 0Adding constants: 0.25 +1 +1 = 2.25Variables: -h -2k -2lQuadratic terms cancel: h² - h² + k² - k² + l² - l² = 0So:2.25 - h -2k -2l = 0 ⇒ -h -2k -2l = -2.25 ⇒ h + 2k + 2l = 2.25 --- (7)Now, so far, we have equations (5), (6), and (7):(5): h + k = 1(6): 2h + k + 2l = 2.25(7): h + 2k + 2l = 2.25Now, we can solve these three equations for h, k, l.First, from equation (5): h = 1 - k. Let's substitute h = 1 - k into equations (6) and (7).Substitute into equation (6):2*(1 - k) + k + 2l = 2.25Compute:2 - 2k + k + 2l = 2.25 ⇒ 2 -k + 2l = 2.25 ⇒ -k + 2l = 0.25 --- (8)Substitute h =1 -k into equation (7):(1 - k) + 2k + 2l = 2.25 ⇒ 1 + k + 2l = 2.25 ⇒ k + 2l = 1.25 --- (9)Now, equations (8) and (9):(8): -k + 2l = 0.25(9): k + 2l = 1.25Let me add equations (8) and (9):(-k + 2l) + (k + 2l) = 0.25 + 1.25 ⇒ 4l = 1.5 ⇒ l = 1.5 /4 ⇒ l = 0.375So l = 3/8? Wait, 1.5 divided by 4 is 0.375, which is 3/8. Wait, 0.375 is 3/8? Wait, 1/2 is 0.5, 1/4 is 0.25, 3/4 is 0.75. So 0.375 is 3/8. Hmm. Let me check 3/8 is 0.375. Yes, because 1/8 is 0.125, so 3/8 is 0.375. So l = 3/8.Now, from equation (9): k + 2*(3/8) = 1.25 ⇒ k + 6/8 = 1.25 ⇒ k + 0.75 = 1.25 ⇒ k = 1.25 - 0.75 = 0.5. So k = 0.5, which is 1/2.Then from equation (5): h =1 -k =1 -0.5=0.5. So h=0.5.Therefore, the center of the sphere is at (h, k, l) = (0.5, 0.5, 0.375). Let me confirm this.So, h=0.5, k=0.5, l=0.375.Now, compute R² from equation (1): h² + k² + l² = (0.5)^2 + (0.5)^2 + (0.375)^2 = 0.25 + 0.25 + 0.140625 = 0.640625Therefore, R = sqrt(0.640625). Let me compute that.0.640625 is equal to 640625/1000000. Let me see:sqrt(0.640625). Let's convert to fraction.0.640625 = 0.64 + 0.000625 = 64/100 + 625/1000000 = 16/25 + 1/1600. Wait, maybe better to write as 640625/1000000.Simplify numerator and denominator by dividing numerator and denominator by 25:640625 ÷25=25625, 1000000÷25=40000.So 25625/40000. Simplify again by dividing numerator and denominator by 25:25625÷25=1025, 40000÷25=1600.1025/1600. Check if divisible by 5: 1025/5=205, 1600/5=320. 205/320. Again by 5: 41/64.So 41/64. Thus sqrt(41/64)=sqrt(41)/8≈6.403/8≈0.800375. But let's see, 41 is prime, so sqrt(41) is irrational. So exact value is sqrt(41)/8.Wait, 0.640625 * 64 = 41. So (sqrt(41)/8)^2 = 41/64 = 0.640625. Yes, correct.Therefore, the radius R is sqrt(41)/8. But let me confirm this.Wait, 0.5 squared is 0.25, another 0.5 squared is 0.25, 0.375 squared is 0.140625. Adding up: 0.25 + 0.25 = 0.5, plus 0.140625 gives 0.640625. Which is 41/64. So sqrt(41)/8. Yes. So R = √41 / 8.But let me verify if this is correct by plugging into another equation, say equation (3) for point F (1, 0.5, 1).Compute (1 - 0.5)^2 + (0.5 - 0.5)^2 + (1 - 0.375)^2 = ?(0.5)^2 + 0 + (0.625)^2 = 0.25 + 0 + 0.390625 = 0.640625. Which is equal to R². Correct.Similarly, for point E (0.5, 1, 1):(0.5 -0.5)^2 + (1 -0.5)^2 + (1 -0.375)^2 = 0 + 0.25 + 0.390625 = 0.640625. Correct.And for point C (1,1,0):(1 -0.5)^2 + (1 -0.5)^2 + (0 -0.375)^2 = 0.25 + 0.25 + 0.140625 = 0.640625. Correct.And point A (0,0,0):0.5² +0.5² +0.375² = same as above. Correct.So all four points satisfy the sphere equation with center (0.5, 0.5, 0.375) and radius √(41)/8. Therefore, the radius R is √41 divided by 8. Simplifying √41/8, which is approximately 0.8...But the question says to present the exact value, so √41 /8 is the exact radius. Let me check if I can simplify further, but 41 is prime, so √41 is irrational and cannot be simplified. Thus, the answer should be √41 /8.Wait, but let me check once again to avoid any mistake in the algebra.So equations (5), (6), (7):(5): h + k = 1(6): 2h + k + 2l = 9/4 (since 2.25 is 9/4)(7): h + 2k + 2l = 9/4Then solving:From (5): h =1 -kSubstitute into (6):2(1 -k) +k +2l = 9/4 ⇒ 2 -2k +k +2l = 9/4 ⇒ 2 -k +2l =9/4Multiply both sides by 4 to eliminate fractions:8 -4k +8l =9 ⇒ -4k +8l=1 ⇒ divide by 4: -k +2l=0.25, which is equation (8)Similarly, substitute h=1 -k into (7):(1 -k) +2k +2l =9/4 ⇒1 +k +2l=9/4 ⇒k +2l=9/4 -1=5/4=1.25, which is equation (9)So equations (8) and (9):- k +2l=0.25k +2l=1.25Adding them: 4l=1.5 ⇒ l=1.5/4=3/8=0.375Then from equation (9): k +2*(3/8)=5/4 ⇒k=5/4 -6/8=5/4 -3/4=2/4=1/2=0.5Then h=1 -0.5=0.5. Thus center (0.5,0.5,0.375). Correct.Then R squared is 0.5² +0.5² + (3/8)²=0.25+0.25+9/64=0.5+0.140625=0.640625=41/64. Hence R=√(41)/8. Correct.Therefore, the radius is √41 divided by 8.But just to be thorough, let me visualize the sphere. The cube has edges of length 1, so the sphere passes through two opposite corners of the base (A and C) and two midpoints on the top face. The center is at (0.5,0.5, 0.375), which is along the vertical centerline of the cube but shifted down towards the base, which makes sense since the sphere passes through points on the base and top. The radius is sqrt(41)/8. Let's compute sqrt(41):sqrt(41) ≈6.403, so 6.403/8≈0.8004. The edge length is 1, so a radius of ~0.8 seems plausible. Since the sphere has to reach from the base to the top, but the center is closer to the base, the radius would be a bit less than the distance from center to the top, which is 1 -0.375=0.625. Wait, but 0.8 is larger than 0.625. Wait, maybe I miscalculated.Wait, the center is at (0.5,0.5,0.375). The top face is at z=1. The distance from center to top face is 1 -0.375=0.625. The radius is ~0.8, which is larger, so the sphere extends above the cube. Wait, but the cube's edge is 1, so from z=0 to z=1. The center is at z=0.375. The radius is ~0.8, so the sphere would extend from z=0.375 -0.8 ≈-0.425 to z=0.375 +0.8≈1.175. So it goes below the cube and above the cube. But the sphere passes through points A (z=0), C (z=0), E and F (z=1). So the sphere must touch those points, which are on the cube's surfaces. So even though the center is inside the cube along x and y, it's at z=0.375, so the distance from center to A (z=0) is sqrt(0.5² +0.5² +0.375²)=sqrt(0.25+0.25+0.140625)=sqrt(0.640625)=~0.8, which is R. Similarly, the distance from center to E (z=1) is sqrt( (0.5 -0.5)^2 + (1 -0.5)^2 + (1 -0.375)^2 )=sqrt(0 +0.25 +0.390625)=sqrt(0.640625)=same R. So even though the sphere extends beyond the cube, the four points lie on it. That's okay. The calculations are correct.Thus, after thorough analysis, the radius is √41 /8.Final AnswerThe radius ( R ) of the sphere is (boxed{dfrac{sqrt{41}}{8}}).

✅:Given a cube ( ABCD A_1 B_1 C_1 D_1 ) with an edge length of 1, we need to find the radius ( R ) of a sphere that passes through vertices ( A ) and ( C ) and the midpoints ( F ) and ( E ) of edges ( B_1 C_1 ) and ( C_1 D_1 ) respectively.First, we determine the coordinates of the points:- ( A = (0, 0, 0) )- ( C = (1, 1, 0) )- ( F ) (midpoint of ( B_1 C_1 )) = ( (1, 0.5, 1) )- ( E ) (midpoint of ( C_1 D_1 )) = ( (0.5, 1, 1) )The equation of a sphere in 3D space is ((x - h)^2 + (y - k)^2 + (z - l)^2 = R^2). We set up the following equations using the coordinates of the points:1. For ( A ): ( h^2 + k^2 + l^2 = R^2 )2. For ( C ): ( (1 - h)^2 + (1 - k)^2 + l^2 = R^2 )3. For ( F ): ( (1 - h)^2 + (0.5 - k)^2 + (1 - l)^2 = R^2 )4. For ( E ): ( (0.5 - h)^2 + (1 - k)^2 + (1 - l)^2 = R^2 )Subtracting these equations to eliminate ( R^2 ):- Subtracting equation 1 from 2: ( 2 - 2h - 2k = 0 Rightarrow h + k = 1 )- Subtracting equation 1 from 3: ( 2.25 - 2h - k - 2l = 0 Rightarrow 2h + k + 2l = 2.25 )- Subtracting equation 1 from 4: ( 2.25 - h - 2k - 2l = 0 Rightarrow h + 2k + 2l = 2.25 )Solving these equations:1. ( h + k = 1 )2. ( 2h + k + 2l = 2.25 )3. ( h + 2k + 2l = 2.25 )Substituting ( h = 1 - k ) into the other equations:- ( 2(1 - k) + k + 2l = 2.25 Rightarrow -k + 2l = 0.25 )- ( (1 - k) + 2k + 2l = 2.25 Rightarrow k + 2l = 1.25 )Solving these:- Adding equations: ( 4l = 1.5 Rightarrow l = 0.375 )- Substituting ( l ): ( k = 0.5 )- Substituting ( k ): ( h = 0.5 )The center of the sphere is ((0.5, 0.5, 0.375)). Calculating ( R ):[ R^2 = (0.5)^2 + (0.5)^2 + (0.375)^2 = 0.25 + 0.25 + 0.140625 = 0.640625 ][ R = sqrt{frac{41}{64}} = frac{sqrt{41}}{8} ]The radius ( R ) of the sphere is (boxed{dfrac{sqrt{41}}{8}}).

🔑:1. Let the cube be denoted as ( ABCDA_1B_1C_1D_1 ), where each edge of the cube has a length of 1. 2. Points ( A ) and ( C ) are vertices of the cube, while ( E ) and ( F ) refer to the midpoints of edges ( C_1D_1 ) and ( B_1C_1 ), respectively. We need to find the radius ( R ) of the sphere passing through these points.3. The center ( M ) of the sphere is equidistant from points ( A ) and ( C ). Therefore, it lies on the plane of symmetry that contains these points. This plane is ( BDD_1D ).4. Similarly, point ( M ) is also equidistant from points ( E ) and ( F ). Hence, it lies on another plane of symmetry passing through the midpoint of edges ( B_1C_1 ) and ( C_1D_1 ), which is the plane ( A_1C_1C ).5. Consequently, the center ( M ) must lie on the intersection line of these two planes ( A_1C_1C ) and ( BDD_1D ).6. Set up a rectangular coordinate system such that ( A=(0,0,0) ), ( B=(1,0,0) ), ( D=(0,1,0) ), etc. Let midpoint ( E = left( frac{1}{2}, 1, 0 right) ) and midpoint ( F = left( 1, frac{1}{2}, 0 right) ).7. Consider the coordinate of point ( M ) as ( left( 0.5, 0.5, z right) ). Since ( M ) lies on the perpendicular bisector of ( A ) and ( C ), distances ( ME ) and ( MC ) must be equal.8. Calculate ( ME ) and ( MC ):[ME = sqrt{(0.5-0.5)^2 + (0.5-1)^2 + (z-1)^2} = sqrt{0 + 0.25 + (z-1)^2} = sqrt{0.25 + (z-1)^2}][MC = sqrt{(0.5-1)^2 + (0.5-1)^2 + (0-z)^2} = sqrt{0.25 + 0.25 + z^2} = sqrt{0.5 + z^2}]9. Set ( ME = MC ):[sqrt{0.25 + (z-1)^2} = sqrt{0.5 + z^2}]10. Square both sides:[0.25 + (z-1)^2 = 0.5 + z^2]11. Expand and simplify:[0.25 + z^2 - 2z + 1 = 0.5 + z^2][1.25 - 2z + z^2 = 0.5 + z^2][1.25 - 2z = 0.5][-2z = -0.75][z = frac{3}{8}]12. Determine the radius ( R ), where ( R = MC ):[R = sqrt{(1-0.5)^2 + (1-0.5)^2 + left(0-frac{3}{8}right)^2}][R = sqrt{0.25 + 0.25 + left(frac{-3}{8}right)^2}][R = sqrt{0.25 + 0.25 + frac{9}{64}}][R = sqrt{frac{16}{64} + frac{16}{64} + frac{9}{64}}][R= sqrt{frac{41}{64}}][R = frac{sqrt{41}}{8}] Conclusion:[boxed{frac{sqrt{41}}{8}}]

🧠:Alright, so I need to find ( (f(x))^2 - (g(x))^2 ) given that ( f(x) + g(x) = sqrt{frac{1 + cos 2x}{1 - sin x}} ) where ( f ) is odd and ( g ) is even. Hmm, okay. Let me start by recalling some trigonometric identities and properties of odd and even functions.First, the expression inside the square root: ( frac{1 + cos 2x}{1 - sin x} ). I remember that ( cos 2x ) can be written in terms of ( 2cos^2 x - 1 ), so substituting that in, the numerator becomes ( 1 + 2cos^2 x - 1 = 2cos^2 x ). So the expression simplifies to ( sqrt{frac{2cos^2 x}{1 - sin x}} ). Let me check that again:( 1 + cos 2x = 1 + 2cos^2 x - 1 = 2cos^2 x ). Yes, that's right. So the numerator is ( 2cos^2 x ), denominator is ( 1 - sin x ). Therefore, the entire expression under the square root is ( frac{2cos^2 x}{1 - sin x} ). Let me see if I can simplify that further.Maybe multiply numerator and denominator by ( 1 + sin x ) to rationalize the denominator? Let's try that:Multiply numerator and denominator by ( 1 + sin x ):Numerator becomes ( 2cos^2 x (1 + sin x) ).Denominator becomes ( (1 - sin x)(1 + sin x) = 1 - sin^2 x = cos^2 x ).So the entire fraction becomes ( frac{2cos^2 x (1 + sin x)}{cos^2 x} = 2(1 + sin x) ).Therefore, the original expression simplifies to ( sqrt{2(1 + sin x)} ). Wait, but hold on, we need to take the square root of that. So the square root of ( 2(1 + sin x) ). Let me verify this step again:Original expression: ( sqrt{frac{2cos^2 x}{1 - sin x}} ).Multiply numerator and denominator by ( 1 + sin x ):Numerator: ( 2cos^2 x (1 + sin x) ).Denominator: ( (1 - sin x)(1 + sin x) = cos^2 x ).So, ( frac{2cos^2 x (1 + sin x)}{cos^2 x} = 2(1 + sin x) ). Therefore, the square root is ( sqrt{2(1 + sin x)} ). That seems right.But wait, the square root of a square is the absolute value, right? So originally, ( sqrt{frac{2cos^2 x}{1 - sin x}} ). The ( cos^2 x ) is non-negative, and the denominator ( 1 - sin x ). Since ( x in (-pi/2, pi/2) ), ( sin x ) ranges from -1 to 1, so ( 1 - sin x ) is positive (since ( sin x < 1 ) except at ( pi/2 ), which isn't included). So denominator is positive, numerator is non-negative, so the entire expression under the square root is non-negative. Then, when we take the square root, we can write it as ( sqrt{2(1 + sin x)} ), but we need to check if there's an absolute value involved. However, ( 1 + sin x ) is non-negative because ( sin x geq -1 ), so ( 1 + sin x geq 0 ). But in the interval ( x in (-pi/2, pi/2) ), ( sin x ) ranges from -1 to 1, so ( 1 + sin x ) ranges from 0 to 2. Therefore, ( sqrt{2(1 + sin x)} ) is always real and non-negative, so we don't need absolute value. Therefore, ( f(x) + g(x) = sqrt{2(1 + sin x)} ).Okay, that simplifies the expression. Now, since ( f(x) ) is odd and ( g(x) ) is even, we can use their properties to find another equation. Remember, for an odd function, ( f(-x) = -f(x) ), and for an even function, ( g(-x) = g(x) ). Let's consider the expression ( f(-x) + g(-x) ). According to the original equation, this should equal ( sqrt{frac{1 + cos(-2x)}{1 - sin(-x)}} ). Let me compute that.First, ( cos(-2x) = cos 2x ) because cosine is even. ( sin(-x) = -sin x ) because sine is odd. Therefore, the expression becomes ( sqrt{frac{1 + cos 2x}{1 - (-sin x)}} = sqrt{frac{1 + cos 2x}{1 + sin x}} ). Let's simplify this as we did before.Again, numerator is ( 1 + cos 2x = 2cos^2 x ). Denominator is ( 1 + sin x ). Therefore, the expression is ( sqrt{frac{2cos^2 x}{1 + sin x}} ). Let's rationalize the denominator by multiplying numerator and denominator by ( 1 - sin x ):Numerator: ( 2cos^2 x (1 - sin x) ).Denominator: ( (1 + sin x)(1 - sin x) = 1 - sin^2 x = cos^2 x ).So, ( frac{2cos^2 x (1 - sin x)}{cos^2 x} = 2(1 - sin x) ).Therefore, the square root of this is ( sqrt{2(1 - sin x)} ). Again, since ( 1 - sin x ) is non-negative (as ( sin x leq 1 )), so the square root is valid. So, ( f(-x) + g(-x) = sqrt{2(1 - sin x)} ).But since ( f(-x) = -f(x) ) (because f is odd) and ( g(-x) = g(x) ) (because g is even), the left-hand side becomes ( -f(x) + g(x) ). Therefore, we have:( -f(x) + g(x) = sqrt{2(1 - sin x)} ).So now we have two equations:1. ( f(x) + g(x) = sqrt{2(1 + sin x)} )2. ( -f(x) + g(x) = sqrt{2(1 - sin x)} )Now, we can solve these two equations for ( f(x) ) and ( g(x) ). Let's write them down:Equation 1: ( f + g = sqrt{2(1 + sin x)} )Equation 2: ( -f + g = sqrt{2(1 - sin x)} )Let me add these two equations:Adding equation 1 and equation 2: ( (f + g) + (-f + g) = sqrt{2(1 + sin x)} + sqrt{2(1 - sin x)} )Simplifies to: ( 2g = sqrt{2(1 + sin x)} + sqrt{2(1 - sin x)} )Similarly, subtract equation 2 from equation 1:( (f + g) - (-f + g) = sqrt{2(1 + sin x)} - sqrt{2(1 - sin x)} )Simplifies to: ( 2f = sqrt{2(1 + sin x)} - sqrt{2(1 - sin x)} )Therefore, we can express ( f(x) ) and ( g(x) ) as:( f(x) = frac{sqrt{2(1 + sin x)} - sqrt{2(1 - sin x)}}{2} )( g(x) = frac{sqrt{2(1 + sin x)} + sqrt{2(1 - sin x)}}{2} )But maybe we can simplify ( f(x) ) and ( g(x) ) further. Let's factor out ( sqrt{2} ) from numerator:( f(x) = frac{sqrt{2}}{2} left( sqrt{1 + sin x} - sqrt{1 - sin x} right) = frac{1}{sqrt{2}} left( sqrt{1 + sin x} - sqrt{1 - sin x} right) )Similarly, ( g(x) = frac{sqrt{2}}{2} left( sqrt{1 + sin x} + sqrt{1 - sin x} right) = frac{1}{sqrt{2}} left( sqrt{1 + sin x} + sqrt{1 - sin x} right) )Hmm, interesting. Maybe we can rationalize ( sqrt{1 + sin x} pm sqrt{1 - sin x} ). Let me recall that ( sqrt{1 + sin x} ) can be expressed as ( sqrt{ left( sin frac{x}{2} + cos frac{x}{2} right)^2 } ), but I need to check. Wait, let's recall that ( 1 + sin x = left( sin frac{x}{2} + cos frac{x}{2} right)^2 ). Let's verify:( left( sin frac{x}{2} + cos frac{x}{2} right)^2 = sin^2 frac{x}{2} + 2sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} = 1 + sin x ). Yes, because ( 2sin frac{x}{2} cos frac{x}{2} = sin x ). Therefore, ( sqrt{1 + sin x} = | sin frac{x}{2} + cos frac{x}{2} | ). Similarly, ( 1 - sin x = left( sin frac{x}{2} - cos frac{x}{2} right)^2 ). Let's check:( left( sin frac{x}{2} - cos frac{x}{2} right)^2 = sin^2 frac{x}{2} - 2sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} = 1 - sin x ). Yes, so ( sqrt{1 - sin x} = | sin frac{x}{2} - cos frac{x}{2} | ).But since ( x in (-pi/2, pi/2) ), then ( frac{x}{2} in (-pi/4, pi/4) ). In that interval, ( cos frac{x}{2} ) is positive and greater than ( sin frac{x}{2} ) in absolute value because ( pi/4 ) is 45 degrees, so for angles between -45 and 45 degrees, cosine is greater than sine in magnitude. Therefore, ( sin frac{x}{2} + cos frac{x}{2} ) is always positive because both terms are positive when ( x ) is positive, and for ( x ) negative, ( sin frac{x}{2} ) is negative but the magnitude of ( cos frac{x}{2} ) is larger. Similarly, ( sin frac{x}{2} - cos frac{x}{2} ) would be negative when ( x ) is positive because ( sin frac{x}{2} < cos frac{x}{2} ), but since we have absolute value, we can write:( sqrt{1 + sin x} = sin frac{x}{2} + cos frac{x}{2} )( sqrt{1 - sin x} = cos frac{x}{2} - sin frac{x}{2} )Therefore, substituting back into ( f(x) ):( f(x) = frac{1}{sqrt{2}} left( (sin frac{x}{2} + cos frac{x}{2}) - (cos frac{x}{2} - sin frac{x}{2}) right) )Simplify inside the brackets:( (sin frac{x}{2} + cos frac{x}{2} - cos frac{x}{2} + sin frac{x}{2}) = 2sin frac{x}{2} )Therefore, ( f(x) = frac{1}{sqrt{2}} times 2sin frac{x}{2} = sqrt{2} sin frac{x}{2} )Similarly, ( g(x) = frac{1}{sqrt{2}} left( (sin frac{x}{2} + cos frac{x}{2}) + (cos frac{x}{2} - sin frac{x}{2}) right) )Simplify inside the brackets:( (sin frac{x}{2} + cos frac{x}{2} + cos frac{x}{2} - sin frac{x}{2}) = 2cos frac{x}{2} )Therefore, ( g(x) = frac{1}{sqrt{2}} times 2cos frac{x}{2} = sqrt{2} cos frac{x}{2} )So now we have expressions for ( f(x) ) and ( g(x) ):( f(x) = sqrt{2} sin frac{x}{2} )( g(x) = sqrt{2} cos frac{x}{2} )Let me check if these satisfy the original conditions. First, ( f(x) ) is odd: ( f(-x) = sqrt{2} sin frac{-x}{2} = -sqrt{2} sin frac{x}{2} = -f(x) ). Good. ( g(x) ) is even: ( g(-x) = sqrt{2} cos frac{-x}{2} = sqrt{2} cos frac{x}{2} = g(x) ). Perfect.Now, let's compute ( (f(x))^2 - (g(x))^2 ). That's ( 2sin^2 frac{x}{2} - 2cos^2 frac{x}{2} = 2(sin^2 frac{x}{2} - cos^2 frac{x}{2}) ). Hmm, recall that ( cos a = cos^2 frac{a}{2} - sin^2 frac{a}{2} ), so ( sin^2 frac{x}{2} - cos^2 frac{x}{2} = -cos x ). Therefore, ( 2(-cos x) = -2cos x ).Wait, let me verify that step. The identity ( cos a = cos^2 frac{a}{2} - sin^2 frac{a}{2} ). So ( cos^2 frac{a}{2} - sin^2 frac{a}{2} = cos a ). Therefore, ( sin^2 frac{a}{2} - cos^2 frac{a}{2} = -cos a ). Therefore, substituting ( a = x ), we get ( sin^2 frac{x}{2} - cos^2 frac{x}{2} = -cos x ). Therefore, ( (f(x))^2 - (g(x))^2 = 2(-cos x) = -2cos x ).But wait, hold on. Let's compute ( (f(x))^2 = (sqrt{2} sin frac{x}{2})^2 = 2sin^2 frac{x}{2} ).Similarly, ( (g(x))^2 = (sqrt{2} cos frac{x}{2})^2 = 2cos^2 frac{x}{2} ).Therefore, ( (f(x))^2 - (g(x))^2 = 2sin^2 frac{x}{2} - 2cos^2 frac{x}{2} = 2(sin^2 frac{x}{2} - cos^2 frac{x}{2}) = 2(-cos x) = -2cos x ).Yes, that's correct. Therefore, the answer should be ( -2cos x ).But let me verify this by another approach. Alternatively, we can note that ( (f(x))^2 - (g(x))^2 = (f(x) - g(x))(f(x) + g(x)) ). From the original problem, we have ( f(x) + g(x) = sqrt{frac{1 + cos 2x}{1 - sin x}} ), which we simplified earlier to ( sqrt{2(1 + sin x)} ). So if we can compute ( f(x) - g(x) ), then multiply by ( f(x) + g(x) ), we can get ( (f(x))^2 - (g(x))^2 ).But from the equations we had earlier:Equation 1: ( f + g = sqrt{2(1 + sin x)} )Equation 2: ( -f + g = sqrt{2(1 - sin x)} )If we subtract equation 2 from equation 1:( (f + g) - (-f + g) = sqrt{2(1 + sin x)} - sqrt{2(1 - sin x)} )Simplifies to ( 2f = sqrt{2(1 + sin x)} - sqrt{2(1 - sin x)} ), which is how we found ( f(x) ).Alternatively, if we rearrange equation 2 to express ( g = sqrt{2(1 - sin x)} + f ), and substitute into equation 1:( f + (sqrt{2(1 - sin x)} + f) = sqrt{2(1 + sin x)} )Which gives ( 2f + sqrt{2(1 - sin x)} = sqrt{2(1 + sin x)} )But this might not help. Alternatively, perhaps express ( f - g ). Wait, from equation 1 and 2:From equation 1: ( f = sqrt{2(1 + sin x)} - g )From equation 2: ( -f = sqrt{2(1 - sin x)} - g )Substitute equation 1 into equation 2:( -(sqrt{2(1 + sin x)} - g) = sqrt{2(1 - sin x)} - g )Simplifies to:( -sqrt{2(1 + sin x)} + g = sqrt{2(1 - sin x)} - g )Bring terms with ( g ) to one side:( 2g = sqrt{2(1 - sin x)} + sqrt{2(1 + sin x)} )Which brings us back to the expression for ( g ). So this approach might not be more helpful.Alternatively, let's compute ( f(x) - g(x) ). From the expressions we found:( f(x) - g(x) = sqrt{2}sin frac{x}{2} - sqrt{2}cos frac{x}{2} = sqrt{2}(sin frac{x}{2} - cos frac{x}{2}) )Multiply this by ( f(x) + g(x) = sqrt{2}(sin frac{x}{2} + cos frac{x}{2}) ):( (f(x))^2 - (g(x))^2 = [sqrt{2}(sin frac{x}{2} - cos frac{x}{2})][sqrt{2}(sin frac{x}{2} + cos frac{x}{2})] )Multiply the terms:( 2 (sin frac{x}{2} - cos frac{x}{2})(sin frac{x}{2} + cos frac{x}{2}) = 2 (sin^2 frac{x}{2} - cos^2 frac{x}{2}) )Which again gives ( 2(-cos x) = -2cos x ). So this confirms the previous result.Therefore, the answer should be ( -2cos x ).But let me check with specific values to ensure there's no mistake. Let's pick ( x = 0 ). Then:Original expression: ( f(0) + g(0) = sqrt{frac{1 + 1}{1 - 0}} = sqrt{2} ).Since ( f ) is odd, ( f(0) = 0 ). Therefore, ( g(0) = sqrt{2} ). Then ( (f(0))^2 - (g(0))^2 = 0 - 2 = -2 ). According to our answer, ( -2cos 0 = -2 times 1 = -2 ). Correct.Another test: ( x = pi/4 ). Then:Original expression: ( f(pi/4) + g(pi/4) = sqrt{frac{1 + cos(pi/2)}{1 - sin(pi/4)}} = sqrt{frac{1 + 0}{1 - frac{sqrt{2}}{2}}} = sqrt{frac{1}{1 - frac{sqrt{2}}{2}}} ).Let me compute this:( sqrt{frac{1}{1 - frac{sqrt{2}}{2}}} = sqrt{frac{2}{2 - sqrt{2}}} = sqrt{frac{2(2 + sqrt{2})}{(2 - sqrt{2})(2 + sqrt{2})}} = sqrt{frac{2(2 + sqrt{2})}{4 - 2}} = sqrt{frac{2(2 + sqrt{2})}{2}} = sqrt{2 + sqrt{2}} ).But according to our expressions for ( f(x) ) and ( g(x) ):( f(pi/4) = sqrt{2} sin frac{pi}{8} ), ( g(pi/4) = sqrt{2} cos frac{pi}{8} ).Then ( f(pi/4)^2 - g(pi/4)^2 = 2sin^2 frac{pi}{8} - 2cos^2 frac{pi}{8} = -2(cos^2 frac{pi}{8} - sin^2 frac{pi}{8}) = -2cos frac{pi}{4} = -2 times frac{sqrt{2}}{2} = -sqrt{2} ).According to our formula, ( -2cos(pi/4) = -2 times frac{sqrt{2}}{2} = -sqrt{2} ). Which matches. Let's compute ( sqrt{2(1 + sin (pi/4))} ):( sqrt{2(1 + frac{sqrt{2}}{2})} = sqrt{2 + sqrt{2}} ), which is what we had earlier. Therefore, ( f + g = sqrt{2 + sqrt{2}} ), which matches. The calculation seems consistent.Another test: ( x = pi/6 ).Original expression: ( f(pi/6) + g(pi/6) = sqrt{frac{1 + cos(pi/3)}{1 - sin(pi/6)}} = sqrt{frac{1 + 0.5}{1 - 0.5}} = sqrt{frac{1.5}{0.5}} = sqrt{3} ).According to our expressions:( f(pi/6) = sqrt{2} sin frac{pi}{12} ), ( g(pi/6) = sqrt{2} cos frac{pi}{12} ).Then ( (f(pi/6))^2 - (g(pi/6))^2 = 2sin^2 frac{pi}{12} - 2cos^2 frac{pi}{12} = -2cos frac{pi}{6} = -2 times (sqrt{3}/2) = -sqrt{3} ).According to the formula ( -2cos(pi/6) = -2 times (sqrt{3}/2) = -sqrt{3} ), which matches. Also, ( f(pi/6) + g(pi/6) = sqrt{2}(sin frac{pi}{12} + cos frac{pi}{12}) ). Let me compute this:( sin frac{pi}{12} = sin 15^circ = frac{sqrt{6} - sqrt{2}}{4} ), ( cos frac{pi}{12} = cos 15^circ = frac{sqrt{6} + sqrt{2}}{4} ).So ( sin frac{pi}{12} + cos frac{pi}{12} = frac{sqrt{6} - sqrt{2} + sqrt{6} + sqrt{2}}{4} = frac{2sqrt{6}}{4} = frac{sqrt{6}}{2} ).Therefore, ( sqrt{2} times frac{sqrt{6}}{2} = frac{sqrt{12}}{2} = frac{2sqrt{3}}{2} = sqrt{3} ). Which matches the original expression. So all tests check out.Therefore, I can be confident that ( (f(x))^2 - (g(x))^2 = -2cos x ).Final AnswerThe value of ( (f(x))^{2} - (g(x))^{2} ) is boxed{-2cos x}.

✅:Given functions ( f(x) ) and ( g(x) ) satisfy ( f(x) + g(x) = sqrt{frac{1 + cos 2x}{1 - sin x}} ) for ( x in left( -frac{pi}{2}, frac{pi}{2} right) ), where ( f(x) ) is an odd function and ( g(x) ) is an even function, we need to find ( (f(x))^2 - (g(x))^2 ).First, we simplify the given expression using trigonometric identities:[frac{1 + cos 2x}{1 - sin x} = frac{2cos^2 x}{1 - sin x}]Rationalizing the denominator by multiplying numerator and denominator by ( 1 + sin x ):[frac{2cos^2 x (1 + sin x)}{(1 - sin x)(1 + sin x)} = frac{2cos^2 x (1 + sin x)}{cos^2 x} = 2(1 + sin x)]Thus, the expression simplifies to:[sqrt{2(1 + sin x)}]Similarly, considering ( f(-x) + g(-x) ), we get:[sqrt{frac{1 + cos(-2x)}{1 - sin(-x)}} = sqrt{frac{2cos^2 x}{1 + sin x}} = sqrt{2(1 - sin x)}]Using the properties of odd and even functions, we have:1. ( f(x) + g(x) = sqrt{2(1 + sin x)} )2. ( -f(x) + g(x) = sqrt{2(1 - sin x)} )Solving these equations, we find:[f(x) = frac{sqrt{2(1 + sin x)} - sqrt{2(1 - sin x)}}{2}][g(x) = frac{sqrt{2(1 + sin x)} + sqrt{2(1 - sin x)}}{2}]Simplifying further using trigonometric identities, we find:[f(x) = sqrt{2} sin frac{x}{2}, quad g(x) = sqrt{2} cos frac{x}{2]Finally, we compute ( (f(x))^2 - (g(x))^2 ):[(f(x))^2 - (g(x))^2 = 2sin^2 frac{x}{2} - 2cos^2 frac{x}{2} = -2(cos^2 frac{x}{2} - sin^2 frac{x}{2}) = -2cos x]Thus, the value of ( (f(x))^2 - (g(x))^2 ) is boxed{-2cos x}.

🔑:1. Given the functions (f(x)) and (g(x)) satisfy the equation: [ f(x) + g(x) = sqrt{frac{1 + cos 2x}{1 - sin x}} quad left(x in left(-frac{pi}{2}, frac{pi}{2}right)right) ] where (f(x)) is an odd function and (g(x)) is an even function.2. Since ( f(x) ) is an odd function, we know ( f(-x) = -f(x) ). Similarly, since ( g(x) ) is an even function, we know ( g(-x) = g(x) ).3. Consider the expression for ( f(-x) + g(-x) ): [ f(-x) + g(-x) = -f(x) + g(x) ]4. Also, we are given: [ f(-x) + g(-x) = sqrt{frac{1 + cos 2x}{1 + sin x}} ]5. Therefore, we have: [ -f(x) + g(x) = sqrt{frac{1 + cos 2x}{1 + sin x}} ]6. To find ( (f(x))^2 - (g(x))^2 ), we use the identity (a^2 - b^2 = (a+b)(a-b)): [ (f(x))^2 - (g(x))^2 = (f(x) + g(x))(f(x) - g(x)) ]7. Substituting the expressions obtained: [ (f(x))^2 - (g(x))^2 = left( sqrt{frac{1 + cos 2x}{1 - sin x}} right) left( -sqrt{frac{1 + cos 2x}{1 + sin x}} right) ]8. Simplify by multiplying the terms inside the square roots: [ (f(x))^2 - (g(x))^2 = sqrt{frac{1 + cos 2x}{1 - sin x}} cdot left(-sqrt{frac{1 + cos 2x}{1 + sin x}}right) = -sqrt{frac{(1 + cos 2x)^2}{(1 - sin x)(1 + sin x)}} ]9. Notice that: [ (1 - sin x)(1 + sin x) = 1 - (sin x)^2 = cos^2 x ]10. Therefore, the expression becomes: [ -sqrt{frac{(1 + cos 2x)^2}{cos^2 x}} = -frac{1 + cos 2x}{cos x} ]11. Finally, using the double-angle identity for cosine: ( cos 2x = 2 cos^2 x -1 ): [ 1 + cos 2x = 1 + (2 cos^2 x - 1) = 2 cos^2 x ]12. Thus: [ -frac{1 + cos 2x}{cos x} = -frac{2 cos^2 x}{cos x} = -2 cos x ] Conclusion:(boxed{-2 cos x})