🧠:Okay, let's try to tackle this problem step by step. It's a bit complex with several conditions, so I need to break it down carefully. Let me start by listing out all the given information and see how I can translate it into mathematical equations.First, there are three problems: A, B, and C. There were 25 competitors who solved at least one of the three. So, the total number of people in the Venn diagram for A, B, and C is 25. That's straightforward.Now, the next condition: Amongst those who did not solve A, twice as many solved B as solved C. Hmm, okay. Let me parse that. The group we're considering here is the set of competitors who did not solve problem A. Within this group, the number of people who solved B is twice the number who solved C. But wait, does "solved B" mean solved only B, or B and possibly other problems? Similarly for C. I think since we're talking about those who did not solve A, solving B would mean solving B only or B and C, but not A. Similarly for C. So, maybe I need to think in terms of the Venn diagram regions outside of A.Let me recall that in a three-set Venn diagram, there are 8 regions: only A, only B, only C, A and B but not C, A and C but not B, B and C but not A, all three A, B, C, and none. But in our case, since all 25 solved at least one, the "none" region is zero.But the problem mentions "those who did not solve A". That would be the regions outside of A: only B, only C, B and C but not A. So, the total number of people who didn't solve A is the sum of these three regions: only B, only C, and B and C. Then, among these, the number who solved B (which would be only B plus B and C) is twice the number who solved C (which would be only C plus B and C). Wait, but that might not be correct. If they solved B, does that include those who solved both B and C? Similarly for C.Wait, the problem says: "twice as many solved B as solved C". So, among those who did not solve A, count how many solved B (regardless of whether they solved C or not) and how many solved C (regardless of whether they solved B or not). Then, the number solving B is twice the number solving C.So, if someone didn't solve A, but solved both B and C, they would be counted in both the B and C groups. Therefore, in the non-A group, the total number of people who solved B is (only B + B and C), and the total number who solved C is (only C + B and C). According to the problem, (only B + B and C) = 2*(only C + B and C).Let me write that as an equation:Let’s denote:- Only B as b- Only C as c- B and C but not A as dThen the number of people who solved B among those who didn't solve A is b + dThe number who solved C among those who didn't solve A is c + dSo, according to the problem: b + d = 2*(c + d)Simplify this equation:b + d = 2c + 2dSubtract d from both sides: b = 2c + dSo, equation (1): b = 2c + dOkay, moving on to the next condition: The number solving only A was one more than the number solving A and at least one other. Let me parse that. The number solving only A is equal to (the number solving A and at least one other) plus one. Let's denote:- Only A as a- A and B but not C as e- A and C but not B as f- All three A, B, C as gThen, the number solving only A is a. The number solving A and at least one other is e + f + g. So, according to the problem:a = (e + f + g) + 1That's equation (2): a = e + f + g + 1Third condition: The number solving just A equalled the number solving just B plus the number solving just C. So,a = b + cThat's equation (3): a = b + cAlso, the total number of competitors is 25, so all regions sum up to 25:a + b + c + d + e + f + g = 25That's equation (4): a + b + c + d + e + f + g = 25Now, let's summarize the equations we have:1. b = 2c + d2. a = e + f + g + 13. a = b + c4. a + b + c + d + e + f + g = 25We need to find the value of b, which is the number of people who solved just B.Now, let's see if we can express variables in terms of others to reduce the number of variables.From equation (3): a = b + c. So, we can replace a with b + c in other equations.From equation (2): a = e + f + g + 1. Since a = b + c, then:b + c = e + f + g + 1 --> equation (2a): e + f + g = b + c - 1From equation (1): b = 2c + d. So, we can express d in terms of b and c: d = b - 2cNow, substitute d = b - 2c into equation (4):a + b + c + (b - 2c) + e + f + g = 25Simplify:a + b + c + b - 2c + e + f + g = 25Combine like terms:a + 2b - c + e + f + g = 25But from equation (2a), e + f + g = b + c - 1, so substitute that in:a + 2b - c + (b + c - 1) = 25Simplify:a + 2b - c + b + c - 1 = 25Combine like terms:a + 3b - 1 = 25Then:a + 3b = 26But from equation (3), a = b + c. Substitute that into the equation:(b + c) + 3b = 26Simplify:4b + c = 26 --> equation (5): 4b + c = 26So now, we have equations (1): b = 2c + d, (5): 4b + c = 26, and we also have d = b - 2c from equation (1).But let's see if we can find another equation. So far, we have:Equation (5): 4b + c = 26We need another equation to solve for b and c. Let's check what variables we have. Let's recall that we also have equation (2a): e + f + g = b + c - 1, but we haven't used that yet. However, e, f, g are other variables that might need to be related. Wait, but do we have any other conditions? The problem didn't mention anything about the overlaps between A and B, A and C, or all three. The only other conditions are the ones we've already used. So maybe we need to see if we can relate other variables or if there are constraints we haven't considered.Wait, perhaps there's a way to express other variables in terms of b and c. Let's recall that we have:From equation (1): d = b - 2cFrom equation (3): a = b + cFrom equation (2a): e + f + g = b + c - 1So, if we can express e + f + g in terms of b and c, but we need more relations. Alternatively, maybe we can note that all variables must be non-negative integers. So, perhaps we can use equation (5): 4b + c = 26, and since b and c are non-negative integers, we can look for integer solutions where c = 26 - 4b. Since c must be non-negative, 26 - 4b ≥ 0 ⇒ 4b ≤ 26 ⇒ b ≤ 6.5. Since b is an integer, b ≤ 6.Also, from equation (1): b = 2c + d. Since d must be non-negative, then b ≥ 2c. But from equation (5): c = 26 - 4b. Substituting that into b ≥ 2c:b ≥ 2*(26 - 4b)b ≥ 52 - 8b9b ≥ 52b ≥ 52/9 ≈ 5.777...Since b must be an integer, so b ≥ 6. But we already found that b ≤ 6. Therefore, b must be 6.So, if b = 6, then from equation (5): 4*6 + c = 26 ⇒ 24 + c = 26 ⇒ c = 2Then, from equation (3): a = b + c = 6 + 2 = 8From equation (1): d = b - 2c = 6 - 2*2 = 6 - 4 = 2From equation (2a): e + f + g = b + c -1 = 6 + 2 -1 = 7Now, let's check if all variables are non-negative. So far:a = 8 (only A)b = 6 (only B)c = 2 (only C)d = 2 (B and C but not A)e + f + g = 7 (A and B, A and C, all three)We also need to ensure that the total sum is 25. Let's check:a + b + c + d + e + f + g = 8 + 6 + 2 + 2 + 7 = 25. Perfect, that adds up.But we need to make sure that all individual variables e, f, g are non-negative. However, the problem doesn't specify the exact values for e, f, g, only their sum. Since the question only asks for the number who solved just B, which is b = 6, maybe we don't need to find e, f, g. However, we should verify if there are any constraints we missed that could affect the validity of the solution.Wait, but the problem might not require e, f, g to be specific values as long as their sum is 7 and they are non-negative integers. Since there are multiple possibilities (e.g., e=7, f=0, g=0; or e=3, f=2, g=2; etc.), but as the question doesn't ask for those, our answer for b=6 seems valid.But just to make sure, let's check all conditions again.1. Among those who did not solve A (which are b + c + d = 6 + 2 + 2 = 10), twice as many solved B as solved C. Solved B in this group is b + d = 6 + 2 = 8. Solved C is c + d = 2 + 2 = 4. Indeed, 8 = 2*4. Check.2. Number solving only A (a=8) is one more than those solving A and at least one other (e + f + g=7). 8 = 7 +1. Check.3. Number solving just A (8) equals number solving just B (6) plus just C (2). 8 = 6 + 2. Check.4. Total is 25. Check.Therefore, all conditions are satisfied. Thus, the number of people who solved just B is 6.Final AnswerThe number of competitors who solved just B is boxed{6}.

✅:Given the problem, we need to determine the number of competitors who solved just problem B. We start by defining the regions of a Venn diagram for problems A, B, and C:- (a): Only A- (b): Only B- (c): Only C- (d): B and C but not A- (e): A and B but not C- (f): A and C but not B- (g): All three (A, B, and C)We are given the following conditions:1. 25 competitors solved at least one of the three problems.2. Among those who did not solve A, twice as many solved B as solved C.3. The number solving only A is one more than the number solving A and at least one other.4. The number solving just A equals the number solving just B plus the number solving just C.Translating these conditions into equations:1. The total number of competitors: (a + b + c + d + e + f + g = 25)2. For those who did not solve A: (b + d = 2(c + d))3. For those solving only A: (a = e + f + g + 1)4. For those solving just A, B, and C: (a = b + c)From condition 2, we derive:[ b + d = 2(c + d) implies b = 2c + d ]From condition 3, substituting (a = b + c) (from condition 4):[ b + c = e + f + g + 1 implies e + f + g = b + c - 1 ]Substituting (a = b + c) and (d = b - 2c) into the total number equation:[ a + b + c + d + e + f + g = 25 ][ (b + c) + b + c + (b - 2c) + (b + c - 1) = 25 ][ 4b + c - 1 = 25 implies 4b + c = 26 ]Solving for (b) and (c) as non-negative integers:- (4b + c = 26)- (b geq 2c) (from (b = 2c + d) and (d geq 0))Testing integer solutions:- (b = 6) and (c = 2) satisfy (4b + c = 26) and (b geq 2c)Substituting (b = 6) and (c = 2):- (a = b + c = 8)- (d = b - 2c = 2)- (e + f + g = 7)Verifying all conditions:1. Total competitors: (8 + 6 + 2 + 2 + 7 = 25)2. Those who did not solve A: (6 + 2 + 2 = 10), with (6 + 2 = 8) solving B and (2 + 2 = 4) solving C (8 = 2 * 4)3. Only A (8) is one more than those solving A and others (7)4. Only A (8) equals only B (6) plus only C (2)Thus, the number of competitors who solved just B is (boxed{6}).

🔑:Let us define the following variables:- ( a ) as the number of competitors who solved just problem A.- ( b ) as the number of competitors who solved just problem B.- ( c ) as the number of competitors who solved just problem C.- ( d ) as the number of competitors who solved problems B and C, but not A.- ( e ) as the number of competitors who solved problems A and B, but not C.- ( f ) as the number of competitors who solved problems A and C, but not B.- ( g ) as the number of competitors who solved all three problems A, B, and C.Given that the total number of competitors is 25, we have:[ a + b + c + d + e + f + g = 25. ]From the problem conditions, we can derive the following equations:1. Among those who did not solve A, twice as many solved B as solved C: [ b + d = 2(c + d). ]2. The number solving only A was one more than the number solving A and at least one other problem: [ a = 1 + (e + f + g). ]3. The number solving only A equalled the number solving just B plus the number solving just C: [ a = b + c. ]Since competitors solving A and at least one other problem can be expressed as:[ e + f + g = 25 - (a + b + c + d), ]we substitute this into equation (2):[ a = 1 + (25 - (a + b + c + d)). ]Simplifying, we get:[ a = 26 - (a + b + c + d). ]Thus,[ 2a + b + c + d = 26. ]We use equation (3):[ a = b + c. ]We now express ( b + d ) and ( c + d ) from equation (1):[ b + d = 2c + 2d, ]so,[ b + d = 2c + 2d, ][ b = 2c + d. ]Now substitute this and ( a = b + c ) into ( 2a + b + c + d = 26 ):[ 2(b + c) + b + c + d = 26. ]Substituting ( b = 2c + d ) into it:[ 2(2c + d + c) + 2c + d + c + d = 26, ][ 2(3c + d) + 3c + d = 26, ][ 6c + 2d + 3c + d = 26, ][ 9c + 3d = 26. ]Since ( b = 2c + d ) and rearranging:Thus,[ 3b + c = 26 - 2d. ][ b - 2c geq 0, ]We solve:[ b = 6, c = 2.]Conclusion:[boxed{6}]

🧠:Okay, so I need to prove that in triangle ABC, the inequality (a/(s-a) + b/(s-b) + c/(s-c)) ≥ 3R/r holds. Hmm, let me start by recalling some basic concepts and formulas related to triangles. First, in a triangle, a, b, c are the lengths of the sides opposite to angles A, B, C respectively. The semiperimeter s is defined as (a + b + c)/2. So, s - a would be (b + c - a)/2, right? Similarly for s - b and s - c. R is the radius of the circumcircle, and r is the radius of the incircle. I remember that there are formulas connecting R, r, and other elements of the triangle. For example, the area of the triangle can be expressed as r*s, and also as (a*b*c)/(4R). Also, there's the formula that relates R, r, and the distances from the inradius to the sides. Let me write down the inequality again:(a/(s - a) + b/(s - b) + c/(s - c)) ≥ 3R/r.I need to show that the sum of these three terms is at least 3 times the ratio of the circumradius to the inradius. I wonder if there are known inequalities involving R and r. The Euler inequality states that R ≥ 2r, with equality if and only if the triangle is equilateral. But here, the inequality is in the opposite direction, multiplied by 3. Wait, 3R/r would be at least 6 if Euler's inequality holds. But the left-hand side of our inequality might be larger than 6? Hmm, maybe. Alternatively, perhaps I can express the left-hand side in terms of R and r. Let me see if I can manipulate the left-hand side. First, let's note that s - a = (b + c - a)/2. Similarly for the others. So, each denominator is half the sum of the other two sides minus the side in the numerator. Alternatively, in terms of the semiperimeter, s - a, s - b, s - c are the lengths related to the exradii. Wait, the exradius opposite to a is given by r_a = Δ/(s - a), where Δ is the area of the triangle. Similarly for the others. But how does that help here?Alternatively, the terms a/(s - a) can be expressed as 2a/(b + c - a). Maybe this is useful? Let me think. Alternatively, perhaps express the left-hand side in terms of trigonometric functions. Since in a triangle, a = 2R sin A, b = 2R sin B, c = 2R sin C. Maybe substituting these into the expression could help. Let me try that.So, substituting a = 2R sin A, b = 2R sin B, c = 2R sin C. Then s = (a + b + c)/2 = R(sin A + sin B + sin C). Therefore, s - a = R(sin B + sin C - sin A). Similarly for s - b and s - c. Therefore, the left-hand side becomes:[2R sin A]/[R(sin B + sin C - sin A)] + [2R sin B]/[R(sin B + sin C - sin A)] + [2R sin C]/[R(sin C + sin A - sin B)]Wait, no, each denominator is different. Let me check again. Wait, s - a is equal to (b + c - a)/2, which would be (2R sin B + 2R sin C - 2R sin A)/2 = R(sin B + sin C - sin A). Similarly, s - b = R(sin A + sin C - sin B), and s - c = R(sin A + sin B - sin C). Therefore, each term in the sum is:a/(s - a) = [2R sin A]/[R(sin B + sin C - sin A)] = 2 sin A / (sin B + sin C - sin A)Similarly for the other terms. So the entire left-hand side becomes:2 [ sin A / (sin B + sin C - sin A ) + sin B / (sin A + sin C - sin B ) + sin C / (sin A + sin B - sin C ) ]Hmm, this seems complicated, but maybe there's a trigonometric identity or inequality that can help here. Alternatively, perhaps there's a substitution that can simplify this expression.Alternatively, perhaps use the formula for r and R. The inradius r is given by Δ/s, where Δ is the area. The circumradius R is given by (a b c)/(4Δ). So, R/r = (a b c)/(4Δ) * (s/Δ) ) = (a b c s)/(4Δ²). Hmm, maybe not directly helpful. Wait, let's compute R/r:R/r = (abc)/(4Δ) / (Δ/s) ) = (abc s)/(4Δ²). Maybe express Δ in terms of r and s: Δ = r s. So substituting that in:R/r = (a b c s)/(4 (r s)²) = (a b c)/(4 r s). So R/r = (a b c)/(4 r s). Hmm, not sure if this helps.Alternatively, perhaps express the left-hand side in terms of r and R. Let me think. Alternatively, note that in terms of the exradii, since r_a = Δ/(s - a), so 1/(s - a) = r_a / Δ. Similarly for the others. Therefore, a/(s - a) = a r_a / Δ. But I don't know if that's helpful. Alternatively, perhaps express a, b, c in terms of r and angles. Wait, there are formulas that connect sides with inradius and angles. For example, a = 2r (cot A/2 + cot B/2 + cot C/2) ? Not sure. Wait, perhaps the formula for the inradius: r = (Δ)/s. Also, the exradius r_a = Δ/(s - a). So, r_a = Δ/(s - a). Therefore, a/(s - a) = a r_a / Δ. Similarly, the left-hand side is (a r_a + b r_b + c r_c)/Δ. But I don't see the connection to R/r. Maybe this path isn't the best.Alternatively, perhaps consider using the Cauchy-Schwarz inequality. Let me recall that Cauchy-Schwarz says that (sum x_i y_i)^2 ≤ (sum x_i²)(sum y_i²). Or in another form, (sum a_i^2)(sum b_i^2) ≥ (sum a_i b_i)^2. Alternatively, the Titu's lemma, which is a form of Cauchy-Schwarz: sum (a_i^2 / b_i) ≥ (sum a_i)^2 / sum b_i.Alternatively, perhaps use the Nesbitt's inequality. Wait, Nesbitt's inequality states that for positive real numbers a, b, c, (a/(b + c) + b/(a + c) + c/(a + b)) ≥ 3/2. But our expression is similar but different. Here, we have a/(s - a) = a/( (b + c - a)/2 ) = 2a/(b + c - a). So the left-hand side is 2(a/(b + c - a) + b/(a + c - b) + c/(a + b - c)). So it's similar but with denominators being (b + c - a) etc. instead of (b + c). Is there a known inequality for this kind of expression? Let me check. If I let x = b + c - a, y = a + c - b, z = a + b - c. Then x, y, z are all positive because of the triangle inequality, and x + y + z = a + b + c. Then our left-hand side becomes 2(a/x + b/y + c/z). So we need to prove that 2(a/x + b/y + c/z) ≥ 3R/r.But x = b + c - a = 2(s - a), similarly y = 2(s - b), z = 2(s - c). Wait, but in the original problem, the denominators are s - a, s - b, s - c, so actually x = 2(s - a), so a/x = a/(2(s - a)). Therefore, the sum is (a/(s - a) + ...) = 2(a/x + b/y + c/z). Wait, no, wait: x = b + c - a = 2(s - a). So a/x = a/(2(s - a)). Therefore, the left-hand side of our original inequality is (a/(s - a) + ...) = (a/(s - a) + b/(s - b) + c/(s - c)) = 2(a/x + b/y + c/z). So the left-hand side is 2 times the sum of a/x + b/y + c/z.But in terms of x, y, z, we have that x + y + z = a + b + c. Also, note that in terms of R and r, perhaps expressing x, y, z in terms of R and r. Alternatively, perhaps using substitution variables.Alternatively, since x = b + c - a, then in terms of the triangle's angles, using the sine formula. Let me try that. Let me write x, y, z in terms of R and angles. Since a = 2R sin A, so x = 2R sin B + 2R sin C - 2R sin A = 2R (sin B + sin C - sin A). Similarly for y and z. Therefore, a/x = (2R sin A)/(2R (sin B + sin C - sin A)) ) = sin A / (sin B + sin C - sin A). Therefore, the sum a/x + b/y + c/z becomes [ sin A / (sin B + sin C - sin A ) + sin B / (sin A + sin C - sin B ) + sin C / (sin A + sin B - sin C ) ].So, the left-hand side of our original inequality is 2 times this sum, and we need to show that this is ≥ 3R/r.Hmm, this seems complicated. Maybe I need to find another approach.Wait, let's recall that R = abc/(4Δ) and r = Δ/s. Therefore, R/r = abc/(4Δ) * s/Δ = (abc s)/(4Δ²). But Δ = r s, so substituting, R/r = (abc s)/(4 r² s²) ) = abc/(4 r s²). Hmm, not sure if that helps.Alternatively, maybe express abc in terms of R. Since a = 2R sin A, b = 2R sin B, c = 2R sin C. Therefore, abc = 8R³ sin A sin B sin C. Then R/r = (8R³ sin A sin B sin C)/(4 r s²) = (2 R³ sin A sin B sin C)/(r s²). Hmm, not sure.Alternatively, perhaps use the formula r = 4R sin(A/2) sin(B/2) sin(C/2). Yes, I think that's a standard formula. So r = 4R sin(A/2) sin(B/2) sin(C/2). Then 3R/r = 3/(4 sin(A/2) sin(B/2) sin(C/2)). So, the inequality becomes:Left-hand side ≥ 3/(4 sin(A/2) sin(B/2) sin(C/2)).Hmm, maybe if we can express the left-hand side in terms of the half-angle functions.Alternatively, let's think about the left-hand side again. Since s - a = (b + c - a)/2, and in terms of the triangle's angles, perhaps using cosine rule? Let me see. For example, b + c - a can be expressed as 2(s - a). Alternatively, express in terms of cosines. Wait, using the law of cosines: a² = b² + c² - 2bc cos A. Maybe not directly helpful.Alternatively, consider using substitution variables. Let me set x = s - a, y = s - b, z = s - c. Then, x + y + z = 3s - (a + b + c) = 3s - 2s = s. So x + y + z = s. Also, a = y + z, b = x + z, c = x + y. Because s - a = (b + c - a)/2 = ((x + z) + (x + y) - (y + z))/2 = (2x)/2 = x. Similarly, s - b = y, s - c = z.Therefore, the left-hand side becomes:(a/(s - a) + b/(s - b) + c/(s - c)) = ( (y + z)/x + (x + z)/y + (x + y)/z ).Ah, this seems more manageable. So, the left-hand side is ( (y + z)/x + (x + z)/y + (x + y)/z ). Let me write this as (y/x + z/x + x/y + z/y + x/z + y/z ). Which is equal to ( (y/x + x/y ) + (z/x + x/z ) + (z/y + y/z ) ). Each pair like y/x + x/y is ≥ 2 by AM-GM inequality. Therefore, the entire sum is ≥ 2 + 2 + 2 = 6. Therefore, the left-hand side is ≥ 6. But the right-hand side is 3R/r. So if we can show that 6 ≥ 3R/r, which would mean 2 ≥ R/r, but Euler's inequality states that R ≥ 2r, so R/r ≥ 2, hence 6 ≥ 3R/r would require 6 ≥ 3*(≥2), which is 6 ≥ 6. But Euler's inequality is R ≥ 2r, so 3R/r ≥ 6. Hence, if our left-hand side is ≥6 and the right-hand side is ≥6, but this approach only gives that LHS ≥6 and RHS ≥6, but the problem states LHS ≥ 3R/r, which if R/r ≥2, then 3R/r ≥6, but we have LHS ≥6. Therefore, if LHS ≥6 and 3R/r ≥6, but this would not necessarily imply LHS ≥3R/r, unless LHS is exactly 6 when R/r=2. Wait, maybe equality holds when the triangle is equilateral. Let me check.In an equilateral triangle, R = a/(√3), r = a/(2√3), so R/r = 2. Then 3R/r = 6. Also, in an equilateral triangle, all sides are equal, so a = b = c. Then s = (3a)/2, so s - a = (3a/2) - a = a/2. Therefore, a/(s - a) = a/(a/2) = 2. So the sum is 2 + 2 + 2 = 6, which equals 3R/r = 6. So equality holds when the triangle is equilateral.But according to the above approach, LHS ≥6 and 3R/r ≥6, but that doesn't necessarily mean LHS ≥3R/r. For example, suppose in some triangle, LHS=7 and 3R/r=6.5, then the inequality holds, but if in another triangle, LHS=6.5 and 3R/r=7, then the inequality would fail. But according to Euler's inequality, R/r ≥2, so 3R/r ≥6. But from our previous step, LHS ≥6, so perhaps LHS is always greater than or equal to 6, and 3R/r is also greater than or equal to 6, but which one is larger? How do we know that LHS ≥3R/r?Wait, but in the equilateral case, they are equal. So maybe this inequality is actually equivalent to LHS ≥6, since 3R/r ≥6. But no, because 3R/r can be larger than 6. For example, in a very skewed triangle where R is large compared to r, then 3R/r could be larger than 6, but according to our previous step, LHS is always at least 6, but if 3R/r can be larger than 6, then the inequality LHS ≥3R/r is not automatically true. So perhaps my earlier approach is insufficient.Wait, perhaps there is a relation between LHS and R/r that can be established through other means. Let me think again.Earlier, I converted the left-hand side to (y + z)/x + (x + z)/y + (x + y)/z, where x = s - a, y = s - b, z = s - c. Then, since x + y + z = s, and a = y + z, b = x + z, c = x + y.Alternatively, perhaps express R and r in terms of x, y, z.Wait, the inradius r = Δ/s, and Δ, the area, can be expressed using Heron's formula: Δ = √(s x y z). So r = √(s x y z)/s = √(x y z/s).The circumradius R = (a b c)/(4Δ) = ( (y + z)(x + z)(x + y) )/(4 √(s x y z) ). Therefore, R/r = ( (y + z)(x + z)(x + y) )/(4 √(s x y z) ) / ( √(x y z/s) ) = ( (y + z)(x + z)(x + y) )/(4 √(s x y z) ) * √(s)/(√(x y z)) ) = ( (y + z)(x + z)(x + y) )/(4 x y z ) * √(s)/√(s) ) = ( (x + y)(y + z)(z + x) )/(4 x y z ).Therefore, R/r = ( (x + y)(y + z)(z + x) )/(4 x y z ).So, 3R/r = 3*( (x + y)(y + z)(z + x) )/(4 x y z ).Our goal is to show that:(y + z)/x + (x + z)/y + (x + y)/z ≥ 3*( (x + y)(y + z)(z + x) )/(4 x y z )Hmm, this seems like a purely algebraic inequality now. Let's denote the left-hand side as S = (y + z)/x + (x + z)/y + (x + y)/z. And the right-hand side as 3*( (x + y)(y + z)(z + x) )/(4 x y z ). So we need to prove S ≥ 3*( (x + y)(y + z)(z + x) )/(4 x y z )Let me compute (x + y)(y + z)(z + x). Expanding this:(x + y)(y + z)(z + x) = (x + y)(y z + y x + z² + x z ) = x(y z + y x + z² + x z ) + y(y z + y x + z² + x z )= x y z + x y x + x z² + x x z + y² z + y² x + y z² + y x z= x² y + x² z + y² x + y² z + z² x + z² y + 2x y zTherefore, (x + y)(y + z)(z + x) = x²(y + z) + y²(x + z) + z²(x + y) + 2x y zTherefore, 3*( (x + y)(y + z)(z + x) )/(4 x y z ) = 3/(4 x y z ) * [ x²(y + z) + y²(x + z) + z²(x + y) + 2x y z ]Let me write this as 3/(4 x y z ) * [ x²(y + z) + y²(x + z) + z²(x + y) ] + 3/(4 x y z ) * 2x y zSimplifying the second term: 3/(4 x y z ) * 2x y z = 3/(4 x y z ) * 2x y z = 3/2.Therefore, 3R/r = 3/(4 x y z ) * [ x²(y + z) + y²(x + z) + z²(x + y) ] + 3/2.Therefore, our inequality S ≥ 3R/r becomes:(y + z)/x + (x + z)/y + (x + y)/z ≥ 3/(4 x y z ) * [ x²(y + z) + y²(x + z) + z²(x + y) ] + 3/2.Let me denote S = (y + z)/x + (x + z)/y + (x + y)/z. Then, we can write S = (y + z)/x + (x + z)/y + (x + y)/z = (y/x + z/x + x/y + z/y + x/z + y/z ). So S = Σ (y/x + x/y) over all pairs. Which is Σ (x/y + y/x) = 3 terms each. Let me think if there's a way to relate S and [ x²(y + z) + y²(x + z) + z²(x + y) ].Alternatively, let's try to manipulate the inequality:S - 3/2 ≥ 3/(4 x y z ) * [ x²(y + z) + y²(x + z) + z²(x + y) ]Multiply both sides by 4 x y z / 3:(4 x y z / 3)(S - 3/2) ≥ x²(y + z) + y²(x + z) + z²(x + y)Hmm, not sure if this helps. Alternatively, perhaps expand S and compare term by term.S = (y + z)/x + (x + z)/y + (x + y)/z = y/x + z/x + x/y + z/y + x/z + y/z.Let me consider multiplying both sides of the inequality by 4xyz/3 to eliminate denominators:Left-hand side * 4xyz/3 = (y + z)/x * 4xyz/3 + similar terms. Wait, perhaps not. Let me think again.Alternatively, let's cross-multiplied to eliminate denominators. But this might get messy.Alternatively, let's consider homogenizing the inequality. Since both sides are homogeneous in x, y, z (if we consider scaling x, y, z by a constant factor, both sides scale similarly), so we can assume that x + y + z = 1. Wait, but earlier we have x + y + z = s. But perhaps this complicates things. Alternatively, use substitution variables.Alternatively, perhaps use the AM-GM inequality on the left-hand side. Since each term like (y + z)/x is equal to (y + z)/x. But since x, y, z are positive real numbers, perhaps there's a way to bound this.Alternatively, note that the left-hand side S can be written as Σ ( (y + z)/x ) = Σ ( (y + z)/x ). Since y + z = a, x = s - a, but not sure.Wait, going back to the original variables. Since x = s - a, y = s - b, z = s - c, and x + y + z = s. So, perhaps using substitution variables with x + y + z = s. But I don't know if that helps.Alternatively, let's try to test the inequality for an equilateral triangle. If x = y = z, since in an equilateral triangle, all sides are equal, so s - a = s - b = s - c = s/3. Therefore, x = y = z = s/3. Then, S = 3*( (s/3 + s/3)/ (s/3) ) = 3*( (2s/3)/(s/3) ) = 3*2 = 6. On the other hand, 3R/r = 6, as we saw earlier. So equality holds. Now, suppose we take a different triangle, say a right-angled triangle with sides 3, 4, 5. Let's compute both sides.For a 3-4-5 triangle: a=3, b=4, c=5. Then s=(3+4+5)/2=6. So s - a=6-3=3, s - b=6-4=2, s - c=6-5=1. Therefore, the left-hand side is 3/3 + 4/2 +5/1 =1 + 2 +5=8.Now, compute R and r. For a 3-4-5 triangle, area Δ=6. r=Δ/s=6/6=1. R= (a b c)/(4Δ)= (3*4*5)/(4*6)=60/24=2.5. So 3R/r=3*(2.5)/1=7.5. So LHS=8 ≥7.5. So the inequality holds here.Another example: take an isoceles triangle with sides 2, 2, 3. Then s=(2+2+3)/2=3.5. s - a=3.5 -2=1.5, s - b=1.5, s - c=3.5 -3=0.5. Then LHS=2/1.5 +2/1.5 +3/0.5= (4/1.5) +6= (8/3) +6≈2.666+6≈8.666.Compute R and r. Area Δ: using Heron's formula, sqrt(3.5*(3.5 -2)*(3.5 -2)*(3.5 -3))=sqrt(3.5*1.5*1.5*0.5)=sqrt(3.5*1.5*0.75)=sqrt( 3.5*1.125 )=sqrt(3.9375)≈1.984. So r=Δ/s≈1.984/3.5≈0.567. R=(a b c)/(4Δ)= (2*2*3)/(4*1.984)=12/(7.936)≈1.511. Therefore, 3R/r≈3*(1.511)/0.567≈7.99. So LHS≈8.666 ≥7.99, which holds.Another test case: a very skewed triangle, say sides 1, 1, 1.999. Then s=(1 +1 +1.999)/2≈1.9995. s - a≈1.9995 -1=0.9995, similarly for s - b≈0.9995, and s - c≈1.9995 -1.999≈0.0005. Then LHS≈1/0.9995 +1/0.9995 +1.999/0.0005≈1 +1 +3998≈4000.Compute R and r. Area Δ: using Heron's formula, sqrt(1.9995*(1.9995 -1)*(1.9995 -1)*(1.9995 -1.999))≈sqrt(1.9995*0.9995*0.9995*0.0005). This is approximately sqrt(1.9995 * (0.9995)^2 * 0.0005)≈sqrt(1.9995 * 0.99900025 * 0.0005)≈sqrt(1.9995 * 0.0004995)≈sqrt(0.0009985)≈0.0316. So r=Δ/s≈0.0316/1.9995≈0.0158. R=(a b c)/(4Δ)= (1*1*1.999)/(4*0.0316)≈1.999/(0.1264)≈15.82. Therefore, 3R/r≈3*15.82/0.0158≈3*1001≈3003. So LHS≈4000 ≥3003, which holds.So in these test cases, the inequality holds, and the LHS is significantly larger when the triangle is very skewed, which makes sense because R/r becomes large, but the LHS also becomes large. However, in the previous approach, we saw that LHS ≥6 and 3R/r ≥6, but how do we formally establish that LHS ≥3R/r?Perhaps another approach is needed. Let me recall some identities or inequalities that relate R, r, and other triangle quantities.There's the identity that in any triangle, (a + b + c)/2 = s, and there are formulas involving R and r. For example, r = 4R sin(A/2) sin(B/2) sin(C/2). Also, there's the formula cos A + cos B + cos C = 1 + r/R. Maybe this can be useful.Alternatively, we can use trigonometric identities. Let me think again about expressing the left-hand side in terms of trigonometric functions.Earlier, we had:Left-hand side = 2 [ sin A / (sin B + sin C - sin A ) + sin B / (sin A + sin C - sin B ) + sin C / (sin A + sin B - sin C ) ]Let me denote the terms as T_A = sin A / (sin B + sin C - sin A ), and similarly for T_B and T_C. So the left-hand side is 2(T_A + T_B + T_C). Perhaps we can relate these terms to R and r. Since r = 4R sin(A/2) sin(B/2) sin(C/2), as mentioned earlier, and R is known in terms of the sides.Alternatively, let's consider the denominators. For example, sin B + sin C - sin A. Using sine addition formulas or other trigonometric identities. Let me recall that in a triangle, A + B + C = π. So, B + C = π - A. Therefore, sin B + sin C = 2 sin( (B + C)/2 ) cos( (B - C)/2 ) = 2 sin( (π - A)/2 ) cos( (B - C)/2 ) = 2 cos(A/2) cos( (B - C)/2 ). Therefore, sin B + sin C - sin A = 2 cos(A/2) cos( (B - C)/2 ) - sin A.But sin A = 2 sin(A/2) cos(A/2). Therefore, sin B + sin C - sin A = 2 cos(A/2) [ cos( (B - C)/2 ) - sin(A/2) ].Hmm, not sure if this helps. Alternatively, perhaps express the denominator in terms of cosines.Alternatively, note that sin B + sin C - sin A = 2 sin( (B + C)/2 ) cos( (B - C)/2 ) - sin A = 2 sin( (π - A)/2 ) cos( (B - C)/2 ) - sin A = 2 cos(A/2) cos( (B - C)/2 ) - 2 sin(A/2) cos(A/2 )= 2 cos(A/2) [ cos( (B - C)/2 ) - sin(A/2) ]This still seems complicated. Maybe there's a substitution. Let me define u = A/2, v = B/2, w = C/2. Then u + v + w = π/2. Then, sin A = 2 sin u cos u, sin B = 2 sin v cos v, sin C = 2 sin w cos w. Then, let's try to express T_A in terms of u, v, w.Denominator sin B + sin C - sin A = 2 sin v cos v + 2 sin w cos w - 2 sin u cos u.Hmm, maybe not helpful. Alternatively, using the formula for r = 4R sin u sin v sin w. So, 3R/r = 3/(4 sin u sin v sin w). So we need to prove that 2(T_A + T_B + T_C) ≥ 3/(4 sin u sin v sin w).Alternatively, let's consider each term T_A = sin A / (sin B + sin C - sin A ). Substituting A = 2u, B = 2v, C = 2w:T_A = sin 2u / (sin 2v + sin 2w - sin 2u ). Similarly for T_B and T_C.Expressing sin 2u = 2 sin u cos u, sin 2v = 2 sin v cos v, sin 2w = 2 sin w cos w. So,T_A = (2 sin u cos u) / (2 sin v cos v + 2 sin w cos w - 2 sin u cos u )= (sin u cos u) / ( sin v cos v + sin w cos w - sin u cos u )Similarly for T_B and T_C.But this seems messy. Maybe there's another approach.Alternatively, let's use the substitution t = s - a, u = s - b, v = s - c. Wait, but we already used x, y, z for that. Maybe this isn't helpful.Alternatively, recall that in terms of the exradii, r_a = Δ/(s - a), so a/(s - a) = a r_a / Δ. Similarly, the sum becomes (a r_a + b r_b + c r_c)/Δ. So LHS = (a r_a + b r_b + c r_c)/Δ. Then, we need to prove that (a r_a + b r_b + c r_c)/Δ ≥ 3R/r.But since Δ = r s, we have LHS = (a r_a + b r_b + c r_c)/(r s). So, (a r_a + b r_b + c r_c)/s ≥ 3R. But I don't know if there's a relation between a r_a + b r_b + c r_c and R.Alternatively, express the exradii in terms of R and angles. Recall that r_a = Δ/(s - a) = (r s)/(s - a). Also, from the formula r = 4R sin(u) sin(v) sin(w), where u, v, w are the half-angles.Alternatively, perhaps express r_a in terms of R and angles. The exradius r_a is given by r_a = Δ/(s - a) = (r s)/(s - a). Also, using formula for exradius: r_a = 4R sin(u) cos(v) cos(w). Wait, is that correct? Let me recall the formula for exradius. The exradius opposite to A is r_a = Δ/(s - a) = 4R sin(A/2) cos(B/2) cos(C/2). Yes, that's correct. Similarly, r_b = 4R sin(B/2) cos(A/2) cos(C/2), r_c = 4R sin(C/2) cos(A/2) cos(B/2).Therefore, a r_a = a * 4R sin(A/2) cos(B/2) cos(C/2). But a = 2R sin A, so:a r_a = 2R sin A * 4R sin(A/2) cos(B/2) cos(C/2) = 8 R² sin A sin(A/2) cos(B/2) cos(C/2)But sin A = 2 sin(A/2) cos(A/2). Therefore,a r_a = 8 R² * 2 sin²(A/2) cos(A/2) cos(B/2) cos(C/2) = 16 R² sin²(A/2) cos(A/2) cos(B/2) cos(C/2)Similarly, b r_b = 16 R² sin²(B/2) cos(B/2) cos(A/2) cos(C/2)And c r_c = 16 R² sin²(C/2) cos(C/2) cos(A/2) cos(B/2)Therefore, the sum a r_a + b r_b + c r_c = 16 R² cos(A/2) cos(B/2) cos(C/2) [ sin²(A/2) + sin²(B/2) + sin²(C/2) ]Therefore, LHS = (a r_a + b r_b + c r_c)/Δ = [ 16 R² cos(A/2) cos(B/2) cos(C/2) ( sin²A/2 + sin²B/2 + sin²C/2 ) ] / ΔBut Δ = r s, and r = 4R sin(A/2) sin(B/2) sin(C/2), s = (a + b + c)/2 = 2R (sin A + sin B + sin C )/2 = R (sin A + sin B + sin C )Therefore, Δ = r s = 4R sin(A/2) sin(B/2) sin(C/2) * R (sin A + sin B + sin C ) = 4 R² sin(A/2) sin(B/2) sin(C/2) (sin A + sin B + sin C )Therefore, substituting back into LHS:LHS = [ 16 R² cos(A/2) cos(B/2) cos(C/2) ( sin²A/2 + sin²B/2 + sin²C/2 ) ] / [ 4 R² sin(A/2) sin(B/2) sin(C/2) (sin A + sin B + sin C ) ]Simplify:= [16 / 4] * [ cos(A/2) cos(B/2) cos(C/2) / sin(A/2) sin(B/2) sin(C/2) ] * [ sin²A/2 + sin²B/2 + sin²C/2 ) ] / (sin A + sin B + sin C )= 4 * [ cot(A/2) cot(B/2) cot(C/2) ] * [ sin²A/2 + sin²B/2 + sin²C/2 ] / (sin A + sin B + sin C )But cot(A/2) cot(B/2) cot(C/2) = (cos(A/2)/sin(A/2))(cos(B/2)/sin(B/2))(cos(C/2)/sin(C/2)) ) = [ cos(A/2) cos(B/2) cos(C/2) ] / [ sin(A/2) sin(B/2) sin(C/2) ] = (1/r) * something? Wait, since r = 4R sin(A/2) sin(B/2) sin(C/2), so cot(A/2) cot(B/2) cot(C/2) = [ cos(A/2) cos(B/2) cos(C/2) ] / [ sin(A/2) sin(B/2) sin(C/2) ] = (1/(4R)) * [ cos(A/2) cos(B/2) cos(C/2) ] / (r/(4R)) )? Wait, maybe not directly helpful.Alternatively, note that cot(A/2) = (s - a)/r, similarly for others. So cot(A/2) cot(B/2) cot(C/2) = [(s - a)(s - b)(s - c)]/r³. But this might not help directly.Alternatively, perhaps express [ sin²A/2 + sin²B/2 + sin²C/2 ] in terms of other trigonometric identities. Let me recall that in a triangle, A + B + C = π, so A/2 + B/2 + C/2 = π/2. Let me set x = A/2, y = B/2, z = C/2, so x + y + z = π/2. Then, sin²x + sin²y + sin²z.There's an identity for sin²x + sin²y + sin²z when x + y + z = π/2. Let me compute:sin²x + sin²y + sin²z = sin²x + sin²y + sin²(π/2 - x - y )= sin²x + sin²y + cos²(x + y )Expand cos²(x + y) = [ cosx cosy - sinx siny ]² = cos²x cos²y - 2 sinx siny cosx cosy + sin²x sin²yTherefore, sin²x + sin²y + cos²x cos²y - 2 sinx siny cosx cosy + sin²x sin²yThis seems complicated. Maybe there's a better way. Alternatively, note that since x + y + z = π/2, we can use the identity:sin²x + sin²y + sin²z + 2 sinx siny sinz = 1 - (cos²x + cos²y + cos²z - 2 sinx siny sinz )Not sure. Alternatively, use specific values. For example, in an equilateral triangle, x = y = z = π/6. Then sin²x = sin²(π/6) = (1/2)^2 = 1/4. So sum is 3*(1/4) = 3/4. Then, LHS in this case would be 4 * [cot(π/6) cot(π/6) cot(π/6)] * (3/4) / (sin π/3 + sin π/3 + sin π/3 )cot(π/6) = √3, so cot^3(π/6) = (√3)^3 = 3√3. Then 4 * 3√3 * (3/4) / (3*(√3/2)) ) = 3√3 * (3/4) * 4/(3√3/2) ) = 3√3 * 3/4 * 4/(3√3/2 ) = 3√3 * 3/4 * 8/(3√3) ) = (3√3 * 3 * 8 ) / (4 * 3√3 ) ) = (72√3)/(12√3) ) = 6. Which matches the equilateral case.But this doesn't help in proving the general inequality. Perhaps another route is necessary. Let me recall that in the initial substitution, we rewrote the left-hand side as S = (y + z)/x + (x + z)/y + (x + y)/z, and the right-hand side is 3R/r. Then, after some manipulation, we saw that 3R/r = 3*( (x + y)(y + z)(z + x) )/(4 x y z ). So the inequality is S ≥ 3*( (x + y)(y + z)(z + x) )/(4 x y z )Let me denote this as S ≥ 3*K/(4 x y z ), where K = (x + y)(y + z)(z + x).But perhaps we can use the AM-GM inequality to relate S and K. Let's see:Note that (x + y)(y + z)(z + x) ≤ 8/27 (x + y + z)^3 by AM-GM, but since x + y + z = s, but not sure. Wait, no, actually for three variables, the product (x + y)(y + z)(z + x) is maximum when x = y = z, but here we need a relation between S and K.Alternatively, let's expand both S and K.We have:S = (y + z)/x + (x + z)/y + (x + y)/z = Σ (y/x + z/x) = Σ (x/y + y/x + x/z + z/x + y/z + z/y )Wait, no, S = Σ ( (y + z)/x ) = (y + z)/x + (x + z)/y + (x + y)/z.Alternatively, expand K:K = (x + y)(y + z)(z + x) = x y (x + y + z) + y z (x + y + z) + z x (x + y + z) - x y z = (x + y + z)(x y + y z + z x) - x y z But x + y + z = s, so K = s (x y + y z + z x) - x y z But not sure if this helps.Alternatively, perhaps use the inequality between S and K. Let me consider that:We need to show that S ≥ 3K/(4 x y z ). Let's rearrange this as:4 x y z S ≥ 3KBut S = (y + z)/x + (x + z)/y + (x + y)/z. Multiplying by 4 x y z:4 x y z * S = 4 x y z [ (y + z)/x + (x + z)/y + (x + y)/z ] = 4 y z (y + z) + 4 x z (x + z) + 4 x y (x + y )= 4 y z (y + z) + 4 x z (x + z) + 4 x y (x + y )We need to show that this is ≥ 3K = 3(x + y)(y + z)(z + x)So, the inequality becomes:4[y z(y + z) + x z(x + z) + x y(x + y)] ≥ 3(x + y)(y + z)(z + x)Let me expand both sides.Left-hand side:4[ y z(y + z) + x z(x + z) + x y(x + y) ] = 4[ y² z + y z² + x² z + x z² + x² y + x y² ].Right-hand side:3(x + y)(y + z)(z + x). Expand this step by step:First, (x + y)(y + z) = x y + x z + y² + y z.Then, multiply by (z + x):= (x y + x z + y² + y z)(z + x)= x y z + x² y + x z² + x² z + y² z + x y² + y z² + x y z= x² y + x² z + y² z + y² x + z² x + z² y + 2x y z.Therefore, 3(x + y)(y + z)(z + x) = 3[ x² y + x² z + y² z + y² x + z² x + z² y + 2x y z ].So, the inequality is:4[ y² z + y z² + x² z + x z² + x² y + x y² ] ≥ 3[ x² y + x² z + y² z + y² x + z² x + z² y + 2x y z ]Let's bring all terms to the left side:4[ y² z + y z² + x² z + x z² + x² y + x y² ] - 3[ x² y + x² z + y² z + y² x + z² x + z² y + 2x y z ] ≥ 0.Simplify term by term:For x² y:4x² y - 3x² y = x² y.Similarly, for x² z:4x² z -3x² z = x² z.For y² z:4y² z -3y² z = y² z.For y² x:4x y² -3x y² = x y².For z² x:4x z² -3z² x = x z².For z² y:4y z² -3y z² = y z².Then, the remaining terms are:4y z² + 4x z² + 4x y² + 4x² z + 4x² y + 4y² z -3[ ... ] which are accounted for above.Wait, no, wait:Wait, the left-hand side after expansion is:4(y² z + y z² + x² z + x z² + x² y + x y²) = 4y² z +4 y z² +4 x² z +4 x z² +4 x² y +4 x y².Subtracting 3(x² y +x² z +y² z +y² x +z² x +z² y +2x y z):= 4y² z +4 y z² +4 x² z +4 x z² +4 x² y +4 x y² -3x² y -3x² z -3y² z -3y² x -3z² x -3z² y -6x y z.Grouping similar terms:x² y: 4x² y -3x² y = x² yx² z: 4x² z -3x² z = x² zy² z:4y² z -3y² z = y² zy² x:4x y² -3x y² = x y²z² x:4x z² -3x z² = x z²z² y:4y z² -3y z² = y z²And the remaining terms:4y z² +4x z² +4x y² +4x² z +4x² y +4y² z minus the terms above.Wait, perhaps I need to re-express the entire left-hand side:After subtraction, we have:x² y +x² z +y² z +y² x +z² x +z² y -6x y z ≥0.So the inequality reduces to:x² y +x² z +y² z +y² x +z² x +z² y -6x y z ≥0.Factorizing, this can be written as:x y (x + y) + y z (y + z) + z x (z + x) -6x y z ≥0.Hmm, not sure. Alternatively, note that the expression is symmetric in x, y, z. Perhaps factor it as a sum of squares or apply AM-GM.Let me consider applying AM-GM to each term. For example, x² y + x² z ≥ 2x² sqrt(y z). Similarly for other terms. But this might not lead directly to the desired inequality.Alternatively, note that x² y + y² z + z² x ≥3x y z by AM-GM. Similarly, x² z + y² x + z² y ≥3x y z. Therefore, adding these two inequalities gives x² y + x² z + y² z + y² x + z² x + z² y ≥6x y z. Therefore, x² y + x² z + y² z + y² x + z² x + z² y -6x y z ≥0, which is exactly the expression we have. Hence, the inequality holds, and equality occurs when x = y = z, i.e., when the triangle is equilateral.Therefore, the original inequality 4[y z(y + z) + x z(x + z) + x y(x + y)] ≥3(x + y)(y + z)(z + x) holds, which implies that S ≥3R/r.Hence, the inequality is proven.

✅:To prove that in triangle ABC:frac{a}{s-a}+frac{b}{s-b}+frac{c}{s-c} geqslant frac{3R}{r},we proceed as follows:Step 1: Substitute Variables Let x = s - a, y = s - b, z = s - c, where s = frac{a+b+c}{2} is the semiperimeter. Then, a = y + z, b = x + z, c = x + y, and x + y + z = s.Step 2: Rewrite the Left-Hand Side (LHS) The LHS becomes:frac{a}{s-a} + frac{b}{s-b} + frac{c}{s-c} = frac{y+z}{x} + frac{x+z}{y} + frac{x+y}{z}.Step 3: Express in Terms of x, y, z This simplifies to:left(frac{y}{x} + frac{z}{x}right) + left(frac{x}{y} + frac{z}{y}right) + left(frac{x}{z} + frac{y}{z}right) = sum_{text{sym}} frac{x}{y}.Step 4: Relate to Circumradius (R) and Inradius (r) Using formulas for R and r:R = frac{abc}{4Delta}, quad r = frac{Delta}{s},where Delta is the area. Express R/r as:frac{R}{r} = frac{(y+z)(x+z)(x+y)}{4xyz}.Thus, frac{3R}{r} = frac{3(y+z)(x+z)(x+y)}{4xyz}.Step 5: Formulate the Inequality We need to show:frac{y+z}{x} + frac{x+z}{y} + frac{x+y}{z} geq frac{3(y+z)(x+z)(x+y)}{4xyz}.Step 6: Expand and Compare Expand both sides:- LHS: 4(yz(y+z) + xz(x+z) + xy(x+y))- RHS: 3(x+y)(y+z)(z+x)Subtracting RHS from LHS and simplifying yields:x^2y + x^2z + y^2x + y^2z + z^2x + z^2y - 6xyz geq 0.Step 7: Apply AM-GM Inequality By AM-GM, x^2y + y^2z + z^2x geq 3xyz and similarly for cyclic permutations. Summing these gives:x^2y + x^2z + y^2x + y^2z + z^2x + z^2y geq 6xyz,thus proving the inequality. Equality holds when x = y = z, i.e., in an equilateral triangle.Conclusion Therefore, the inequality frac{a}{s-a}+frac{b}{s-b}+frac{c}{s-c} geqslant frac{3R}{r} holds in any triangle ABC, with equality if and only if the triangle is equilateral. boxed{frac{a}{s-a}+frac{b}{s-b}+frac{c}{s-c} geqslant frac{3R}{r}}

🔑:To prove: frac{a}{s-a} + frac{b}{s-b} + frac{c}{s-c} geq frac{3R}{r}, where (a, b, c) are the sides of ( triangle ABC ), (s) is the semi-perimeter of the triangle, (R) is the circumradius, and (r) is the inradius.1. Set-Up with Substitution: Let (a = y + z), (b = z + x), and (c = x + y) where (x, y, z > 0). Thus: [ s = frac{a + b + c}{2} = frac{(y + z) + (z + x) + (x + y)}{2} = x + y + z. ] So, ( s - a = (x + y + z) - (y + z) = x ), ( s - b = (x + y + z) - (z + x) = y ), and ( s - c = (x + y + z) - (x + y) = z ). 2. Transform the Original Expresssion: Substitute the values into the inequality: [ frac{y+z}{x} + frac{z+x}{y} + frac{x+y}{z} geq frac{3R}{r}. ] 3. Reformulating Using AM-GM Inequality: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM) inequality: [ frac{y+z}{x} + frac{z+x}{y} + frac{x+y}{z} geq 3 sqrt[3]{frac{(y+z)(z+x)(x+y)}{xyz}}. ] 4. Check Product Simplifications: Simplify the right-hand side product: [ (y+z)(z+x)(x+y) quad text{and} quad xyz. ] - Notice that: [ (y+z)(z+x)(x+y) = sum_{text{sym}} yzx + yz^2 + y^2z + xz^2 + x^2z + xy^2 + x^2y. ]5. Compare With the Original Products: The expanded form becomes: [ 4[x y(x+y) + y z (y + z) + z x (z + x)] geq 3(x+y)(y+z)(z+x). ]6. Simplify and Constrain: Simplify further to: [ (x+y)(y+z)(z+x) geq 8 xyz. ]This completes the proof. Since:[(x+y)(y+z)(z+x) geq 8xyz,]it confirms that:[frac{a}{s-a}+frac{b}{s-b}+frac{c}{s-c} geq frac{3R}{r}.]Hence, the initial proposition holds true. [boxed{frac{a}{s-a} + frac{b}{s-b} + frac{c}{s-c} geq frac{3R}{r}}.]

🧠:Okay, let's see. The problem says: "The number which is 6 less than twice the square of 4 is..." with options from A to E. Hmm, I need to figure out which one is correct. Let me break it down step by step.First, let's parse the sentence. The main components here are "twice the square of 4" and then "6 less than" that. So the structure is: take the square of 4, multiply it by 2, and then subtract 6 from the result. Let me write that down in mathematical terms to be clear.The square of 4 is 4 squared, which is 4². Then twice that would be 2 times 4². Then subtract 6. So the expression would be 2*(4²) - 6. Let me compute that step by step to avoid mistakes.First, compute 4 squared. 4 times 4 is 16. So 4² = 16. Then twice that is 2*16. Let me do that multiplication. 2 times 16 is 32. So now, we have 32. Then subtract 6 from 32. 32 minus 6 is 26. So the result should be 26. Let me check the answer choices. Option C is 26, so that's probably the answer. Wait, but let me double-check to make sure I didn't misinterpret the wording.The problem says "6 less than twice the square of 4." In math terms, when something is "A less than B," it translates to B - A. So "6 less than twice the square of 4" would be (twice the square of 4) minus 6. So yes, that's exactly what I did: 2*(4^2) -6. So 2*16=32, 32-6=26. That seems correct.Let me check if any of the other options could be confused here. For example, if someone misplaces the order and does 4 squared minus 6 first, then multiplies by 2, that would be different. Let's test that. If it was 2*(4² - 6), then that would be 2*(16 -6)=2*10=20, but 20 isn't one of the options. So that's not the case here. The wording specifies "twice the square of 4," so the square is computed first, then multiplied by 2, then subtract 6. So the order is correct as per my initial calculation.Another possible mistake could be squaring (4*2) first instead of squaring 4 and then multiplying by 2. Let's see: if someone does (2*4)^2 -6, that would be 8 squared, which is 64, minus 6, which is 58. But 58 is option E. However, that's not what the problem states. The problem says "twice the square of 4," which is 2*(4²), not (2*4)². The phrase "twice the square of 4" implies that the squaring happens first, and then the multiplication by two. So the correct interpretation is 2*(4²), not (2*4)². Therefore, option E would be incorrect here.Let me check the answer options again. The options are (A) -26, (B) 10, (C) 26, (D) 38, (E) 58. Our calculation leads us to 26, which is option C. The other options might come from different misinterpretations, but according to the correct order of operations and the phrasing of the problem, 26 is the right answer.Just to recap: square of 4 is 16, multiply by 2 to get 32, subtract 6 to get 26. All steps are in the correct order based on the problem's wording. Therefore, I'm confident that the answer is C) 26.

✅:The problem asks for a number that is 6 less than twice the square of 4. Breaking this down:1. Square of 4: (4^2 = 16).2. Twice the square of 4: (2 times 16 = 32).3. 6 less than the result: (32 - 6 = 26).Key checks ensure no misinterpretation: - Order of operations confirms squaring precedes multiplication and subtraction. - Misreading as ((2 times 4)^2 - 6) leads to (58) (option E), but phrasing explicitly specifies "twice the square of 4," not "square of twice 4."Final Answer: (boxed{C})

🔑:1. Find the square of 4: First, calculate (4^2): [ 4^2 = 16 ]2. Multiply the square by 2: Next, multiply the result by 2: [ 2 times 16 = 32 ]3. Subtract 6 from the product: Now, subtract 6 from the product obtained in the previous step: [ 32 - 6 = 26 ] ConclusionFrom the steps above, the number which is 6 less than twice the square of 4 is:[boxed{26}]

🧠:Okay, let's tackle this problem step by step. So, we have a regular hexagon named ABCDEF. A frog starts at vertex A, and each time it can jump to one of the two adjacent vertices. The frog stops if it reaches vertex D within 5 jumps; otherwise, it stops after 5 jumps. We need to find the number of different ways the frog can jump from A until it stops.First, let me visualize the hexagon. Since it's regular, all sides and angles are equal. The vertices are labeled in order: A, B, C, D, E, F, and back to A. So, each vertex is connected to two adjacent vertices. For example, A is connected to B and F. Similarly, D is connected to C and E.The frog starts at A. Each jump can be either clockwise or counter-clockwise. The problem is similar to counting the number of paths the frog can take, considering that it might stop earlier if it reaches D within 5 jumps. So, we need to consider both possibilities: paths that reach D before or at the 5th jump, and paths that don't reach D at all within 5 jumps.But the key here is that once the frog reaches D, it stops. That means if the frog gets to D on, say, the 3rd jump, it won't make any more jumps. Therefore, any path that reaches D before the 5th jump is terminated early. So, we need to count all possible paths starting at A, making up to 5 jumps, but some of them might end early at D.This seems like a problem that can be approached using recursion or dynamic programming. Let's think about states. Let's define the state as the current vertex and the number of jumps made so far. We need to count the number of ways to reach each vertex at each jump count, considering that once the frog reaches D, it stops.Alternatively, since the frog stops either when it reaches D or after 5 jumps, whichever comes first, we need to calculate all possible paths from A, considering that once D is reached, the path is terminated.To formalize this, let's define f(n, v) as the number of ways to reach vertex v in exactly n jumps without having reached D before. Then, the total number of paths would be the sum over all n from 1 to 5 of f(n, D) plus the number of paths that don't reach D even after 5 jumps. Wait, actually, no. Because once the frog reaches D, it stops. So, the total number of paths is the sum over all n from 1 to 5 of the number of paths that reach D for the first time at the nth jump, plus the number of paths that never reach D in 5 jumps.But perhaps a better way is to model the process step by step, keeping track of the current position and the number of jumps, making sure that once D is reached, we don't consider further jumps from D.Let me structure this as follows:- The frog starts at A (jump 0). At each step, it can jump to the left or right adjacent vertex, except if it's already at D, in which case it stops.We can model this using recursion with memoization. Let's define a function count_jumps(current_vertex, jumps_made) that returns the number of ways to continue jumping from current_vertex with jumps_made so far. The base cases are:- If current_vertex is D, return 1 (since the frog stops here, so this is one valid path).- If jumps_made == 5, return 1 (since the frog must stop regardless of position).Otherwise, from the current vertex, the frog can jump to the two adjacent vertices. So, for each vertex adjacent to current_vertex, we add the number of ways from those vertices. But we need to make sure that if jumping to D, we count that path as terminating, so we don't continue jumping from D.Wait, actually, in the problem statement, the frog stops jumping when it reaches D. So, once it jumps to D, it cannot make any more jumps. Therefore, when building the recursion, if the frog is at a vertex other than D, it can jump to two adjacent vertices, but if it jumps to D, that path is terminated.Alternatively, perhaps it's better to think of each path as a sequence of jumps that either ends when it reaches D or ends after 5 jumps. Therefore, each path can be of length 1 to 5, but if it reaches D before 5 jumps, the path is shorter.But counting such paths is a bit tricky. Let me think of it as a tree where each node is a state (current vertex, number of jumps so far). The root is (A, 0). From each node (v, n), if v is not D and n < 5, then it branches to two nodes: (left adjacent, n+1) and (right adjacent, n+1). If v is D, then it's a leaf node. Similarly, if n=5, it's a leaf node regardless of v.Therefore, the total number of paths is the number of leaf nodes in this tree. Each leaf node corresponds to a path that either ended at D or ended at some other vertex after 5 jumps.Therefore, we need to count all such leaf nodes.Alternatively, since each jump is a binary choice (left or right), except when at D, the total number of possible paths without considering D would be 2^5 = 32. But since some paths might reach D before 5 jumps and terminate early, those paths would have fewer jumps, but we still need to count them as distinct. However, the problem states that the frog stops jumping when it reaches D, so those paths are shorter. Therefore, the total number of paths is not just 32, because some paths are shorter, but all paths (including the shorter ones) need to be counted.Wait, but how exactly does the counting work? For example, if the frog reaches D in 3 jumps, then the path is 3 jumps long, and we don't consider the remaining 2 jumps. So, each path is uniquely determined by the sequence of jumps until it either stops at D or completes 5 jumps.Therefore, the problem is equivalent to counting all possible sequences of jumps starting at A, where each jump is to an adjacent vertex, the sequence stops when reaching D, and the maximum length is 5 jumps.Hence, the total number of different ways is the number of such sequences.To model this, perhaps we can use dynamic programming where we track, for each vertex and each number of jumps, the number of ways to reach that vertex at that step without having previously reached D.Let me formalize this.Let dp[n][v] be the number of ways to reach vertex v in exactly n jumps without having reached D in any of the previous jumps.Then, the initial condition is dp[0][A] = 1, since the frog starts at A with 0 jumps.For each step from n = 0 to 4:- For each vertex v (excluding D), if dp[n][v] > 0, then from v, the frog can jump to two adjacent vertices, say v_prev and v_next.But in a hexagon, each vertex has two neighbors. For example, A is connected to B and F. Similarly, B is connected to A and C, etc.We need to be careful about the adjacency. Let's label the vertices in order: A (0), B (1), C (2), D (3), E (4), F (5). So each vertex i is adjacent to (i-1) mod 6 and (i+1) mod 6.Therefore, for vertex v, the two adjacent vertices are (v-1) mod 6 and (v+1) mod 6.But since the frog cannot jump to D once it's there, but D is a stopping point. Wait, actually, once the frog jumps to D, it stops, so in the DP, once we reach D at any step, we don't consider further jumps from D.But in our DP table, dp[n][v] counts the number of ways to reach v in exactly n jumps without having reached D before. Therefore, once we reach D, we can add those paths to our total count and not consider them in future steps.Therefore, the total number of paths is the sum over all n from 1 to 5 of dp[n][D] (the number of paths that reach D for the first time at step n) plus the number of paths that after 5 jumps haven't reached D (sum of dp[5][v] for v ≠ D).Therefore, we need to compute dp[n][v] for all n from 0 to 5 and v from A to F, excluding D once it's reached.Let's structure this step by step.Initialize dp[0][A] = 1.For each step n from 0 to 4:- For each vertex v (A, B, C, D, E, F): - If v is D, then dp[n][D] is added to the total count for D at step n, and we don't process further jumps from D. - If v is not D, then from v, the frog can jump to two adjacent vertices. So, for each adjacent vertex u of v: - If u is not D, then dp[n+1][u] += dp[n][v] - If u is D, then we add dp[n][v] to the total count for D at step n+1, but we don't propagate further from D.Wait, perhaps another way: when computing dp[n+1][u], if u is not D, then it's part of the ongoing paths. If u is D, then those paths terminate at D, so we need to accumulate them into the total.Alternatively, we can separate the counts into two parts: the number of paths that have already stopped (by reaching D) and the number of paths that are still ongoing (haven't reached D yet).But maybe a better approach is to model the DP as follows:At each step, we have a current state which is the current vertex and whether we have already stopped. But since once we reach D, we stop, the state is either "active" (haven't reached D yet) or "stopped" (have reached D). However, since we need to count all possible paths, whether they stopped early or continued until 5 jumps, we can model the DP as follows:Let active[n][v] be the number of active paths at step n that are at vertex v without having reached D yet.Let stopped[n] be the number of paths that have stopped by reaching D at step n.Initially, active[0][A] = 1, and stopped[0] = 0.For each step from n = 0 to 4: For each vertex v: If active[n][v] > 0: For each adjacent vertex u of v: If u == D: stopped[n+1] += active[n][v] Else: active[n+1][u] += active[n][v] Then, after processing all vertices, move to n+1.After 5 steps, the total number of paths is the sum over stopped[1] to stopped[5] plus the sum over active[5][v] for all v ≠ D.This seems like a solid approach.Let's formalize this.We need to track active paths and stopped paths.Initialize:active = [ [0]*6 for _ in range(6) ] # active[n][v], n from 0 to 5, v from 0 to 5 (A=0, B=1, ..., F=5)active[0][0] = 1 # Start at A (vertex 0)stopped = [0]*6 # stopped[n], n from 0 to 5Then, for each n from 0 to 4: For each vertex v in 0 to 5: If active[n][v] > 0: left = (v - 1) % 6 right = (v + 1) % 6 for u in [left, right]: if u == 3: # D is vertex 3 stopped[n+1] += active[n][v] else: active[n+1][u] += active[n][v]After processing all n from 0 to 4, the total stopped paths are sum(stopped[1..5]), and the total active paths after 5 jumps are sum(active[5][v] for v in 0..5 if v != 3). So total paths = sum(stopped[1..5]) + sum(active[5][v] for v != 3).Let's compute this step by step.First, label the vertices:A=0, B=1, C=2, D=3, E=4, F=5.Initialize active[0][0] = 1.Now, let's compute for each n from 0 to 4.n = 0:active[0][0] = 1.Process vertex 0 (A):left = (0 - 1) % 6 = 5 (F)right = (0 + 1) % 6 = 1 (B)Check u=5 (F) and u=1 (B):Neither is D (3), so:active[1][5] += 1active[1][1] += 1So after n=0:active[1][5] = 1active[1][1] = 1stopped remains [0,0,0,0,0,0]n=1:Process active[1][1] and active[1][5]First, active[1][1] = 1.Left of 1 is 0 (A), right is 2 (C).u=0 and u=2.Neither is D (3), so:active[2][0] += 1active[2][2] += 1Next, active[1][5] = 1.Left of 5 is 4 (E), right is 0 (A).u=4 and u=0.Neither is D, so:active[2][4] += 1active[2][0] += 1Now, active[2][0] = 1 (from 1) +1 (from 5) = 2active[2][2] = 1active[2][4] =1stopped[2] = 0n=1 complete.n=2:Process active[2][0], active[2][2], active[2][4]First, active[2][0] = 2.Left of 0 is 5, right is 1.u=5 and u=1.Neither is D, so:active[3][5] += 2active[3][1] += 2Next, active[2][2] = 1.Left of 2 is 1, right is 3.u=1 and u=3.u=3 is D, so:stopped[3] += 1u=1 is not D, so:active[3][1] += 1Then, active[2][4] =1.Left of 4 is 3 (D), right is 5.u=3 and u=5.u=3 is D, so:stopped[3] +=1u=5 is not D, so:active[3][5] +=1So after processing n=2:active[3][5] = 2 (from 0) +1 (from 4) = 3active[3][1] = 2 (from 0) +1 (from 2) = 3stopped[3] =1 (from 2) +1 (from4) = 2Other active entries at n=3 are 0.n=3:Process active[3][5] =3, active[3][1] =3First, active[3][1] =3.Left of 1 is 0, right is 2.u=0 and u=2.Neither is D, so:active[4][0] +=3active[4][2] +=3Next, active[3][5] =3.Left of 5 is 4, right is 0.u=4 and u=0.Neither is D, so:active[4][4] +=3active[4][0] +=3So active[4][0] =3 +3=6active[4][2] =3active[4][4] =3Also, check if any jumps from active[3][v] to D:But in n=3, we process active[3][v], which is at step 3. Jumps to D would be added to stopped[4].But in this case, from active[3][1], jumps to 0 and 2, neither is D. From active[3][5], jumps to 4 and 0, neither is D. So no additions to stopped[4].Thus, stopped[4] remains 0.n=3 complete.n=4:Process active[4][0] =6, active[4][2]=3, active[4][4]=3First, active[4][0] =6.Left is 5, right is 1.u=5 and u=1.Neither is D, so:active[5][5] +=6active[5][1] +=6Next, active[4][2] =3.Left is 1, right is 3.u=1 and u=3.u=3 is D, so:stopped[5] +=3u=1 is not D, so:active[5][1] +=3Then, active[4][4] =3.Left is 3, right is 5.u=3 is D, so:stopped[5] +=3u=5 is not D, so:active[5][5] +=3So after processing n=4:active[5][5] =6 (from 0) +3 (from4) =9active[5][1] =6 (from0) +3 (from2) =9stopped[5] =3 (from2) +3 (from4) =6Now, we've completed n=4. Now, after n=5, we don't process further since the frog stops at 5 jumps.Now, total stopped paths: sum(stopped[1..5]) = stopped[1] + stopped[2] + stopped[3] + stopped[4] + stopped[5]Looking back:stopped[1] =0stopped[2]=0stopped[3]=2stopped[4]=0stopped[5]=6Total stopped = 2 +6 =8Now, active[5][v] for v ≠3:active[5][0] =0active[5][1]=9active[5][2]=0active[5][4]=0active[5][5]=9So sum(active[5][v] for v≠3) =9 +9 =18Total paths = stopped_total + active_total =8 +18=26Wait, but according to this calculation, the answer is 26. But let's verify step by step to check for errors.Let me retrace the steps.Starting with n=0:active[0][0] =1n=0:jump to 1 and 5.active[1][1]=1, active[1][5]=1n=1:From 1: jump to 0 and 2. active[2][0]=1, active[2][2]=1From 5: jump to 4 and 0. active[2][4]=1, active[2][0]=1+1=2n=2:active[2][0]=2, active[2][2]=1, active[2][4]=1From 0: jump to 5 and 1. active[3][5]=2, active[3][1]=2From 2: jump to 1 and 3. u=1: active[3][1] +=1, u=3: stopped[3] +=1From 4: jump to 3 and 5. u=3: stopped[3] +=1, u=5: active[3][5] +=1Thus, active[3][5]=2+1=3, active[3][1]=2+1=3stopped[3]=1+1=2n=3:active[3][1]=3, active[3][5]=3From 1: jump to 0 and 2. active[4][0]=3, active[4][2]=3From 5: jump to 4 and 0. active[4][4]=3, active[4][0]=3+3=6n=4:active[4][0]=6, active[4][2]=3, active[4][4]=3From 0: jump to 5 and 1. active[5][5]=6, active[5][1]=6From 2: jump to 1 and 3. u=1: active[5][1] +=3, u=3: stopped[5] +=3From 4: jump to 3 and 5. u=3: stopped[5] +=3, u=5: active[5][5] +=3Thus, active[5][5]=6+3=9, active[5][1]=6+3=9stopped[5]=3+3=6Total stopped paths: stopped[3] + stopped[5] =2 +6=8Active paths after 5 jumps: active[5][1] + active[5][5]=9 +9=18Total=8 +18=26Hmm, but I have a feeling that the answer might be different. Let me think.Wait, in the hexagon, the distance from A to D is 3 steps (A to B to C to D), or 3 steps in the other direction (A to F to E to D). So the shortest path to D is 3 jumps. Therefore, the frog can reach D in 3, 4, or 5 jumps.But according to our calculation, stopped[3]=2, stopped[5]=6, but stopped[4]=0. Is that correct?Wait, in our calculation, stopped[3] is the number of paths that reach D at step 3. Similarly, stopped[5] is the number of paths that reach D at step 5. But wait, in step 4, active[4][v] are the active paths, and when processing n=4, we add to stopped[5] when jumping to D. So stopped[5] counts the paths that reach D at step 5.But according to our calculation, stopped[3]=2, stopped[5]=6. So total stopped paths are 8. Then the active paths after 5 jumps are 18. Total is 26.But let's think: without any restrictions, the total number of paths in 5 jumps would be 2^5=32. But since some paths are stopped early, those paths are less than 5 jumps. However, in our problem, the frog stops either when it reaches D or after 5 jumps. Therefore, the total number of paths should be equal to the number of all possible paths of length up to 5, where paths that reach D earlier are terminated. So, each path is uniquely determined by its sequence of jumps until it reaches D or until 5 jumps.So, according to our calculation, the answer is 26, which is less than 32. But let's see if that's correct.Alternatively, let's try to compute it manually for small steps.At step 1: The frog is at B or F. Neither is D. So active paths: 2.At step 2: From B, can go to A or C; from F, can go to E or A. So positions: A, C, E. Each has 1 path? Wait, no:Wait, from step 1:At step 1, positions are B and F.From B, step 2 can be to A or C.From F, step 2 can be to E or A.Thus, at step 2, positions: A (from B and F: 2 paths), C (1 path), E (1 path). Total active paths: 4.But in our calculation earlier, at n=2, active[2][0]=2, active[2][2]=1, active[2][4]=1. That matches.From step 2, positions A, C, E.From A: step 3 can go to B or F.From C: step 3 can go to B or D.From E: step 3 can go to D or F.So, from A: B and F (2 paths)From C: B and D. Jumping to D stops, so those are 1 path to B and 1 path stopped.From E: D and F. Similarly, 1 path to D (stopped) and 1 path to F.Thus, at step 3:Active paths: B (from A and C: 2+1=3), F (from A and E: 2+1=3). Stopped paths: 2 (from C and E).Which matches our DP: active[3][1]=3 (B is vertex1), active[3][5]=3 (F is vertex5), stopped[3]=2.Then step 4:From B (active[3][1]=3):Can go to A or C.From F (active[3][5]=3):Can go to E or A.Thus, step 4:From B: A and C (each 3 paths)From F: E and A (each 3 paths)So positions at step4:A: 3 (from B) +3 (from F) =6C: 3 (from B)E: 3 (from F)Thus, active[4][0]=6 (A=0), active[4][2]=3 (C=2), active[4][4]=3 (E=4). Which matches our DP.Then step5:From A (6 paths):Go to B and F. So 6 paths to B and 6 to F.From C (3 paths):Go to B and D. 3 paths to B, 3 stopped at D.From E (3 paths):Go to D and F. 3 paths to D stopped, 3 paths to F.Thus, active[5][1]=6 (from A) +3 (from C) =9active[5][5]=6 (from A) +3 (from E) =9stopped[5]=3 (from C) +3 (from E) =6Total active at step5: 18, stopped total:8 (2 at step3 and 6 at step5). Total 26.Therefore, the calculation seems consistent. Therefore, the answer should be 26.But I recall that in some similar problems, the answer might be different. Let me think again.Alternatively, perhaps I made a mistake in the labels. Let's check if D is vertex 3. Yes, A=0, B=1, C=2, D=3, E=4, F=5.Another way: since the frog can reach D in 3 jumps. There are two paths of length 3: A-B-C-D and A-F-E-D. But wait, actually, those are two different paths. However, according to our calculation, stopped[3]=2, which corresponds to these two paths. Then, in step5, we have 6 paths that reach D at step5. How?Let's see. To reach D in 5 jumps, the frog must make a round trip. For example, starting at A, going to B, then back to A, then to B, C, D. But the exact number of such paths is 6.Similarly, from the other direction: A-F-E-D, but bouncing back and forth.Alternatively, let's think recursively. Let’s define f(n, v) as the number of ways to reach vertex v in n jumps without having reached D before.The total number of paths is sum_{n=1}^5 f(n, D) + sum_{v≠D} f(5, v)We can compute f(n, D) for n=3,4,5.As calculated earlier:f(3, D)=2 (paths: ABC D and AFE D)For f(4, D):To reach D in 4 jumps. The frog must be at C or E at step3, then jump to D. However, in our previous calculation, we didn't have any stopped[4], which suggests that no paths reach D at step4. But is that correct?Wait, if the frog is at step3 at C or E, it can jump to D. But in our previous DP steps, when processing n=3 (step3), we have active[3][1] and active[3][5]. Wait, no. At step3, active[3][1]=3 (B) and active[3][5]=3 (F). So to reach D at step4, the frog must jump from C or E at step3. But in our case, at step3, the frog is at B or F, not C or E. Therefore, it cannot reach D in step4. Hence, stopped[4]=0. Which aligns with our previous result.Therefore, to reach D in step4, the frog must be at C or E at step3. But since at step3, the frog is at B or F, it's impossible. Therefore, no paths reach D at step4. Hence, stopped[4]=0.Similarly, to reach D at step5, the frog must be at C or E at step4.Looking at step4, active[4][0]=6 (A), active[4][2]=3 (C), active[4][4]=3 (E)From A, jumps to B and F; from C, jumps to B and D; from E, jumps to D and F.Therefore, from C: 3 paths to B and 3 to D (stopped)From E: 3 paths to D (stopped) and 3 to FFrom A: 6 to B and 6 to FTherefore, at step5, active[5][1]=6 (from A) +3 (from C) =9active[5][5]=6 (from A) +3 (from E) =9stopped[5]=3 (from C) +3 (from E)=6Thus, the total is indeed 26.Therefore, after careful step-by-step calculation, the total number of different ways is 26.Final Answerboxed{26}

✅:Given a regular hexagon (ABCDEF) with a frog starting at vertex (A), the frog can jump to adjacent vertices. The frog stops if it reaches vertex (D) within 5 jumps; otherwise, it stops after 5 jumps. We need to count the number of different ways the frog can jump from (A) until it stops.We use dynamic programming to track the number of ways the frog can reach each vertex without having reached (D) previously. We define two arrays: `active` to track the number of active paths at each step and `stopped` to track the number of paths that have stopped at (D).1. Initialization: - `active[0][A] = 1` (starting at (A)) - `stopped` array initialized to 0.2. Dynamic Programming Steps: - For each step (n) from 0 to 4: - For each vertex (v): - If `active[n][v] > 0`, the frog jumps to adjacent vertices. - If the jump reaches (D), update `stopped[n+1]`. - Otherwise, update `active[n+1][u]` for the adjacent vertex (u).3. Final Calculation: - Total stopped paths: sum of `stopped[1]` to `stopped[5]`. - Total active paths after 5 jumps: sum of `active[5][v]` for all (v neq D).Following the detailed steps:- After 5 jumps, the total stopped paths are (2 + 6 = 8).- The total active paths after 5 jumps are (9 + 9 = 18).Summing these results gives the total number of different ways the frog can jump from (A) until it stops:[boxed{26}]

🔑:1. Calculate Total Possible Jump Patterns if Jumping 5 Times: [ text{Number of different ways to jump 5 times} = 2^5 = 32. ]2. Check Jumps to Point ( D ) Within 3 Jumps: If the frog reaches point ( D ) in exactly 3 jumps, there are (2) possible jump patterns to achieve this. This is because from point ( A ) (starting point), the frog can only follow certain patterns to reach ( D ) in 3 jumps. - Let’s detail the patterns: - Suppose the points adjacent to ( D ) are ( C ) and ( E ). - Starting at ( A ), the frog could jump ( A rightarrow B rightarrow C rightarrow D ). - Another pattern could be ( A rightarrow F rightarrow E rightarrow D ). 3. Consider Subsequent 2 Jumps After Reaching ( D ): After reaching point ( D ) in 3 jumps, the subsequent 2 jumps should technically be irrelevant because the frog has already reached and should stop at ( D ). Thus, each of these sequences corresponds to ( 2 ) choices for each of the remaining ( 2 ) jumps, giving (2^2 = 4) choices.4. Correct Calculation of Unwanted Valid Patterns: Indeed, such sequences where frog intentionally continues jumping after reaching (D) are not valid, and we need to exclude such invalid sequences. So the number of invalid patterns is: [ 2 times 2^2 = 2 times 4 = 8 text{ invalid patterns} ]5. Recount Valid Jump Patterns: Since we have (32) total patterns. The (8) invalid patterns (where the frog reached (D) in 3 jumps but continued jumping for the remaining (2) jumps) should be excluded. Adding back the (2) valid patterns where the frog jumps directly to (D) (in the correct manner as detailed earlier): [ 32 - 8 + 2 = 32 - 6 = 26 text{ valid jump patterns} ] Conclusion:The number of different possible jump patterns from the beginning to the point when the frog stops is: [ boxed{26} ]

🧠:Okay, so I need to find the units digit of (a_{128,1}) in this integer sequence defined recursively. Let me start by understanding the problem step by step.First, let's parse the definition of the sequence. The sequence is two-dimensional, with indices (i) and (j), both natural numbers. The initial condition is given for (i=1): (a_{1, n} = n^n) where (n) is a positive integer. Then, for (i, j geq 1), each term (a_{i, j}) is the sum of the term above it and the term diagonally above and to the right, that is, (a_{i-1, j} + a_{i-1, j+1}). Our target is (a_{128, 1}), so we need to compute this term and find its units digit. But computing this directly by building the entire table up to the 128th row seems impractical. There must be a pattern or a mathematical property that can help us shortcut this process, especially since we only care about the units digit.Let me recall that when dealing with units digits, modular arithmetic, specifically modulo 10, is very useful. The units digit of a number is equivalent to the number modulo 10. So perhaps we can compute all terms modulo 10, which would simplify calculations.Given that, let's consider building the sequence modulo 10. Starting with the first row, where (a_{1, n} = n^n mod 10). Then, each subsequent term is the sum of the two terms above it, modulo 10.But even building a table up to 128 rows would be time-consuming. There must be a pattern or periodicity in the units digits that we can exploit. Let's explore the problem step by step.First, let's understand the first row: (a_{1, n} = n^n mod 10). The units digit of (n^n) depends on the units digit of (n) itself, due to the properties of modular arithmetic. For example, if (n) ends with 2, then (2^n) has a cycle of 4 in its units digit (2, 4, 8, 6, 2, ...). However, since the exponent here is equal to (n), not a linear function, the cycles might vary. Let's try to compute the units digit of (n^n) for the first few (n) to see if we can find a pattern.Let's compute (n^n mod 10) for (n = 1) to, say, 20:- (1^1 = 1 mod 10 = 1)- (2^2 = 4 mod 10 = 4)- (3^3 = 27 mod 10 = 7)- (4^4 = 256 mod 10 = 6)- (5^5 = 3125 mod 10 = 5)- (6^6 = 46656 mod 10 = 6)- (7^7 = 823543 mod 10 = 3)- (8^8 = 16777216 mod 10 = 6)- (9^9 = 387420489 mod 10 = 9)- (10^{10} = 10000000000 mod 10 = 0)- (11^{11}): The units digit of 11 is 1, so 1^any power is 1, so 1.- (12^{12}): Units digit of 12 is 2. The cycle for 2 is 2,4,8,6. Since the exponent is 12, which is a multiple of 4 (cycle length 4), 2^12 mod 10 = 6.- (13^{13}): Units digit 3. The cycle for 3 is 3,9,7,1. Exponent 13 mod 4 = 1, so 3^1 mod 10 = 3. Wait, 3^1 is 3, but 3^4 is 1. So 3^13 = 3^(4*3 + 1) = (3^4)^3 * 3^1 ≡ 1^3 * 3 ≡ 3 mod 10. So units digit 3.- (14^{14}): Units digit 4. The cycle for 4 is 4,6. Exponent 14. 14 mod 2 = 0, so 4^14 mod 10 = 6.- (15^{15}): Units digit 5. Any power of 5 ends with 5. So 5.- (16^{16}): Units digit 6. Any power of 6 ends with 6. So 6.- (17^{17}): Units digit 7. The cycle for 7 is 7,9,3,1. Exponent 17 mod 4 = 1. So 7^17 ≡ 7^1 ≡ 7 mod 10.- (18^{18}): Units digit 8. The cycle for 8 is 8,4,2,6. Exponent 18 mod 4 = 2. 8^2 = 64 ≡ 4 mod 10.- (19^{19}): Units digit 9. The cycle for 9 is 9,1. Exponent 19 is odd, so 9^19 ≡ 9 mod 10.- (20^{20}): Units digit 0. Any power of 0 ends with 0. So 0.Hm, interesting. So the units digits of (n^n) mod 10 for n starting at 1 are:1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, 6, 7, 4, 9, 0,...Is there a pattern here? It seems like after n=10, the pattern starts repeating some similar digits, but it's not immediately clear. However, for the purpose of constructing the first row, we can note that for each n, the units digit of (n^n) is equal to the units digit of (n mod 10)^(n mod φ(10) + φ(10)) due to Euler's theorem, but since 10 is composite, Euler's theorem applies only when n and 10 are coprime. Wait, maybe this is getting too complicated. Alternatively, perhaps we can use the fact that the units digit cycles with period depending on the base.But maybe instead of computing (a_{1,n}) for all n, we need to recognize that the recursive relation (a_{i,j} = a_{i-1,j} + a_{i-1,j+1}) is similar to Pascal's triangle, but in two dimensions. However, in Pascal's triangle, each term is the sum of the two above it. Here, each term is the sum of the term above and the term above and to the right, which is similar but not exactly the same. Wait, actually, in this case, the recurrence is (a_{i,j} = a_{i-1,j} + a_{i-1,j+1}). So it's a downward recurrence, where each row is built from the previous row by adding adjacent terms. So it's similar to a moving average or convolution. Hmm.Wait, perhaps there's a generating function approach here. Let me think. If we fix i and consider the generating function for row i, maybe we can express the generating function recursively.Alternatively, let's note that this recursion resembles the way differences or sums are computed in sequences. For example, in finite differences, each term is the difference of consecutive terms. Here, it's a sum, so perhaps each subsequent row is a cumulative sum or something similar. Let me check with a small example.Let's compute the first few rows manually modulo 10 to see if a pattern emerges.Starting with row 1: (a_{1,n} = n^n mod 10). Let's list the units digits for n starting at 1:Row 1 (i=1):n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...a_{1,n}: 1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, ...But for the purposes of building the triangle, since each row i has terms a_{i,1}, a_{i,2}, a_{i,3}, ... each computed as a_{i-1,1} + a_{i-1,2}, a_{i-1,2} + a_{i-1,3}, etc.So row i=2 would be:a_{2,1} = a_{1,1} + a_{1,2} = 1 + 4 = 5a_{2,2} = a_{1,2} + a_{1,3} = 4 + 7 = 11 ≡ 1 mod 10a_{2,3} = a_{1,3} + a_{1,4} = 7 + 6 = 13 ≡ 3 mod 10a_{2,4} = a_{1,4} + a_{1,5} = 6 + 5 = 11 ≡ 1 mod 10a_{2,5} = a_{1,5} + a_{1,6} = 5 + 6 = 11 ≡ 1 mod 10a_{2,6} = a_{1,6} + a_{1,7} = 6 + 3 = 9a_{2,7} = a_{1,7} + a_{1,8} = 3 + 6 = 9a_{2,8} = a_{1,8} + a_{1,9} = 6 + 9 = 15 ≡ 5 mod 10a_{2,9} = a_{1,9} + a_{1,10} = 9 + 0 = 9a_{2,10} = a_{1,10} + a_{1,11} = 0 + 1 = 1a_{2,11} = a_{1,11} + a_{1,12} = 1 + 6 = 7a_{2,12} = a_{1,12} + a_{1,13} = 6 + 3 = 9a_{2,13} = a_{1,13} + a_{1,14} = 3 + 6 = 9a_{2,14} = a_{1,14} + a_{1,15} = 6 + 5 = 11 ≡ 1 mod 10... and so on.So row 2 modulo 10 is: 5, 1, 3, 1, 1, 9, 9, 5, 9, 1, 7, 9, 9, 1, ...Row 3 would be built from row 2:a_{3,1} = a_{2,1} + a_{2,2} = 5 + 1 = 6a_{3,2} = a_{2,2} + a_{2,3} = 1 + 3 = 4a_{3,3} = a_{2,3} + a_{2,4} = 3 + 1 = 4a_{3,4} = a_{2,4} + a_{2,5} = 1 + 1 = 2a_{3,5} = a_{2,5} + a_{2,6} = 1 + 9 = 10 ≡ 0 mod 10a_{3,6} = a_{2,6} + a_{2,7} = 9 + 9 = 18 ≡ 8 mod 10a_{3,7} = a_{2,7} + a_{2,8} = 9 + 5 = 14 ≡ 4 mod 10a_{3,8} = a_{2,8} + a_{2,9} = 5 + 9 = 14 ≡ 4 mod 10a_{3,9} = a_{2,9} + a_{2,10} = 9 + 1 = 10 ≡ 0 mod 10a_{3,10} = a_{2,10} + a_{2,11} = 1 + 7 = 8a_{3,11} = a_{2,11} + a_{2,12} = 7 + 9 = 16 ≡ 6 mod 10a_{3,12} = a_{2,12} + a_{2,13} = 9 + 9 = 18 ≡ 8 mod 10a_{3,13} = a_{2,13} + a_{2,14} = 9 + 1 = 10 ≡ 0 mod 10... and so on.So row 3 modulo 10: 6, 4, 4, 2, 0, 8, 4, 4, 0, 8, 6, 8, 0, ...This is getting a bit tedious, but maybe if we continue a few more rows, we can see a pattern.Row 4:a_{4,1} = 6 + 4 = 10 ≡ 0 mod 10a_{4,2} = 4 + 4 = 8a_{4,3} = 4 + 2 = 6a_{4,4} = 2 + 0 = 2a_{4,5} = 0 + 8 = 8a_{4,6} = 8 + 4 = 12 ≡ 2 mod 10a_{4,7} = 4 + 4 = 8a_{4,8} = 4 + 0 = 4a_{4,9} = 0 + 8 = 8a_{4,10} = 8 + 6 = 14 ≡ 4 mod 10a_{4,11} = 6 + 8 = 14 ≡ 4 mod 10a_{4,12} = 8 + 0 = 8... So row 4: 0, 8, 6, 2, 8, 2, 8, 4, 8, 4, 4, 8, ...Row 5:a_{5,1} = 0 + 8 = 8a_{5,2} = 8 + 6 = 14 ≡ 4a_{5,3} = 6 + 2 = 8a_{5,4} = 2 + 8 = 10 ≡ 0a_{5,5} = 8 + 2 = 10 ≡ 0a_{5,6} = 2 + 8 = 10 ≡ 0a_{5,7} = 8 + 4 = 12 ≡ 2a_{5,8} = 4 + 8 = 12 ≡ 2a_{5,9} = 8 + 4 = 12 ≡ 2a_{5,10} = 4 + 4 = 8a_{5,11} = 4 + 8 = 12 ≡ 2... So row 5: 8, 4, 8, 0, 0, 0, 2, 2, 2, 8, 2, ...Row 6:a_{6,1} = 8 + 4 = 12 ≡ 2a_{6,2} = 4 + 8 = 12 ≡ 2a_{6,3} = 8 + 0 = 8a_{6,4} = 0 + 0 = 0a_{6,5} = 0 + 0 = 0a_{6,6} = 0 + 2 = 2a_{6,7} = 2 + 2 = 4a_{6,8} = 2 + 2 = 4a_{6,9} = 2 + 8 = 10 ≡ 0a_{6,10} = 8 + 2 = 10 ≡ 0... Row 6: 2, 2, 8, 0, 0, 2, 4, 4, 0, 0, ...Row 7:a_{7,1} = 2 + 2 = 4a_{7,2} = 2 + 8 = 10 ≡ 0a_{7,3} = 8 + 0 = 8a_{7,4} = 0 + 0 = 0a_{7,5} = 0 + 2 = 2a_{7,6} = 2 + 4 = 6a_{7,7} = 4 + 4 = 8a_{7,8} = 4 + 0 = 4a_{7,9} = 0 + 0 = 0... Row 7: 4, 0, 8, 0, 2, 6, 8, 4, 0, ...Row 8:a_{8,1} = 4 + 0 = 4a_{8,2} = 0 + 8 = 8a_{8,3} = 8 + 0 = 8a_{8,4} = 0 + 2 = 2a_{8,5} = 2 + 6 = 8a_{8,6} = 6 + 8 = 14 ≡ 4a_{8,7} = 8 + 4 = 12 ≡ 2a_{8,8} = 4 + 0 = 4... Row 8: 4, 8, 8, 2, 8, 4, 2, 4, ...Row 9:a_{9,1} = 4 + 8 = 12 ≡ 2a_{9,2} = 8 + 8 = 16 ≡ 6a_{9,3} = 8 + 2 = 10 ≡ 0a_{9,4} = 2 + 8 = 10 ≡ 0a_{9,5} = 8 + 4 = 12 ≡ 2a_{9,6} = 4 + 2 = 6a_{9,7} = 2 + 4 = 6... Row 9: 2, 6, 0, 0, 2, 6, 6, ...Row 10:a_{10,1} = 2 + 6 = 8a_{10,2} = 6 + 0 = 6a_{10,3} = 0 + 0 = 0a_{10,4} = 0 + 2 = 2a_{10,5} = 2 + 6 = 8a_{10,6} = 6 + 6 = 12 ≡ 2... Row 10: 8, 6, 0, 2, 8, 2, ...Row 11:a_{11,1} = 8 + 6 = 14 ≡ 4a_{11,2} = 6 + 0 = 6a_{11,3} = 0 + 2 = 2a_{11,4} = 2 + 8 = 10 ≡ 0a_{11,5} = 8 + 2 = 10 ≡ 0... Row 11: 4, 6, 2, 0, 0, ...Row 12:a_{12,1} = 4 + 6 = 10 ≡ 0a_{12,2} = 6 + 2 = 8a_{12,3} = 2 + 0 = 2a_{12,4} = 0 + 0 = 0... Row 12: 0, 8, 2, 0, ...Row 13:a_{13,1} = 0 + 8 = 8a_{13,2} = 8 + 2 = 10 ≡ 0a_{13,3} = 2 + 0 = 2... Row 13: 8, 0, 2, ...Row 14:a_{14,1} = 8 + 0 = 8a_{14,2} = 0 + 2 = 2... Row 14: 8, 2, ...Row 15:a_{15,1} = 8 + 2 = 10 ≡ 0... Row 15: 0, ...Wait, this seems like the sequence is collapsing. From row 15, the first term is 0. But then, if we follow the recursion, the next row would be:Row 16:a_{16,1} = 0 + a_{15,2} but a_{15,2} doesn't exist because row 15 only has one term? Wait, no. Wait, actually, each row i has terms from j=1 to j=... as many as needed. Wait, when building the rows, each row i has as many terms as row i-1 minus 1, because each term is a sum of two adjacent terms. Wait, no, actually, if row i-1 has m terms, then row i will have m - 1 terms. So, starting from row 1, which has infinitely many terms (since j can be any natural number), each subsequent row will also have infinitely many terms. But in our manual computation above, we stopped at some j, but in reality, each row continues indefinitely. However, since we're only interested in a_{128,1}, which is on row 128, and each subsequent row has one fewer term in the sense that to compute a_{i,1}, we only need a_{i-1,1} and a_{i-1,2}, which in turn depend on a_{i-2,1}, a_{i-2,2}, a_{i-2,2}, a_{i-2,3}, etc. So effectively, to compute a_{128,1}, we need a triangular section of the initial rows up to j=128 in the first row. But even so, computing this manually is impractical.However, observing the rows we computed above, the units digits seem to eventually reach zero and then propagate in some periodic fashion. But maybe there's a cycle or period in the units digits of the terms as we go down the rows. For example, looking at a_{i,1} for i from 1 to 15:Row 1: 1Row 2: 5Row 3: 6Row 4: 0Row 5: 8Row 6: 2Row 7: 4Row 8: 4Row 9: 2Row 10: 8Row 11: 4Row 12: 0Row 13: 8Row 14: 8Row 15: 0Hmm, not obvious. Let's list the units digits of a_{i,1}:1 (i=1), 5, 6, 0, 8, 2, 4, 4, 2, 8, 4, 0, 8, 8, 0,...Is there a cycle here? Let's see:From i=1 to i=15: 1,5,6,0,8,2,4,4,2,8,4,0,8,8,0Looking for repetition. Let's check from i=5 onwards:i=5:8, i=6:2, i=7:4, i=8:4, i=9:2, i=10:8, i=11:4, i=12:0, i=13:8, i=14:8, i=15:0So from i=5 to i=15: 8,2,4,4,2,8,4,0,8,8,0Hmm, not a clear cycle. Maybe we need to compute more terms. But this is tedious manually. Alternatively, maybe there's a mathematical approach here.Another thought: the process of generating each subsequent row is equivalent to applying a linear transformation. Since we're working modulo 10, this transformation is a linear operator over the integers modulo 10. If we can diagonalize this operator or find its period, we might find that the sequence of units digits cycles with a certain period, allowing us to compute 128 modulo the period and find the answer.Alternatively, note that the operation from row i-1 to row i is similar to a linear filter or a difference operator. Specifically, each term in row i is the sum of two consecutive terms from row i-1. This is equivalent to convolving row i-1 with the kernel [1, 1]. Repeated applications of this convolution would correspond to convolving with [1,1]^k, which is similar to the binomial coefficients. Wait a minute! This seems related to the binomial coefficients. Let's explore this.Suppose that we model the transformation from the first row to the i-th row as a series of convolutions with [1,1]. Then, the value of (a_{i,1}) would be the sum of the first row terms multiplied by the binomial coefficients from the i-th row. Wait, actually, in such a recurrence, each term in the i-th row is a combination of the terms in the first row with coefficients given by the number of paths from the original term to the target term, which in this case is the binomial coefficient.Specifically, in such a recursive setup, each term (a_{i,j}) can be expressed as the sum over k from j to j+i-1 of (a_{1,k}) multiplied by the number of paths from (a_{1,k}) to (a_{i,j}), which is (binom{i-1}{k - j}). Wait, let's verify this.For example, when i=2, j=1:a_{2,1} = a_{1,1} + a_{1,2} = 1 + 4 = 5. According to the binomial coefficient idea, it would be (binom{1}{0}a_{1,1} + binom{1}{1}a_{1,2}), which is 1*1 + 1*4 = 5. Correct.Similarly, for i=3, j=1:a_{3,1} = a_{2,1} + a_{2,2} = 5 + 1 = 6. According to the binomial coefficients, it should be (binom{2}{0}a_{1,1} + binom{2}{1}a_{1,2} + binom{2}{2}a_{1,3}) = 1*1 + 2*4 + 1*7 = 1 + 8 + 7 = 16. But 16 mod 10 is 6, which matches. Wait, so actually, the coefficients are binomial coefficients, but we need to take modulo 10 at each step? Or perhaps, since we are working modulo 10, the coefficients themselves can be considered modulo 10 as well?Wait, but in reality, the recurrence builds up each term as a sum of two previous terms, so each term in row i is a linear combination of the terms in row 1 with coefficients equal to the number of paths, which are binomial coefficients. Therefore, (a_{i,1} = sum_{k=1}^{i} binom{i-1}{k-1} a_{1,k}). Let's verify this.For i=1: (a_{1,1} = binom{0}{0}a_{1,1} = 1*1 = 1). Correct.For i=2: (binom{1}{0}a_{1,1} + binom{1}{1}a_{1,2}) = 1*1 + 1*4 = 5. Correct.For i=3: (binom{2}{0}a_{1,1} + binom{2}{1}a_{1,2} + binom{2}{2}a_{1,3}) = 1*1 + 2*4 + 1*7 = 1 + 8 + 7 = 16 ≡ 6 mod 10. Correct.For i=4: (binom{3}{0}a_{1,1} + binom{3}{1}a_{1,2} + binom{3}{2}a_{1,3} + binom{3}{3}a_{1,4}) = 1*1 + 3*4 + 3*7 + 1*6 = 1 + 12 + 21 + 6 = 40 ≡ 0 mod 10. Which matches our earlier computation of row 4, a_{4,1} = 0. Correct.Great! So this formula seems to hold. Therefore, in general,[a_{i,1} = sum_{k=1}^{i} binom{i-1}{k-1} a_{1,k} mod 10]Therefore, to compute (a_{128,1} mod 10), we need to compute the sum:[sum_{k=1}^{128} binom{127}{k-1} a_{1,k} mod 10]Where (a_{1,k} = k^k mod 10).So the problem reduces to calculating this sum modulo 10. However, computing this directly for k up to 128 is still quite intensive. However, we can look for patterns or properties that allow us to simplify this sum.First, note that the binomial coefficients (binom{127}{k-1}) modulo 10 can be broken down using Lucas' theorem, since 10 is a composite modulus. Lucas' theorem allows us to compute binomial coefficients modulo a prime by considering their digits in base p. However, 10 is not prime, but we can factor it into primes 2 and 5 and use the Chinese Remainder Theorem. But maybe this is getting too complex.Alternatively, note that the binomial coefficients modulo 10 repeat with some period. For example, binomial coefficients modulo 2 have a known structure related to Sierpiński triangles, but modulo 10, which factors into 2 and 5, the period might be more complicated.Alternatively, since 10 = 2 * 5, we can compute the sum modulo 2 and modulo 5 separately, then combine the results using the Chinese Remainder Theorem. Let's explore this approach.First, compute (S = sum_{k=1}^{128} binom{127}{k-1} a_{1,k})Compute S mod 2 and S mod 5, then combine them.Compute S mod 2:Each term in the sum is (binom{127}{k-1} a_{1,k} mod 2).First, note that (a_{1,k} = k^k mod 10), so (a_{1,k} mod 2) is the parity of (k^k). Since any even k raised to any power is even, and any odd k raised to any power is odd. Therefore:- If k is even, (a_{1,k} mod 2 = 0)- If k is odd, (a_{1,k} mod 2 = 1)Therefore, in the sum S mod 2, only the terms where k is odd contribute 1 * (binom{127}{k-1} mod 2), and the rest contribute 0.So S mod 2 = (sum_{k text{ odd}, 1 leq k leq 128} binom{127}{k-1} mod 2)But note that k is odd, so k-1 is even. Let m = k-1, so m ranges from 0 to 127, even m. So S mod 2 = (sum_{m=0}^{63} binom{127}{2m} mod 2)But by Lucas' theorem, (binom{n}{k} mod 2) is 1 if and only if the binary representation of k is a subset of the binary representation of n. Let's write 127 in binary:127 in binary is 64 + 32 + 16 + 8 + 4 + 2 + 1 = 1111111.So the binary representation of 127 is seven 1s. Then, (binom{127}{m} mod 2) is 1 for all m from 0 to 127, because every bit of m (which has up to 7 bits) is a subset of 127's bits. Wait, but that can't be right, since the binomial coefficients modulo 2 are 1 only when the binary digits of m are less than or equal to those of n. But since n=127 has all bits set to 1, any m from 0 to 127 will have a subset of bits, so indeed, (binom{127}{m} mod 2 = 1) for all m. Therefore, S mod 2 = number of terms where k is odd, which is 64 terms (from k=1 to 128, half are odd). Each term contributes 1, so S mod 2 = 64 mod 2 = 0.Therefore, S ≡ 0 mod 2.Now compute S mod 5:This is more complicated. Let's proceed step by step.First, S mod 5 = (sum_{k=1}^{128} binom{127}{k-1} a_{1,k} mod 5)We can write this as (sum_{m=0}^{127} binom{127}{m} a_{1,m+1} mod 5), where m = k-1.So S mod 5 = (sum_{m=0}^{127} binom{127}{m} (m+1)^{m+1} mod 5)This sum seems difficult to compute directly, but perhaps we can find a generating function or exploit periodicity.Note that (a_{1,m+1} = (m+1)^{m+1}), so the term is (binom{127}{m} (m+1)^{m+1}). To compute this modulo 5.We can use the fact that for modulus 5, the terms may repeat every certain period. Let's consider the periodicity of ( (m+1)^{m+1} mod 5 ).The units digit modulo 5 of (n^n) where n = m+1. Wait, actually, (n^n mod 5) can be simplified using Euler's theorem. Since 5 is prime, φ(5) = 4. For n coprime to 5, (n^4 ≡ 1 mod 5), so (n^n ≡ n^{n mod 4} mod 5). However, when n is a multiple of 5, (n^n ≡ 0 mod 5).Therefore, for n = m+1:- If n ≡ 0 mod 5, then (n^n ≡ 0 mod 5)- If n ≡ 1 mod 5, then (n^n ≡ 1^n ≡ 1 mod 5)- If n ≡ 2 mod 5, then (n^n ≡ 2^{n mod 4} mod 5)- If n ≡ 3 mod 5, then (n^n ≡ 3^{n mod 4} mod 5)- If n ≡ 4 mod 5, then (n^n ≡ 4^{n mod 2} mod 5) (since 4^2 ≡ 1 mod 5, so period 2)This complicates things, but maybe we can split the sum into cases based on n mod 5 or m mod 5.Let me define n = m + 1, so m = n - 1, and n ranges from 1 to 128.Therefore, S mod 5 = (sum_{n=1}^{128} binom{127}{n-1} n^n mod 5)We can split this sum into parts where n ≡ 0,1,2,3,4 mod 5.Let me consider each residue class separately.First, let's handle n ≡ 0 mod 5: n = 5,10,...,125. There are 25 such terms (since 125 = 5*25, but 128 is the upper limit, so 5*25=125, then 130 would be next, which is beyond 128). So n=5,10,...,125: total 25 terms.For each such n, (n^n ≡ 0 mod 5). Therefore, the contribution of these terms to S mod 5 is 0.Next, n ≡ 1 mod 5: n =1,6,11,...,126. Let's compute how many terms there are. The first term is n=1, then n=6, ..., up to n=126 (since 126 = 5*25 + 1 = 125 +1 = 126). The sequence is 1,6,...,126. The number of terms is floor((126 -1)/5) +1 = floor(125/5) +1=25 +1=26 terms.Similarly, n ≡2 mod5: n=2,7,...,127. The last term would be 127=5*25 + 2=127, so number of terms: floor((127-2)/5)+1= floor(125/5)+1=25+1=26 terms.n≡3 mod5: n=3,8,...,128. Wait, 128=5*25 +3=125 +3=128. So n=3,8,...,128. Number of terms: floor((128-3)/5)+1=floor(125/5)+1=25+1=26 terms.n≡4 mod5: n=4,9,...,124. The last term is 124=5*24 +4=120 +4=124. So number of terms: floor((124-4)/5)+1=floor(120/5)+1=24 +1=25 terms.So total terms: 25 (0 mod5) +26 +26 +26 +25=128. Correct.Now, for each residue class:1. For n ≡1 mod5: n=5k+1, where k=0 to25 (since 5*25 +1=126). Compute the contribution:Sum_{k=0}^{25} binom{127}{5k} (5k+1)^{5k+1} mod5But note that (5k+1)^{5k+1} mod5. Since 5k+1 ≡1 mod5, so (5k+1)^{5k+1} ≡1^{5k+1} ≡1 mod5. Therefore, each term in this sum is binom{127}{5k} *1 mod5. Therefore, the sum is Sum_{k=0}^{25} binom{127}{5k} mod5.2. For n ≡2 mod5: n=5k+2, k=0 to25 (n=2 to127). Compute the contribution:Sum_{k=0}^{25} binom{127}{5k+1} (5k+2)^{5k+2} mod5.Since 5k+2 ≡2 mod5, so (5k+2)^{5k+2} ≡2^{5k+2} mod5. Now, 2 has period 4 modulo5: 2^1=2, 2^2=4, 2^3=3, 2^4=1, then repeats.Thus, 2^{5k+2} ≡2^{(5k+2) mod4} mod5.Compute exponent (5k +2) mod4:5k mod4 = (since 5≡1 mod4) so 5k ≡k mod4. Thus, 5k +2 ≡k +2 mod4.Therefore, 2^{5k+2} ≡2^{(k +2) mod4} mod5.So the term becomes binom{127}{5k+1} *2^{(k +2) mod4} mod5.3. For n ≡3 mod5: n=5k+3, k=0 to25 (n=3 to128). Contribution:Sum_{k=0}^{25} binom{127}{5k+2} (5k+3)^{5k+3} mod5.Similarly, 5k+3 ≡3 mod5, so (5k+3)^{5k+3} ≡3^{5k+3} mod5.3 has period 4 modulo5: 3^1=3,3^2=4,3^3=2,3^4=1.Thus, 3^{5k+3} ≡3^{(5k+3) mod4} mod5.5k mod4 =k mod4, so 5k+3 ≡k +3 mod4.Therefore, 3^{5k+3} ≡3^{(k +3) mod4} mod5.So each term is binom{127}{5k+2}*3^{(k +3) mod4} mod5.4. For n ≡4 mod5: n=5k+4, k=0 to24 (n=4 to124). Contribution:Sum_{k=0}^{24} binom{127}{5k+3} (5k+4)^{5k+4} mod5.5k+4 ≡4 mod5, so (5k+4)^{5k+4} ≡4^{5k+4} mod5.4 has period 2 modulo5: 4^1=4,4^2=1.Thus, 4^{5k+4} ≡4^{(5k+4) mod2} mod5.Since 5k mod2 =k mod2, so 5k+4 ≡k +0 mod2 (since 4≡0 mod2). Wait, 5k +4 ≡k +0 mod2 (because 5k ≡k mod2, 4≡0 mod2), so 5k+4 ≡k +0 mod2.Therefore, 4^{5k+4} ≡4^{k mod2} mod5.Which is 4^0=1 if k even, 4^1=4 if k odd.Therefore, each term is binom{127}{5k+3}*4^{k mod2} mod5.So putting it all together, S mod5 is:Sum_{k=0}^{25} binom{127}{5k} + Sum_{k=0}^{25} binom{127}{5k+1}*2^{(k+2) mod4} + Sum_{k=0}^{25} binom{127}{5k+2}*3^{(k+3) mod4} + Sum_{k=0}^{24} binom{127}{5k+3}*4^{k mod2} mod5.This seems very complex, but maybe we can find generating functions or use properties of binomial coefficients modulo primes.Alternatively, perhaps we can use Lucas's theorem to compute the binomial coefficients modulo5.Lucas's theorem states that for primes p, and non-negative integers m and n with base p expansions m = m0 + m1 p + ... + mn p^n and n = n0 + n1 p + ... + nn p^n, then:[binom{m}{n} equiv prod_{i=0}^{k} binom{m_i}{n_i} mod p]Where if ni > mi for any i, the binomial coefficient is zero.Given that 5 is prime, we can apply Lucas's theorem here.First, express 127 in base5:5^0 =1, 5^1=5,5^2=25,5^3=125.127 divided by 125 is 1 with remainder 2. 2 divided by 25 is 0 with remainder 2. 2 divided by 5 is 0 with remainder 2. So 127 in base5 is 1*5^3 + 0*5^2 + 0*5^1 + 2*5^0, which is written as 1002_5.Similarly, any m in the binomial coefficient (binom{127}{m}) can be expressed in base5.To compute (binom{127}{m} mod5), we need to consider the digits of m in base5.For example, consider m =5k. In base5, m would have a digit of 0 in the units place. Similarly, m=5k+1 has a units digit of1, etc.But since 127 in base5 is 1002, which has digits [1,0,0,2] (from highest to lowest power), then according to Lucas's theorem, to compute (binom{127}{m} mod5), we need to break m into its base5 digits m = m3 m2 m1 m0, and ensure that each digit of m is less than or equal to the corresponding digit in 127.However, since the digits of 127 in base5 are [1,0,0,2], we have constraints:- For the 5^3 digit (leftmost digit), m3 must be ≤1- For the 5^2 digit, m2 must be ≤0, so m2=0- For the 5^1 digit, m1 must be ≤0, so m1=0- For the 5^0 digit, m0 must be ≤2Therefore, any m such that in base5 has m3 ≤1, m2=0, m1=0, and m0 ≤2. Otherwise, the binomial coefficient is 0 mod5.Thus, the binomial coefficient (binom{127}{m} mod5) is non-zero only if m, in base5, has digits [m3,0,0,m0], where m3 ∈ {0,1} and m0 ∈ {0,1,2}.Therefore, m can be written as m = m3*5^3 + m0*5^0, where m3 ∈ {0,1}, m0 ∈ {0,1,2}.So, m can take the following values:- If m3=0: m = 0,1,2- If m3=1: m = 125,126,127Therefore, the only m for which (binom{127}{m} notequiv0 mod5) are m=0,1,2,125,126,127.This is a crucial observation! Therefore, in all the sums over m, only these values of m contribute. Therefore, in the expression for S mod5, most terms are zero modulo5, except when m is 0,1,2,125,126,127. Therefore, we can simplify each sum by considering only these values of m.Let's verify this. For example, consider (binom{127}{5k}). If 5k is not one of 0,1,2,125,126,127, then (binom{127}{5k} equiv0 mod5). Similarly for other residues.Therefore, the first sum (n≡1 mod5): Sum_{k=0}^{25} binom{127}{5k} mod5. However, 5k must be one of 0,1,2,125,126,127. But 5k is a multiple of5. Among 0,1,2,125,126,127, the multiples of5 are 0 and125. Therefore, only k=0 (5k=0) and k=25 (5k=125) contribute. Therefore:Sum_{k=0}^{25} binom{127}{5k} ≡ binom{127}{0} + binom{127}{125} mod5.But (binom{127}{125} = binom{127}{2} = (127*126)/2 = 8001. 8001 mod5. 127 mod5=2, 126 mod5=1, so (2*1)/2=1 mod5. Wait, but division by 2 in modulo5 is multiplication by 3, since 2*3=6≡1 mod5. So (127*126)/2 ≡ (2*1)/2 ≡ (2*1)*3 ≡6≡1 mod5. Therefore, binom{127}{125} ≡1 mod5. And (binom{127}{0}=1). Therefore, the sum is 1 +1=2 mod5.Next, second sum (n≡2 mod5): Sum_{k=0}^{25} binom{127}{5k+1}*2^{(k+2) mod4} mod5.Here, we need 5k+1 to be one of the non-zero residues. The possible values of m=5k+1 must be in {1,126}. Wait, the non-zero m in {0,1,2,125,126,127} that are ≡1 mod5 are m=1 and m=126 (since 126 mod5=1). Therefore, possible m=1 and126.So 5k+1=1 =>k=0, and5k+1=126 =>k=25 (since5*25 +1=125 +1=126). Therefore, the sum reduces to:k=0: binom{127}{1}*2^{(0 +2) mod4}=127*2^2 mod5. 127 mod5=2, 2^2=4, so 2*4=8≡3 mod5.k=25: binom{127}{126}*2^{(25 +2) mod4}. binom{127}{126}=127≡2 mod5. (25 +2)=27 mod4=3, so 2^3=8≡3 mod5. So 2*3=6≡1 mod5.Thus, total sum=3 +1=4 mod5.Third sum (n≡3 mod5): Sum_{k=0}^{25} binom{127}{5k+2}*3^{(k+3) mod4} mod5.Similarly, 5k+2 must be in {2,127}, since those are the residues ≡2 mod5 in {0,1,2,125,126,127}. So m=2 and m=127.For m=2:5k+2=2 =>k=0.For m=127:5k+2=127 =>k=(127 -2)/5=125/5=25.Therefore, sum is:k=0: binom{127}{2}*3^{(0 +3) mod4}= binom{127}{2}*3^3 mod5. As before, binom{127}{2}=8001≡1 mod5. 3^3=27≡2 mod5. So 1*2=2 mod5.k=25: binom{127}{127}*3^{(25 +3) mod4}=1*3^(28 mod4)=3^0=1 mod5. So contribution=1*1=1 mod5.Total sum=2 +1=3 mod5.Fourth sum (n≡4 mod5): Sum_{k=0}^{24} binom{127}{5k+3}*4^{k mod2} mod5.Here, 5k+3 must be in {0,1,2,125,126,127} and ≡3 mod5. Checking these values, none of {0,1,2,125,126,127} are ≡3 mod5. Therefore, all terms in this sum are zero mod5.Therefore, the total S mod5 is 2 (from first sum) +4 (second sum)+3 (third sum)+0=2+4+3=9≡4 mod5.Thus, S ≡0 mod2 and ≡4 mod5. We need to find a number x such that x≡0 mod2 and x≡4 mod5. Solving these congruences:x≡4 mod5 =>x=5k+4. Also x≡0 mod2 =>5k+4≡0 mod2 =>5k≡-4≡0 mod2 =>k≡0 mod2. So k=2m. Therefore, x=5*(2m)+4=10m +4. The smallest non-negative solution is x=4, then 14,24,... Since we are working modulo10, the solutions modulo10 are x≡4 mod10.Therefore, the units digit of (a_{128,1}) is 4.Final AnswerThe unit digit of (a_{128,1}) is boxed{4}.

✅:To find the unit digit of (a_{128,1}) in the given integer sequence, we start by understanding the recursive definition and utilizing modular arithmetic. The sequence is defined as follows:- (a_{1, n} = n^n mod 10) for (n in mathbf{Z}_{+})- (a_{i, j} = a_{i-1, j} + a_{i-1, j+1} mod 10) for (i, j geq 1)We need to compute (a_{128,1} mod 10). The key insight is recognizing that each term (a_{i,j}) can be expressed as a sum of binomial coefficients multiplied by the initial terms modulo 10. Specifically, (a_{i,1}) is given by:[a_{i,1} = sum_{k=1}^{i} binom{i-1}{k-1} a_{1,k} mod 10]For (i = 128), this becomes:[a_{128,1} = sum_{k=1}^{128} binom{127}{k-1} a_{1,k} mod 10]We compute this sum modulo 2 and modulo 5 separately, then combine the results using the Chinese Remainder Theorem.1. Modulo 2 Analysis: - (a_{1,k} mod 2) is 1 if (k) is odd, and 0 otherwise. - Using Lucas' theorem, (binom{127}{k-1} mod 2 = 1) for all (k), leading to: [ S mod 2 = sum_{k text{ odd}} 1 mod 2 = 64 mod 2 = 0 ]2. Modulo 5 Analysis: - We split the sum into residue classes modulo 5. - Using Lucas' theorem, we find that only specific terms contribute non-zero values modulo 5. - After detailed calculations, we find: [ S mod 5 = 2 + 4 + 3 = 9 equiv 4 mod 5 ]Combining results using the Chinese Remainder Theorem:- (S equiv 0 mod 2)- (S equiv 4 mod 5)The solution is (x equiv 4 mod 10). Therefore, the unit digit of (a_{128,1}) is (boxed{4}).

🔑:Given the integer sequences ( {a_{i, j}} (i, j in mathbb{N}) ), where:[begin{array}{l}a_{1, n} = n^n quad (n in mathbb{Z}_+), a_{i, j} = a_{i-1, j} + a_{i-1, j+1} quad (i, j geq 1).end{array}]To find the value of ( a_{128,1} ):1. By repeatedly applying the recurrence relation, we find:[begin{array}{l}a_{1,1} = 1, a_{2, n} = n^n + (n+1)^{n+1}, a_{3, n} = n^n + 2(n+1)^{n+1} + (n+2)^{n+2}.end{array}]2. Following this pattern, we can generalize to:[a_{n, m} = sum_{k=0}^{m-1} binom{m-1}{k}(n+k)^{n+k}.]3. For verification, when ( k = 0 ), ( binom{m}{0} = binom{m-1}{0} ) holds true. Additionally, by the binomial coefficient property, ( binom{m}{k} = binom{m-1}{k} + binom{m-1}{k-1} ) for ( k geq 1 ), confirming the recurrence relation.4. For ( m = 128 ), using Lucas' theorem:[begin{array}{l}binom{127}{k} equiv 1 , (text{mod} , 2) quad (1 leq k leq 127), binom{127}{k} equiv 0 , (text{mod} , 5) quad (3 leq k leq 124).end{array}]5. Thus, we calculate ( a_{128,1} ):[a_{128,1} = sum_{k=0}^{127} binom{127}{k} (k+1)^{k+1}.]6. Modulo 2, we have:[sum_{k=0}^{127} (k+1)^{k+1} equiv 0 , (text{mod} , 2),]7. And modulo 5, considering only ( k in {0, 1, 2} ):[a_{128,1} = sum_{k in {0,1,2}} (k+1)^{k+1} equiv 4 , (text{mod} , 5).]Thus,[ a_{128,1} equiv 4 , (text{mod} , 10). ]Therefore, the final value is:(boxed{4}).

🧠:Alright, let's try to figure out this geometry problem. So, we have two circles, k₁ and k₂, centered at points A and B with radii r₁ and r₂ respectively. There's another circle, k, that is tangent to both k₁ and k₂ from the inside, and we need to find the ratio r₃/r₁ as point C approaches A, which means r₁ is getting smaller and approaching zero. Hmm, okay, let's break this down.First, let me visualize the problem. We have two fixed points A and B. There's a circle around A (k₁) with radius r₁ and a circle around B (k₂) with radius r₂. Then there's a circle k that is tangent to both k₁ and k₂ from the inside. So, circle k is inside both k₁ and k₂ and touches each of them exactly once. As C approaches A, which probably means that the position of circle k is moving towards A, but since A and B are fixed, maybe the configuration is changing such that the radius r₁ is decreasing towards zero. Wait, the problem says "fixing A and B, the point C approaches A". Hmm, point C is probably the center of the circle k? Because the circle k is tangent to k₁ and k₂, so its center must be somewhere relative to A and B. Maybe C is the center of circle k. If that's the case, then as C approaches A, the radius r₁ of circle k₁ (around A) is approaching zero? Wait, the problem states: "Let k be a circle that is tangent to k₁ and k₂ from the inside, with radius r₃. What does the ratio r₃ / r₁ approach when, fixing A and B, the point C approaches A, i.e., as r₁ approaches 0?"Wait, so maybe the circle k is tangent to both k₁ and k₂ from the inside. That means circle k lies inside both k₁ and k₂ and touches each of them. But if k is inside both, then the centers of k₁ and k₂ must be such that the distance between A and B is greater than the sum of their radii? Wait, no. If k is tangent to k₁ and k₂ from the inside, then k must be inside k₁ and inside k₂. Therefore, the centers A and B must be outside of circle k, and the distance from A to the center of k must be equal to r₁ - r₃, since it's tangent from the inside. Similarly, the distance from B to the center of k must be r₂ - r₃. So, the center of circle k, let's call it point C, must satisfy:AC = r₁ - r₃BC = r₂ - r₃Since points A and B are fixed, the distance AB is a constant. Let's denote AB as d. Then, the distance between A and B is d, so in the triangle ABC, we have:AC + BC ≥ AB (triangle inequality). But since C is moving towards A, as r₁ approaches zero, maybe BC is approaching AB? Wait, let's think again.Wait, if we fix points A and B, then AB is fixed. The circle k is tangent to k₁ and k₂ from the inside. As point C (the center of k) approaches A, then AC is getting smaller. Since AC = r₁ - r₃, and if C is approaching A, then AC approaches zero. Therefore, r₁ - r₃ approaches zero, so r₃ approaches r₁. But the problem says as r₁ approaches zero, what happens to the ratio r₃/r₁? If r₃ approaches r₁, then the ratio would approach 1. But that seems too straightforward. Maybe I'm missing something here.Wait, but hold on. If we fix A and B, and C approaches A, then the position of circle k is moving towards A. But the circle k is also tangent to k₂, which is fixed at B with radius r₂. So, even as C approaches A, the circle k still has to maintain tangency with k₂. Therefore, the distance from C to B must equal r₂ - r₃. But if C is approaching A, then the distance from C to B is approaching AB (since A and B are fixed). Let me denote AB as d. So, as C approaches A, BC approaches d. Therefore, we have BC = d - AC. But since AC = r₁ - r₃, as C approaches A, AC approaches zero, so BC approaches d. But BC is also equal to r₂ - r₃. So, as C approaches A, BC approaches d, which would mean r₂ - r₃ approaches d. But since r₂ is fixed (because we're only changing r₁ and keeping A and B fixed), then as d is fixed, how does that reconcile?Wait, hold on. Maybe I need to clarify the setup. The problem says: fixing A and B, the point C approaches A, i.e., as r₁ approaches 0. So, when they fix A and B, but they let the radius r₁ approach zero. So, circle k₁ is around A with radius r₁ approaching zero, and circle k₂ is around B with radius r₂ (fixed?). Then circle k is tangent to both k₁ and k₂ from the inside, so it's inside both. But as r₁ approaches zero, the circle k₁ is shrinking to a point at A. So, the circle k must be tangent to this shrinking circle k₁ and also tangent to k₂. So, the center of circle k is at a distance of r₁ - r₃ from A, but as r₁ approaches zero, this distance approaches -r₃. Wait, but distance can't be negative. Therefore, maybe the formula is different. Wait, if circle k is tangent to k₁ from the inside, then the distance between centers A and C (center of k) should be equal to r₁ - r₃. But if r₁ is approaching zero, and the distance AC is approaching zero (since C is approaching A), then:AC = r₁ - r₃ => 0 = 0 - r₃ => r₃ = 0. But that can't be right because then the ratio r₃/r₁ would be 0/0, which is undefined. Hmm, maybe my initial assumption is wrong.Wait, perhaps the circle k is tangent to k₁ externally? But the problem says "tangent from the inside". So, k is inside k₁ and k₂, so the tangent is internal. Therefore, the distance between centers is equal to the difference of radii. So, for two circles tangent internally, the distance between centers is equal to the difference of their radii. So, if k is inside k₁, then the distance between A and C is r₁ - r₃. Similarly, the distance between B and C is r₂ - r₃. But as r₁ approaches zero, the center C is at distance r₁ - r₃ from A. If r₁ approaches zero, then C is approaching A, but if r₃ is positive, then r₁ - r₃ would be negative unless r₃ < r₁. But since r₁ is approaching zero, r₃ must be smaller than r₁, but we need to find the ratio r₃/r₁ as r₁ approaches zero. Hmm.Alternatively, maybe the formula is different. If k is tangent to k₁ from the inside, then the distance between centers is r₁ - r₃. But if r₁ is approaching zero, then r₃ must be approaching zero as well, but how? Let me try to write the equations.Let’s denote the centers of the circles k₁, k₂, and k as A, B, and C respectively. The circle k is tangent to both k₁ and k₂ internally. Therefore:AC = r₁ - r₃BC = r₂ - r₃Also, the distance AB is fixed, let's call it d. So, points A, B, and C form a triangle where:AC = r₁ - r₃BC = r₂ - r₃AB = dTherefore, by the triangle inequality, AC + BC ≥ AB. So:(r₁ - r₃) + (r₂ - r₃) ≥ dr₁ + r₂ - 2r₃ ≥ dBut as r₁ approaches 0, and since A and B are fixed (so d is fixed), and r₂ is fixed, we can write:0 + r₂ - 2r₃ ≥ dWhich implies:r₂ - d ≥ 2r₃But since the circle k is inside both k₁ and k₂, the radii must satisfy r₃ < r₁ and r₃ < r₂. But as r₁ approaches 0, this condition becomes r₃ < 0, which is impossible. Therefore, perhaps my understanding is incorrect.Wait, maybe when the problem says "fixing A and B", the position of A and B are fixed, but the radii r₁ and r₂ can vary? Wait, the problem states: "the ratio r₃ / r₁ approach when, fixing A and B, the point C approaches A, i.e., as r₁ approaches 0". So, fixing A and B, meaning their positions are fixed, so AB is fixed. Then, as point C approaches A, which is equivalent to r₁ approaching 0. Therefore, in this scenario, r₁ is approaching 0, but r₂ is fixed? Or is r₂ also changing?Wait, the problem says "Let k be a circle that is tangent to k₁ and k₂ from the inside, with radius r₃". So, k₁ and k₂ are given circles around A and B with radii r₁ and r₂. Then, as we fix A and B, and move point C (the center of k) towards A, which causes r₁ to approach 0. Wait, maybe the circle k₁ is being shrunk towards A, while circle k₂ remains fixed? So, in this case, r₂ is fixed, and r₁ is approaching 0. Then, circle k must be tangent to both k₁ and k₂ from the inside. So, as k₁ shrinks, the circle k must adjust its position and radius to stay tangent to both k₁ and k₂.Let me try to model this. Let’s consider coordinates. Let’s place point A at (0,0) and point B at (d, 0), where d is the fixed distance between A and B. The circle k₁ is centered at A with radius r₁, and k₂ is centered at B with radius r₂ (fixed). Circle k is tangent to both k₁ and k₂ from the inside, so it must be inside both. Let’s denote the center of circle k as point C with coordinates (x, 0) since the problem is symmetric along the line AB. The radius of circle k is r₃.Since k is tangent to k₁ internally, the distance between A and C is r₁ - r₃. So:x = r₁ - r₃Similarly, since k is tangent to k₂ internally, the distance between B and C is r₂ - r₃. The coordinates of B are (d, 0), so the distance from C to B is |d - x|. Therefore:d - x = r₂ - r₃Substituting x from the first equation:d - (r₁ - r₃) = r₂ - r₃Simplify:d - r₁ + r₃ = r₂ - r₃Bring like terms together:d - r₁ - r₂ = -2r₃Multiply both sides by -1:r₁ + r₂ - d = 2r₃Therefore:r₃ = (r₁ + r₂ - d)/2Wait, but this is a key equation. So, the radius r₃ of the circle tangent to both k₁ and k₂ from the inside is (r₁ + r₂ - d)/2. But hold on, if we have this formula, then for r₃ to be positive, we need r₁ + r₂ - d > 0, so r₁ + r₂ > d. But if A and B are fixed, and d is fixed, then if we let r₁ approach 0, and r₂ is fixed, then r₁ + r₂ - d would approach r₂ - d. If r₂ is fixed and d is fixed, and assuming that in the original problem setup, there exists such a circle k, then r₂ must be greater than d, because when r₁ approaches zero, r₃ would be (0 + r₂ - d)/2 = (r₂ - d)/2. For r₃ to be positive, r₂ must be greater than d. But that contradicts the fact that circle k₂ is around B with radius r₂, and the distance between A and B is d. If r₂ > d, then circle k₂ would encompass point A, but the problem states that circle k is tangent to both k₁ and k₂ from the inside. If k is inside k₂, then k must be entirely within k₂. But if k is also inside k₁, which is around A with radius approaching zero, then k has to be near A, inside k₁. However, if k₂ has radius r₂ > d, then k could be near A and still inside k₂. But this seems possible.Wait, but let's check the initial conditions. The formula r₃ = (r₁ + r₂ - d)/2. If we need r₃ positive, then r₁ + r₂ > d. So, for the circle k to exist, the sum of radii r₁ and r₂ must exceed the distance between A and B. But in the problem statement, circles k₁ and k₂ are around A and B, and circle k is tangent to both from the inside. So, perhaps this formula is correct. Let me verify with a simple case.Suppose d = 10 units. Let r₁ = 3, r₂ = 8. Then r₃ = (3 + 8 - 10)/2 = 1/2. Then the distance from A to C is 3 - 0.5 = 2.5, and from B to C is 10 - 2.5 = 7.5, which should be equal to 8 - 0.5 = 7.5. Correct. So the formula works here. So, yes, the radius of the inner tangent circle is (r₁ + r₂ - d)/2. Therefore, in this problem, we have:r₃ = (r₁ + r₂ - AB)/2But AB is fixed, let's denote AB = d. So, as r₁ approaches 0, r₃ approaches (0 + r₂ - d)/2 = (r₂ - d)/2. Therefore, the ratio r₃/r₁ approaches ((r₂ - d)/2)/r₁ as r₁ approaches 0. Wait, but (r₂ - d)/2 is a constant if r₂ is fixed. Therefore, as r₁ approaches 0, r₃ approaches (r₂ - d)/2, so the ratio r₃/r₁ approaches ((r₂ - d)/2)/r₁, which would go to infinity if (r₂ - d)/2 is positive. But that can't be, because we need r₃ to be positive. Wait, but if r₂ > d, then (r₂ - d)/2 is positive, but as r₁ approaches 0, the ratio r₃/r₁ tends to infinity. But that seems problematic.Wait, but in reality, when r₁ approaches 0, how does r₃ behave? According to the formula, r₃ = (r₁ + r₂ - d)/2. If we fix r₂ and d, then as r₁ approaches 0, r₃ approaches (r₂ - d)/2. So, if r₂ > d, then r₃ approaches a positive constant, so r₃/r₁ tends to infinity. If r₂ = d, then r₃ approaches 0, but then the ratio would be 0/r₁ = 0. If r₂ < d, then r₃ would be negative, which is impossible. So, the existence of circle k requires that r₁ + r₂ > d. So, as r₁ approaches 0, we need r₂ > d to have r₃ positive. Therefore, in the limit as r₁ approaches 0, r₃ approaches (r₂ - d)/2, which is a constant. Therefore, the ratio r₃/r₁ would approach infinity. But the problem says "fixing A and B", so AB = d is fixed. If r₂ is also fixed, then yes, as r₁ approaches 0, the ratio tends to infinity. But this contradicts the problem statement's implication that such a limit exists and is finite. Therefore, perhaps my formula is incorrect.Wait, maybe I made a wrong assumption in setting up the coordinates. I assumed the center C is along the line AB, which is correct because the problem is symmetric. But the key equation I got was r₃ = (r₁ + r₂ - d)/2. However, when I consider the circle k being tangent to both k₁ and k₂ from the inside, maybe the formula should be different. Wait, when two circles are tangent internally, the distance between centers is equal to the difference of radii. So, for circle k and k₁: distance AC = r₁ - r₃. For circle k and k₂: distance BC = r₂ - r₃. Then, since points A, B, C are colinear (along AB), then AC + BC = AB if C is between A and B. But in our case, if C is approaching A, then BC would be AB - AC = d - AC. So, substituting:AC = r₁ - r₃BC = d - AC = d - (r₁ - r₃) = d - r₁ + r₃But BC should also equal r₂ - r₃. Therefore:d - r₁ + r₃ = r₂ - r₃Then:d - r₁ + r₃ = r₂ - r₃Bring terms with r₃ to one side:r₃ + r₃ = r₂ - d + r₁2r₃ = r₁ + r₂ - dSo:r₃ = (r₁ + r₂ - d)/2Same result as before. So, the formula seems correct. Therefore, if we fix d and r₂, as r₁ approaches 0, then r₃ approaches (r₂ - d)/2. If r₂ > d, then r₃ approaches a positive constant, so r₃/r₁ tends to infinity. But the problem states "the ratio r₃ / r₁ approach when, fixing A and B, the point C approaches A, i.e., as r₁ approaches 0". Hmm, but according to this, the ratio would go to infinity. But maybe there's a different interpretation.Wait, the problem says "fixing A and B", which could mean fixing their positions, but not necessarily the radii r₁ and r₂. Wait, no, the problem mentions "Denote the circles around A and B in problem 1448 by k₁ and k₂, respectively, with radii r₁ and r₂". So, perhaps in the original problem 1448, the circles k₁ and k₂ have fixed radii? Or maybe in this problem, we are to consider varying r₁ while keeping A and B fixed (i.e., fixed distance AB = d), and r₂ is fixed? The problem says "fixing A and B, the point C approaches A, i.e., as r₁ approaches 0". So, perhaps when they fix A and B, they mean keeping AB = d fixed, and varying r₁ and r₂? Wait, but the problem statement doesn't mention varying r₂. It only mentions r₁ approaching 0. Therefore, I think r₂ is fixed. Therefore, if r₂ is fixed and d is fixed, then as r₁ approaches 0, r₃ approaches (0 + r₂ - d)/2 = (r₂ - d)/2. Therefore, the ratio r₃ / r₁ approaches (r₂ - d)/(2r₁), which as r₁ approaches 0, if r₂ > d, this ratio goes to infinity. If r₂ = d, then the ratio approaches 0. If r₂ < d, the ratio is negative, which is impossible. Therefore, unless r₂ is approaching d as r₁ approaches 0, we can't have a finite limit.But the problem says "fixing A and B", so AB = d is fixed, and presumably r₂ is fixed as well. Therefore, unless r₂ = d + something, but if r₂ is fixed greater than d, then the ratio would go to infinity. But the problem asks "What does the ratio r₃ / r₁ approach...", implying that the limit is a finite number. Therefore, my previous analysis must be missing something.Wait, maybe I misunderstood the problem's configuration. Let me read it again:"Denote the circles around A and B in problem 1448 by k₁ and k₂, respectively, with radii r₁ and r₂. Let k be a circle that is tangent to k₁ and k₂ from the inside, with radius r₃. What does the ratio r₃ / r₁ approach when, fixing A and B, the point C approaches A, i.e., as r₁ approaches 0?"Since it refers to problem 1448, which I don't have access to, but the key is that k is tangent to k₁ and k₂ from the inside. So, k is inside both k₁ and k₂. If we are to fix A and B, and move C towards A, which is equivalent to r₁ approaching 0. But if k is inside k₁, which is itself shrinking to a point, how does k remain tangent to k₂?Alternatively, maybe the circles k₁ and k₂ are not fixed, but the positions of A and B are fixed. As C approaches A, the radius r₁ is getting smaller. But how are k₁ and k₂ related? Maybe in the original problem 1448, circles k₁ and k₂ are tangent to each other or something else, but without that context, it's hard to tell.Wait, perhaps the problem is about the Soddy circles or Apollonius circles. When you have two circles, there are two circles tangent to both from the inside. The radius can be found using Descartes' Circle Theorem. Maybe applying that here.Descartes' Circle Theorem states that for four mutually tangent circles, the curvatures satisfy a certain equation. But here we have three circles: k₁, k₂, and k. Wait, Descartes' Theorem for three circles. If we have three circles tangent to each other, we can find the radius of the fourth. But in our case, we have two circles and a circle tangent to both from the inside. So maybe using the formula for the radius of the inner tangent circle.Alternatively, recall that for two circles, the radius of a circle tangent to both from the inside can be found by the formula I derived earlier: r₃ = (r₁ + r₂ - d)/2. But according to that formula, if we fix d and r₂, as r₁ approaches 0, r₃ approaches (r₂ - d)/2. Therefore, unless r₂ = d, which would make r₃ = (0 + d - d)/2 = 0, which would give a ratio of 0. But if r₂ = d, then k₂ is a circle around B with radius equal to the distance AB, so point A is on the circumference of k₂. Therefore, circle k, which is inside k₂ and tangent to it, would have to be within k₂. If we also have circle k₁ around A with r₁ approaching 0, then the circle k must be tangent to k₁ (which is a tiny circle around A) and inside k₂. So, the center of k is very close to A, with radius r₃ approximately equal to r₁, but also needs to be inside k₂.Wait, but if r₂ = d, then the circle k₂ has radius equal to AB, so B is the center, and the circle k₂ extends to A. So, the maximum distance from B to any point in k₂ is d, so A is on the circumference of k₂. Then, a circle k inside k₂ and tangent to k₂ would have its center at a distance of r₂ - r₃ = d - r₃ from B. Also, being tangent to k₁ (radius r₁) from the inside, the distance from A to the center of k is r₁ - r₃. So, if the center of k is at distance r₁ - r₃ from A and d - r₃ from B. But AB = d, so:The distance from A to center of k is x = r₁ - r₃The distance from B to center of k is d - x = d - (r₁ - r₃) = d - r₁ + r₃But this must equal r₂ - r₃ = d - r₃ (since r₂ = d)Therefore:d - r₁ + r₃ = d - r₃Simplify:-r₁ + r₃ = -r₃Then:-r₁ = -2r₃Multiply both sides by -1:r₁ = 2r₃Therefore:r₃ = r₁ / 2Therefore, the ratio r₃ / r₁ = 1/2.Wait, so if r₂ = d, then as r₁ approaches 0, the ratio approaches 1/2. But in the previous case where r₂ was fixed greater than d, the ratio went to infinity. So, maybe in the problem, the circle k₂ is such that when we fix A and B, and move C towards A (i.e., r₁ approaches 0), the radius r₂ is not fixed but is adjusted such that circle k remains tangent to both. Wait, but the problem says "fixing A and B", which usually would mean their positions and radii are fixed. But maybe in problem 1448, the context is different. Perhaps in the original problem, circles k₁ and k₂ are tangent to each other, or something else.Alternatively, perhaps when point C approaches A, the circle k₂ must also be adjusted. But the problem states "fixing A and B", which probably means their positions and radii. Wait, no, fixing A and B might just mean their positions, not their radii. So, if we fix the positions of A and B, but allow their radii r₁ and r₂ to vary, then as C approaches A, which implies r₁ approaches 0, and we might need to adjust r₂ accordingly to maintain the tangency of circle k. But the problem doesn't mention r₂ changing; it only mentions fixing A and B. Hmm.Wait, the problem says: "the ratio r₃ / r₁ approach when, fixing A and B, the point C approaches A, i.e., as r₁ approaches 0". So, fixing A and B probably means fixing their positions (so distance AB = d is fixed), but the radii r₁ and r₂ could be variables. However, in the problem statement, k₁ and k₂ are circles around A and B with radii r₁ and r₂, and k is tangent to both. So, as r₁ approaches 0, perhaps r₂ is also changing? But the problem doesn't specify how r₂ changes. It just says "fixing A and B", which might mean their radii are fixed as well. But if radii are fixed, then as r₁ approaches 0, the circle k must satisfy the equations:r₃ = (r₁ + r₂ - d)/2If r₂ and d are fixed, then as r₁ approaches 0, r₃ approaches (r₂ - d)/2. If r₂ > d, ratio approaches infinity; if r₂ = d, ratio approaches 0; if r₂ < d, r₃ negative which is impossible. Therefore, unless the problem implies that r₂ is a function of r₁ as r₁ approaches 0, perhaps in such a way that (r₂ - d)/2 is proportional to r₁, making the ratio finite.Alternatively, maybe there's a different configuration. Suppose the circle k is tangent to k₁ and k₂ externally. But the problem says "from the inside", so it must be tangent internally. Alternatively, maybe one is inside and one is outside. But the problem states "tangent to k₁ and k₂ from the inside", so both are internal tangents.Wait, perhaps the formula is different if the circles are tangent in a different configuration. Let me double-check. For two circles, if they are tangent internally, the distance between centers is difference of radii. If tangent externally, the distance is sum of radii. So, if circle k is inside both k₁ and k₂, then distances from A and B to C are r₁ - r₃ and r₂ - r₃, respectively.But given that, in the earlier analysis, we get r₃ = (r₁ + r₂ - d)/2. If we fix d, and let r₁ approach 0, then unless r₂ approaches d + 2r₃, but since r₃ is (r₁ + r₂ - d)/2, substituting gives r₃ = (0 + r₂ - d)/2, so r₂ = d + 2r₃. Therefore, if we assume that as r₁ approaches 0, r₂ is approaching d, then r₃ approaches (0 + d - d)/2 = 0. But how does r₂ approach d? The problem states "fixing A and B", which probably means fixing their radii as well. If not, then maybe in the problem's original context (problem 1448), there is a specific relation between r₁, r₂, and d.Alternatively, maybe the problem is in a configuration where the two circles k₁ and k₂ are tangent to each other. If k₁ and k₂ are tangent, then the distance AB = r₁ + r₂. But in that case, d = r₁ + r₂. Then, substituting into the formula for r₃:r₃ = (r₁ + r₂ - d)/2 = (r₁ + r₂ - (r₁ + r₂))/2 = 0, which doesn't make sense. Therefore, perhaps not tangent.Alternatively, maybe in problem 1448, the circles k₁ and k₂ are the circumcircle and incircle of a triangle or something else. Without the original problem, it's hard to tell, but perhaps we can infer.Alternatively, maybe there is a misinterpretation of the problem's geometric configuration. Let's consider that point C is a third point, not the center of circle k. Wait, the problem says "Let k be a circle that is tangent to k₁ and k₂ from the inside, with radius r₃. What does the ratio r₃ / r₁ approach when, fixing A and B, the point C approaches A, i.e., as r₁ approaches 0?"Wait, the mention of point C approaching A suggests that in the original problem 1448, point C is defined somewhere, perhaps as the center of circle k or another point. Since the problem here references circles around A and B from problem 1448, maybe in that problem, point C is another point related to the configuration, and as it approaches A, the radii change accordingly. Without the original problem, it's challenging, but given the current information, we need to work with what we have.Assuming that point C is the center of circle k, as I did earlier, and given the formula r₃ = (r₁ + r₂ - d)/2, then as r₁ approaches 0, r₃ approaches (r₂ - d)/2. For this to be positive, r₂ must be greater than d. If that's the case, then the ratio r₃/r₁ approaches ((r₂ - d)/2)/r₁, which tends to infinity as r₁ approaches 0. But the problem likely expects a finite answer, so this suggests that my assumption is incorrect.Alternatively, maybe the circle k is tangent to k₁ externally and k₂ internally. Wait, but the problem says "tangent to k₁ and k₂ from the inside", so both are internal tangents. Therefore, my original analysis holds. However, the problem might be in a different configuration where AB is not the line along which the centers lie, but that seems unlikely due to the problem's symmetry.Wait, maybe the circle k is tangent to k₁ and k₂ externally instead of internally. Let's explore this possibility. If circle k is tangent to both k₁ and k₂ externally, then the distance between centers would be r₁ + r₃ and r₂ + r₃. Then, the distance between A and B would be AC + BC = (r₁ + r₃) + (r₂ + r₃) = r₁ + r₂ + 2r₃. But AB is fixed as d, so:r₁ + r₂ + 2r₃ = dTherefore:r₃ = (d - r₁ - r₂)/2In this case, as r₁ approaches 0, r₃ approaches (d - r₂)/2. For r₃ to be positive, d - r₂ must be positive, so d > r₂. If d and r₂ are fixed, then as r₁ approaches 0, r₃ approaches (d - r₂)/2, so the ratio r₃/r₁ approaches (d - r₂)/(2r₁), which also tends to infinity. Still problematic.But again, this contradicts the problem's implication that the ratio approaches a finite limit. Therefore, perhaps the problem's original configuration in problem 1448 has a specific relation that makes this ratio finite. Given that the user refers to problem 1448, which might be from a specific source, perhaps a textbook, where the circles k₁ and k₂ are related in a certain way.Alternatively, perhaps there's an error in assuming the center C is colinear with A and B. Maybe in the original problem, point C is not on the line AB, leading to a different configuration. Let's consider that.If circle k is tangent to both k₁ and k₂ from the inside, but its center C is not on the line AB, then we can model the problem in 2D space. Let’s denote the coordinates:Let’s place point A at (0,0) and point B at (d,0). Let the center of circle k be at (x,y) with radius r₃. The distance from (x,y) to A must be r₁ - r₃, and the distance from (x,y) to B must be r₂ - r₃. Therefore:√(x² + y²) = r₁ - r₃√((x - d)² + y²) = r₂ - r₃Squaring both equations:x² + y² = (r₁ - r₃)²(x - d)² + y² = (r₂ - r₃)²Subtract the first equation from the second:(x - d)² + y² - x² - y² = (r₂ - r₃)² - (r₁ - r₃)²Expand left side:x² - 2dx + d² - x² = -2dx + d²Right side:(r₂ - r₃)² - (r₁ - r₃)² = [r₂² - 2r₂r₃ + r₃²] - [r₁² - 2r₁r₃ + r₃²] = r₂² - 2r₂r₃ - r₁² + 2r₁r₃Therefore:-2dx + d² = r₂² - 2r₂r₃ - r₁² + 2r₁r₃Solve for x:-2dx = r₂² - 2r₂r₃ - r₁² + 2r₁r₃ - d²Therefore:x = [d² + r₁² - r₂² + 2r₂r₃ - 2r₁r₃]/(2d)But from the first equation, we also have:x² + y² = (r₁ - r₃)²Therefore, we can write y² = (r₁ - r₃)² - x²Substituting x from above:y² = (r₁ - r₃)² - [ (d² + r₁² - r₂² + 2r₂r₃ - 2r₁r₃ )/(2d) ]²This seems complicated, but perhaps in the limit as r₁ approaches 0, we can find the leading behavior of r₃.Assume that as r₁ → 0, r₃ also approaches 0. Let’s set r₃ = k * r₁, where k is a constant we need to find. So, r₃ = k r₁. Substitute this into the equations.First, from the previous result:r₃ = (r₁ + r₂ - d)/2But if r₃ = k r₁, then:k r₁ = (r₁ + r₂ - d)/2Multiply both sides by 2:2k r₁ = r₁ + r₂ - dRearrange:(2k - 1)r₁ = r₂ - dAs r₁ approaches 0, the left-hand side approaches 0. Therefore, we must have r₂ - d = 0, i.e., r₂ = d. Otherwise, the equation can't hold as r₁ → 0. Therefore, this suggests that r₂ must equal d in the limit as r₁ approaches 0. Therefore, if r₂ = d, then the equation becomes:2k r₁ = r₁ + d - d = r₁Therefore, 2k r₁ = r₁ ⇒ 2k = 1 ⇒ k = 1/2Therefore, the ratio r₃/r₁ approaches 1/2 as r₁ approaches 0, provided that r₂ approaches d as r₁ approaches 0. However, the problem states "fixing A and B", which likely means fixing their positions and radii. If r₂ is fixed and equal to d, then the earlier analysis when r₂ = d gives us r₃ = r₁/2, so the ratio is always 1/2, regardless of r₁. Wait, let's check that.If r₂ = d, then according to the formula:r₃ = (r₁ + d - d)/2 = r₁ / 2Therefore, r₃/r₁ = 1/2 for any r₁. Therefore, as r₁ approaches 0, the ratio remains 1/2. Therefore, the limit is 1/2.But this requires that r₂ = d. However, the problem didn't specify that r₂ = d, but if we are to fix A and B, which are centers of circles k₁ and k₂, perhaps in the original problem 1448, the radii r₁ and r₂ are such that the circle k exists, which requires r₁ + r₂ ≥ d for internal tangent. But if we fix A and B (i.e., fix d), and vary r₁ approaching 0, then for the circle k to exist, we must have r₂ ≥ d - r₁. As r₁ approaches 0, this requires r₂ ≥ d. Therefore, if r₂ is exactly d, then as r₁ approaches 0, the ratio r₃/r₁ is always 1/2. If r₂ > d, then as r₁ approaches 0, r₃ approaches (r₂ - d)/2, so the ratio tends to infinity. But the problem states "fixing A and B", which might imply that r₂ is adjusted to maintain the existence of circle k. However, typically, fixing A and B would mean their positions and radii are fixed. But given that the problem allows r₁ to approach 0, it's likely that in this configuration, r₂ is set to d, making the ratio 1/2. Therefore, the answer is 1/2.But how do we reconcile this with the earlier analysis? If r₂ is fixed at d, then k₂ is a circle with radius d centered at B, so A lies on the circumference of k₂. Then, circle k, which is inside both k₁ (radius r₁) and k₂ (radius d), has to be tangent to both. The center of k is at distance r₁ - r₃ from A and d - r₃ from B. But since AB = d, the distance from B to the center of k is d - (r₁ - r₃) = d - r₁ + r₃. Setting this equal to d - r₃ (because it's inside k₂), we get:d - r₁ + r₃ = d - r₃ ⇒ -r₁ + r₃ = -r₃ ⇒ r₁ = 2r₃ ⇒ r₃ = r₁/2.Therefore, the ratio r₃/r₁ is always 1/2, regardless of r₁, as long as r₂ = d. Therefore, if in the problem's context, fixing A and B implies that k₂ has radius equal to the distance AB, then the ratio approaches 1/2. However, if r₂ is fixed at a different value, the limit could be different or nonexistent.But since the problem references problem 1448, which likely establishes a specific relationship between A, B, and C, and the circles k₁ and k₂, we might need to infer that in the original problem, k₂ is the circle with radius equal to AB. Alternatively, it's possible that in the original problem, the circles k₁ and k₂ are orthogonal or have another relation that makes r₂ = d.Given that the problem asks for the limit as r₁ approaches 0, and given that without additional constraints, the ratio could be 1/2 or infinity depending on r₂, but the problem expects a unique answer, it's plausible that the correct answer is 1/2. This is because if we assume that the problem requires the circle k to exist for all r₁ approaching 0, then r₂ must be exactly d, leading to the ratio 1/2. Otherwise, the circle k would cease to exist or the ratio would tend to infinity.Therefore, after considering various configurations and the likely intentions of the problem, the ratio r₃/r₁ approaches 1/2 as r₁ approaches 0.

✅:To determine the ratio ( frac{r_3}{r_1} ) as ( r_1 to 0 ), we analyze the geometric configuration of the circles. Let the distance between centers ( A ) and ( B ) be ( d ). The circle ( k ) (radius ( r_3 )) is tangent to both ( k_1 ) (radius ( r_1 )) and ( k_2 ) (radius ( r_2 )) internally. The distances between their centers are:[AC = r_1 - r_3 quad text{and} quad BC = r_2 - r_3.]Using the triangle inequality for ( triangle ABC ), we derive:[r_3 = frac{r_1 + r_2 - d}{2}.]As ( r_1 to 0 ), for ( k ) to exist, ( r_2 ) must equal ( d ) (since ( r_3 > 0 )). Substituting ( r_2 = d ):[r_3 = frac{0 + d - d}{2} = frac{r_1}{2} quad Rightarrow quad frac{r_3}{r_1} = frac{1}{2}.]Thus, the ratio approaches ( frac{1}{2} ).Final Answer:boxed{dfrac{1}{2}}

🔑:We are asked to find the limit of the ratio ( frac{r_3}{r_1} ) as ( C ) approaches ( A ), which implies ( r_1 ) approaches 0. Below is the step-by-step solution to the problem.1. Express the Given Quantities: From the reference solution, we know the following representations: - Distance ( OC = t ) where ( 0 leq t < 1 ). - The radius of the circle tangent from inside to ( k ), ( k_1 ), and ( k_2 ): [ r_3 = frac{1 - t^2}{4} = frac{(1 - t)(1 + t)}{4} ] - The radius ( r_1 ) of the circle shrinking to a point, i.e., distance ( AC ): [ r_1 = AC = 1 - t ]2. Determine the Ratio ( frac{r_3}{r_1} ): Calculate the ratio as follows: [ frac{r_3}{r_1} = frac{frac{(1 - t)(1 + t)}{4}}{1 - t} = frac{1 + t}{4} ]3. Analyze the Behavior as ( C to A ): As ( C ) approaches ( A ), the distance ( CA = 1 - t ) tends to 0 which means ( t ) approaches 1. Substituting ( t = 1 ) into the ratio: [ t to 1 Rightarrow frac{1 + t}{4} to frac{1 + 1}{4} = frac{2}{4} = frac{1}{2} ]4. Conclusion: The limit of the ratio ( frac{r_3}{r_1} ) as ( r_1 ) tends to 0 is: [ boxed{frac{1}{2}} ]