💡Okay, so I have this problem about two similar right triangles. The smaller one has an area of 12 square inches, and the larger one has an area of 300 square inches. The hypotenuse of the smaller triangle is 10 inches. I need to find the sum of the lengths of the legs of the larger triangle.First, since the triangles are similar, their corresponding sides are proportional. That means the ratio of their areas will be the square of the ratio of their corresponding sides. So, if I can find the ratio of their areas, I can find the scale factor for their sides.The area of the smaller triangle is 12, and the area of the larger one is 300. So, the ratio of their areas is 300 divided by 12, which is 25. That means the area ratio is 25:1. Since area scales with the square of the side lengths, the ratio of their side lengths should be the square root of 25, which is 5. So, each side of the larger triangle is 5 times the corresponding side of the smaller triangle.Now, I need to find the lengths of the legs of the smaller triangle. Let's denote the legs as 'a' and 'b', and the hypotenuse as 'c'. For the smaller triangle, we know the area is 12, so (1/2)*a*b = 12. That simplifies to a*b = 24.We also know that the hypotenuse is 10 inches, so by the Pythagorean theorem, a² + b² = 10² = 100.So, I have two equations:1. a*b = 242. a² + b² = 100I need to solve for 'a' and 'b'. Maybe I can express one variable in terms of the other from the first equation and substitute into the second equation.From the first equation, b = 24/a. Substitute that into the second equation:a² + (24/a)² = 100Simplify that:a² + 576/a² = 100Multiply both sides by a² to eliminate the denominator:a⁴ + 576 = 100a²Bring all terms to one side:a⁴ - 100a² + 576 = 0This looks like a quadratic equation in terms of a². Let me set x = a²:x² - 100x + 576 = 0Now, I can use the quadratic formula to solve for x:x = [100 ± sqrt(100² - 4*1*576)] / 2Calculate the discriminant:sqrt(10000 - 2304) = sqrt(7696)Hmm, sqrt(7696) is approximately 87.73, but let me check that. Wait, 87² is 7569, and 88² is 7744. So, sqrt(7696) is between 87 and 88. Let me calculate it more accurately.7696 - 7569 = 127, so sqrt(7696) ≈ 87 + 127/(2*87) ≈ 87 + 127/174 ≈ 87 + 0.73 ≈ 87.73So, x = [100 ± 87.73]/2Calculating both possibilities:x = (100 + 87.73)/2 ≈ 187.73/2 ≈ 93.865x = (100 - 87.73)/2 ≈ 12.27/2 ≈ 6.135So, x ≈ 93.865 or x ≈ 6.135Since x = a², then a = sqrt(x). So,a ≈ sqrt(93.865) ≈ 9.69 inchesora ≈ sqrt(6.135) ≈ 2.48 inchesSo, the legs of the smaller triangle are approximately 2.48 inches and 9.69 inches.Now, since the larger triangle is similar and scaled by a factor of 5, each leg will be 5 times longer.So, the legs of the larger triangle are approximately 2.48*5 ≈ 12.4 inches and 9.69*5 ≈ 48.45 inches.Therefore, the sum of the lengths of the legs of the larger triangle is approximately 12.4 + 48.45 ≈ 60.85 inches.Wait, that's approximately 60.85, but in the initial calculation, I thought it was 61.3. Maybe I made a rounding error somewhere. Let me check my calculations again.Starting from the quadratic equation:x² - 100x + 576 = 0Using the quadratic formula:x = [100 ± sqrt(10000 - 2304)] / 2sqrt(10000 - 2304) = sqrt(7696). Let me calculate sqrt(7696) more precisely.I know that 87² = 7569 and 88² = 7744.7696 - 7569 = 127So, sqrt(7696) = 87 + 127/(2*87) = 87 + 127/174 ≈ 87 + 0.73 ≈ 87.73So, x = (100 ± 87.73)/2x1 = (100 + 87.73)/2 ≈ 187.73/2 ≈ 93.865x2 = (100 - 87.73)/2 ≈ 12.27/2 ≈ 6.135So, a² = 93.865 or 6.135Thus, a = sqrt(93.865) ≈ 9.69 inchesand a = sqrt(6.135) ≈ 2.48 inchesSo, the legs are approximately 2.48 and 9.69 inches.Scaling by 5, we get 12.4 and 48.45 inches.Sum is 12.4 + 48.45 ≈ 60.85 inches.Wait, but in the initial thought process, I thought it was 61.3. Maybe I approximated sqrt(7696) as 87.73, but actually, 87.73² is approximately 7696. Let me check:87.73 * 87.73:Calculate 87 * 87 = 756987 * 0.73 = 63.510.73 * 87 = 63.510.73 * 0.73 ≈ 0.5329So, total is 7569 + 63.51 + 63.51 + 0.5329 ≈ 7569 + 127.02 + 0.5329 ≈ 7696.5529, which is very close to 7696. So, sqrt(7696) ≈ 87.73 is accurate.Therefore, the sum is approximately 60.85 inches. But the initial thought process had 61.3. Maybe I made a mistake in the initial calculation.Wait, in the initial thought process, I had:x = 93.2 or 6.8So, a = sqrt(6.8) ≈ 2.61, b = sqrt(93.2) ≈ 9.65Then, 5*(2.61 + 9.65) = 5*12.26 = 61.3But in my detailed calculation here, I have a = 2.48 and b = 9.69, which gives 5*(2.48 + 9.69) = 5*12.17 ≈ 60.85So, there's a slight discrepancy due to rounding. Maybe in the initial thought process, they approximated sqrt(7696) as 87.73, but in reality, it's slightly less.Alternatively, perhaps I should carry out the calculations more precisely without rounding too early.Let me try solving the quadratic equation without approximating:x² - 100x + 576 = 0Using quadratic formula:x = [100 ± sqrt(10000 - 2304)] / 2sqrt(7696) = let's see, 87² = 7569, 88²=7744So, sqrt(7696) = 87 + (7696 - 7569)/ (2*87 + 1) = 87 + 127/175 ≈ 87 + 0.7257 ≈ 87.7257So, x = (100 ± 87.7257)/2x1 = (100 + 87.7257)/2 = 187.7257/2 = 93.86285x2 = (100 - 87.7257)/2 = 12.2743/2 = 6.13715So, a² = 93.86285 or 6.13715Thus, a = sqrt(93.86285) ≈ 9.69 inchesand a = sqrt(6.13715) ≈ 2.48 inchesSo, legs are approximately 2.48 and 9.69 inches.Scaling by 5, we get 12.4 and 48.45 inches.Sum is 12.4 + 48.45 ≈ 60.85 inches.But in the initial thought process, the sum was 61.3. So, perhaps the initial thought process had a slight rounding difference.Alternatively, maybe I should solve the quadratic equation more accurately.Let me try to calculate sqrt(7696) more precisely.We have 87² = 756987.7² = (87 + 0.7)² = 87² + 2*87*0.7 + 0.7² = 7569 + 121.8 + 0.49 = 7691.2987.7² = 7691.29But we need sqrt(7696). So, 7696 - 7691.29 = 4.71So, 87.7 + d)^2 = 7696(87.7 + d)^2 = 87.7² + 2*87.7*d + d² = 7691.29 + 175.4d + d² = 7696So, 175.4d + d² = 4.71Assuming d is small, d² is negligible, so 175.4d ≈ 4.71Thus, d ≈ 4.71 / 175.4 ≈ 0.02686So, sqrt(7696) ≈ 87.7 + 0.02686 ≈ 87.72686So, x = [100 ± 87.72686]/2x1 = (100 + 87.72686)/2 ≈ 187.72686/2 ≈ 93.86343x2 = (100 - 87.72686)/2 ≈ 12.27314/2 ≈ 6.13657So, a² = 93.86343 or 6.13657Thus, a = sqrt(93.86343) ≈ 9.69 inchesand a = sqrt(6.13657) ≈ 2.48 inchesSo, legs are approximately 2.48 and 9.69 inches.Scaling by 5, we get 12.4 and 48.45 inches.Sum is 12.4 + 48.45 ≈ 60.85 inches.So, approximately 60.85 inches.But the initial thought process had 61.3. Maybe I should check if I made a mistake in the initial thought process.Wait, in the initial thought process, I had:x = 93.2 or 6.8So, a = sqrt(6.8) ≈ 2.61, b = sqrt(93.2) ≈ 9.65Then, 5*(2.61 + 9.65) = 5*12.26 = 61.3But in reality, x was 93.86343 and 6.13657, not 93.2 and 6.8. So, the initial thought process had a rounding error.Therefore, the correct sum should be approximately 60.85 inches.But let's see, maybe I can find the exact value without approximating.Let me try to solve the quadratic equation exactly.We have x² - 100x + 576 = 0Using quadratic formula:x = [100 ± sqrt(10000 - 2304)] / 2sqrt(7696) = sqrt(16*481) = 4*sqrt(481)So, x = [100 ± 4*sqrt(481)] / 2 = 50 ± 2*sqrt(481)So, x = 50 + 2*sqrt(481) or x = 50 - 2*sqrt(481)Therefore, a² = 50 + 2*sqrt(481) or a² = 50 - 2*sqrt(481)Thus, a = sqrt(50 + 2*sqrt(481)) or a = sqrt(50 - 2*sqrt(481))Similarly, b = sqrt(50 - 2*sqrt(481)) or b = sqrt(50 + 2*sqrt(481))So, the legs are sqrt(50 + 2*sqrt(481)) and sqrt(50 - 2*sqrt(481))Now, let's compute sqrt(50 + 2*sqrt(481)) and sqrt(50 - 2*sqrt(481))First, compute sqrt(481):sqrt(481) ≈ 21.9317So, 2*sqrt(481) ≈ 43.8634Thus,sqrt(50 + 43.8634) = sqrt(93.8634) ≈ 9.69 inchesandsqrt(50 - 43.8634) = sqrt(6.1366) ≈ 2.48 inchesSo, same as before.Therefore, the legs of the smaller triangle are approximately 2.48 and 9.69 inches.Scaling by 5, the legs of the larger triangle are approximately 12.4 and 48.45 inches.Sum is 12.4 + 48.45 ≈ 60.85 inches.But the problem asks for the sum of the lengths of the legs of the larger triangle. So, 60.85 inches is the approximate value.But maybe we can express it exactly.Let me see, the legs of the smaller triangle are sqrt(50 + 2*sqrt(481)) and sqrt(50 - 2*sqrt(481))So, the sum of the legs is sqrt(50 + 2*sqrt(481)) + sqrt(50 - 2*sqrt(481))Let me denote S = sqrt(50 + 2*sqrt(481)) + sqrt(50 - 2*sqrt(481))Let me compute S²:S² = [sqrt(50 + 2*sqrt(481)) + sqrt(50 - 2*sqrt(481))]²= (50 + 2*sqrt(481)) + (50 - 2*sqrt(481)) + 2*sqrt{(50 + 2*sqrt(481))(50 - 2*sqrt(481))}Simplify:= 100 + 2*sqrt{(50)² - (2*sqrt(481))²}= 100 + 2*sqrt{2500 - 4*481}= 100 + 2*sqrt{2500 - 1924}= 100 + 2*sqrt{576}= 100 + 2*24= 100 + 48= 148So, S² = 148Therefore, S = sqrt(148) = 2*sqrt(37)So, the sum of the legs of the smaller triangle is 2*sqrt(37) inches.Therefore, the sum of the legs of the larger triangle is 5 times that, which is 10*sqrt(37) inches.But 10*sqrt(37) is approximately 10*6.08276 ≈ 60.8276 inches, which is approximately 60.83 inches.So, the exact value is 10*sqrt(37) inches, which is approximately 60.83 inches.But in the initial thought process, I had 61.3, which is a bit off due to rounding.Therefore, the correct sum is 10*sqrt(37) inches, which is approximately 60.83 inches.But let me check if 10*sqrt(37) is indeed the exact value.Yes, because S = sqrt(50 + 2*sqrt(481)) + sqrt(50 - 2*sqrt(481)) = 2*sqrt(37)Wait, how?Wait, S² = 148, so S = sqrt(148) = 2*sqrt(37)Yes, because 148 = 4*37, so sqrt(148) = 2*sqrt(37)Therefore, the sum of the legs of the smaller triangle is 2*sqrt(37), and the larger triangle is 5 times that, so 10*sqrt(37)Therefore, the exact value is 10*sqrt(37) inches.So, the sum of the lengths of the legs of the larger triangle is 10*sqrt(37) inches, which is approximately 60.83 inches.But the problem didn't specify whether to give an exact value or an approximate decimal. Since sqrt(37) is irrational, it's better to give the exact value.Therefore, the answer is 10*sqrt(37) inches.But let me check if 10*sqrt(37) is indeed the correct exact value.Yes, because:Sum of legs of smaller triangle = 2*sqrt(37)Therefore, larger triangle's sum = 5*2*sqrt(37) = 10*sqrt(37)Yes, that's correct.So, the exact value is 10*sqrt(37) inches.Alternatively, if we rationalize or approximate, it's approximately 60.83 inches.But since the problem didn't specify, I think the exact value is preferred.Therefore, the sum of the lengths of the legs of the larger triangle is 10*sqrt(37) inches.But let me double-check my steps to ensure I didn't make any mistakes.1. The ratio of areas is 300/12 = 25, so the scale factor is 5.2. For the smaller triangle, area = 12 = (1/2)*a*b => a*b = 243. Hypotenuse c = 10, so a² + b² = 1004. Solved for a and b by setting up the quadratic equation, found a² = 50 ± 2*sqrt(481)5. Sum of legs S = sqrt(50 + 2*sqrt(481)) + sqrt(50 - 2*sqrt(481)) = 2*sqrt(37)6. Therefore, larger triangle's sum = 5*S = 10*sqrt(37)Yes, that seems correct.Alternatively, perhaps there's a simpler way to find the sum of the legs without solving for a and b individually.Let me think.We have a*b = 24 and a² + b² = 100We need to find a + b.We know that (a + b)² = a² + 2ab + b² = (a² + b²) + 2ab = 100 + 48 = 148Therefore, (a + b)² = 148 => a + b = sqrt(148) = 2*sqrt(37)Yes, that's a simpler way.So, the sum of the legs of the smaller triangle is 2*sqrt(37), and the larger triangle is 5 times that, so 10*sqrt(37)Therefore, the answer is 10*sqrt(37) inches.This method avoids solving the quadratic equation and directly finds the sum of the legs.So, that's a more efficient approach.Therefore, the sum of the lengths of the legs of the larger triangle is 10*sqrt(37) inches.

💡Okay, so I need to prove that for any set ( Y subseteq mathbf{N}_{+} ) with ( n ) elements, there exists a subset ( B subseteq Y ) such that ( |B| > frac{n}{3} ) and if ( u, v in B ), then ( u + v notin B ). Hmm, interesting. Let me try to break this down.First, I should understand what the problem is asking. We have a set of positive integers, and we need to find a subset that's larger than a third of the original set. Moreover, this subset has the property that the sum of any two elements isn't in the subset. So, it's like avoiding any pairs whose sum is also in the subset. That reminds me a bit of sum-free sets, which are sets where no two elements add up to another element in the set. Maybe that concept can help here.I remember that in additive combinatorics, sum-free sets have been studied extensively. One classic result is that any set of integers has a sum-free subset of size at least ( frac{n}{3} ). Wait, that seems related. But the problem here is slightly different because it requires the subset to be larger than ( frac{n}{3} ). So, maybe I can use some similar techniques or build upon that result.Let me think about how sum-free sets are constructed. One common method is to consider the upper third of the set. For example, if you take the largest ( lceil frac{n}{3} rceil ) elements of ( Y ), then the sum of any two elements will be larger than the largest element, hence not in the subset. But in this case, we need a subset larger than ( frac{n}{3} ). So, taking the upper third might not be sufficient because it only gives us size ( frac{n}{3} ). Hmm.Wait, maybe there's a way to partition the set into three parts and then select the appropriate part. If I can partition ( Y ) into three classes such that one class is sum-free and larger than ( frac{n}{3} ), that would solve the problem. But how?Alternatively, perhaps using modular arithmetic could help. I've heard of using residues modulo some number to construct sum-free sets. Let me explore that idea. If I choose a modulus ( p ) such that ( p ) is relatively prime to all elements of ( Y ), then multiplying each element by a suitable number modulo ( p ) might spread them out in a way that allows me to pick a sum-free subset.Let me try to formalize this. Suppose I choose a prime ( p ) larger than ( n ) and such that ( p ) is coprime to all elements of ( Y ). Then, for each element ( y in Y ), I can consider its residue modulo ( p ). If I can find a range of residues such that adding any two residues in that range doesn't fall back into the same range, then the corresponding elements would form a sum-free set.For example, if I take the middle third of the residues modulo ( p ), say from ( k ) to ( 2k-1 ) where ( p = 3k - 1 ), then adding any two numbers in this range would result in a number either in the lower third or the upper third, but not in the middle third. That way, the subset corresponding to the middle third residues would be sum-free.But wait, how do I ensure that such a construction gives me a subset larger than ( frac{n}{3} )? If I can show that, on average, each residue class has about ( frac{n}{p} ) elements, and since ( p = 3k - 1 ), which is roughly ( 3k ), the number of elements in each third would be roughly ( frac{n}{3} ). But I need more than that. Maybe by choosing ( p ) appropriately, I can ensure that one of the residue classes has more than ( frac{n}{3} ) elements.Alternatively, perhaps I can use the probabilistic method. If I randomly select elements from ( Y ), what's the probability that the subset I get is sum-free and has size greater than ( frac{n}{3} )? But I'm not sure if that approach would directly give me the desired result, and it might be more complicated.Going back to the modular arithmetic idea, let me try to make it precise. Let me choose ( p = 3k - 1 ) where ( k ) is chosen such that ( p ) is prime and greater than ( n ). Also, ( p ) should be coprime to all elements of ( Y ). Then, define ( A = {k, k+1, ldots, 2k-1} ). If I take any two elements ( x, y in A ), their sum ( x + y ) will be between ( 2k ) and ( 4k - 2 ). Since ( p = 3k - 1 ), the residues modulo ( p ) of these sums will either be in ( {2k, ldots, 3k - 1} ) or wrap around to ( {1, ldots, k - 1} ). Therefore, ( x + y ) modulo ( p ) is not in ( A ), so the corresponding elements in ( Y ) won't have their sum in the subset.Now, to construct the subset ( B ), I can consider each element ( y in Y ) and look at its residue modulo ( p ). If the residue is in ( A ), include ( y ) in ( B ). Since ( p ) is coprime to all elements of ( Y ), the residues are uniformly distributed, so on average, each residue class has about ( frac{n}{p} ) elements. Since ( p = 3k - 1 approx 3k ), the size of ( B ) would be roughly ( frac{n}{3} ). But I need more than that.Wait, maybe I can use the pigeonhole principle here. If I consider all possible residues modulo ( p ), and since each element of ( Y ) falls into exactly one residue class, the total number of elements is ( n ). If I partition the residues into three intervals: ( {1, ldots, k-1} ), ( {k, ldots, 2k-1} ), and ( {2k, ldots, 3k-1} ), then each interval has ( k-1 ), ( k ), and ( k-1 ) residues respectively. Therefore, the total number of elements in the middle interval ( A ) is ( k times frac{n}{p} approx frac{n}{3} ). But since ( p = 3k - 1 ), ( frac{n}{p} approx frac{n}{3k} ), so the total size is ( k times frac{n}{3k} = frac{n}{3} ). Hmm, that's still just ( frac{n}{3} ).But the problem requires ( |B| > frac{n}{3} ). Maybe I need to adjust the construction slightly. Perhaps instead of taking exactly the middle third, I can take a slightly larger interval. Let me think. If I take ( A = {k, k+1, ldots, 2k} ), then the size of ( A ) is ( k + 1 ), which is slightly larger. Then, the sum of two elements in ( A ) would be between ( 2k ) and ( 4k ). Modulo ( p = 3k + 1 ), the sums would either be in ( {2k, ldots, 3k} ) or wrap around to ( {1, ldots, k} ). But ( A ) is ( {k, ldots, 2k} ), so the sums modulo ( p ) would not fall back into ( A ). Therefore, this ( A ) would still be sum-free.Now, the size of ( A ) is ( k + 1 ), which is ( frac{p + 1}{3} ) since ( p = 3k + 1 ). Therefore, the expected number of elements in ( B ) would be ( frac{n}{p} times (k + 1) = frac{n}{3k + 1} times (k + 1) approx frac{n}{3} times frac{k + 1}{k} approx frac{n}{3} times (1 + frac{1}{k}) ). If ( k ) is large enough, this could be slightly larger than ( frac{n}{3} ). But I'm not sure if this guarantees ( |B| > frac{n}{3} ).Alternatively, maybe I can use a more combinatorial approach. Let's consider the graph where each vertex is an element of ( Y ), and there is an edge between two vertices if their sum is also in ( Y ). Then, finding a subset ( B ) with the desired properties is equivalent to finding an independent set in this graph of size greater than ( frac{n}{3} ). But I'm not sure if this perspective helps directly, as the structure of the graph could be complicated.Wait, another idea: perhaps using the concept of Schur numbers. Schur numbers deal with partitioning sets to avoid monochromatic solutions to ( a + b = c ). The Schur number ( S(k) ) is the largest number such that the set ( {1, 2, ldots, S(k)} ) can be partitioned into ( k ) sum-free sets. But I'm not sure if this directly applies here since we're dealing with an arbitrary set ( Y ), not necessarily the first ( n ) integers.Let me go back to the modular arithmetic approach. Suppose I fix a modulus ( p ) and define ( A ) as a sum-free set modulo ( p ). Then, for each element ( y in Y ), if ( y mod p in A ), include ( y ) in ( B ). The key is to choose ( p ) such that ( A ) is sum-free and the number of elements in ( B ) is more than ( frac{n}{3} ).To ensure that ( A ) is sum-free, I can take ( A ) to be the set of quadratic residues or some other structured set, but that might complicate things. Alternatively, taking an interval as before seems simpler. Let me formalize this:Let ( p = 3k - 1 ) be a prime such that ( p > n ) and ( p ) is coprime to all elements of ( Y ). Define ( A = {k, k+1, ldots, 2k - 1} ). For any ( x, y in A ), ( x + y ) modulo ( p ) is either in ( {2k, ldots, 3k - 2} ) or wraps around to ( {1, ldots, k - 1} ), neither of which is in ( A ). Therefore, ( A ) is sum-free modulo ( p ).Now, for each ( y in Y ), compute ( overline{y} = y mod p ). If ( overline{y} in A ), include ( y ) in ( B ). Since ( p ) is coprime to all ( y ), the mapping ( y mapsto overline{y} ) is a bijection on the residues. Therefore, the number of elements in ( B ) is equal to the number of elements in ( Y ) whose residues fall into ( A ).The size of ( A ) is ( k - 1 ), and since ( p = 3k - 1 ), the density of ( A ) is ( frac{k - 1}{3k - 1} approx frac{1}{3} ). Therefore, the expected number of elements in ( B ) is ( frac{k - 1}{3k - 1} n approx frac{n}{3} ). But we need more than ( frac{n}{3} ).Wait, maybe by choosing ( p ) slightly larger, say ( p = 3k ), but then ( p ) might not be prime. Alternatively, perhaps I can adjust the interval ( A ) to be slightly larger. Let me try ( A = {k, k+1, ldots, 2k} ), which has size ( k + 1 ). Then, ( p = 3k + 1 ), which is prime for some ( k ). The density of ( A ) is ( frac{k + 1}{3k + 1} approx frac{1}{3} + frac{2}{3(3k + 1)} ), which is slightly more than ( frac{1}{3} ). Therefore, the expected size of ( B ) is slightly more than ( frac{n}{3} ).But how do I ensure that at least one such ( B ) exists with size greater than ( frac{n}{3} )? Maybe by considering all possible shifts of ( A ) modulo ( p ) and using the pigeonhole principle. If I consider all possible starting points for ( A ), then the total number of elements covered by all such ( A ) would be ( n times text{size of } A ). Since each element is covered by exactly one ( A ) in each shift, the average size of ( B ) across all shifts is ( frac{n times text{size of } A}{p} ). If this average is greater than ( frac{n}{3} ), then by the pigeonhole principle, at least one shift must have ( |B| > frac{n}{3} ).Let me calculate this. The size of ( A ) is ( k + 1 ), and ( p = 3k + 1 ). Therefore, the average size is ( frac{n(k + 1)}{3k + 1} ). Simplifying, ( frac{k + 1}{3k + 1} = frac{1}{3} + frac{2}{3(3k + 1)} ), which is indeed slightly more than ( frac{1}{3} ). Therefore, the average size is slightly more than ( frac{n}{3} ), so by the pigeonhole principle, there exists at least one shift where ( |B| > frac{n}{3} ).Putting it all together, here's the plan:1. Choose a prime ( p = 3k + 1 ) such that ( p > n ) and ( p ) is coprime to all elements of ( Y ).2. Define ( A = {k, k+1, ldots, 2k} subseteq mathbb{Z}_p ), which is sum-free.3. For each element ( y in Y ), compute ( overline{y} = y mod p ).4. Define ( B = { y in Y mid overline{y} in A } ).5. By the pigeonhole principle, there exists a shift of ( A ) such that ( |B| > frac{n}{3} ).6. Since ( A ) is sum-free modulo ( p ), ( B ) satisfies the condition that ( u + v notin B ) for any ( u, v in B ).I think this approach works. The key steps are choosing the right modulus, defining a sum-free interval, and using the pigeonhole principle to ensure the subset is large enough. I should double-check the modular arithmetic part to make sure that the sums indeed don't fall back into ( A ).Let me verify that. If ( x, y in A ), then ( x + y ) modulo ( p ) is in ( {2k, ldots, 4k} ). Since ( p = 3k + 1 ), the residues wrap around at ( 3k + 1 ). So, ( 2k leq x + y leq 4k ). If ( x + y leq 3k ), then ( overline{x + y} = x + y ), which is in ( {2k, ldots, 3k} ). If ( x + y > 3k ), then ( overline{x + y} = x + y - (3k + 1) ), which is in ( { -1, ldots, k - 1} ), but since we take the smallest positive residue, it becomes ( {1, ldots, k - 1} ). Therefore, in both cases, ( overline{x + y} notin A ), which confirms that ( B ) is sum-free.Alright, I think I've got a solid plan. Now, let me write out the step-by-step explanation to formalize this.

💡Okay, so I need to figure out what numeral is in the 150th decimal place of the fraction 13 over 23. Hmm, that sounds a bit tricky, but I think I can handle it by breaking it down step by step.First, I remember that when you divide a number by another, especially when it doesn't divide evenly, you get a decimal that either terminates or repeats. Since 23 is a prime number and 13 isn't a multiple of 23, I suspect the decimal will repeat. So, I think the first thing I should do is perform the division of 13 by 23 to see what the decimal looks like.Alright, let's try dividing 13 by 23. I know that 23 goes into 13 zero times, so I'll write that down as 0. and then proceed with the division. I'll add a decimal point and some zeros to 13 to continue the division process.23 into 130... Hmm, 23 times 5 is 115, which is less than 130. If I try 6, that's 138, which is too big. So, 5 it is. 5 times 23 is 115. Subtracting that from 130 gives me 15. Bring down the next zero, making it 150.Now, 23 into 150. Let's see, 23 times 6 is 138, which is less than 150. 23 times 7 is 161, which is too big. So, 6 again. 6 times 23 is 138. Subtracting that from 150 gives me 12. Bring down another zero, making it 120.23 into 120. 23 times 5 is 115, which is less than 120. 23 times 6 is 138, too big. So, 5. 5 times 23 is 115. Subtracting from 120 gives me 5. Bring down another zero, making it 50.23 into 50. 23 times 2 is 46, which is less than 50. 23 times 3 is 69, too big. So, 2. 2 times 23 is 46. Subtracting from 50 gives me 4. Bring down another zero, making it 40.23 into 40. 23 times 1 is 23, which is less than 40. 23 times 2 is 46, too big. So, 1. 1 times 23 is 23. Subtracting from 40 gives me 17. Bring down another zero, making it 170.23 into 170. 23 times 7 is 161, which is less than 170. 23 times 8 is 184, too big. So, 7. 7 times 23 is 161. Subtracting from 170 gives me 9. Bring down another zero, making it 90.23 into 90. 23 times 3 is 69, which is less than 90. 23 times 4 is 92, too big. So, 3. 3 times 23 is 69. Subtracting from 90 gives me 21. Bring down another zero, making it 210.23 into 210. 23 times 9 is 207, which is less than 210. 23 times 10 is 230, too big. So, 9. 9 times 23 is 207. Subtracting from 210 gives me 3. Bring down another zero, making it 30.23 into 30. 23 times 1 is 23, which is less than 30. 23 times 2 is 46, too big. So, 1. 1 times 23 is 23. Subtracting from 30 gives me 7. Bring down another zero, making it 70.23 into 70. 23 times 3 is 69, which is less than 70. 23 times 4 is 92, too big. So, 3. 3 times 23 is 69. Subtracting from 70 gives me 1. Bring down another zero, making it 10.23 into 10. 23 times 0 is 0, which is less than 10. So, 0. Bring down another zero, making it 100.23 into 100. 23 times 4 is 92, which is less than 100. 23 times 5 is 115, too big. So, 4. 4 times 23 is 92. Subtracting from 100 gives me 8. Bring down another zero, making it 80.23 into 80. 23 times 3 is 69, which is less than 80. 23 times 4 is 92, too big. So, 3. 3 times 23 is 69. Subtracting from 80 gives me 11. Bring down another zero, making it 110.23 into 110. 23 times 4 is 92, which is less than 110. 23 times 5 is 115, too big. So, 4. 4 times 23 is 92. Subtracting from 110 gives me 18. Bring down another zero, making it 180.23 into 180. 23 times 7 is 161, which is less than 180. 23 times 8 is 184, too big. So, 7. 7 times 23 is 161. Subtracting from 180 gives me 19. Bring down another zero, making it 190.23 into 190. 23 times 8 is 184, which is less than 190. 23 times 9 is 207, too big. So, 8. 8 times 23 is 184. Subtracting from 190 gives me 6. Bring down another zero, making it 60.23 into 60. 23 times 2 is 46, which is less than 60. 23 times 3 is 69, too big. So, 2. 2 times 23 is 46. Subtracting from 60 gives me 14. Bring down another zero, making it 140.23 into 140. 23 times 6 is 138, which is less than 140. 23 times 7 is 161, too big. So, 6. 6 times 23 is 138. Subtracting from 140 gives me 2. Bring down another zero, making it 20.23 into 20. 23 times 0 is 0, which is less than 20. So, 0. Bring down another zero, making it 200.23 into 200. 23 times 8 is 184, which is less than 200. 23 times 9 is 207, too big. So, 8. 8 times 23 is 184. Subtracting from 200 gives me 16. Bring down another zero, making it 160.23 into 160. 23 times 6 is 138, which is less than 160. 23 times 7 is 161, too big. So, 6. 6 times 23 is 138. Subtracting from 160 gives me 22. Bring down another zero, making it 220.23 into 220. 23 times 9 is 207, which is less than 220. 23 times 10 is 230, too big. So, 9. 9 times 23 is 207. Subtracting from 220 gives me 13. Bring down another zero, making it 130.Wait a minute, I've seen this before. Earlier, when I started the division, I had 130 after the decimal point. So, it seems like the decimals are starting to repeat now. So, the decimal expansion of 13/23 is repeating every certain number of digits. Let me check how many digits are in this repeating cycle.Looking back, the decimal started repeating at 130, which was after the first few digits. Let me count the number of digits in the repeating sequence. From the division steps, I can see that the repeating cycle is 565217391304347826086956, which is 23 digits long. So, the decimal repeats every 23 digits.Now, I need to find the 150th decimal place. Since the decimal repeats every 23 digits, I can use modular arithmetic to find out which digit in the cycle corresponds to the 150th place.To do this, I'll divide 150 by 23 and find the remainder. The remainder will tell me the position within the repeating cycle.Let's calculate 150 divided by 23. 23 times 6 is 138, and 150 minus 138 is 12. So, the remainder is 12. That means the 150th decimal place corresponds to the 12th digit in the repeating sequence.Now, I need to look at the repeating sequence: 565217391304347826086956. Let's count to the 12th digit.1: 52: 63: 54: 25: 16: 77: 38: 99: 110: 311: 012: 4Wait, the 12th digit is 4? But earlier, I thought it was 3. Did I make a mistake somewhere?Let me double-check the counting. Starting from the first digit after the decimal:1: 52: 63: 54: 25: 16: 77: 38: 99: 110: 311: 012: 4Yes, it seems like the 12th digit is indeed 4. But earlier, I thought it was 3. Maybe I miscounted earlier. Let me check the repeating sequence again.Wait, the repeating sequence is 565217391304347826086956. Let me write it out with positions:1:52:63:54:25:16:77:38:99:110:311:012:413:314:415:716:817:218:619:020:821:622:923:524:6 (and then it repeats)Wait, so the 12th digit is 4, not 3. So, earlier conclusion was wrong. It should be 4.But in the initial problem, the user mentioned the repeating block as 565217391304347826086956, which is 23 digits. So, the 12th digit is indeed 4.But in the initial thought process, I thought it was 3. So, there must be a mistake in the initial thought process.Wait, let me recount the positions carefully:1:52:63:54:25:16:77:38:99:110:311:012:4Yes, 12th is 4. So, the correct answer should be 4, not 3.But the initial answer was 3. So, I must have made a mistake in the initial thought process.Wait, maybe I miscounted the digits in the repeating sequence. Let me write out the repeating sequence again:565217391304347826086956Let me count the digits:1:52:63:54:25:16:77:38:99:110:311:012:413:314:415:716:817:218:619:020:821:622:923:524:6 (and then it repeats)Yes, so the 12th digit is 4. Therefore, the correct numeral in the 150th decimal place is 4.So, I think the initial answer was incorrect because of a miscount. The correct answer should be 4.

💡Okay, so I have this problem about a regular decagon with area n, and I need to find the ratio m/n where m is the area of quadrilateral ACEG. Hmm, let me think about how to approach this.First, I remember that a regular decagon has 10 sides, all equal in length, and all internal angles equal. The area of a regular polygon can be calculated using the formula involving the number of sides and the side length, but I'm not sure if I need that here.The quadrilateral ACEG is formed by connecting every other vertex of the decagon. So starting from A, then skipping B to connect to C, then skipping D to connect to E, and so on until G. I think this should form a regular quadrilateral, maybe a square or a rectangle, but I need to confirm that.Wait, in a regular decagon, connecting every other vertex doesn't necessarily form a square. Let me visualize a regular decagon. Each internal angle is 144 degrees, right? So, if I connect A to C, that's two sides apart. The angle between AC and the side AB would be half of the internal angle, maybe? Hmm, not sure.Maybe I should consider the central angles. In a regular decagon, each central angle is 36 degrees because 360/10 = 36. So, if I connect every other vertex, the central angle between A and C would be 72 degrees, right? Because from A to B is 36, and B to C is another 36, so A to C is 72 degrees.If I connect A to C, C to E, E to G, and G back to A, each of these connections spans 72 degrees at the center. So, the quadrilateral ACEG is actually a regular quadrilateral inscribed in the same circle as the decagon. Wait, but a regular quadrilateral with each central angle 72 degrees? That doesn't sound right because a regular quadrilateral should have central angles of 90 degrees each.Hmm, maybe it's not a regular quadrilateral. Maybe it's a different type of quadrilateral. Let me think. If each side of the quadrilateral spans 72 degrees, then the shape might be a rectangle or something else.Alternatively, maybe I can use coordinates to figure this out. Let me place the decagon on a coordinate system with the center at the origin. Let's assume the decagon is unit for simplicity, so each vertex is at a distance of 1 from the center.The coordinates of the vertices can be given by (cos θ, sin θ), where θ is the angle from the positive x-axis. For a decagon, the angles would be 0°, 36°, 72°, 108°, 144°, 180°, 216°, 252°, 288°, and 324°.So, vertex A is at (1, 0). Vertex B is at (cos 36°, sin 36°), vertex C is at (cos 72°, sin 72°), vertex D is at (cos 108°, sin 108°), and so on.So, quadrilateral ACEG would have vertices at A(1,0), C(cos 72°, sin 72°), E(cos 144°, sin 144°), G(cos 216°, sin 216°), and back to A.Wait, that's five points, but a quadrilateral should have four vertices. Oh, I see, ACEG is four points: A, C, E, G.So, A(1,0), C(cos 72°, sin 72°), E(cos 144°, sin 144°), G(cos 216°, sin 216°).Now, to find the area of quadrilateral ACEG, I can use the shoelace formula. Let me recall how that works. For a polygon with vertices (x1,y1), (x2,y2), ..., (xn,yn), the area is given by:1/2 |sum from 1 to n of (xi*yi+1 - xi+1*yi)|, where xn+1 = x1 and yn+1 = y1.So, let me compute the coordinates:A: (1, 0)C: (cos 72°, sin 72°)E: (cos 144°, sin 144°)G: (cos 216°, sin 216°)Let me compute these values numerically to make it easier.First, cos 72° ≈ 0.3090, sin 72° ≈ 0.9511cos 144° ≈ -0.8090, sin 144° ≈ 0.5878cos 216° ≈ -0.8090, sin 216° ≈ -0.5878So, the coordinates are approximately:A: (1, 0)C: (0.3090, 0.9511)E: (-0.8090, 0.5878)G: (-0.8090, -0.5878)Now, applying the shoelace formula:List the coordinates in order and repeat the first at the end:(1, 0), (0.3090, 0.9511), (-0.8090, 0.5878), (-0.8090, -0.5878), (1, 0)Compute the sum of xi*yi+1:1*0.9511 + 0.3090*0.5878 + (-0.8090)*(-0.5878) + (-0.8090)*0= 0.9511 + 0.1816 + 0.4755 + 0= 0.9511 + 0.1816 = 1.1327; 1.1327 + 0.4755 = 1.6082Now compute the sum of yi*xi+1:0*0.3090 + 0.9511*(-0.8090) + 0.5878*(-0.8090) + (-0.5878)*1= 0 - 0.7705 - 0.4755 - 0.5878= -0.7705 - 0.4755 = -1.246; -1.246 - 0.5878 = -1.8338Now subtract the two sums:1.6082 - (-1.8338) = 1.6082 + 1.8338 = 3.442Take the absolute value and multiply by 1/2:Area = (1/2)*|3.442| ≈ 1.721So, the area of quadrilateral ACEG is approximately 1.721 when the decagon is unit.Now, I need to find the area of the entire decagon. The formula for the area of a regular decagon with side length a is:(5/2) * a^2 * (1 + sqrt(5)) / 2Wait, actually, I think the area can also be calculated using the formula:(10 * a^2) / (4 * tan(π/10))But since I assumed the decagon is unit, meaning the radius is 1, the side length a can be calculated as 2 * sin(π/10) ≈ 2 * 0.3090 ≈ 0.6180.But maybe it's easier to calculate the area using the formula for a regular polygon with radius R:Area = (1/2) * n * R^2 * sin(2π/n)For a decagon, n = 10, R = 1:Area = (1/2) * 10 * 1^2 * sin(2π/10) = 5 * sin(π/5) ≈ 5 * 0.5878 ≈ 2.9389Wait, but earlier I calculated the area of quadrilateral ACEG as approximately 1.721, and the area of the decagon as approximately 2.9389. So, the ratio m/n would be 1.721 / 2.9389 ≈ 0.586, which is roughly 1/1.705, so about 0.586, which is approximately 1/1.705 ≈ 0.586, which is roughly 1/1.705, so about 0.586.But none of the answer choices are near 0.586. The options are 1/3, 1/4, 1/5, 1/6, 1/7. So, clearly, I must have made a mistake somewhere.Wait, maybe I messed up the shoelace formula. Let me double-check the calculations.Coordinates:A: (1, 0)C: (0.3090, 0.9511)E: (-0.8090, 0.5878)G: (-0.8090, -0.5878)Compute xi*yi+1:1*0.9511 = 0.95110.3090*0.5878 ≈ 0.1816-0.8090*(-0.5878) ≈ 0.4755-0.8090*0 = 0Sum: 0.9511 + 0.1816 + 0.4755 ≈ 1.6082Compute yi*xi+1:0*0.3090 = 00.9511*(-0.8090) ≈ -0.77050.5878*(-0.8090) ≈ -0.4755-0.5878*1 ≈ -0.5878Sum: 0 - 0.7705 - 0.4755 - 0.5878 ≈ -1.8338Difference: 1.6082 - (-1.8338) = 3.442Area: 1/2 * 3.442 ≈ 1.721Hmm, that seems correct. So, the area of ACEG is about 1.721, and the area of the decagon is about 2.9389. So, the ratio is approximately 1.721 / 2.9389 ≈ 0.586, which is roughly 1/1.705.But the answer choices are 1/3, 1/4, 1/5, 1/6, 1/7. So, maybe my assumption that the decagon is unit is causing confusion. Maybe I should consider the decagon with side length a, and then express both areas in terms of a.Alternatively, perhaps I should use vectors or complex numbers to find the area ratio more accurately.Wait, another approach: in a regular decagon, the area can be divided into 10 congruent isosceles triangles, each with a central angle of 36 degrees. The area of each triangle is (1/2)*R^2*sin(36°), so total area is 10*(1/2)*R^2*sin(36°) = 5*R^2*sin(36°).Similarly, quadrilateral ACEG can be divided into triangles or other shapes whose areas can be related to the decagon's area.Alternatively, maybe I can use symmetry. The decagon has rotational symmetry of order 10, and the quadrilateral ACEG is symmetric as well. Maybe the ratio can be found by considering how many times ACEG fits into the decagon.Wait, another thought: in a regular decagon, the diagonals that skip two vertices (like AC, CE, etc.) are longer than the sides. The ratio of the diagonal to the side in a regular decagon is the golden ratio, φ = (1 + sqrt(5))/2 ≈ 1.618.So, if the side length is a, then the diagonal AC is φ*a.Now, quadrilateral ACEG is a rectangle? Or maybe a square? Wait, in a regular decagon, the diagonals that skip two vertices are equal, but the angles between them might not be 90 degrees. So, maybe it's a rectangle, but not necessarily a square.Wait, actually, in a regular decagon, the angle between two adjacent diagonals like AC and CE is 72 degrees, because each central angle is 36 degrees, so the angle between two diagonals would be 2*36 = 72 degrees.So, quadrilateral ACEG is a kite? Or maybe a rhombus? Wait, no, because the sides are equal but the angles are not necessarily 90 degrees.Alternatively, maybe it's a cyclic quadrilateral, since all vertices lie on the circumcircle of the decagon.Wait, if it's cyclic, then maybe I can use Brahmagupta's formula for the area of a cyclic quadrilateral. The formula is sqrt[(s-a)(s-b)(s-c)(s-d)], where s is the semi-perimeter, and a, b, c, d are the sides.But I need to know the lengths of the sides of quadrilateral ACEG. Since ACEG is formed by diagonals of the decagon, each side of ACEG is equal to the length of the diagonal AC, which is φ*a.So, all sides of ACEG are equal, making it a rhombus. But since it's cyclic, a rhombus that is cyclic must be a square. Wait, is that true? A rhombus is cyclic only if it's a square because all sides are equal and all angles are equal (90 degrees). So, if ACEG is a cyclic rhombus, then it must be a square.Wait, but in a regular decagon, the central angles between consecutive vertices of ACEG are 72 degrees, so the angle subtended by each side of ACEG at the center is 72 degrees. Therefore, the sides of ACEG are chords of the circle subtending 72 degrees, so their length is 2*R*sin(36°), where R is the radius of the circumscribed circle.Wait, but earlier I thought the side length of the decagon is a = 2*R*sin(18°), because each side subtends 36 degrees at the center, so half of that is 18 degrees.So, a = 2*R*sin(18°), and the diagonal AC = 2*R*sin(36°). Therefore, AC = a*(sin(36°)/sin(18°)).Using the identity sin(36°) = 2*sin(18°)*cos(18°), so AC = a*(2*sin(18°)*cos(18°)/sin(18°)) = 2*a*cos(18°).Since cos(18°) ≈ 0.9511, so AC ≈ 2*a*0.9511 ≈ 1.902*a.But the golden ratio φ ≈ 1.618, so 2*cos(36°) ≈ 1.618, which is φ. Wait, cos(36°) ≈ 0.8090, so 2*cos(36°) ≈ 1.618, which is φ.Wait, so AC = 2*R*sin(36°) = 2*R*sin(36°). But a = 2*R*sin(18°), so AC = a*(sin(36°)/sin(18°)).Using the double-angle identity, sin(36°) = 2*sin(18°)*cos(18°), so AC = a*(2*sin(18°)*cos(18°)/sin(18°)) = 2*a*cos(18°).But cos(18°) ≈ 0.9511, so AC ≈ 1.902*a, which is more than φ*a. Hmm, maybe I confused the angle.Wait, actually, in a regular decagon, the length of the diagonal that skips one vertex (like AC) is φ*a, where φ is the golden ratio. So, AC = φ*a.So, if a is the side length, then AC = φ*a.Therefore, quadrilateral ACEG has sides of length φ*a, and since it's a square (as it's a cyclic rhombus), its area is (φ*a)^2.But wait, earlier I thought it's a square, but if the central angles are 72 degrees, then the sides of ACEG are chords of 72 degrees, so their length is 2*R*sin(36°). But a = 2*R*sin(18°), so AC = 2*R*sin(36°) = 2*R*2*sin(18°)*cos(18°) = 4*R*sin(18°)*cos(18°) = 2*a*cos(18°).But since a = 2*R*sin(18°), then AC = 2*(2*R*sin(18°))*cos(18°) = 4*R*sin(18°)*cos(18°).Wait, I'm getting confused. Maybe I should express everything in terms of R.Let me denote R as the radius of the circumscribed circle.Then, the side length a = 2*R*sin(π/10) = 2*R*sin(18°).The diagonal AC = 2*R*sin(2π/10) = 2*R*sin(36°).So, AC = 2*R*sin(36°).Therefore, the side length of quadrilateral ACEG is AC = 2*R*sin(36°).Since ACEG is a square, its area is (AC)^2 = (2*R*sin(36°))^2 = 4*R^2*sin^2(36°).Now, the area of the decagon is (10/2)*R^2*sin(2π/10) = 5*R^2*sin(36°).So, the ratio m/n = (4*R^2*sin^2(36°)) / (5*R^2*sin(36°)) = (4*sin(36°))/5.Now, sin(36°) ≈ 0.5878, so m/n ≈ (4*0.5878)/5 ≈ 2.351/5 ≈ 0.4702.Hmm, that's approximately 0.47, which is roughly 1/2.127, so about 0.47, which is still not matching the answer choices.Wait, but maybe I made a mistake in assuming that ACEG is a square. Earlier, I thought it's a square because it's a cyclic rhombus, but maybe that's not correct.Wait, in a regular decagon, the central angles between consecutive vertices of ACEG are 72 degrees, so the angle between two adjacent vertices as seen from the center is 72 degrees. Therefore, the sides of ACEG are chords subtending 72 degrees, so their length is 2*R*sin(36°).But for a square inscribed in a circle, each side subtends 90 degrees at the center, so the chord length would be 2*R*sin(45°). But in our case, it's 72 degrees, so it's not a square.Therefore, ACEG is not a square, but a cyclic quadrilateral with all sides equal (a rhombus) but angles not 90 degrees. So, it's a rhombus, but not a square.Therefore, the area of a rhombus is base*height, but since it's cyclic, the area can also be calculated as (d1*d2)/2, where d1 and d2 are the diagonals.Wait, but I don't know the lengths of the diagonals of ACEG. Alternatively, since it's cyclic, I can use Brahmagupta's formula.Brahmagupta's formula states that the area of a cyclic quadrilateral is sqrt[(s-a)(s-b)(s-c)(s-d)], where s is the semi-perimeter, and a, b, c, d are the sides.But in our case, all sides are equal, so it's a rhombus. Therefore, the area is also equal to a^2*sin(theta), where theta is one of the internal angles.But I need to find theta. Since the quadrilateral is cyclic, the sum of opposite angles is 180 degrees. Also, the central angles corresponding to each side are 72 degrees, so the inscribed angles would be half of that, which is 36 degrees.Wait, no. The central angle is 72 degrees, so the inscribed angle subtended by the same arc would be half, which is 36 degrees. But in a cyclic quadrilateral, the internal angle is equal to half the sum of the central angles of the opposite arcs.Wait, maybe I should think differently. The internal angle at each vertex of the quadrilateral ACEG can be found by considering the arcs between the vertices.From A to C is 72 degrees, C to E is another 72 degrees, E to G is another 72 degrees, and G back to A is another 72 degrees. Wait, but that's 4*72 = 288 degrees, which is less than 360, so that can't be right.Wait, no, the central angles between consecutive vertices of the decagon are 36 degrees each. So, from A to B is 36, B to C is another 36, so A to C is 72 degrees. Similarly, C to E is another 72 degrees, E to G is another 72 degrees, and G back to A is another 72 degrees, totaling 4*72 = 288 degrees, which is less than 360, so the remaining 72 degrees is the arc from A back to G, but that's not part of the quadrilateral.Wait, maybe I'm overcomplicating. Let me try to find the internal angles of quadrilateral ACEG.In a cyclic quadrilateral, the internal angle is equal to half the measure of the opposite arc. So, for example, the internal angle at A is half the measure of the arc opposite to A, which is the arc from C to E to G. Wait, that's 72 + 72 = 144 degrees. So, the internal angle at A is half of 144, which is 72 degrees.Similarly, the internal angle at C would be half the arc from E to G to A, which is also 144 degrees, so 72 degrees. Same for E and G. So, all internal angles of quadrilateral ACEG are 72 degrees, making it a rectangle? Wait, no, because in a rectangle, all angles are 90 degrees. So, if all angles are 72 degrees, it's not a rectangle, but a different type of cyclic quadrilateral.Wait, but if all internal angles are 72 degrees, then it's a rectangle only if all angles are 90 degrees. So, it's not a rectangle, but a cyclic quadrilateral with all angles equal to 72 degrees.Therefore, the area can be calculated as a^2*sin(theta), where a is the side length, and theta is one of the internal angles.But since it's cyclic, another formula is to use the product of the diagonals and the sine of the angle between them divided by 2. But I don't know the lengths of the diagonals.Alternatively, since all sides are equal and it's cyclic, it's a special case called a "cyclic rhombus," which is a square only if all angles are 90 degrees. Since our angles are 72 degrees, it's not a square, but a different cyclic rhombus.Wait, maybe I can use the formula for the area of a cyclic quadrilateral with equal sides. Since all sides are equal, it's a rhombus, and the area is a^2*sin(theta), where theta is one of the internal angles.But I need to find theta. Earlier, I thought theta is 72 degrees, but let me confirm.In a cyclic quadrilateral, the internal angle is equal to half the measure of the opposite arc. So, for quadrilateral ACEG, the arc opposite to angle A is the arc from C to E to G, which is 72 + 72 = 144 degrees. Therefore, angle A is half of 144, which is 72 degrees. Similarly, all other angles are 72 degrees.Therefore, the area of ACEG is a^2*sin(72°), where a is the side length of the quadrilateral, which is AC = 2*R*sin(36°).But the area of the decagon is 5*R^2*sin(36°).So, m = (2*R*sin(36°))^2 * sin(72°) = 4*R^2*sin^2(36°)*sin(72°).And n = 5*R^2*sin(36°).Therefore, m/n = (4*R^2*sin^2(36°)*sin(72°)) / (5*R^2*sin(36°)) = (4*sin(36°)*sin(72°))/5.Now, let's compute sin(36°)*sin(72°). Using the identity sin(A)*sin(B) = [cos(A-B) - cos(A+B)]/2.So, sin(36°)*sin(72°) = [cos(36°) - cos(108°)]/2.We know that cos(108°) = cos(180°-72°) = -cos(72°).Therefore, sin(36°)*sin(72°) = [cos(36°) - (-cos(72°))]/2 = [cos(36°) + cos(72°)]/2.Now, cos(36°) ≈ 0.8090, cos(72°) ≈ 0.3090.So, [0.8090 + 0.3090]/2 = 1.118/2 ≈ 0.559.Therefore, m/n ≈ (4*0.559)/5 ≈ 2.236/5 ≈ 0.4472.Hmm, that's approximately 0.4472, which is roughly 1/2.236, which is approximately 0.4472, which is close to sqrt(2)/2 ≈ 0.7071, but not exactly.Wait, but 0.4472 is approximately 1/sqrt(5) ≈ 0.4472, so m/n ≈ 1/sqrt(5). But sqrt(5) ≈ 2.236, so 1/sqrt(5) ≈ 0.4472.But the answer choices are 1/3, 1/4, 1/5, 1/6, 1/7. None of these is 1/sqrt(5). So, perhaps I made a mistake in the approach.Wait, maybe I should consider that the area of the decagon can be divided into 10 congruent isosceles triangles, each with a central angle of 36 degrees. The area of each triangle is (1/2)*R^2*sin(36°), so total area n = 10*(1/2)*R^2*sin(36°) = 5*R^2*sin(36°).Quadrilateral ACEG can be divided into 4 congruent triangles, each with a central angle of 72 degrees. Wait, no, because ACEG has 4 vertices, each separated by 72 degrees. So, the area of ACEG would be 4*(1/2)*R^2*sin(72°) = 2*R^2*sin(72°).Therefore, m = 2*R^2*sin(72°).So, m/n = (2*R^2*sin(72°)) / (5*R^2*sin(36°)) = (2*sin(72°))/(5*sin(36°)).Now, sin(72°) = 2*sin(36°)*cos(36°), using the double-angle identity.So, m/n = (2*(2*sin(36°)*cos(36°)))/(5*sin(36°)) = (4*sin(36°)*cos(36°))/(5*sin(36°)) = (4*cos(36°))/5.Now, cos(36°) ≈ 0.8090, so m/n ≈ (4*0.8090)/5 ≈ 3.236/5 ≈ 0.6472.Wait, that's approximately 0.6472, which is roughly 2/3, but the answer choices don't have that either.I'm getting conflicting results, so maybe I need to approach this differently.Another approach: use complex numbers to represent the vertices and calculate the area.Let me place the decagon on the unit circle, so R = 1. The vertices are at angles 0°, 36°, 72°, ..., 324°.So, the coordinates are:A: (1, 0)C: (cos 72°, sin 72°)E: (cos 144°, sin 144°)G: (cos 216°, sin 216°)Using these coordinates, I can apply the shoelace formula again, but this time with exact values.But exact trigonometric values might be messy, but perhaps I can express them in terms of radicals.Alternatively, I can use the formula for the area of a polygon given its vertices in the complex plane.The area can be calculated using the formula:Area = (1/2)*|sum from k=1 to n of (x_k*y_{k+1} - x_{k+1}*y_k)|So, let's compute this exactly.First, let's denote the points:A: (1, 0)C: (cos 72°, sin 72°)E: (cos 144°, sin 144°)G: (cos 216°, sin 216°)Let me compute the terms:Term1: x_A*y_C - x_C*y_A = 1*sin72 - cos72*0 = sin72Term2: x_C*y_E - x_E*y_C = cos72*sin144 - cos144*sin72Term3: x_E*y_G - x_G*y_E = cos144*sin216 - cos216*sin144Term4: x_G*y_A - x_A*y_G = cos216*0 - 1*sin216 = -sin216Now, let's compute each term:Term1: sin72 ≈ 0.9511Term2: cos72*sin144 - cos144*sin72Using the identity sin(A - B) = sinA*cosB - cosA*sinB, but here it's cos72*sin144 - cos144*sin72 = sin(144 - 72) = sin72 ≈ 0.9511Wait, that's interesting. So, Term2 = sin72 ≈ 0.9511Term3: cos144*sin216 - cos216*sin144Similarly, this is sin(216 - 144) = sin72 ≈ 0.9511Term4: -sin216 = -sin(180+36) = -(-sin36) = sin36 ≈ 0.5878So, summing all terms:Term1 + Term2 + Term3 + Term4 ≈ 0.9511 + 0.9511 + 0.9511 + 0.5878 ≈ 3.4411Therefore, the area is (1/2)*|3.4411| ≈ 1.7205Now, the area of the decagon with R=1 is 5*sin36 ≈ 5*0.5878 ≈ 2.939So, m/n ≈ 1.7205 / 2.939 ≈ 0.585, which is approximately 1/1.707, which is roughly 0.585.But the answer choices are 1/3, 1/4, 1/5, 1/6, 1/7. So, none of these match. Therefore, I must have made a mistake in my approach.Wait, perhaps the quadrilateral ACEG is not a four-sided figure but a five-sided figure? Wait, no, the problem says quadrilateral ACEG, which has four vertices: A, C, E, G.Wait, maybe I should consider that in a regular decagon, the area of quadrilateral ACEG is exactly 1/4 of the total area. But how?Wait, another thought: in a regular decagon, the number of triangles formed by connecting all diagonals is complex, but maybe the quadrilateral ACEG divides the decagon into regions whose areas can be compared.Alternatively, perhaps using symmetry, the decagon can be divided into 10 congruent triangles, and quadrilateral ACEG covers a certain number of these triangles.But I'm not sure. Maybe I should look for a pattern or a known ratio.Wait, I recall that in a regular decagon, the ratio of the area of the square formed by connecting every other vertex to the total area is 1/5. But I'm not sure.Alternatively, maybe it's 1/4. Let me think.Wait, if I consider that the decagon can be divided into 10 triangles, and quadrilateral ACEG covers 2.5 of them, but that doesn't make sense because you can't have half a triangle.Alternatively, maybe it's 2 triangles, so 2/10 = 1/5. But earlier calculations suggested a higher ratio.Wait, perhaps I should consider that the area of ACEG is 1/5 of the decagon's area. But I'm not sure.Wait, another approach: use the fact that in a regular decagon, the area can be expressed in terms of the golden ratio. Maybe the ratio m/n is related to φ.But I'm not sure. Alternatively, maybe the ratio is 1/5 because there are 5 such quadrilaterals in the decagon, but that's just a guess.Wait, I think I need to find a better approach. Let me try to calculate the area of ACEG using vectors.Let me place the decagon on the unit circle, so each vertex is a complex number of magnitude 1.The coordinates are:A: 1 (angle 0°)C: e^(i*72°)E: e^(i*144°)G: e^(i*216°)The area of quadrilateral ACEG can be calculated using the shoelace formula in the complex plane.The formula for the area is (1/2)*|Im(sum_{k=1}^{n} z_k overline{z_{k+1}})|, where z_k are the vertices in order.So, let's compute this.First, write the vertices as complex numbers:A: 1C: e^(i*72°) = cos72 + i sin72E: e^(i*144°) = cos144 + i sin144G: e^(i*216°) = cos216 + i sin216Now, compute the sum:Sum = A*C̄ + C*Ē + E*Ḡ + G*ĀWhere C̄ is the complex conjugate of C, etc.Compute each term:A*C̄ = 1*(cos72 - i sin72) = cos72 - i sin72C*Ē = (cos72 + i sin72)*(cos144 - i sin144) = cos72*cos144 + sin72*sin144 + i*(sin72*cos144 - cos72*sin144)Using the identity cos(A - B) = cosA cosB + sinA sinB and sin(A - B) = sinA cosB - cosA sinB.So, C*Ē = cos(144 - 72) + i sin(144 - 72) = cos72 + i sin72Similarly, E*Ḡ = (cos144 + i sin144)*(cos216 - i sin216) = cos144*cos216 + sin144*sin216 + i*(sin144*cos216 - cos144*sin216)Again, using the identity:E*Ḡ = cos(216 - 144) + i sin(216 - 144) = cos72 + i sin72G*Ā = (cos216 + i sin216)*1 = cos216 + i sin216Now, sum all terms:Sum = (cos72 - i sin72) + (cos72 + i sin72) + (cos72 + i sin72) + (cos216 + i sin216)Simplify:Sum = cos72 - i sin72 + cos72 + i sin72 + cos72 + i sin72 + cos216 + i sin216Combine like terms:Real parts: cos72 + cos72 + cos72 + cos216 = 3*cos72 + cos216Imaginary parts: (-i sin72) + (i sin72) + (i sin72) + (i sin216) = 0 + i sin72 + i sin216Now, compute the real and imaginary parts:Real part: 3*cos72 + cos216cos72 ≈ 0.3090, so 3*0.3090 ≈ 0.9270cos216 = cos(180+36) = -cos36 ≈ -0.8090So, real part ≈ 0.9270 - 0.8090 ≈ 0.1180Imaginary part: sin72 + sin216sin72 ≈ 0.9511sin216 = sin(180+36) = -sin36 ≈ -0.5878So, imaginary part ≈ 0.9511 - 0.5878 ≈ 0.3633Therefore, Sum ≈ 0.1180 + i*0.3633Now, the area is (1/2)*|Im(Sum)| = (1/2)*|0.3633| ≈ 0.18165Wait, that can't be right because earlier calculations gave a much larger area. I must have made a mistake in the calculation.Wait, no, actually, the shoelace formula in the complex plane requires the vertices to be ordered either clockwise or counterclockwise without crossing. Maybe I messed up the order or the conjugation.Wait, let me double-check the formula. The area is (1/2)*|Im(sum_{k=1}^{n} z_k overline{z_{k+1}})|.So, I think I did it correctly, but the result seems too small. Maybe I made a mistake in the calculation.Wait, let's compute the imaginary part again:Imaginary part: (-i sin72) + (i sin72) + (i sin72) + (i sin216) = 0 + i sin72 + i sin216So, it's i*(sin72 + sin216) = i*(sin72 - sin36) ≈ i*(0.9511 - 0.5878) ≈ i*0.3633So, Im(Sum) ≈ 0.3633Therefore, area ≈ (1/2)*0.3633 ≈ 0.18165But earlier, using coordinates, I got an area of approximately 1.721 when R=1. So, clearly, there's a mistake here.Wait, maybe I messed up the order of the vertices. In the complex plane, the order should be either clockwise or counterclockwise without crossing. Let me check the order:A(1,0), C(cos72, sin72), E(cos144, sin144), G(cos216, sin216)Plotting these points, they are in counterclockwise order, so the shoelace formula should work.Wait, but when I applied the shoelace formula earlier with coordinates, I got a much larger area. So, why is the complex plane method giving a smaller area?Wait, perhaps I made a mistake in the complex plane calculation. Let me try again.Compute Sum = A*C̄ + C*Ē + E*Ḡ + G*ĀA = 1, C = e^(i72°), E = e^(i144°), G = e^(i216°)C̄ = e^(-i72°), Ē = e^(-i144°), Ḡ = e^(-i216°), Ā = 1So,A*C̄ = 1*e^(-i72°) = e^(-i72°)C*Ē = e^(i72°)*e^(-i144°) = e^(-i72°)E*Ḡ = e^(i144°)*e^(-i216°) = e^(-i72°)G*Ā = e^(i216°)*1 = e^(i216°)So, Sum = e^(-i72°) + e^(-i72°) + e^(-i72°) + e^(i216°)= 3*e^(-i72°) + e^(i216°)Now, e^(i216°) = e^(i(180°+36°)) = -e^(i36°)Similarly, e^(-i72°) = cos72 - i sin72So, Sum = 3*(cos72 - i sin72) - (cos36 + i sin36)= 3cos72 - 3i sin72 - cos36 - i sin36Now, compute the imaginary part:Im(Sum) = -3 sin72 - sin36= -3*0.9511 - 0.5878 ≈ -2.8533 - 0.5878 ≈ -3.4411Therefore, the area is (1/2)*|Im(Sum)| ≈ (1/2)*3.4411 ≈ 1.7205Ah, that matches the earlier shoelace result. So, the area of ACEG is approximately 1.7205 when R=1.The area of the decagon is 5*sin36 ≈ 5*0.5878 ≈ 2.939Therefore, m/n ≈ 1.7205 / 2.939 ≈ 0.585, which is approximately 1/1.707.But the answer choices are 1/3, 1/4, 1/5, 1/6, 1/7. So, none of these match. Therefore, I must have made a mistake in my approach.Wait, perhaps the decagon is not unit, but has side length a, and I need to express the areas in terms of a.Let me denote a as the side length of the decagon. Then, the radius R of the circumscribed circle is R = a / (2*sin(π/10)) = a / (2*sin18°) ≈ a / (2*0.3090) ≈ a / 0.618 ≈ 1.618*a.So, R ≈ 1.618*a.Now, the area of the decagon is 5*R^2*sin36 ≈ 5*(1.618*a)^2*0.5878 ≈ 5*(2.618*a^2)*0.5878 ≈ 5*1.538*a^2 ≈ 7.69*a^2.Wait, but the area of a regular decagon is also given by (5/2)*a^2*(1 + sqrt(5)) ≈ (5/2)*a^2*(3.236) ≈ 5*1.618*a^2 ≈ 8.09*a^2, which is close to the previous calculation.Now, the area of quadrilateral ACEG is 1.7205 when R=1. But since R ≈ 1.618*a, then the area scales with R^2.So, m = 1.7205*(R)^2 ≈ 1.7205*(1.618*a)^2 ≈ 1.7205*2.618*a^2 ≈ 4.508*a^2.Therefore, m/n ≈ 4.508*a^2 / 8.09*a^2 ≈ 0.557, which is approximately 1/1.795, still not matching the answer choices.Wait, maybe I should consider that the area of ACEG is 1/5 of the decagon's area because there are 5 such quadrilaterals in the decagon. But I'm not sure.Alternatively, perhaps the ratio is 1/5 because the decagon can be divided into 5 such quadrilaterals, but I'm not certain.Wait, another thought: in a regular decagon, the number of triangles formed by connecting all diagonals is complex, but maybe the quadrilateral ACEG covers exactly 2 of the 10 congruent triangles, making the ratio 2/10 = 1/5.But earlier calculations suggested a higher ratio, so I'm confused.Wait, perhaps I should look for a pattern or a known ratio. I recall that in a regular decagon, the ratio of the area of the square formed by connecting every other vertex to the total area is 1/5. But I'm not sure.Alternatively, maybe it's 1/4. Let me think.Wait, I think I need to find a better approach. Let me try to calculate the area of ACEG using the formula for a cyclic quadrilateral.The formula is:Area = sqrt[(s - a)(s - b)(s - c)(s - d)]where s is the semi-perimeter, and a, b, c, d are the sides.But in our case, all sides are equal, so a = b = c = d = AC = 2*R*sin36.So, s = (4*a)/2 = 2*a.Therefore, Area = sqrt[(2a - a)^4] = sqrt[a^4] = a^2.Wait, that can't be right because the area should depend on the angles as well.Wait, no, Brahmagupta's formula for a cyclic quadrilateral with equal sides (a rhombus) is:Area = a^2*sin(theta), where theta is one of the internal angles.Earlier, we found that theta = 72 degrees.So, Area = a^2*sin72 ≈ a^2*0.9511.But the area of the decagon is 5*R^2*sin36 ≈ 5*(1.618*a)^2*0.5878 ≈ 5*2.618*a^2*0.5878 ≈ 5*1.538*a^2 ≈ 7.69*a^2.Therefore, m/n ≈ (a^2*0.9511) / (7.69*a^2) ≈ 0.9511 / 7.69 ≈ 0.1236, which is approximately 1/8.09, which is about 1/8, but that's not one of the options either.I'm clearly missing something here. Maybe I should consider that the quadrilateral ACEG is not a rhombus but another shape.Wait, another approach: use the fact that in a regular decagon, the area can be divided into 10 congruent isosceles triangles, each with a central angle of 36 degrees. The area of each triangle is (1/2)*R^2*sin36.Quadrilateral ACEG can be seen as covering 4 of these triangles, but not entirely, because the diagonals overlap.Wait, no, because ACEG is a four-sided figure, it might cover more than 4 triangles.Alternatively, maybe it's better to use the fact that the area of ACEG is 1/5 of the decagon's area because of the symmetry and the number of sides.But I'm not sure. Given the time I've spent and the conflicting results, I think the answer is likely 1/5, which is option C.But earlier calculations suggested a higher ratio, so I'm not entirely confident. However, given the answer choices, I'll go with 1/5.

💡Okay, so I have this problem where I need to make an open-top water tank from a square sheet of iron that's 120 cm on each side. The process involves cutting out small squares from each of the four corners and then folding up the sides to form the tank. I need to figure out the maximum volume this tank can hold. Hmm, let's break this down step by step.First, I visualize the square sheet. It's 120 cm by 120 cm. If I cut out squares from each corner, the size of these squares will determine the height of the tank. Let's say the side length of each small square I cut out is 'x' centimeters. So, from each corner, I'm removing a square that's x cm by x cm.After cutting out these squares, the remaining part of the sheet will have flaps on each side that I can fold up to form the walls of the tank. The original length of each side is 120 cm, but after cutting out two squares (one from each end), the length of the base of the tank will be 120 cm minus 2 times x cm. So, the base length becomes (120 - 2x) cm.Similarly, the width of the base will also be (120 - 2x) cm because the sheet is square, so both length and width are the same. The height of the tank, which is determined by how much we fold up the sides, will be equal to the side length of the squares we cut out, which is x cm.Now, the volume of the tank can be calculated using the formula for the volume of a rectangular prism, which is length × width × height. In this case, the length and width are both (120 - 2x) cm, and the height is x cm. So, the volume V is:V = (120 - 2x) × (120 - 2x) × xLet me write that out more neatly:V = x × (120 - 2x)^2Okay, so now I have an expression for the volume in terms of x. To find the maximum volume, I need to find the value of x that maximizes V. This is an optimization problem, and I think calculus is the way to go here. Specifically, I'll need to take the derivative of V with respect to x, set it equal to zero, and solve for x. That should give me the critical points, and then I can determine which one gives the maximum volume.Before I proceed, let me make sure I understand the domain of x. Since I'm cutting out squares from each corner, x has to be greater than zero, otherwise, I wouldn't be cutting anything. Also, x can't be so large that the length of the base becomes negative. So, (120 - 2x) must be greater than zero. Solving for x:120 - 2x > 0120 > 2x60 > xSo, x must be less than 60 cm. Therefore, the domain of x is 0 < x < 60 cm.Alright, now back to the volume equation:V = x × (120 - 2x)^2Let me expand this to make differentiation easier. First, let me compute (120 - 2x)^2:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, that's not quite right. Let me recall the formula for squaring a binomial: (a - b)^2 = a^2 - 2ab + b^2.So, applying that:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, hold on, that's not correct. The middle term should be 2×a×b, but since it's (a - b)^2, it's -2ab. So, let's correct that:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, no, that's still not right. The middle term is -2ab, where a is 120 and b is 2x. So:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, no, that's incorrect. The middle term is -2ab, so:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, no, that's not correct. Let me clarify:(a - b)^2 = a^2 - 2ab + b^2So, here, a = 120, b = 2x.Therefore:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, that's still not correct. The middle term is -2ab, so:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, no, that's incorrect. The middle term is -2ab, so:(120 - 2x)^2 = 120^2 - 2×120×2x + (2x)^2Wait, no, that's not correct. Let me do it step by step:(a - b)^2 = a^2 - 2ab + b^2So, a = 120, b = 2xTherefore:(120 - 2x)^2 = (120)^2 - 2*(120)*(2x) + (2x)^2= 14400 - 480x + 4x^2Okay, that's correct. So, (120 - 2x)^2 = 14400 - 480x + 4x^2Now, going back to the volume:V = x × (14400 - 480x + 4x^2)Let me multiply this out:V = x*14400 - x*480x + x*4x^2= 14400x - 480x^2 + 4x^3So, V = 4x^3 - 480x^2 + 14400xNow, to find the maximum volume, I need to take the derivative of V with respect to x, set it equal to zero, and solve for x.So, let's compute dV/dx:dV/dx = d/dx (4x^3 - 480x^2 + 14400x)= 12x^2 - 960x + 14400Okay, so the derivative is 12x^2 - 960x + 14400.Now, set this equal to zero to find critical points:12x^2 - 960x + 14400 = 0Hmm, that's a quadratic equation. Let's see if we can simplify it before solving.First, notice that all coefficients are divisible by 12. Let's divide each term by 12:(12x^2)/12 - (960x)/12 + 14400/12 = 0Simplifying:x^2 - 80x + 1200 = 0Okay, that's simpler. Now, let's solve for x using the quadratic formula.The quadratic equation is:x^2 - 80x + 1200 = 0The quadratic formula is:x = [80 ± sqrt(80^2 - 4*1*1200)] / 2*1Let's compute the discriminant first:Discriminant D = 80^2 - 4*1*1200= 6400 - 4800= 1600Okay, so sqrt(D) = sqrt(1600) = 40So, the solutions are:x = [80 ± 40]/2So, two solutions:x = (80 + 40)/2 = 120/2 = 60x = (80 - 40)/2 = 40/2 = 20So, x = 60 cm or x = 20 cmBut wait, earlier we established that x must be less than 60 cm because (120 - 2x) must be positive. So, x = 60 cm would make (120 - 2x) = 0, which would mean the base of the tank has zero length, which doesn't make sense. Therefore, x = 60 cm is not a feasible solution.So, the critical point we're interested in is x = 20 cm.Now, to confirm whether this critical point is a maximum, we can use the second derivative test.First, let's find the second derivative of V.We had the first derivative:dV/dx = 12x^2 - 960x + 14400So, the second derivative is:d^2V/dx^2 = 24x - 960Now, evaluate the second derivative at x = 20 cm:d^2V/dx^2 = 24*20 - 960= 480 - 960= -480Since the second derivative is negative (-480 < 0), this means that the function V has a local maximum at x = 20 cm.Therefore, the maximum volume occurs when x = 20 cm.Now, let's compute the maximum volume.Recall that V = 4x^3 - 480x^2 + 14400xPlugging in x = 20:V = 4*(20)^3 - 480*(20)^2 + 14400*(20)First, compute each term:4*(20)^3 = 4*8000 = 32,000480*(20)^2 = 480*400 = 192,00014400*(20) = 288,000Now, plug these back into the equation:V = 32,000 - 192,000 + 288,000Compute step by step:32,000 - 192,000 = -160,000-160,000 + 288,000 = 128,000So, V = 128,000 cm³Alternatively, I can compute the volume using the original expression:V = x*(120 - 2x)^2Plugging in x = 20:V = 20*(120 - 2*20)^2= 20*(120 - 40)^2= 20*(80)^2= 20*6400= 128,000 cm³Same result.Therefore, the maximum volume of the tank is 128,000 cubic centimeters.Just to recap, the steps were:1. Define the variable x as the side length of the squares cut out from each corner.2. Express the volume V in terms of x.3. Expand and simplify the volume equation.4. Take the derivative of V with respect to x to find critical points.5. Solve for x, considering the domain constraints.6. Use the second derivative test to confirm that the critical point is a maximum.7. Calculate the maximum volume by plugging x back into the volume equation.This makes sense because when you cut out larger squares, the height increases, but the base area decreases. There's a balance between these two factors that leads to the maximum volume. In this case, cutting out 20 cm squares from each corner gives the optimal dimensions for the tank.

💡To determine the perimeter of the "monster" in the arcade game, we need to consider both the arc length of the sector and the two radii that form the mouth.1. **Calculate the length of the radii**: - Each radius is 1 cm. - There are two radii, so their combined length is: [ 1 , text{cm} + 1 , text{cm} = 2 , text{cm} ]2. **Determine the fraction of the circle forming the arc**: - The central angle of the mouth is (60^circ). - The remaining angle of the sector is: [ 360^circ - 60^circ = 300^circ ] - The fraction of the circle that this arc represents is: [ frac{300^circ}{360^circ} = frac{5}{6} ]3. **Calculate the circumference of the full circle**: - The radius is 1 cm, so the full circumference is: [ 2pi times 1 = 2pi , text{cm} ]4. **Calculate the length of the arc**: - The arc length is (frac{5}{6}) of the full circumference: [ frac{5}{6} times 2pi = frac{5}{3}pi , text{cm} ]5. **Calculate the total perimeter**: - The total perimeter is the sum of the arc length and the two radii: [ frac{5}{3}pi , text{cm} + 2 , text{cm} = frac{5}{3}pi + 2 , text{cm} ]Thus, the perimeter of the monster is (frac{5}{3}pi + 2) cm. The correct answer is:[boxed{E}]