🧠:Okay, so I need to figure out the maximum number of pieces a round pancake can be divided into with three straight cuts. Hmm, let's start by visualizing. If I have a pancake, which is basically a circle, and I want to cut it into as many pieces as possible using straight lines. Each cut is a straight line, right? First, maybe I should start with fewer cuts and see the pattern. Let's see:- With 0 cuts, there's obviously just 1 piece, the whole pancake.- With 1 cut, you can divide it into 2 pieces. That's straightforward.- With 2 cuts, if I make the second cut intersect the first one, then each new cut can intersect the previous ones to create more regions. So, if the second cut crosses the first one, that would make 4 pieces? Wait, no. Wait, if the second cut is parallel, you only get 3 pieces. But if you cross them, you get 4. So the maximum for 2 cuts is 4.Wait, let me verify that. First cut: 2 pieces. Second cut: if it's not parallel and intersects the first, then each time a new line crosses all previous lines, it adds as many regions as the number of times it crosses. So for the second cut, if it crosses the first once, then the number of regions increases by 2. Wait, maybe there's a formula here.I recall there's a formula for the maximum number of regions created by n lines in a plane: R(n) = (n^2 + n + 2)/2. Let me check that. For n=1: (1 +1 +2)/2 = 4/2 = 2. Correct. For n=2: (4 +2 +2)/2 = 8/2 =4. That works. For n=3: (9 +3 +2)/2=14/2=7. Wait, but if we have three lines, the maximum number of regions is 7? Let me visualize three lines. Each new line intersects all previous ones. So first line: 2 regions. Second line crosses the first, creating 4 regions. Third line crosses both previous lines, creating two more intersections, which would divide the third line into three segments, each adding a new region. So the third line would add 3 regions, making total 4 +3 =7. Yeah, that seems right. So the formula gives 7 regions for 3 lines.But wait, the question is about a pancake, which is a circle. Does that make a difference? Because lines can go to infinity, but here we have a bounded circle. Hmm. Wait, maybe in the case of a circle, the maximum number of regions with n cuts is different? Let me think.If the cuts are chords of the circle, then each new chord can intersect existing chords. The maximum number of regions would be similar to the planar case, but maybe adjusted because chords are finite. Let me check.Actually, the formula for the maximum number of regions created by n chords in a circle is given by the lazy caterer's sequence: R(n) = (n^2 + n + 2)/2. Wait, that's the same as the planar case? That can't be right. Wait, no, maybe not. Wait, let's check with n=3.In the planar case with 3 lines, you get 7 regions. For a circle with 3 chords, if each chord intersects all others, how many regions? Let me draw it out mentally. First chord: splits circle into 2. Second chord, if it intersects the first, splits it into 4. Third chord, if it intersects both previous chords, each intersection creates a new region. So each intersection with a previous chord adds a region. So the third chord would cross two existing chords, creating two intersection points, thereby dividing the third chord into three segments, each adding a new region. So the third chord would add 3 regions. So total regions would be 4 + 3 =7. Huh, same as the planar case. So maybe for both lines and chords, the maximum regions are the same? That seems possible.But wait, in the planar case, lines can extend infinitely, but in the circle case, chords are finite. However, if each new chord intersects all previous chords, then the number of regions added is the same as with lines. So perhaps the formula is the same. Therefore, the maximum number of regions with n chords is the same as with n lines, which is R(n) = (n(n + 1))/2 + 1. Let me check that formula. For n=1: (1*2)/2 +1 =1 +1=2. Correct. For n=2: (2*3)/2 +1=3 +1=4. Correct. For n=3: (3*4)/2 +1=6 +1=7. Yes, so that formula works. So for three cuts, the maximum number of pieces is 7.Wait, but I need to confirm this. Let me think through each step.First cut: divides the circle into 2.Second cut: if it intersects the first cut, then each region is split, so 4 regions. If it doesn't intersect, it's parallel, so only 3 regions. So maximum at 4.Third cut: to maximize regions, it should intersect both previous cuts. Each intersection with a previous cut creates an additional region. So the third cut crosses two existing cuts, creating two intersection points. Each time it crosses a cut, it enters a new region, thereby splitting that region. So each time it crosses a cut, it adds one more region than before. For the third cut, crossing two existing cuts, it will pass through three regions, thereby splitting each into two, so adding three regions. So starting from 4, adding 3 gives 7.Yes, that seems right. So three cuts can give up to 7 pieces. Is that the case? Let me think if there's any way to get more. Suppose the third cut is arranged such that it intersects all previous cuts, but maybe not in the same way. Wait, but with three cuts, each subsequent cut can intersect all previous ones. Wait, but in two-dimensional space, each new line can intersect all previous lines only once each. So for three cuts, the third cut can intersect the first two cuts at two distinct points, which would divide the third cut into three segments, each creating a new region. So that's three new regions. So yes, 4 +3=7.Alternatively, maybe if the cuts are arranged differently, but I don't think you can get more than seven. Let me try to imagine. First cut: vertical. Second cut: horizontal, crossing the first, making four quadrants. Third cut: diagonal, crossing both the vertical and horizontal, but not passing through the center. So each intersection with previous cuts, so two intersections, and the diagonal line would pass through three regions, splitting each into two. So adding three regions. So total of 4 +3=7. Yes.Alternatively, if the third cut is arranged to pass through the intersection point of the first two cuts, but then it would only intersect each of the first two cuts once, but overlapping at the center. Wait, if the third cut passes through the center, where the first two cuts intersect, then it would intersect each of the first two cuts once, but only at a single point. So does that count as two intersections? Wait, no. If the third line passes through the common intersection point of the first two lines, then it intersects both lines at that single point. So in terms of regions, how does that affect?If the first two cuts are crossed at the center, making four regions. Then the third cut is a line passing through the center, so it's a diameter. Then that third cut would intersect each of the first two cuts at the center. So the third cut would be divided into two segments by the center, each going through two existing regions. So each half of the third cut would divide two regions into two each. So the third cut would split four regions? Wait, no. Wait, the third cut is a diameter. If the first two cuts were perpendicular diameters, making four quadrants. Then adding a third diameter at, say, 45 degrees, would pass through two opposite quadrants, splitting each into two. So total regions would be 6. But that's less than 7. Hence, arranging the third cut to pass through the intersection point of the first two cuts actually results in fewer regions. Therefore, to maximize regions, the third cut should not pass through any existing intersection points, thereby intersecting each previous cut at a distinct point. Thus, creating two new intersections and three new regions.Therefore, the maximum number of regions with three cuts is indeed 7. So the answer should be 7.But let me cross-verify with the formula. The formula for the maximum number of regions created by n cuts in a circle is the same as the planar case: R(n) = (n^2 + n + 2)/2. For n=3: (9 +3 +2)/2 =14/2=7. Yep, that matches.Alternatively, the lazy caterer's sequence is defined as the maximum number of pieces formed by n planar cuts, and it's given by R(n) = (n^2 + n + 2)/2. So that's the same as the formula above. Therefore, for three cuts, the maximum is 7 pieces.So I think that's the answer. But just to make sure, let's think of another way. Imagine three cuts, each intersecting the others at different points. First cut: splits into 2. Second cut: intersects the first, making 4. Third cut: crosses both cuts at two different points, not at the center. So entering the circle, crossing the first cut, then crossing the second cut, exiting. So each time it crosses a cut, it goes into a new region, thereby increasing the count by one each time. So for the third cut, it crosses two existing cuts, so it adds 2 +1 =3 regions. Therefore, 4 +3=7. That seems right.Alternatively, if the third cut is arranged to cross one existing cut twice? Wait, but in a circle, a straight cut is a chord, and two chords can intersect at most once. So each pair of cuts can intersect at most once. So with three cuts, each pair can intersect once, but only if they are not parallel and not concurrent. So maximum intersections for three cuts is three points (each pair intersecting once). But wait, three cuts can have up to three intersection points: each pair intersects once. So if all three cuts intersect each other at different points, then you have three intersection points. How does that affect the number of regions?Let me try to imagine. First cut: horizontal. Second cut: vertical, crossing the horizontal at the center. Third cut: a diagonal that crosses the horizontal cut off-center and crosses the vertical cut off-center as well. So three intersection points: horizontal-vertical at center, horizontal-diagonal somewhere left, vertical-diagonal somewhere up. Then, how many regions does that create?Let's count. First cut: 2. Second cut: crosses first, making 4. Third cut: crosses first at point A and second at point B, both not at the center. So the third cut enters the circle, crosses the first cut (horizontal) at A, then proceeds to cross the vertical cut at B, then exits. Each crossing splits a region. So the third cut would pass through three regions: starting in one region, crossing into another at A, then crossing into another at B. Each time it enters a new region, it splits it. So adding three regions. So total regions 4 +3=7. So even with three intersection points, it's still seven regions. Because the third cut crosses two previous cuts, so two intersection points, leading to three segments on the third cut, each adding a region. So yeah, regardless of where the intersections are, as long as each new cut intersects all previous ones, the number of regions added is equal to the number of intersections plus one. So each new cut can intersect n-1 previous cuts, adding n regions. Therefore, for the third cut, intersecting two previous cuts, adding three regions. So total 7.Therefore, I'm confident the answer is 7.

✅:The maximum number of pieces a round pancake can be divided into using three straight cuts is determined by analyzing how each new cut intersects existing ones to maximize regions. Here's the breakdown:1. First Cut: Divides the pancake into 2 pieces.2. Second Cut: Intersects the first cut, creating 4 pieces.3. Third Cut: Intersects both previous cuts at distinct points, passing through three regions and adding 3 new pieces.Using the formula for the maximum number of regions created by ( n ) cuts in a plane or circle: [R(n) = frac{n^2 + n + 2}{2}] For ( n = 3 ): [R(3) = frac{3^2 + 3 + 2}{2} = frac{14}{2} = 7]Answer: The maximum number of pieces is (boxed{7}).

🔑:To solve this problem, let's analyze how the number of pieces a circle can be divided into increases with each additional straight cut. 1. Base Case: - 0 cuts: A circle with zero cuts remains a single piece. [ P(0) = 1 ] 2. 1 cut: - 1 cut: A single straight line divides the circle into 2 pieces. [ P(1) = 2 ] 3. 2 cuts: - 2 cuts: The maximum number of pieces with two cuts occurs when the two lines intersect inside the circle, providing 4 pieces. [ P(2) = 4 ]4. 3 cuts: - 3 cuts: The goal is to determine the maximum number of pieces with 3 cuts. Based on combinatorial geometry: 1. Each new line can intersect all previous lines, creating new pieces. 2. Consider the general formula for the maximum number of pieces: [ P(n) = frac{n(n+1)}{2} + 1 ] Let's calculate ( P(3) ): [ P(3) = frac{3(3+1)}{2} + 1 = frac{3 times 4}{2} + 1 = 6 + 1 = 7 ] Therefore, the maximum number of pieces obtained with 3 cuts happens when each pair of cuts intersects exactly once within the circle. Conclusion:The maximum number of pieces you can divide a circle into with three straight line cuts is: [ boxed{7} ]

🧠:Alright, so I need to prove that for positive real numbers a, b, and c, the sum of a/(b+c) + b/(c+a) + c/(a+b) is at least 3/2. Hmm, okay. Let me start by recalling some inequalities that might be useful here. Since the expression involves fractions with variables in the denominator, maybe the Cauchy-Schwarz inequality or the AM-HM inequality could come into play. Or perhaps the Nesbitt's inequality? Wait, isn't this exactly Nesbitt's inequality? But even if it is, I should still work through the proof.Let me try the Cauchy-Schwarz approach first. The Cauchy-Schwarz inequality states that for any real numbers, (sum u_i v_i)^2 ≤ (sum u_i^2)(sum v_i^2). Alternatively, in the form (sum a_i^2)(sum b_i^2) ≥ (sum a_i b_i)^2. But how can I apply this here? Maybe if I consider the numerators and denominators appropriately.Alternatively, the Titu's lemma, which is a specific case of Cauchy-Schwarz, might be useful here. Titu's lemma states that sum(a_i^2 / b_i) ≥ (sum a_i)^2 / sum b_i. Let me see if I can apply that here. If I look at each term, like a/(b+c), maybe I can write it as a^2 / [a(b+c)]. Then the entire sum would be sum(a^2 / [a(b+c)]). So that's sum(a / (b + c)). Then applying Titu's lemma, this would be greater than or equal to (a + b + c)^2 / [a(b + c) + b(c + a) + c(a + b)]. Let me compute that denominator.Denominator becomes a(b + c) + b(c + a) + c(a + b) = ab + ac + bc + ba + ca + cb = 2(ab + bc + ca). So the expression from Titu's lemma would be (a + b + c)^2 / [2(ab + bc + ca)]. So according to Titu's lemma, the original sum is ≥ (a + b + c)^2 / [2(ab + bc + ca)].Now, I need to show that this is ≥ 3/2. Therefore, we have:(a + b + c)^2 / [2(ab + bc + ca)] ≥ 3/2Multiply both sides by 2(ab + bc + ca):(a + b + c)^2 ≥ 3(ab + bc + ca)Expanding the left-hand side: a² + b² + c² + 2ab + 2bc + 2ca ≥ 3ab + 3bc + 3caSubtract 3ab + 3bc + 3ca from both sides:a² + b² + c² - ab - bc - ca ≥ 0Which can be written as (1/2)[(a - b)^2 + (b - c)^2 + (c - a)^2] ≥ 0. Since squares are always non-negative, this is true. Therefore, the inequality holds, and thus the original expression is indeed ≥ 3/2. So that works!But let me check if there's another way to approach this, maybe using the AM-GM inequality. Let's see. AM-GM states that the arithmetic mean is ≥ the geometric mean. However, the terms here are fractions, so maybe combining them in a clever way. Alternatively, using substitution variables. Let me try substituting x = b + c, y = c + a, z = a + b. Then the denominators become x, y, z. But the numerators would need to be expressed in terms of x, y, z. Let's see:Since x = b + c, then a = (y + z - x)/2. Similarly, b = (x + z - y)/2, c = (x + y - z)/2. Hmm, this might complicate things. Maybe not the best approach here.Alternatively, consider homogenizing the inequality. Since the inequality is homogeneous, meaning that if we scale a, b, c by a constant factor, both sides remain the same. Let's check: if we replace a with ka, b with kb, c with kc, then each term becomes (ka)/(kb + kc) = a/(b + c), so the left side remains the same. The right side is 3/2, which is a constant. Therefore, the inequality is homogeneous. So we can assume without loss of generality that a + b + c = 1. Let me try that.Assume a + b + c = 1. Then the left-hand side becomes a/(1 - a) + b/(1 - b) + c/(1 - c). Hmm, not sure if that helps. Alternatively, since a, b, c are positive and sum to 1, maybe using Jensen's inequality. The function f(x) = x/(1 - x) is convex on (0, 1) because its second derivative is positive. Let me check:First derivative f'(x) = [1*(1 - x) - x*(-1)]/(1 - x)^2 = [1 - x + x]/(1 - x)^2 = 1/(1 - x)^2.Second derivative f''(x) = 2/(1 - x)^3, which is positive for x in (0, 1). Therefore, f is convex. By Jensen's inequality, we have:f(a) + f(b) + f(c) ≥ 3f((a + b + c)/3) = 3f(1/3) = 3*( (1/3)/(1 - 1/3) ) = 3*( (1/3)/(2/3) ) = 3*(1/2) = 3/2. So this gives the result. Wait, but this requires that a + b + c = 1. However, since the inequality is homogeneous, we can normalize a + b + c = 1. So this approach works as well. Therefore, another way to prove it is by using Jensen's inequality.Alternatively, there's the classic proof using the AM-HM inequality. Let's see. For each term a/(b + c), if we consider adding them up, maybe we can pair them with (b + c). But I need to think more carefully. Let me try another angle.Another approach: Let's denote S = a + b + c. Then each term a/(b + c) can be written as a/(S - a). So the sum becomes:a/(S - a) + b/(S - b) + c/(S - c).Maybe consider substituting variables such that x = S - a, y = S - b, z = S - c. Then x = b + c, y = a + c, z = a + b. Then we have the sum as (S - x)/x + (S - y)/y + (S - z)/z = S(1/x + 1/y + 1/z) - 3.But not sure if that helps. Alternatively, using the substitution x = b + c, y = a + c, z = a + b, we can write the sum as (y + z - x)/2x + (x + z - y)/2y + (x + y - z)/2z. Simplifying each term: (y)/(2x) + z/(2x) - x/(2x) + similar terms. But this might complicate things further.Wait, perhaps another way: Let's consider the sum:sum(a/(b + c)) = sum(a/(b + c)).Let me try to use the Cauchy-Schwarz inequality in the following form:sum(a/(b + c)) ≥ (a + b + c)^2 / [a(b + c) + b(c + a) + c(a + b)].But this is exactly the same as Titu's lemma applied here. Then the denominator is 2(ab + bc + ca). So we get (a + b + c)^2 / [2(ab + bc + ca)] ≥ 3/2, which simplifies to (a + b + c)^2 ≥ 3(ab + bc + ca), which is equivalent to a² + b² + c² ≥ ab + bc + ca, which is true. Therefore, the same as before.So both approaches via Titu's lemma (Cauchy-Schwarz) and Jensen's inequality work. Alternatively, another method might be to use the AM-GM inequality on the denominators. Let's see.For each term a/(b + c), since b + c ≤ (b + c + a)/ something? Not sure. Alternatively, notice that by AM-GM, (b + c) ≤ (a + b + c)/ something. Wait, but that might not directly help. Let me think.Alternatively, using the Nesbitt's inequality which directly states that for positive a, b, c, the sum a/(b + c) + b/(a + c) + c/(a + b) ≥ 3/2. So essentially, this is Nesbitt's inequality. The proof is what we've done above via Cauchy-Schwarz or Jensen.Alternatively, another method is to use the AM-HM inequality on each term. Let's see:For each term a/(b + c), we can write it as a/(b + c) = (a^2)/(a(b + c)). Then if we apply the Cauchy-Schwarz inequality on the numerator and denominator:(sum a/(b + c)) = sum (a^2)/(a(b + c)) ≥ (a + b + c)^2 / [a(b + c) + b(c + a) + c(a + b)].Which is the same as before. So maybe that's the same approach.Alternatively, perhaps rearrangement inequality? But since the terms are cyclic, not sure.Alternatively, let's try adding 3 to both sides and see if that helps. Wait, not sure. Let's consider expanding the terms.Alternatively, let's consider the function f(a, b, c) = a/(b + c) + b/(a + c) + c/(a + b). To find its minimum under the constraint that a, b, c > 0. By symmetry, the minimum might occur when a = b = c. Let's test that. If a = b = c, then each term is a/(a + a) = 1/2, so the sum is 3*(1/2) = 3/2. So the minimum is indeed 3/2 when a = b = c. Therefore, the inequality holds. But to make this rigorous, we need to show that the function attains its minimum when a = b = c. Since the function is symmetric and convex, perhaps we can use the method of Lagrange multipliers or consider using the method of symmetry.Alternatively, assume without loss of generality that a ≥ b ≥ c. Then maybe some substitution or majorization can be applied. But I think the earlier methods are more straightforward.So, summarizing, the two main approaches are:1. Apply Titu's lemma (Cauchy-Schwarz) to get the sum ≥ (a + b + c)^2 / [2(ab + bc + ca)], then show that (a + b + c)^2 ≥ 3(ab + bc + ca), which simplifies to a² + b² + c² ≥ ab + bc + ca, which is a known inequality.2. Use Jensen's inequality on the convex function f(x) = x/(1 - x) after normalizing a + b + c = 1.Either method leads to the conclusion that the sum is at least 3/2. Therefore, the inequality is proven.I should check if all steps are valid. For Titu's lemma: Yes, since all denominators are positive, Titu's lemma applies. For the Jensen's approach: We normalized a + b + c = 1, which is allowed because the inequality is homogeneous, and the function f(x) = x/(1 - x) is indeed convex on (0,1). Therefore, both methods are valid.Another check: Let me test with a = 1, b = 1, c = 1. Then the sum is 1/(1+1) * 3 = 3/2, which matches the equality case. If I take a different case, say a = 2, b = 1, c = 1. Then the sum is 2/(1+1) + 1/(2+1) + 1/(2+1) = 1 + 1/3 + 1/3 = 1.666..., which is 5/3 ≈ 1.666..., which is greater than 3/2. So the inequality holds here. Another test case: a = 3, b = 1, c = 1. Then sum is 3/(1+1) + 1/(3+1) + 1/(3+1) = 1.5 + 0.25 + 0.25 = 2, which is greater than 1.5. So seems to hold.What if one variable approaches zero? Let’s say c approaches 0, then the sum becomes a/(b + 0) + b/(a + 0) + 0/(a + b) = a/b + b/a + 0. By AM-GM, a/b + b/a ≥ 2. Which is greater than 3/2. So the inequality still holds. Therefore, even in the edge cases, the inequality is valid.Therefore, the proof is solid.

✅:To prove that for positive real numbers (a), (b), and (c), the inequality (frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{3}{2}) holds, we can use Titu's Lemma (a special case of the Cauchy-Schwarz inequality) and algebraic manipulation.Step-by-Step Proof:1. Apply Titu's Lemma: [ frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} = frac{a^2}{a(b+c)} + frac{b^2}{b(c+a)} + frac{c^2}{c(a+b)} ] By Titu's Lemma, this is greater than or equal to: [ frac{(a + b + c)^2}{a(b+c) + b(c+a) + c(a+b)} ]2. Simplify the Denominator: [ a(b+c) + b(c+a) + c(a+b) = 2(ab + bc + ca) ]3. Formulate the Inequality: [ frac{(a + b + c)^2}{2(ab + bc + ca)} geq frac{3}{2} ] Multiply both sides by (2(ab + bc + ca)): [ (a + b + c)^2 geq 3(ab + bc + ca) ]4. Expand and Simplify: [ a^2 + b^2 + c^2 + 2(ab + bc + ca) geq 3(ab + bc + ca) ] Subtract (2(ab + bc + ca)) from both sides: [ a^2 + b^2 + c^2 geq ab + bc + ca ]5. Recognize a Known Inequality: The inequality (a^2 + b^2 + c^2 geq ab + bc + ca) is always true since: [ a^2 + b^2 + c^2 - ab - bc - ca = frac{1}{2}left[(a - b)^2 + (b - c)^2 + (c - a)^2right] geq 0 ]Conclusion:Thus, (frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{3}{2}) holds for all positive real numbers (a), (b), and (c), with equality when (a = b = c).Final Answer:boxed{frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{3}{2}}

🔑:1. Assume Symmetry and Reordering: Assume that a geq b geq c. The variables a, b, and c can be reordered because they play symmetric roles in the inequality. Hence, we have: - (b + c leq a + c leq a + b).2. Ordering of Reciprocals: Given the above reordering, the reciprocals will also follow the reverse order: [ frac{1}{b+c} geq frac{1}{c+a} geq frac{1}{a+b} ]3. Applying the Reordering Inequality: - By the Reordering Inequality, we can state that: [ frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{b}{b+c} + frac{c}{c+a} + frac{a}{a+b} ] - We can also state: [ frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{c}{b+c} + frac{a}{c+a} + frac{b}{a+b} ]4. Summing the Two Inequalities: - Adding the two inequalities together: [ 2 left(frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b}right) geq frac{b}{b+c} + frac{c}{c+a} + frac{a}{a+b} + frac{c}{b+c} + frac{a}{c+a} + frac{b}{a+b} ] - Since (frac{b}{b+c} + frac{c}{c+a} + frac{a}{a+b} + frac{c}{b+c} + frac{a}{c+a} + frac{b}{a+b} = 3), we get: [ 2 left( frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} right) geq 3 ]5. Dividing by 2: - Dividing both sides by 2: [ frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{3}{2} ]6. Harmonic Mean-Arithmetic Mean Inequality: - Using the fact that the harmonic mean is less than or equal to the arithmetic mean: [ frac{1}{3} left( frac{1}{b+c} + frac{1}{c+a} + frac{1}{a+b} right) geq frac{3}{(b+c) + (c+a) + (a+b)} ] - This simplifies to: [ frac{1}{3} left( frac{1}{b+c} + frac{1}{c+a} + frac{1}{a+b} right) geq frac{3}{2(a+b+c)} ]7. Multiplication by (6(a+b+c)): - Multiplying both sides by (6(a+b+c)): [ 2 left( frac{a + (b+c)}{b+c} + frac{b + (c+a)}{c+a} + frac{c + (a+b)}{a+b} right) geq 9 ]8. Simplification: - This can be rewritten as: [ 2 left( frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} + 1 + 1 + 1 right) geq 9 ] - Subtracting 6 from both sides: [ 2 left( frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} right) geq 3 ] - Dividing by 2: [ frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{3}{2} ] Conclusion:The inequality is proved:[boxed{frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{3}{2}}]

🧠:Okay, let me try to figure this out. The question is asking what type of triangle it is if the sum of any two of its angles is greater than 90 degrees. Hmm, triangles are classified based on their angles as acute, right, or obtuse. Let me recall: an acute triangle has all angles less than 90 degrees, a right triangle has one angle exactly 90 degrees, and an obtuse triangle has one angle greater than 90 degrees.So the given condition here is that for any two angles in the triangle, their sum is greater than 90 degrees. Let me denote the three angles of the triangle as A, B, and C. The condition is then A + B > 90°, B + C > 90°, and A + C > 90°. Since in any triangle, the sum of all three angles is 180°, maybe I can use that to find out more about each individual angle.Let me start by considering the first condition: A + B > 90°. Since the total sum is 180°, if A + B > 90°, then the third angle C must be less than 90°, because C = 180° - (A + B). If A + B > 90°, then 180° - (A + B) = C < 90°. Similarly, applying the same logic to the other conditions:- If B + C > 90°, then A = 180° - (B + C) < 90°- If A + C > 90°, then B = 180° - (A + C) < 90°So, from all three conditions, each angle A, B, and C must be less than 90°. Wait, that suggests that all three angles are acute. So does that mean the triangle is acute?But hold on, let me verify this. If all three angles are less than 90°, then each pair of angles would sum to more than 90°, right? Let's check. Suppose each angle is 80°, then 80° + 80° = 160°, which is greater than 90°. But what if one angle is smaller? Let's take another example: suppose angles are 70°, 70°, and 40°. Then 70° + 70° = 140° > 90°, 70° + 40° = 110° > 90°, and 70° + 40° = 110° > 90°. So even if one angle is 40°, the sum of any two is still greater than 90°. Wait, but in this case, the triangle is still acute because all angles are less than 90°. However, in this example, even with a 40° angle, the sums are still above 90°. So maybe the condition given in the problem actually ensures that all angles are acute?But let's test another case. Suppose we have an obtuse triangle, say angles 100°, 40°, 40°. Let's check the sums: 100° + 40° = 140° > 90°, 40° + 40° = 80° < 90°, so here the sum of the two smaller angles is less than 90°, which violates the problem's condition. So in an obtuse triangle, the sum of the two acute angles would be less than 90°, right? Because the obtuse angle is greater than 90°, so the sum of the other two angles is 180° - (obtuse angle) < 90°. Hence, in such a case, the sum of those two angles would be less than 90°, which doesn't satisfy the problem's condition. Therefore, an obtuse triangle can't satisfy the given condition.Similarly, let's check a right triangle. Suppose angles are 90°, 45°, 45°. Then, 90° + 45° = 135° > 90°, 45° + 45° = 90°, which is not greater than 90°, so that also violates the condition. Therefore, a right triangle doesn't satisfy the condition either.So both right and obtuse triangles have at least one pair of angles whose sum is not greater than 90°. Hence, only acute triangles can satisfy the condition where all pairs of angles sum to more than 90°. But wait, is every acute triangle satisfying this condition? Let me check with an example. Suppose an acute triangle with angles 80°, 60°, 40°. Then, 80° + 60° = 140° > 90°, 80° + 40° = 120° > 90°, 60° + 40° = 100° > 90°. All sums are greater than 90°, so that's good. But what if we have an acute triangle with smaller angles? Let's try 50°, 50°, 80°. Wait, but 50° + 50° = 100° > 90°, 50° + 80° = 130° > 90°, and 50° + 80° = 130° > 90°. Still, all sums are greater than 90°. Wait, maybe all acute triangles satisfy this condition?Wait, hold on. If all angles are less than 90°, then the sum of any two angles must be greater than 90°? Let me test with a very small angle. Suppose angles are 30°, 60°, 90°... but that's a right triangle. So not allowed. Let me take angles 30°, 60°, 90° – no, that's a right angle. Let me take an acute triangle with angles 20°, 70°, 90° – no, again 90° is a right angle. Wait, maybe 20°, 80°, 80°. Wait, 20 + 80 + 80 = 180. Then, the sums: 20 + 80 = 100 > 90, 80 + 80 = 160 > 90, 20 + 80 = 100 > 90. So even with a 20° angle, the sums are all above 90°. Wait, is that true?Wait, let's take a more extreme case. Suppose angles are 10°, 85°, 85°. Then 10 + 85 = 95 > 90, 85 + 85 = 170 > 90, and 10 + 85 = 95 > 90. Still all sums are above 90. Wait, so even if one angle is very small, as long as the other two angles are large enough to compensate, the sum of any two angles will still be greater than 90. Hmm, but in that case, all angles are still acute. Wait, but in such a triangle, even with a 10° angle, the other two angles are 85°, which are still acute. So maybe any acute triangle, regardless of how small one angle is, as long as all angles are less than 90°, the sum of any two angles will always be greater than 90°? Let me check.Suppose we have angles A, B, C, all less than 90°. Then, the sum of any two angles, say A + B. Since all angles are positive and A + B + C = 180°, then A + B = 180° - C. Since C is less than 90°, 180° - C is greater than 90°, so A + B > 90°. Similarly, B + C = 180° - A > 90°, and A + C = 180° - B > 90°, since A and B are each less than 90°. Therefore, in any acute triangle, the sum of any two angles must be greater than 90°, because each angle is less than 90°, so 180° minus any angle is greater than 90°. Therefore, all acute triangles satisfy the condition that the sum of any two angles is greater than 90°, and conversely, if a triangle has all pairs of angles summing to more than 90°, then all angles must be acute. Therefore, the triangle must be acute.But wait, let me verify this converse. Suppose we have a triangle where the sum of any two angles is greater than 90°, does that imply all angles are acute? Let's see. Suppose one angle is 90°, then the sum of the other two angles is 90°, so their sum is exactly 90°, which is not greater. So, if the triangle has a right angle, then one pair of angles sums to exactly 90°, violating the condition. Similarly, if there's an obtuse angle, say 100°, then the other two angles sum to 80°, which is less than 90°, so again violating the condition. Hence, if all pairs of angles sum to more than 90°, then all angles must be less than 90°, making it an acute triangle.Therefore, the type of triangle is acute.But just to be thorough, let me consider if there's any other possibility. For example, an equilateral triangle has all angles 60°, so sums are 120°, which is greater than 90°, so that's acute. A triangle with angles 80°, 60°, 40° as before – all sums are above 90°. So yes, seems like all acute triangles satisfy this condition, and only acute triangles do.Therefore, the answer is that the triangle is acute.

✅:The triangle is acute.Step-by-Step Explanation:1. Understand Triangle Types by Angles: - Acute Triangle: All angles < 90°. - Right Triangle: One angle = 90°. - Obtuse Triangle: One angle > 90°.2. Given Condition: - For angles ( A, B, C ), the sum of any two angles > 90°: [ A + B > 90°, quad B + C > 90°, quad A + C > 90°. ]3. Analyze Implications: - Since ( A + B + C = 180° ), substitute into each inequality: - ( A + B > 90° implies C < 90° ), - ( B + C > 90° implies A < 90° ), - ( A + C > 90° implies B < 90° ). - All angles ( A, B, C ) must individually be < 90°, satisfying the definition of an acute triangle.4. Verify Converse: - If all angles are acute ((< 90°)), then ( 180° - text{any angle} > 90° ), meaning the sum of any two angles (= 180° - text{third angle} > 90°). Thus, all acute triangles meet the condition. - In contrast, right/obtuse triangles fail the condition because: - Right triangle: Two angles sum to 90° (violates > 90°). - Obtuse triangle: Two angles sum to (< 90°) (since the obtuse angle > 90°).Conclusion: A triangle where the sum of any two angles exceeds ( 90° ) must have all angles acute. Thus, it is an acute triangle.

🔑:1. Consider a triangle, and denote its three angles as ( alpha ), ( beta ), and ( gamma ).2. According to the problem, the sum of any two angles is greater than ( 90^circ ). That is: [ alpha + beta > 90^circ ] [ alpha + gamma > 90^circ ] [ beta + gamma > 90^circ ]3. First, let us examine the consequences of the sum of the angles in a triangle. The sum of the interior angles in any triangle is always ( 180^circ ): [ alpha + beta + gamma = 180^circ ]4. If we take the inequality ( alpha + beta > 90^circ ): [ gamma = 180^circ - (alpha + beta) ] Substituting the inequality: [ gamma < 180^circ - 90^circ = 90^circ ] Thus, ( gamma ) is less than ( 90^circ ).5. Applying the same reasoning to the other inequalities: [ alpha + gamma > 90^circ implies beta = 180^circ - (alpha + gamma) < 90^circ ] [ beta + gamma > 90^circ implies alpha = 180^circ - (beta + gamma) < 90^circ ]6. Consequently, each of the angles ( alpha ), ( beta ), and ( gamma ) is less than ( 90^circ ).Conclusion: Given that all three angles of the triangle are acute angles (i.e., less than ( 90^circ )), the triangle is an acute triangle.[boxed{text{The triangle is acute.}}]

🧠:Okay, let's try to figure out the cost price of the toy car. Hmm, the problem says that the car was initially priced with a 100% profit, which means the first day's price is double the cost price. Then, on the second day, the price was reduced by 10%, and it still didn't sell. On the third day, they reduced the price by another 360 yuan, and then it sold. The final selling price is 1.44 times the cost price. So we need to find the original cost price.Let me break this down step by step. Let's denote the cost price as C. Then, the first day's price would be C + 100% of C, which is 2C. Right? Because 100% profit means the selling price is cost plus the same amount as the cost.On the second day, the price is reduced by 10%. So the new price is 90% of the first day's price. That would be 0.9 times 2C, which is 1.8C. But still, no one bought it. Then on the third day, they reduce the price by another 360 yuan. So the final selling price is 1.8C minus 360 yuan. And according to the problem, this final price is 1.44 times the cost price. So 1.8C - 360 = 1.44C.Wait, let me check that equation again. The final selling price after the third day is 1.44C. So the equation should be:First day: 2CSecond day: 2C * 0.9 = 1.8CThird day: 1.8C - 360 = 1.44CYes, that seems right. Now, we can solve for C.So, subtract 1.44C from both sides:1.8C - 1.44C = 3600.36C = 360Then, C = 360 / 0.36 = 1000 yuan.Wait, that seems straightforward, but let me verify again to make sure I didn't make a mistake. Let's plug C = 1000 back into the original steps.First day price: 2 * 1000 = 2000 yuan.Second day price: 2000 - 10% of 2000 = 2000 - 200 = 1800 yuan.Third day price: 1800 - 360 = 1440 yuan.And 1.44 times the cost price is 1.44 * 1000 = 1440 yuan. That matches. So the cost price is indeed 1000 yuan.Wait, but I need to make sure there are no other interpretations of the problem. For example, when it says "reduced by 10%", does that mean 10% of the original price (which was 2C) or 10% of the current price each day? In this case, the problem states "the price was reduced by 10%", which is likely referring to the previous day's price. So on the second day, it's 10% off from the first day's price, which is 2C, so 2C * 0.9 = 1.8C. Then the third day, it's a flat reduction of 360 yuan. So the steps are correct.Alternatively, if someone misreads the problem and thinks that each day's reduction is based on the previous day's price, but the second day is 10% off the first day, and third day is 360 off the second day. But that seems to be exactly what we did.Yes, so the answer should be 1000 yuan. Let me check once again with another approach.Let’s assume the cost price is C. The final selling price is 1.44C. The third day's price is 1.44C. Before that, on the third day, they reduced the price by 360 yuan, so the second day's price was 1.44C + 360. The second day's price is 90% of the first day's price, which was 2C. So:Second day's price = 0.9 * 2C = 1.8CBut we also have that second day's price is equal to 1.44C + 360. Therefore:1.8C = 1.44C + 360Subtract 1.44C from both sides:0.36C = 360So C = 360 / 0.36 = 1000. Yep, same result. So that's consistent. Therefore, the cost price is 1000 yuan.I think that's solid. All steps check out, and the verification confirms it.

✅:The cost price of the toy car is determined as follows:1. Define Variables: Let the cost price be ( C ) yuan.2. First Day Price: Priced at 100% profit, so ( 2C ).3. Second Day Price: Reduced by 10%, resulting in ( 0.9 times 2C = 1.8C ).4. Third Day Price: Reduced by 360 yuan, so ( 1.8C - 360 ).5. Final Selling Price: Equals ( 1.44C ) (given condition).Set up the equation:[1.8C - 360 = 1.44C]Solve for ( C ):[1.8C - 1.44C = 360 implies 0.36C = 360 implies C = frac{360}{0.36} = 1000]Verification:- First day: ( 2 times 1000 = 2000 ) yuan.- Second day: ( 2000 times 0.9 = 1800 ) yuan.- Third day: ( 1800 - 360 = 1440 ) yuan.- ( 1.44 times 1000 = 1440 ), which matches the final price.Answer: The cost price of the toy car is (boxed{1000}) yuan.

🔑:Given: A toy car is sold on the third day at a price reduced by 10% on the second day and further reduced by 360 yuan on the third day, resulting in sale.Let the initial price of the toy car be ( P ) (100% price). We are told:1. The sale price on the third day after reducing by 10% and then 360 yuan equals 1.44 times the cost price.Assuming the cost price is ( x ), we formulate the following steps:1. Initial price after 10% reduction: [ P_{text{second day}} = P times (1 - 0.10) = 0.90P ]2. Price after an additional 360 yuan reduction: [ P_{text{third day}} = 0.90P - 360 ]3. Price relation given in the problem: This price is equal to 1.44 times the cost price: [ 0.90P - 360 = 1.44x ]4. Relate ( P ) and ( x ): Since ( P ) is the 100% initial price and does not exceed ( x ), set ( P ) in terms of ( x ): [ P = 100% = 1.00x ]5. Substitute ( P = x ) into the equation: [ 0.90x - 360 = 1.44x ]6. Isolate ( x ): Combine like terms to solve for ( x ): [ 0.90x - 1.44x = 360 ] [ -0.54x = 360 ] [ x = frac{360}{-0.54} ]7. Calculate the value of ( x ): [ x = frac{360}{-0.54} = frac{36000}{54} = 1000 ] Conclusion:Thus, the cost price of the toy car is:[boxed{1000 text{ yuan}}]

🧠:Okay, so I need to prove that the set of points M satisfying the equation k|AM|² + l|BM|² = d is either a circle with its center on line AB, a single point, or an empty set. Let me start by recalling some coordinate geometry concepts. Maybe if I place points A and B on a coordinate system, it would simplify things. Let me assign coordinates to A and B. Let's say point A is at (0,0) and point B is at (b,0) on the x-axis. That way, the line AB is just the x-axis, which might make calculations easier.So, let me define coordinates:- Let A = (0, 0)- Let B = (b, 0) where b is the distance between A and B.Now, let M be a point (x, y). Then, the distances squared from M to A and B are:- |AM|² = x² + y²- |BM|² = (x - b)² + y²The given equation is k|AM|² + l|BM|² = d. Plugging in the coordinates, that becomes:k(x² + y²) + l[(x - b)² + y²] = dLet me expand this equation:k(x² + y²) + l(x² - 2bx + b² + y²) = dMultiply out the terms:kx² + ky² + lx² - 2blx + lb² + ly² = dCombine like terms:(k + l)x² + (k + l)y² - 2blx + lb² = dHmm, so this simplifies to:(k + l)x² + (k + l)y² - 2blx + (lb² - d) = 0Let me factor out (k + l) from the x² and y² terms:(k + l)(x² + y²) - 2blx + (lb² - d) = 0Wait, if I want to write this in the standard form of a circle equation, which is (x - h)² + (y - k)² = r², I need to complete the square for the x terms. Let me rearrange the equation:(k + l)(x² - (2bl/(k + l))x) + (k + l)y² = d - lb²Hmm, actually, let's divide both sides by (k + l) to make it easier, assuming k + l ≠ 0, which is given. So:x² + y² - (2bl/(k + l))x + (lb² - d)/(k + l) = 0Now, let's group the x terms to complete the square:x² - (2bl/(k + l))x + y² = (d - lb²)/(k + l)Completing the square for x:Take the coefficient of x, which is -2bl/(k + l), divide by 2, square it: [bl/(k + l)]²Add and subtract this term on the left side:[x² - (2bl/(k + l))x + (bl/(k + l))²] + y² = (d - lb²)/(k + l) + (bl/(k + l))²This simplifies to:(x - bl/(k + l))² + y² = [ (d - lb²) + (b² l²)/(k + l) ] / (k + l)Let me compute the right-hand side:First, write (d - lb²) as [d(k + l) - lb²(k + l)] / (k + l), but wait, that might not be helpful. Let me instead compute:(d - lb²)/(k + l) + (b² l²)/(k + l)²Factor out 1/(k + l):[ (d - lb²)(k + l) + b² l² ] / (k + l)²Let me expand the numerator:d(k + l) - lb²(k + l) + b² l²= d(k + l) - lb²k - lb² l + l² b²= d(k + l) - lb²k - lb² l + l² b²Note that - lb²k - lb² l = -lb²(k + l), and l² b² is + l² b². So:= d(k + l) - lb²(k + l) + l² b²Factor out (k + l) from the first two terms:= (k + l)(d - lb²) + l² b²Hmm, maybe there's a better way to simplify. Let me compute:Numerator: d(k + l) - lb²(k + l) + l² b²= d(k + l) - lb²k - lb² l + l² b²= d(k + l) - lb²k - lb² l + l² b²Notice that the last two terms: - lb² l + l² b² = -l² b² + l² b² = 0. Wait, that can't be right. Wait:Wait, the original expression is:d(k + l) - lb²k - lb² l + l² b²Let me rearrange:= d(k + l) - lb²(k + l) + l² b²Factor (k + l) from the first two terms? Wait:Wait, -lb²k - lb² l = -lb²(k + l). So the numerator is:d(k + l) - lb²(k + l) + l² b²= (k + l)(d - lb²) + l² b²Alternatively, maybe factor something else. Let me compute:= d(k + l) - lb²(k + l) + l² b²= [d - lb²](k + l) + l² b²Hmm, not sure if that helps. Let me just keep going.So the right-hand side is:[ (k + l)(d - lb²) + l² b² ] / (k + l)^2= [ (k + l)(d - lb²) + l² b² ] / (k + l)^2Let me factor (k + l) in the numerator:= (k + l)(d - lb²) + l² b² / (k + l)^2Wait, perhaps expanding (k + l)(d - lb²):= k d + l d - k l b² - l² b² + l² b²= k d + l d - k l b²So the numerator becomes k d + l d - k l b²Hence, the right-hand side is (k d + l d - k l b²) / (k + l)^2So putting it all together, the equation becomes:(x - (bl)/(k + l))² + y² = (k d + l d - k l b²) / (k + l)^2Simplify the numerator:Factor d(k + l) - k l b²So:= [d(k + l) - k l b²] / (k + l)^2Therefore, the equation is:(x - (bl)/(k + l))² + y² = [d(k + l) - k l b²] / (k + l)^2Let me denote the right-hand side as R², where R is the radius.So, R² = [d(k + l) - k l b²] / (k + l)^2Therefore, the equation represents a circle with center at ( bl/(k + l), 0 ) and radius sqrt( [d(k + l) - k l b²] / (k + l)^2 )Simplify the radius expression:sqrt( [d(k + l) - k l b²] ) / (k + l)So, sqrt( d(k + l) - k l b² ) divided by (k + l)Therefore, the radius is sqrt( [d(k + l) - k l b²] ) / (k + l)But since the radius squared must be non-negative, the expression inside the square root must be non-negative:d(k + l) - k l b² ≥ 0So, if d(k + l) - k l b² > 0, then the radius is real and positive, so the locus is a circle.If d(k + l) - k l b² = 0, then the radius is zero, so the locus is a single point (the center).If d(k + l) - k l b² < 0, then there's no real solution, so the locus is empty.Therefore, the locus is a circle (when the radius squared is positive), a single point (radius zero), or empty (radius squared negative).Moreover, the center of the circle is at ( bl/(k + l), 0 ), which lies on the x-axis, i.e., the line AB, since points A and B are on the x-axis. Therefore, the center is on line AB.Therefore, we have proved that the locus is either a circle with center on AB, a point, or empty.Wait, but in the problem statement, they mention "given numbers k, l, d", but in the coordinate system I chose, I set AB as the x-axis with A at (0,0) and B at (b,0). However, the problem is general for any points A and B. But since the problem statement mentions the line AB, the coordinate system can be chosen such that A and B lie on the x-axis without loss of generality, because any line can be rotated or translated to the x-axis. Therefore, the conclusion about the center being on the x-axis (i.e., line AB) in this coordinate system translates to the original problem's line AB.Therefore, the center of the circle is indeed on line AB.Thus, the locus is as described.Alternatively, another approach without coordinates could involve vectors. Let me think about that.Let me denote vectors with position vectors. Let the position vectors of A, B, M be a, b, m respectively.Then |AM|² = |m - a|² and |BM|² = |m - b|².The equation is k|m - a|² + l|m - b|² = dExpanding both terms:k[ |m|² - 2m·a + |a|² ] + l[ |m|² - 2m·b + |b|² ] = dCombine like terms:(k + l)|m|² - 2m·(ka + lb) + (k|a|² + l|b|²) = dBring the constant term to the right:(k + l)|m|² - 2m·(ka + lb) = d - (k|a|² + l|b|²)Divide both sides by (k + l) assuming k + l ≠ 0:|m|² - 2m·[ (ka + lb)/(k + l) ] = [ d - (k|a|² + l|b|²) ] / (k + l )Let me denote c = (ka + lb)/(k + l). Then, the equation becomes:|m|² - 2m·c = [ d - k|a|² - l|b|² ] / (k + l )Complete the square on the left side by adding |c|² to both sides:|m - c|² = [ d - k|a|² - l|b|² ] / (k + l ) + |c|²Compute |c|²:|c|² = |(ka + lb)/(k + l)|² = (k²|a|² + 2kla·b + l²|b|²)/(k + l)^2Therefore, the right-hand side becomes:[ d - k|a|² - l|b|² ] / (k + l ) + (k²|a|² + 2kla·b + l²|b|²)/(k + l)^2Combine the terms over a common denominator (k + l)^2:= [ (d - k|a|² - l|b|²)(k + l) + k²|a|² + 2kla·b + l²|b|² ] / (k + l)^2Expand the numerator:d(k + l) - k(k + l)|a|² - l(k + l)|b|² + k²|a|² + 2kla·b + l²|b|²Simplify term by term:- For the |a|² terms:- k(k + l)|a|² is k²|a|² + kl|a|²- Then subtract that gives -k²|a|² - kl|a|²- Then adding k²|a|² cancels the first term, leaving -kl|a|²Similarly, for |b|² terms:- l(k + l)|b|² is kl|b|² + l²|b|²- Subtract that gives -kl|b|² - l²|b|²- Then adding l²|b|² cancels the second term, leaving -kl|b|²So the numerator simplifies to:d(k + l) - kl|a|² - kl|b|² + 2kla·bFactor out kl:= d(k + l) + kl(-|a|² - |b|² + 2a·b)Notice that -|a|² - |b|² + 2a·b = -(|a|² + |b|² - 2a·b) = -|a - b|²Therefore, the numerator becomes:d(k + l) - kl|a - b|²Therefore, the equation is:|m - c|² = [ d(k + l) - kl|a - b|² ] / (k + l)^2Which is similar to the coordinate approach result. So, the radius squared is [ d(k + l) - kl|AB|² ] / (k + l)^2, since |a - b| is the distance between A and B, which is |AB|.Therefore, the locus is a circle with center at c = (ka + lb)/(k + l), which is a weighted average of points A and B, so it lies on the line AB. If the numerator is positive, it's a circle; zero, a point; negative, empty set.Hence, regardless of the coordinate system, the center is on line AB, and the locus is a circle, point, or empty.Therefore, the proof is complete.

✅:To determine the locus of points ( M ) satisfying ( k|AM|^2 + l|BM|^2 = d ), we analyze the equation using coordinate geometry and vector algebra.1. Coordinate Geometry Approach:- Place ( A ) at ((0,0)) and ( B ) at ((b,0)).- Let ( M = (x,y) ). The distances are ( |AM|^2 = x^2 + y^2 ) and ( |BM|^2 = (x - b)^2 + y^2 ).- Substitute into the equation: [ k(x^2 + y^2) + l[(x - b)^2 + y^2] = d ]- Expand and simplify: [ (k + l)x^2 + (k + l)y^2 - 2blx + lb^2 = d ]- Divide by ( k + l ) and complete the square for ( x ): [ left(x - frac{bl}{k + l}right)^2 + y^2 = frac{d(k + l) - klb^2}{(k + l)^2} ]- Let ( R^2 = frac{d(k + l) - kl|AB|^2}{(k + l)^2} ). The locus is: - A circle with center (left(frac{bl}{k + l}, 0right)) (on line ( AB )) if ( R^2 > 0 ). - A single point (the center) if ( R^2 = 0 ). - Empty set if ( R^2 < 0 ).2. Vector Algebra Approach:- Let ( mathbf{a}, mathbf{b}, mathbf{m} ) be position vectors of ( A, B, M ).- Expand the given equation: [ (k + l)|mathbf{m}|^2 - 2mathbf{m} cdot (kmathbf{a} + lmathbf{b}) + k|mathbf{a}|^2 + l|mathbf{b}|^2 = d ]- Complete the square and simplify: [ |mathbf{m} - mathbf{c}|^2 = frac{d(k + l) - kl|mathbf{a} - mathbf{b}|^2}{(k + l)^2}, quad mathbf{c} = frac{kmathbf{a} + lmathbf{b}}{k + l} ]- The center ( mathbf{c} ) lies on line ( AB ), confirming the geometric interpretation.Conclusion:The locus is a circle centered on ( AB ), a point, or empty based on the value of ( d(k + l) - kl|AB|^2 ). Thus, the locus satisfies the given conditions.Final AnswerThe locus is a circle with its center on line ( AB ), a single point, or an empty set. Therefore, the statement is proven as boxed{text{True}}.

🔑:Given two points (A) and (B). We need to prove that the geometric locus of points (M) such that (k|AM|^{2} + l|BM|^{2} = d) (where (k, l, d) are given constants, (k + l neq 0)) is either a circle with its center on the line (AB), a point, or an empty set.We will follow the argument based on the contexts mentioned in Problem II.7.1. Rewrite the Equation in Analytical Form: Let the coordinates of points (A) and (B) be ((a_1, a_2)) and ((b_1, b_2)), respectively. Suppose the coordinates of point (M) are ((x, y)). The squared distances (|AM|^2) and (|BM|^2) can be expressed as: [ |AM|^2 = (x - a_1)^2 + (y - a_2)^2 ] [ |BM|^2 = (x - b_1)^2 + (y - b_2)^2 ]2. Substitute into the Given Equation: Substitute these squared distances into the equation: [ k((x - a_1)^2 + (y - a_2)^2) + l((x - b_1)^2 + (y - b_2)^2) = d ]3. Expand and Group Like Terms: Expand the equation to group the terms involving (x) and (y): [ k(x^2 - 2a_1x + a_1^2 + y^2 - 2a_2y + a_2^2) + l(x^2 - 2b_1x + b_1^2 + y^2 - 2b_2y + b_2^2) = d ] [ kx^2 - 2ka_1x + ka_1^2 + ky^2 - 2ka_2y + ka_2^2 + lx^2 - 2lb_1x + lb_1^2 + ly^2 - 2lb_2y + lb_2^2 = d ] Simplify to: [ (k + l)x^2 + (k + l)y^2 - 2(ka_1 + lb_1)x - 2(ka_2 + lb_2)y + (ka_1^2 + la_2^2 + kb_1^2 + lb_2^2) = d ]4. Consider Different Cases for the Coefficients: - Case 1: (k + l neq 0): [ (k + l)(x^2 + y^2) - 2(ka_1 + lb_1)x - 2(ka_2 + lb_2)y + (ka_1^2 + la_2^2 + kb_1^2 + lb_2^2) = d ] Divide the whole equation by (k + l): [ x^2 + y^2 - frac{2(ka_1 + lb_1)}{k + l}x - frac{2(ka_2 + lb_2)}{k + l}y + frac{ka_1^2 + la_2^2 + kb_1^2 + lb_2^2 - d}{k + l} = 0 ] This equation is a general second-degree equation in (x) and (y) representing a circle (if the discriminant does not result in a degenerate conic). - Case 2: (k + l = 0): In this scenario, if (k + l = 0), then the general equation simplifies to: [ -2(ka_1 + lb_1)x - 2(ka_2 + lb_2)y + (ka_1^2 + la_2^2 - d) = 0 ] This can either represent a point if consistent, or an empty set if non-consistent.5. Conclusion: Based on the results, the geometric locus of points (M) is either a circle centered on the line (AB), a point, or an empty set. [ boxed{text{A circle, a point, or an empty set}} ]

🧠:Okay, so I need to find all positive integers n such that n² + 5n + 13 is a perfect square. Hmm, let's start by understanding what the problem is asking. A perfect square means that the expression n² + 5n + 13 should equal some integer squared, say k². So, we have the equation:n² + 5n + 13 = k²My goal is to find all positive integers n for which this equation holds. Let's think about how to approach this. Since both n and k are integers, maybe I can manipulate the equation to find possible values of n.First, perhaps I can rearrange the equation to see if I can express it in a form that allows me to complete the square or factor it. Let's try completing the square for the quadratic in n.Starting with n² + 5n + 13. To complete the square, take the coefficient of n, which is 5, divide by 2 to get 5/2, then square it to get 25/4. So:n² + 5n + 25/4 - 25/4 + 13 = (n + 5/2)² - 25/4 + 52/4 = (n + 5/2)² + 27/4So, the equation becomes:(n + 5/2)² + 27/4 = k²But since n and k are integers, (n + 5/2) is a half-integer, which might complicate things. Maybe instead, I can multiply the entire equation by 4 to eliminate the fractions. Let's try that:4n² + 20n + 52 = 4k²Now, notice that 4n² + 20n + 25 is a perfect square: (2n + 5)². Let's see:4n² + 20n + 25 = (2n + 5)²So, subtracting this from 4k² gives:4k² - (2n + 5)² = 52 - 25 = 27So, 4k² - (2n + 5)² = 27Hmm, this looks like a difference of squares. Let me write it as:(2k - (2n + 5))(2k + (2n + 5)) = 27Because (a² - b²) = (a - b)(a + b), so here a = 2k and b = (2n + 5). Therefore:(2k - (2n + 5))(2k + (2n + 5)) = 27Let me denote the two factors as x and y, where x = 2k - (2n + 5) and y = 2k + (2n + 5). Then, x * y = 27. Also, since k and n are positive integers, both x and y must be positive integers as well. Moreover, since y = (2k + 2n +5) and x = (2k - 2n -5), we can see that y > x because adding (2n +5) to 2k is greater than subtracting it.So, we need to find pairs of positive integers (x, y) such that x * y = 27 and y > x. Also, since x and y are both integers, the possible factor pairs of 27 are:1 * 273 * 9That's it, since 27 is 3³, so the only factor pairs are (1,27) and (3,9). Let's check each pair.First pair: x = 1, y = 27Then we have:x = 2k - 2n -5 = 1y = 2k + 2n +5 = 27So, we can set up these two equations:1) 2k - 2n -5 = 12) 2k + 2n +5 = 27Let's solve these equations. Let's add both equations:(2k -2n -5) + (2k +2n +5) = 1 +27Simplify:4k = 28 => k = 7Substitute k =7 into equation 1:2*7 -2n -5 =114 -5 -2n =19 -2n =1 => -2n = -8 => n=4So, n=4 is a possible solution.Now check the second pair: x=3, y=9So, x=3, y=9Set up the equations:1) 2k -2n -5=32) 2k +2n +5=9Again, add the two equations:(2k -2n -5) + (2k +2n +5) =3 +94k =12 =>k=3Substitute k=3 into equation 1:2*3 -2n -5=36 -5 -2n=31 -2n=3 => -2n=2 =>n= -1But n must be a positive integer, so discard this solution.Therefore, the only possible solution from the factor pairs is n=4.Wait, but let me check if there are other factor pairs. Since 27 is positive and we considered positive factors, but maybe negative factors? Let's see.But since x and y are both positive integers (as 2k +2n +5 is definitely positive, and x =2k -2n -5 must also be positive because x * y =27 which is positive, so both x and y have to be positive. Therefore, negative factors aren't considered here.Therefore, only n=4 is the solution? Let me check.Wait, let's test n=4 in the original equation to confirm:n² +5n +13 = 16 +20 +13=49=7², which is correct. So that's good.But maybe there are other solutions beyond these factor pairs. Let's consider if there are more possibilities. Since 27 is a small number, perhaps not. But let me check for other possible values of n.Alternatively, perhaps I can approach this by bounding the expression n² +5n +13 between two consecutive squares and see if it can be a perfect square.Let me consider that for large n, n² +5n +13 is between n² and (n +3)². Let's compute (n + a)² and see for which a the expression lies in between.Compute (n +2)^2 =n² +4n +4Compare with n² +5n +13. For n >0, 5n +13 >4n +4 when n > -9, which is always true. So n² +5n +13 > (n +2)^2Now, compute (n +3)^2 =n² +6n +9Compare with n² +5n +13. The difference is (n² +6n +9) - (n² +5n +13) =n -4So, when n -4 >0, i.e., n>4, then (n +3)^2 >n² +5n +13. So, for n>4, the expression lies between (n +2)^2 and (n +3)^2. Since it's a perfect square, it would have to be equal to one of these squares, but for n>4, n² +5n +13 is between (n +2)^2 and (n +3)^2. Therefore, if n>4, then n² +5n +13 cannot be a perfect square because it's strictly between two consecutive squares. Hence, only possible n≤4 need to be checked.Wait, but let's verify this reasoning.For n>4, as above, (n +2)^2 <n² +5n +13 <(n +3)^2. Therefore, n² +5n +13 is between two consecutive squares, so it can't be a perfect square. Therefore, n must be ≤4.But wait, let's check for n=5:n=5: 25 +25 +13=63, which is not a perfect square.n=6:36 +30 +13=79, not a square.n=7:49 +35 +13=97, not a square.n=8:64 +40 +13=117, not a square.n=9:81 +45 +13=139, not a square.n=10:100 +50 +13=163, not a square. So, yeah, seems like beyond n=4, it doesn't hit a square.But wait, the previous argument says for n>4, the expression is between (n +2)^2 and (n +3)^2. Let's confirm that with n=5:(n +2)^2=7²=49, n² +5n +13=25 +25 +13=63, (n +3)^2=8²=64. So 63 is between 49 and 64? Wait, no. Wait, for n=5:Wait, (n +3)^2=8²=64, and (n +2)^2=7²=49. But 63 is between 49 and 64. Wait, but 63 is just 1 less than 64. So, in this case, for n=5, the expression is 63, which is 64 -1. Similarly, for n=6: 36 +30 +13=79. (n +3)^2=9²=81, so 79 is 81 -2.For n=7:49 +35 +13=97. (n +3)^2=10²=100. So 97 is 100 -3. So, as n increases, the expression n² +5n +13 is getting closer to (n +3)^2 but still less by n -4. Wait, earlier difference was (n +3)^2 - (n² +5n +13)=n -4. So for n >4, the difference is positive, meaning n² +5n +13 is (n +3)^2 - (n -4). Therefore, as n increases beyond 4, the difference (n -4) increases, so the expression is (n +3)^2 minus a growing number, so it's getting further away from the next square. So, for example, when n=5, it's 64 -1=63, n=6:81 -2=79, n=7:100 -3=97, etc. So, these are not perfect squares.Therefore, for n>4, the expression is stuck between (n +2)^2 and (n +3)^2, but closer to (n +3)^2, but not a perfect square. Therefore, the only possible values of n to check are n=1,2,3,4.But wait, in our initial analysis, we found n=4 as a solution. Let's check the others:n=1:1 +5 +13=19, not a square.n=2:4 +10 +13=27, not a square.n=3:9 +15 +13=37, not a square.n=4:16 +20 +13=49=7², which is a square.Therefore, only n=4 works. Therefore, the answer is n=4.But wait, let me check n=0 just in case, even though the problem states n is a positive integer. n=0:0 +0 +13=13, not a square.But since the problem specifies n is a positive integer, n=0 is not considered. So, the only solution is n=4.But let me think again if there could be another approach where maybe there's a larger n that satisfies the equation. For example, maybe for some larger n, the expression could coincide with a perfect square further away?Wait, perhaps the equation n² +5n +13 =k² can be rearranged as k² -n² -5n =13. Which can be written as k² =n² +5n +13. Maybe using Pell's equation or other Diophantine equation techniques?Alternatively, maybe consider this as a quadratic in n:n² +5n + (13 -k²)=0For this quadratic equation to have integer solutions, the discriminant must be a perfect square. The discriminant D is:D=25 -4*(1)*(13 -k²)=25 -52 +4k²=4k² -27So, for n to be an integer, 4k² -27 must be a perfect square. Let's denote m²=4k² -27. Then, we have:4k² -m²=27Which can be written as:(2k -m)(2k +m)=27Again, similar to the previous approach. So, this is the same equation as before. Therefore, the possible factor pairs of 27 are (1,27) and (3,9). Therefore, solving for 2k -m=1 and 2k +m=27, which gives 4k=28 =>k=7, m=13. Then, 4k² -27=4*49 -27=196 -27=169=13², so m=13.Alternatively, 2k -m=3 and 2k +m=9. Adding gives 4k=12 =>k=3, m=3. Then, 4k² -27=36 -27=9=3², which is valid. Then, solving for n in the quadratic equation:n² +5n +13 -k²=0. For k=7, this becomes n² +5n +13 -49= n² +5n -36=0. Solving this quadratic: n=(-5 ±√(25 +144))/2=(-5 ±13)/2. Positive solution: (8)/2=4. So n=4. For k=3: n² +5n +13 -9= n² +5n +4=0. Solutions: n=(-5 ±√(25 -16))/2=(-5 ±3)/2. So n=(-5 +3)/2=-1 or n=(-5 -3)/2=-4. Both negative, so no positive solutions. Hence, only n=4. So this approach also leads to n=4.Therefore, seems like n=4 is the only solution.Alternatively, maybe I can model the equation as k² -n²=5n +13. Which factors as (k -n)(k +n)=5n +13. Let me set a=k -n and b=k +n, so that ab=5n +13. Also, since k>n (because k² =n² +5n +13 >n²), a and b are positive integers with b >a. Also, a and b have the same parity because k -n and k +n are both integers, and their sum and difference must be even. Because (k +n) + (k -n)=2k, which is even, and (k +n) - (k -n)=2n, which is even. Therefore, a and b must both be even or both be odd. Since ab=5n +13, let's see.But 5n +13 is equal to ab. Also, from a=k -n and b=k +n, we have:a + b =2kb -a=2nSo, solving these equations, we can express k and n in terms of a and b:k=(a + b)/2n=(b -a)/2Since n must be a positive integer, (b -a)/2 must be a positive integer. Therefore, b -a must be even and positive. Hence, b >a, and b -a is even.Given that ab=5n +13, and n=(b -a)/2, substitute into ab:ab=5*( (b -a)/2 ) +13Multiply both sides by 2:2ab=5(b -a) +26Rearrange:2ab -5b +5a -26=0Hmm, not sure if this helps. Maybe rearrange terms:2ab +5a -5b =26Factor:a(2b +5) -5b=26Hmm, maybe factor differently. Let's see:a(2b +5) =5b +26Thus,a=(5b +26)/(2b +5)Since a must be a positive integer, the numerator must be divisible by the denominator. So, (5b +26) divided by (2b +5) must be an integer. Let's denote c=2b +5, then b=(c -5)/2. Substitute into numerator:5*((c -5)/2) +26= (5c -25)/2 +26= (5c -25 +52)/2= (5c +27)/2Therefore, a=(5c +27)/(2c). But a must be an integer. Therefore, (5c +27) must be divisible by 2c. Let's write this as:(5c +27)/2c =5/2 +27/(2c). For this to be an integer, 27/(2c) must be a number that, when added to 5/2, results in an integer. Let's denote d=27/(2c). Then, 5/2 +d must be integer. Therefore, d must be of the form integer -5/2. Let me think differently.Alternatively, since a=(5b +26)/(2b +5) must be an integer, let's perform the division:Divide 5b +26 by 2b +5.Express 5b +26 = q*(2b +5) + r, where q is the quotient and r is the remainder.Let me compute q. The leading term is 5b divided by 2b, which is 5/2. But since we need integer division, q=2, because 2*(2b +5)=4b +10. Subtract that from 5b +26: (5b +26) - (4b +10)=b +16. Now, the remainder is b +16. Then, the next term would be dividing b +16 by 2b +5. Since 2b +5 >b +16 for b>11, but let's check for b:Wait, for b=1: 2*1 +5=7, remainder=1 +16=17, not divisible.But maybe this approach isn't the most straightforward. Alternatively, set a=(5b +26)/(2b +5) and look for integer values of b where this fraction is integer.Let me denote t=2b +5, so that b=(t -5)/2. Substitute into a:a=(5*(t -5)/2 +26)/t=( (5t -25)/2 +52/2 )/t=(5t +27)/2tThus, a=(5t +27)/(2t)=5/2 +27/(2t)Since a must be an integer, 27/(2t) must be a half-integer. Wait, but 27/(2t) added to 5/2 must be integer. Let me denote 27/(2t)=k -5/2, where k is integer. Then:27/(2t)=k -5/2Multiply both sides by 2t:27=2t(k -5/2)=2tk -5tRearranged:2tk -5t -27=0Hmm, but this seems a bit convoluted. Let me try plugging in small values of t. Since t=2b +5, and b must be at least (since n=(b -a)/2 is positive, and a is positive, so b >a. Also, a=(5b +26)/(2b +5). Since a must be positive, and b is positive, let's try small integer values of b.Let's test b from 1 upwards and see if a is integer.b=1:a=(5*1 +26)/(2*1 +5)=31/7≈4.428... not integer.b=2:a=(10 +26)/(4 +5)=36/9=4. So a=4. Then check if a and b satisfy the conditions. So a=4, b=2. Wait, but b must be greater than a, since b=k +n and a=k -n, so k +n >k -n =>b >a. But here, b=2 and a=4, which violates b >a. So invalid.Wait, but if b=2, then from n=(b -a)/2=(2 -4)/2=-1, which is negative. Disregarded.b=3:a=(15 +26)/(6 +5)=41/11≈3.727... not integer.b=4:a=(20 +26)/(8 +5)=46/13≈3.538... not integer.b=5:a=(25 +26)/(10 +5)=51/15=3.4, not integer.b=6:a=(30 +26)/(12 +5)=56/17≈3.294... nope.b=7:a=(35 +26)/(14 +5)=61/19≈3.210... nope.b=8:a=(40 +26)/(16 +5)=66/21=22/7≈3.142... nope.b=9:a=(45 +26)/(18 +5)=71/23≈3.087... nope.b=10:a=(50 +26)/(20 +5)=76/25=3.04, nope.b=11:a=(55 +26)/(22 +5)=81/27=3. So a=3. Check if b >a: 11 >3, yes. Then, compute n=(b -a)/2=(11 -3)/2=8/2=4. So n=4. Which is the solution we found earlier.Wait, so b=11, a=3. Then k=(a +b)/2=(3 +11)/2=7, which matches the previous result. So, that's consistent.Continuing to check higher b:b=12:a=(60 +26)/(24 +5)=86/29≈2.965... not integer.b=13:a=(65 +26)/(26 +5)=91/31≈2.935... nope.b=14:a=(70 +26)/(28 +5)=96/33=32/11≈2.909... nope.b=15:a=(75 +26)/(30 +5)=101/35≈2.885... nope.b=16:a=(80 +26)/(32 +5)=106/37≈2.864... nope.And as b increases, the value of a approaches 5/2=2.5 from above, so it's decreasing but never reaching an integer. Hence, the only solution here is when b=11, which gives n=4.Therefore, this method also leads to the same solution, n=4. Hence, confirms the result.Alternatively, perhaps we can use inequalities to bound possible k values. For example, since k² =n² +5n +13, we can say that k² >n² +5n. Also, k² <n² +5n +13 +something? Wait, let's see.Alternatively, notice that for n ≥1, n² +5n +13 <n² +6n +9=(n +3)^2 when n >4, as established earlier. But for n ≤4, we can check individually.Wait, but we already did that. For n=1,2,3,4, only n=4 gives a perfect square.Therefore, all methods point towards n=4 being the only solution.Another check: Let's suppose there exists another solution for some larger n. Suppose n=4 +m, where m is a positive integer. Let's substitute n=4 +m into the equation:(4 +m)² +5(4 +m) +13 =16 +8m +m² +20 +5m +13= m² +13m +49So, the equation becomes m² +13m +49 =k²Then, k² -m² -13m =49(k -m)(k +m)=13m +49This seems similar to the previous equations, but now with coefficients 13 and 49. This might not yield solutions easily, but let's check small m:m=1:1 +13 +49=63, not square.m=2:4 +26 +49=79, nope.m=3:9 +39 +49=97, nope.m=4:16 +52 +49=117, nope.Continuing up, no squares found here. So, this approach doesn't yield new solutions.Alternatively, maybe use Pell's equation. Let me rearrange the original equation:k² -n² -5n =13But not sure how to convert this into Pell's equation form. Pell's equation usually is of the form x² - Dy² =N. Here, we have mixed terms with n and k.Alternatively, write the equation as k² =n² +5n +13. Let me express this as:k² = (n + (5/2))² + 13 -25/4 = (n + 5/2)² + 27/4Multiply both sides by 4:4k² = (2n +5)² +27So,4k² - (2n +5)² =27Which is the same as before, leading to the difference of squares.Thus, this method also leads us back to the original factorization. Hence, confirming that n=4 is the only solution.In conclusion, through multiple approaches—completing the square, factoring into difference of squares, checking small values, bounding the expression between consecutive squares, and even considering the equation as a quadratic in n—we consistently find that the only positive integer solution is n=4. All other potential solutions either result in negative numbers or non-integers, which are invalid per the problem's constraints. Therefore, the answer is n=4.

✅:To determine the positive integer ( n ) such that ( n^2 + 5n + 13 ) is a perfect square, we start by setting the expression equal to ( k^2 ), where ( k ) is an integer:[n^2 + 5n + 13 = k^2]Rearranging and completing the square:[4k^2 - (2n + 5)^2 = 27]This can be factored as a difference of squares:[(2k - (2n + 5))(2k + (2n + 5)) = 27]The factor pairs of 27 are ((1, 27)) and ((3, 9)). Solving for each pair:1. Pair (1, 27): [ begin{cases} 2k - (2n + 5) = 1 2k + (2n + 5) = 27 end{cases} ] Adding these equations: [ 4k = 28 implies k = 7 ] Substituting ( k = 7 ): [ 2(7) - 2n - 5 = 1 implies 14 - 2n - 5 = 1 implies n = 4 ]2. Pair (3, 9): [ begin{cases} 2k - (2n + 5) = 3 2k + (2n + 5) = 9 end{cases} ] Adding these equations: [ 4k = 12 implies k = 3 ] Substituting ( k = 3 ): [ 2(3) - 2n - 5 = 3 implies 6 - 2n - 5 = 3 implies n = -1 quad (text{invalid, as } n > 0) ]Testing small values of ( n ) (1, 2, 3, 4) confirms only ( n = 4 ) yields a perfect square (( 49 = 7^2 )). Bounding arguments show for ( n > 4 ), ( n^2 + 5n + 13 ) lies between consecutive squares and cannot be a perfect square.Final Answer: (boxed{4})

🔑:1. Let ( n ) be a positive integer such that ( n^2 + 5n + 13 ) is a perfect square.2. Suppose ( n^2 + 5n + 13 = m^2 ), where ( m ) is an integer. This can be rewritten as: [ n^2 + 5n + 13 = m^2 ] Rearrange terms to isolate the perfect square form: [ n^2 + 5n + 2.5^2 - 2.5^2 + 13 = m^2 ] Simplify inside the parentheses: [ (n + 2.5)^2 + 6.75 = m^2 ]3. To match both sides as a perfect square, set up the equation: [ m^2 - (n + 2.5)^2 = 6.75 ] Recognize this as a difference of squares, giving: [ (m + n + 2.5)(m - n - 2.5) = 6.75 ]4. Multiply through by 4 to avoid dealing with fractions: [ 4(m + n + 2.5)(m - n - 2.5) = 4 times 6.75 = 27 ] So: [ (2m + 2n + 5)(2m - 2n - 5) = 27 ]5. Evaluate possible factor pairs of 27. These pairs should consider the forms of ( (2m + 2n + 5) ) and ( (2m - 2n - 5) ): Two possible factorizations: begin{align*} (a) quad 2m + 2n + 5 = 27, quad 2m - 2n - 5 = 1 & (b) quad 2m + 2n + 5 = 9, quad 2m - 2n - 5 = 3 & end{align*}6. Solve each case: For (a): [ left{ begin{array}{l} 2m + 2n + 5 = 27 2m - 2n - 5 = 1 end{array} right. ] Adding these two equations: [ (2m + 2n + 5) + (2m - 2n - 5) = 27 + 1 implies 4m = 28 implies m = 7 ] Substituting ( m = 7 ) back: [ 2m + 2n + 5 = 27 implies 14 + 2n + 5 = 27 implies 2n = 8 implies n = 4 ] For (b): [ left{ begin{array}{l} 2m + 2n + 5 = 9 2m - 2n - 5 = 3 end{array} right. ] Adding these two equations: [ (2m + 2n + 5) + (2m - 2n - 5) = 9 + 3 implies 4m = 12 implies m = 3 ] Substituting ( m = 3 ) back: [ 2m + 2n + 5 = 9 implies 6 + 2n + 5 = 9 implies 2n = -2 implies n = -1 ] Since ( n ) is positive, ( n = -1 ) is rejected.7. Confirm the solution is correct: [ boxed{4} ]