💡Alright, so I have this problem about an ellipse and a moving circle. The ellipse is given by the equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) where (a > b > 0). The circle is given by (x^2 + y^2 = R^2) with (b < R < a). Points (A) and (B) are on the ellipse and the circle respectively, and the line (AB) is tangent to both the ellipse and the circle. I need to find the maximum distance (|AB|).Hmm, okay. So, let me try to visualize this. There's an ellipse and a circle inside it, but the circle isn't too small because (R > b). Points (A) and (B) are on the ellipse and circle, and the line connecting them is tangent to both. So, this line just touches the ellipse at (A) and the circle at (B). I need to find the maximum possible distance between such points (A) and (B).I think I should start by recalling the condition for a line to be tangent to an ellipse and a circle. For a circle, the tangent condition is straightforward: the distance from the center to the line equals the radius. For an ellipse, the condition is a bit more involved, but I remember it involves the slope of the tangent line and the ellipse's parameters.Let me denote the equation of line (AB) as (y = kx + m). Since it's tangent to both the ellipse and the circle, I can use the tangent conditions for both.First, for the circle (x^2 + y^2 = R^2), the condition for the line (y = kx + m) to be tangent is that the distance from the center (which is the origin) to the line equals (R). The distance from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is (frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}). So, for the line (y = kx + m), rewriting it as (kx - y + m = 0), the distance from the origin is (frac{|m|}{sqrt{k^2 + 1}}). This should equal (R), so:[frac{|m|}{sqrt{k^2 + 1}} = R implies |m| = R sqrt{k^2 + 1}]So, (m = pm R sqrt{k^2 + 1}). I'll keep this in mind.Now, for the ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), the condition for the line (y = kx + m) to be tangent is that the discriminant of the quadratic equation formed by substituting (y) into the ellipse equation is zero.Substituting (y = kx + m) into the ellipse equation:[frac{x^2}{a^2} + frac{(kx + m)^2}{b^2} = 1]Expanding this:[frac{x^2}{a^2} + frac{k^2x^2 + 2k m x + m^2}{b^2} = 1]Combine like terms:[left(frac{1}{a^2} + frac{k^2}{b^2}right)x^2 + frac{2k m}{b^2}x + left(frac{m^2}{b^2} - 1right) = 0]For this quadratic equation in (x) to have exactly one solution (since the line is tangent), the discriminant must be zero. The discriminant (Delta) is given by:[Delta = left(frac{2k m}{b^2}right)^2 - 4 left(frac{1}{a^2} + frac{k^2}{b^2}right)left(frac{m^2}{b^2} - 1right) = 0]Simplify this expression:First, compute each part:[left(frac{2k m}{b^2}right)^2 = frac{4k^2 m^2}{b^4}]And,[4 left(frac{1}{a^2} + frac{k^2}{b^2}right)left(frac{m^2}{b^2} - 1right) = 4 left(frac{1}{a^2} + frac{k^2}{b^2}right)left(frac{m^2 - b^2}{b^2}right)]So, putting it all together:[frac{4k^2 m^2}{b^4} - 4 left(frac{1}{a^2} + frac{k^2}{b^2}right)left(frac{m^2 - b^2}{b^2}right) = 0]Factor out the 4:[4 left[ frac{k^2 m^2}{b^4} - left(frac{1}{a^2} + frac{k^2}{b^2}right)left(frac{m^2 - b^2}{b^2}right) right] = 0]Divide both sides by 4:[frac{k^2 m^2}{b^4} - left(frac{1}{a^2} + frac{k^2}{b^2}right)left(frac{m^2 - b^2}{b^2}right) = 0]Let me simplify the second term:[left(frac{1}{a^2} + frac{k^2}{b^2}right)left(frac{m^2 - b^2}{b^2}right) = frac{m^2 - b^2}{a^2 b^2} + frac{k^2 (m^2 - b^2)}{b^4}]So, substituting back into the equation:[frac{k^2 m^2}{b^4} - left( frac{m^2 - b^2}{a^2 b^2} + frac{k^2 (m^2 - b^2)}{b^4} right) = 0]Multiply through by (b^4) to eliminate denominators:[k^2 m^2 - left( frac{(m^2 - b^2) b^2}{a^2} + k^2 (m^2 - b^2) right) = 0]Distribute the negative sign:[k^2 m^2 - frac{(m^2 - b^2) b^2}{a^2} - k^2 (m^2 - b^2) = 0]Factor terms:First, group the (k^2) terms:[k^2 m^2 - k^2 (m^2 - b^2) = k^2 (m^2 - m^2 + b^2) = k^2 b^2]Then, the remaining term:[- frac{(m^2 - b^2) b^2}{a^2}]So, putting it all together:[k^2 b^2 - frac{(m^2 - b^2) b^2}{a^2} = 0]Factor out (b^2):[b^2 left( k^2 - frac{m^2 - b^2}{a^2} right) = 0]Since (b^2 neq 0), we have:[k^2 - frac{m^2 - b^2}{a^2} = 0 implies k^2 = frac{m^2 - b^2}{a^2}]So, (k^2 = frac{m^2 - b^2}{a^2}).But earlier, from the circle's tangent condition, we had (m = pm R sqrt{k^2 + 1}). Let's substitute (k^2) from the ellipse condition into this.Let me write (k^2 = frac{m^2 - b^2}{a^2}). Then, substitute into (m = pm R sqrt{k^2 + 1}):First, compute (k^2 + 1):[k^2 + 1 = frac{m^2 - b^2}{a^2} + 1 = frac{m^2 - b^2 + a^2}{a^2} = frac{m^2 + (a^2 - b^2)}{a^2}]So,[m = pm R sqrt{frac{m^2 + (a^2 - b^2)}{a^2}} = pm frac{R}{a} sqrt{m^2 + (a^2 - b^2)}]Let me square both sides to eliminate the square root:[m^2 = frac{R^2}{a^2} (m^2 + a^2 - b^2)]Multiply both sides by (a^2):[a^2 m^2 = R^2 (m^2 + a^2 - b^2)]Expand the right side:[a^2 m^2 = R^2 m^2 + R^2 a^2 - R^2 b^2]Bring all terms to the left:[a^2 m^2 - R^2 m^2 - R^2 a^2 + R^2 b^2 = 0]Factor (m^2):[m^2 (a^2 - R^2) - R^2 (a^2 - b^2) = 0]Solve for (m^2):[m^2 (a^2 - R^2) = R^2 (a^2 - b^2)][m^2 = frac{R^2 (a^2 - b^2)}{a^2 - R^2}]So, (m = pm sqrt{frac{R^2 (a^2 - b^2)}{a^2 - R^2}}).Now, recall that (k^2 = frac{m^2 - b^2}{a^2}). Let's compute (k^2):[k^2 = frac{frac{R^2 (a^2 - b^2)}{a^2 - R^2} - b^2}{a^2}]Simplify the numerator:[frac{R^2 (a^2 - b^2) - b^2 (a^2 - R^2)}{a^2 - R^2}][= frac{R^2 a^2 - R^2 b^2 - a^2 b^2 + b^2 R^2}{a^2 - R^2}][= frac{R^2 a^2 - a^2 b^2}{a^2 - R^2}][= frac{a^2 (R^2 - b^2)}{a^2 - R^2}]So,[k^2 = frac{frac{a^2 (R^2 - b^2)}{a^2 - R^2}}{a^2} = frac{R^2 - b^2}{a^2 - R^2}]Therefore, (k = pm sqrt{frac{R^2 - b^2}{a^2 - R^2}}).Now, with (k) and (m) expressed in terms of (R), I can find the coordinates of points (A) and (B).For point (A) on the ellipse, since the tangent line is (y = kx + m), the point of tangency can be found using the formula for the point of contact of a tangent to an ellipse. The coordinates are given by:[left( -frac{a^2 k}{m}, frac{b^2}{m} right)]Wait, let me verify this formula. For an ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), the tangent at point ((x_1, y_1)) is (frac{x x_1}{a^2} + frac{y y_1}{b^2} = 1). Comparing this with (y = kx + m), we can find (x_1) and (y_1).Let me write the tangent equation in both forms:From the ellipse: (frac{x x_1}{a^2} + frac{y y_1}{b^2} = 1).From the line: (y = kx + m).Let me express the ellipse tangent equation in slope-intercept form:[frac{x x_1}{a^2} + frac{y y_1}{b^2} = 1 implies y = left( -frac{x_1 b^2}{a^2 y_1} right) x + frac{b^2}{y_1}]Comparing with (y = kx + m), we have:[k = -frac{x_1 b^2}{a^2 y_1}][m = frac{b^2}{y_1}]From the second equation, (y_1 = frac{b^2}{m}). Substituting into the first equation:[k = -frac{x_1 b^2}{a^2 cdot frac{b^2}{m}} = -frac{x_1 m}{a^2}][x_1 = -frac{a^2 k}{m}]So, the coordinates of point (A) are:[A left( -frac{a^2 k}{m}, frac{b^2}{m} right)]Similarly, for point (B) on the circle, the tangent line (y = kx + m) touches the circle at point (B). The formula for the point of contact of a tangent to a circle (x^2 + y^2 = R^2) is (left( -frac{R^2 k}{m}, frac{R^2}{m} right)). Let me verify this.The tangent to the circle at point ((x_2, y_2)) is (x x_2 + y y_2 = R^2). Comparing with (y = kx + m), we can express the tangent equation as:[x x_2 + y y_2 = R^2]Express this in slope-intercept form:[y = -frac{x_2}{y_2} x + frac{R^2}{y_2}]Comparing with (y = kx + m), we get:[k = -frac{x_2}{y_2}][m = frac{R^2}{y_2}]From the second equation, (y_2 = frac{R^2}{m}). Substituting into the first equation:[k = -frac{x_2}{frac{R^2}{m}} = -frac{x_2 m}{R^2}][x_2 = -frac{R^2 k}{m}]So, the coordinates of point (B) are:[B left( -frac{R^2 k}{m}, frac{R^2}{m} right)]Now, I have coordinates for both (A) and (B). Let me write them again:[A left( -frac{a^2 k}{m}, frac{b^2}{m} right)][B left( -frac{R^2 k}{m}, frac{R^2}{m} right)]Now, I need to find the distance (|AB|). Let's compute the distance between these two points.First, compute the differences in coordinates:[Delta x = -frac{a^2 k}{m} - left( -frac{R^2 k}{m} right) = -frac{a^2 k}{m} + frac{R^2 k}{m} = frac{(R^2 - a^2)k}{m}][Delta y = frac{b^2}{m} - frac{R^2}{m} = frac{b^2 - R^2}{m}]So, the distance squared is:[|AB|^2 = left( frac{(R^2 - a^2)k}{m} right)^2 + left( frac{b^2 - R^2}{m} right)^2]Factor out (frac{(R^2 - a^2)^2}{m^2}) from the first term and (frac{(b^2 - R^2)^2}{m^2}) from the second term:Wait, actually, let me compute each term:First term:[left( frac{(R^2 - a^2)k}{m} right)^2 = frac{(a^2 - R^2)^2 k^2}{m^2}]Second term:[left( frac{b^2 - R^2}{m} right)^2 = frac{(R^2 - b^2)^2}{m^2}]So, total distance squared:[|AB|^2 = frac{(a^2 - R^2)^2 k^2 + (R^2 - b^2)^2}{m^2}]But I already have expressions for (k^2) and (m^2) in terms of (R). Let me substitute those in.From earlier, we have:[k^2 = frac{R^2 - b^2}{a^2 - R^2}][m^2 = frac{R^2 (a^2 - b^2)}{a^2 - R^2}]So, substitute (k^2) into the distance squared:[|AB|^2 = frac{(a^2 - R^2)^2 cdot frac{R^2 - b^2}{a^2 - R^2} + (R^2 - b^2)^2}{frac{R^2 (a^2 - b^2)}{a^2 - R^2}}]Simplify numerator:First term:[(a^2 - R^2)^2 cdot frac{R^2 - b^2}{a^2 - R^2} = (a^2 - R^2)(R^2 - b^2)]Second term:[(R^2 - b^2)^2]So, numerator becomes:[(a^2 - R^2)(R^2 - b^2) + (R^2 - b^2)^2 = (R^2 - b^2)[(a^2 - R^2) + (R^2 - b^2)] = (R^2 - b^2)(a^2 - b^2)]So, numerator is ((R^2 - b^2)(a^2 - b^2)).Denominator is (frac{R^2 (a^2 - b^2)}{a^2 - R^2}).So, putting it together:[|AB|^2 = frac{(R^2 - b^2)(a^2 - b^2)}{frac{R^2 (a^2 - b^2)}{a^2 - R^2}} = frac{(R^2 - b^2)(a^2 - b^2)(a^2 - R^2)}{R^2 (a^2 - b^2)}]Cancel out ((a^2 - b^2)):[|AB|^2 = frac{(R^2 - b^2)(a^2 - R^2)}{R^2}]So,[|AB| = sqrt{frac{(R^2 - b^2)(a^2 - R^2)}{R^2}} = frac{sqrt{(R^2 - b^2)(a^2 - R^2)}}{R}]Now, I need to find the maximum value of (|AB|) with respect to (R), where (b < R < a).So, let me denote (f(R) = frac{sqrt{(R^2 - b^2)(a^2 - R^2)}}{R}). I need to maximize (f(R)).Alternatively, since the square root and the denominator are positive, I can maximize (f(R)^2 = frac{(R^2 - b^2)(a^2 - R^2)}{R^2}).Let me set (g(R) = frac{(R^2 - b^2)(a^2 - R^2)}{R^2}). I need to find the maximum of (g(R)).Let me expand the numerator:[(R^2 - b^2)(a^2 - R^2) = R^2 a^2 - R^4 - a^2 b^2 + b^2 R^2 = (a^2 + b^2) R^2 - R^4 - a^2 b^2]So,[g(R) = frac{(a^2 + b^2) R^2 - R^4 - a^2 b^2}{R^2} = (a^2 + b^2) - R^2 - frac{a^2 b^2}{R^2}]So,[g(R) = a^2 + b^2 - R^2 - frac{a^2 b^2}{R^2}]Now, to find the maximum of (g(R)), I can take the derivative with respect to (R) and set it to zero.Compute (g'(R)):[g'(R) = -2R + frac{2 a^2 b^2}{R^3}]Set (g'(R) = 0):[-2R + frac{2 a^2 b^2}{R^3} = 0][-2R = - frac{2 a^2 b^2}{R^3}][2R = frac{2 a^2 b^2}{R^3}][R^4 = a^2 b^2][R = sqrt{a b}]Since (R > 0), we take the positive root.So, the critical point is at (R = sqrt{a b}).Now, we need to check if this is a maximum. Let's compute the second derivative or analyze the behavior.Compute (g''(R)):[g''(R) = -2 - frac{6 a^2 b^2}{R^4}]At (R = sqrt{a b}), (g''(R) = -2 - frac{6 a^2 b^2}{(a b)^2} = -2 - 6 = -8 < 0). So, it's a maximum.Therefore, the maximum of (g(R)) occurs at (R = sqrt{a b}).Compute (g(sqrt{a b})):[g(sqrt{a b}) = a^2 + b^2 - (sqrt{a b})^2 - frac{a^2 b^2}{(sqrt{a b})^2}][= a^2 + b^2 - a b - frac{a^2 b^2}{a b}][= a^2 + b^2 - a b - a b][= a^2 + b^2 - 2 a b][= (a - b)^2]So, (g(R)) reaches a maximum of ((a - b)^2) at (R = sqrt{a b}).Therefore, the maximum (|AB|) is:[|AB| = sqrt{g(R)} = sqrt{(a - b)^2} = |a - b|]Since (a > b), this simplifies to (a - b).So, the maximum distance (|AB|) is (a - b).**Final Answer**The maximum distance (|AB|) is boxed{a - b}.

💡Okay, so I have this inequality to prove: (frac{9r}{2S} leq frac{1}{a} + frac{1}{b} + frac{1}{c} leq frac{9R}{4S}). Hmm, let me think about what each of these terms represents in a triangle. First, (a), (b), and (c) are the lengths of the sides of the triangle. (S) is the area of the triangle, (r) is the inradius, and (R) is the circumradius. So, I need to relate the reciprocals of the side lengths to these radii and the area.I remember that the area (S) of a triangle can be expressed in terms of the inradius and the semiperimeter (s): (S = r cdot s), where (s = frac{a + b + c}{2}). Also, the area can be expressed using the circumradius: (S = frac{a b c}{4 R}). Maybe these relationships can help me connect the terms in the inequality.Looking at the left side of the inequality, (frac{9r}{2S}), if I substitute (S = r cdot s), it becomes (frac{9r}{2 r s} = frac{9}{2 s}). So, the left side simplifies to (frac{9}{2 s}). Similarly, the right side is (frac{9R}{4S}). Substituting (S = frac{a b c}{4 R}), it becomes (frac{9R}{4 cdot frac{a b c}{4 R}} = frac{9R^2}{a b c}). Hmm, not sure if that helps directly.Maybe I should think about the reciprocals of the sides. (frac{1}{a} + frac{1}{b} + frac{1}{c}). I wonder if there's a known inequality involving the reciprocals of the sides of a triangle. I recall that in a triangle, the reciprocals of the sides are related to the heights. Specifically, the height corresponding to side (a) is (h_a = frac{2S}{a}), so (frac{1}{a} = frac{h_a}{2S}). Similarly, (frac{1}{b} = frac{h_b}{2S}) and (frac{1}{c} = frac{h_c}{2S}). Therefore, (frac{1}{a} + frac{1}{b} + frac{1}{c} = frac{h_a + h_b + h_c}{2S}). So, the sum of the reciprocals of the sides is equal to the sum of the heights divided by twice the area.Now, I need to find bounds for (h_a + h_b + h_c). I remember that the heights are related to the inradius and the circumradius. Maybe I can find inequalities involving the sum of the heights.I think there's a result that says the sum of the heights is at least (9r). Let me check that. If (h_a + h_b + h_c geq 9r), then substituting back, we get (frac{h_a + h_b + h_c}{2S} geq frac{9r}{2S}), which is the left side of the inequality we need to prove. So, that would establish the lower bound.For the upper bound, I need to find an upper limit for (h_a + h_b + h_c). I recall that the sum of the heights is less than or equal to the sum of the medians, and the sum of the medians is less than or equal to (frac{9R}{2}). Let me verify that.The medians (m_a), (m_b), (m_c) of a triangle satisfy (m_a + m_b + m_c leq frac{9R}{2}). Since (h_a + h_b + h_c leq m_a + m_b + m_c), combining these gives (h_a + h_b + h_c leq frac{9R}{2}). Substituting back, we get (frac{h_a + h_b + h_c}{2S} leq frac{9R}{4S}), which is the right side of the inequality.Wait, let me make sure I didn't mix up any steps. I used the fact that heights are less than or equal to medians, and then the sum of medians is bounded by (frac{9R}{2}). That seems correct. So, putting it all together, I have:1. (frac{1}{a} + frac{1}{b} + frac{1}{c} = frac{h_a + h_b + h_c}{2S})2. (h_a + h_b + h_c geq 9r) implies (frac{1}{a} + frac{1}{b} + frac{1}{c} geq frac{9r}{2S})3. (h_a + h_b + h_c leq frac{9R}{2}) implies (frac{1}{a} + frac{1}{b} + frac{1}{c} leq frac{9R}{4S})Therefore, the inequality (frac{9r}{2S} leq frac{1}{a} + frac{1}{b} + frac{1}{c} leq frac{9R}{4S}) holds.I think that covers both the lower and upper bounds. I should double-check if the relationships between heights, medians, and radii are correctly applied. Also, ensuring that the substitutions for the area (S) in terms of (r) and (R) are accurate. Yes, (S = r cdot s) and (S = frac{a b c}{4 R}) are standard formulas, so substituting them was appropriate. The key step was expressing the reciprocals of the sides in terms of the heights, which allowed me to use known inequalities involving heights, medians, inradius, and circumradius.Overall, the proof seems solid. I just need to present it step-by-step to make sure each part is clear.

💡Alright, so I've got this problem here: If (a, b, c) are integers with (a < c) and they satisfy the equation (a^2 + c^2 = 2b^2), then I need to show that (c^2 - a^2) is divisible by 48. Hmm, okay. Let's try to unpack this step by step.First off, the equation (a^2 + c^2 = 2b^2) reminds me of something related to Pythagorean triples, but it's not exactly the same because of the factor of 2 on the right side. Maybe I can manipulate this equation to make it look more familiar.Let me try multiplying both sides by 2. That gives me:[2a^2 + 2c^2 = 4b^2]Hmm, okay. Now, I wonder if I can factor this or find some relationship between (a) and (c). Maybe I can express this in terms of ((a + c)^2) and ((a - c)^2). Let's see:I know that ((a + c)^2 = a^2 + 2ac + c^2) and ((a - c)^2 = a^2 - 2ac + c^2). If I add these two together, I get:[(a + c)^2 + (a - c)^2 = 2a^2 + 2c^2]Oh, that's exactly the left side of my equation after multiplying by 2! So, substituting back in, I have:[(a + c)^2 + (a - c)^2 = 4b^2]Interesting. So, the sum of these two squares equals (4b^2). That means both ((a + c)^2) and ((a - c)^2) must be even because their sum is divisible by 4. Therefore, both (a + c) and (a - c) must be even numbers. Let me write that down:Let (a + c = 2p) and (a - c = 2q), where (p) and (q) are integers. Then, substituting back, we have:[(2p)^2 + (2q)^2 = 4b^2][4p^2 + 4q^2 = 4b^2]Divide both sides by 4:[p^2 + q^2 = b^2]Okay, so now we have a Pythagorean triple: (p, q, b). That simplifies things a bit. Now, the original question is about (c^2 - a^2). Let me express that in terms of (p) and (q):We know that (c^2 - a^2 = (c - a)(c + a)). From our earlier substitutions, (c + a = 2p) and (c - a = 2q). So:[c^2 - a^2 = (2q)(2p) = 4pq]So, (c^2 - a^2 = 4pq). Now, I need to show that this is divisible by 48. That means I need to show that (pq) is divisible by 12 because (4 times 12 = 48).Alright, so let's focus on showing that (pq) is divisible by 12. To do this, I need to show that (pq) is divisible by both 3 and 4 because 12 is the least common multiple of 3 and 4.First, let's check divisibility by 3. Since (p) and (q) are part of a Pythagorean triple, they must satisfy certain properties. In any Pythagorean triple, at least one of the legs must be divisible by 3. Let me recall that in a primitive Pythagorean triple, one leg is even, and the other is odd, and one of them is divisible by 3.So, in our case, since (p^2 + q^2 = b^2), either (p) or (q) must be divisible by 3. If both were not divisible by 3, then their squares would be congruent to 1 modulo 3, and their sum would be 2 modulo 3, which cannot be a perfect square because squares modulo 3 are either 0 or 1. Therefore, at least one of (p) or (q) must be divisible by 3.Next, let's check divisibility by 4. Since (p) and (q) are integers, and from our earlier substitution, (a + c = 2p) and (a - c = 2q), we can infer that both (p) and (q) must be integers. Moreover, since (a) and (c) are integers, (p) and (q) must also be integers.Now, looking back at the equation (p^2 + q^2 = b^2), we can analyze the parity of (p) and (q). If both (p) and (q) were odd, then their squares would both be 1 modulo 4, and their sum would be 2 modulo 4. However, (b^2) must be either 0 or 1 modulo 4 because squares modulo 4 are either 0 or 1. Therefore, having both (p) and (q) odd leads to a contradiction because (p^2 + q^2) would be 2 modulo 4, which isn't a square.Hence, at least one of (p) or (q) must be even. But since (p) and (q) are part of a Pythagorean triple, and in a primitive triple, exactly one of the legs is even. However, our triple might not necessarily be primitive. But regardless, we can say that at least one of (p) or (q) is even, making (pq) divisible by 2.But we need (pq) to be divisible by 4. So, let's dig deeper. Suppose (p) is even and (q) is odd. Then, (p = 2k) for some integer (k), and (q = 2m + 1) for some integer (m). Then, (pq = 2k(2m + 1) = 4km + 2k), which is divisible by 2 but not necessarily by 4 unless (k) is even.Wait, maybe I need a different approach. Let's consider the fact that (p^2 + q^2 = b^2). If (p) is even, then (p = 2k), so (p^2 = 4k^2). Then, (q^2 = b^2 - 4k^2). For (q^2) to be a perfect square, (b^2 - 4k^2) must be a perfect square. Similarly, if (q) is even, (q = 2m), then (p^2 = b^2 - 4m^2), which must also be a perfect square.But perhaps a better way is to consider the properties of Pythagorean triples. In any Pythagorean triple, one leg is divisible by 4. Wait, is that true? Let me think. In a primitive Pythagorean triple, one leg is even, and the other is odd. The even leg is divisible by 4 because in the formula for generating triples, (m^2 - n^2) and (2mn), the even term (2mn) is divisible by 4 if either (m) or (n) is even.Wait, actually, in a primitive triple, exactly one of the legs is even, and that even leg is divisible by 4. So, if our triple is primitive, then one of (p) or (q) is divisible by 4, and the other is odd. If the triple is not primitive, then both (p) and (q) could be even, making (pq) divisible by 4.But in our case, (p) and (q) are derived from (a) and (c), which are integers with (a < c). So, depending on whether the original triple is primitive or not, (p) and (q) could have different properties.However, regardless of whether the triple is primitive or not, we can say that at least one of (p) or (q) is divisible by 4, and the other is even or odd. Wait, no, that might not necessarily be the case. Let me think again.If the triple is primitive, then one leg is divisible by 4, and the other is odd. If the triple is not primitive, then both legs could be even, meaning both (p) and (q) are even, making (pq) divisible by 4.But in our case, since (p) and (q) are derived from (a) and (c), which are integers, and (a < c), we can assume that (p) and (q) are positive integers with (p > q) because (a + c > a - c) when (c > a).Wait, actually, (a + c = 2p) and (a - c = 2q). Since (a < c), (a - c) is negative, so (q) must be negative. But since we're dealing with squares, the sign doesn't matter. So, we can consider (q) as positive without loss of generality.But maybe I'm overcomplicating. Let's get back to the main point. We need to show that (pq) is divisible by 12, which means it must be divisible by both 3 and 4.We've already established that (pq) is divisible by 3 because in the Pythagorean triple, at least one of (p) or (q) must be divisible by 3.Now, for divisibility by 4, let's consider the parity of (p) and (q). If both (p) and (q) are even, then (pq) is divisible by 4. If one is even and the other is odd, then (pq) is divisible by 2 but not necessarily by 4. However, in our case, since (p^2 + q^2 = b^2), we can analyze the possible parities.If both (p) and (q) are odd, then (p^2 + q^2) would be (1 + 1 = 2) modulo 4, which cannot be a perfect square because squares modulo 4 are 0 or 1. Therefore, both (p) and (q) cannot be odd. Hence, at least one of them must be even.But in a Pythagorean triple, exactly one of the legs is even in a primitive triple. However, if the triple is not primitive, both legs could be even. So, in our case, regardless of whether the triple is primitive or not, we can say that at least one of (p) or (q) is even, making (pq) divisible by 2.But we need (pq) to be divisible by 4. So, let's consider the case where one is even and the other is odd. If (p) is even and (q) is odd, then (p = 2k) and (q = 2m + 1). Then, (pq = 2k(2m + 1) = 4km + 2k). This is divisible by 2, but not necessarily by 4 unless (k) is even.Wait, but in the Pythagorean triple (p^2 + q^2 = b^2), if (p) is even and (q) is odd, then (p) must be divisible by 4. Because in a primitive triple, the even leg is divisible by 4. So, if (p) is even, it's actually divisible by 4, making (p = 4k). Then, (pq = 4k times q), which is divisible by 4.Similarly, if (q) is even, then (q) must be divisible by 4, making (pq) divisible by 4.Therefore, in either case, whether (p) or (q) is even, (pq) is divisible by 4.So, combining both results, (pq) is divisible by both 3 and 4, hence divisible by 12. Therefore, (c^2 - a^2 = 4pq) is divisible by (4 times 12 = 48).Wait, let me just double-check that. If (pq) is divisible by 12, then (4pq) is divisible by 48. Yes, that makes sense.So, putting it all together:1. Start with (a^2 + c^2 = 2b^2).2. Multiply both sides by 2 to get (2a^2 + 2c^2 = 4b^2).3. Recognize that ((a + c)^2 + (a - c)^2 = 4b^2).4. Let (a + c = 2p) and (a - c = 2q), leading to (p^2 + q^2 = b^2).5. Express (c^2 - a^2 = 4pq).6. Show that (pq) is divisible by 12 by proving it's divisible by both 3 and 4.7. Conclude that (4pq) is divisible by 48.Therefore, (c^2 - a^2) is indeed divisible by 48.

💡Okay, so Krista is starting a new savings routine, and she begins by depositing 2 cents on a Sunday morning. Each day after that, she doubles the amount she deposited the previous day. She keeps doing this for two weeks, and we need to figure out on which day of the week the total amount in her bank will first exceed 5.Alright, let me break this down. First, she starts with 2 cents on Sunday. Then, each day, she deposits double the previous day's amount. So, on Monday, she'll deposit 4 cents, Tuesday 8 cents, Wednesday 16 cents, and so on. This seems like a geometric sequence where each term is double the previous one.I remember that the sum of a geometric series can be calculated using the formula:[ S_n = a times frac{r^n - 1}{r - 1} ]where:- ( S_n ) is the sum of the first ( n ) terms,- ( a ) is the first term,- ( r ) is the common ratio,- ( n ) is the number of terms.In this case, the first term ( a ) is 2 cents, which is 0.02. The common ratio ( r ) is 2 because she's doubling each day. We need to find the smallest ( n ) such that the total sum ( S_n ) exceeds 5.So, plugging the values into the formula:[ S_n = 0.02 times frac{2^n - 1}{2 - 1} ][ S_n = 0.02 times (2^n - 1) ]We want this sum to be greater than 5:[ 0.02 times (2^n - 1) > 5 ]Let me solve for ( n ):First, divide both sides by 0.02 to get rid of the coefficient:[ 2^n - 1 > frac{5}{0.02} ][ 2^n - 1 > 250 ][ 2^n > 251 ]Now, I need to find the smallest integer ( n ) such that ( 2^n > 251 ). Let's calculate some powers of 2:- ( 2^7 = 128 )- ( 2^8 = 256 )So, ( 2^8 = 256 ) is the first power of 2 that exceeds 251. Therefore, ( n = 8 ).Since she started on a Sunday, we need to figure out what day of the week it will be 8 days later. Let's count:1. Sunday (day 1)2. Monday (day 2)3. Tuesday (day 3)4. Wednesday (day 4)5. Thursday (day 5)6. Friday (day 6)7. Saturday (day 7)8. Sunday (day 8)So, 8 days later is the following Sunday.Wait, let me double-check. If day 1 is Sunday, then day 8 is also a Sunday because 7 days later would be the next Sunday, and day 8 is the day after that, which is Monday? Hmm, no, that doesn't seem right.Wait, no, actually, if you start counting day 1 as Sunday, then:- Day 1: Sunday- Day 2: Monday- Day 3: Tuesday- Day 4: Wednesday- Day 5: Thursday- Day 6: Friday- Day 7: Saturday- Day 8: SundayYes, that's correct. So, day 8 is the next Sunday.But wait, let me make sure about the total amount. On day 8, the total sum is:[ S_8 = 0.02 times (2^8 - 1) = 0.02 times (256 - 1) = 0.02 times 255 = 5.1 ]So, 5.10, which is indeed more than 5. So, the total exceeds 5 on the 8th day, which is a Sunday.But just to be thorough, let me check the total on day 7:[ S_7 = 0.02 times (2^7 - 1) = 0.02 times (128 - 1) = 0.02 times 127 = 2.54 ]That's only 2.54, which is less than 5. So, day 7 is Saturday, and the total is still below 5. Therefore, the first day the total exceeds 5 is day 8, which is Sunday.So, the answer should be Sunday.

💡Okay, so I have this problem where I need to arrange the numbers 1, 2, ..., n around a circle. The condition is that the difference between any two adjacent numbers doesn't exceed 2. And I need to show that this can be done in only one way, considering rotations and reflections as the same arrangement. Hmm, interesting.Let me start by understanding the problem better. If I arrange numbers in a circle, each number has two neighbors. The difference between each pair of neighbors should be at most 2. So, for example, if I have the number 1, its neighbors can only be 2 or 3 because 1 + 2 = 3. Similarly, if I have the number 2, its neighbors can be 1, 3, or 4, and so on.I think the key here is to figure out a specific arrangement that satisfies this condition and then show that any other arrangement would either violate the difference condition or would be equivalent to the first one under rotation or reflection.Maybe I can start by arranging the numbers in order: 1, 2, 3, ..., n. Let's see if this works. In this case, each adjacent pair has a difference of 1, which is definitely less than or equal to 2. So this arrangement satisfies the condition.But the problem says that this is the only way. So I need to show that no other arrangement is possible. How can I approach this?Perhaps I can assume that there's another arrangement and then show that it must actually be the same as the ordered arrangement. Let's suppose there's a different arrangement where the numbers aren't in order. Maybe somewhere, a number is followed by a number that's more than 2 greater or less than it.Wait, but if I have a number k, the next number can only be k-2, k-1, k+1, or k+2. So if I start at 1, the next number can only be 2 or 3. If I choose 3, then the next number can be 1, 2, 4, or 5. But 1 is already used, so it can be 2, 4, or 5. If I choose 2, then the next number can be 1, 3, or 4. But 1 and 3 are already used, so it must be 4. Hmm, this seems to be forcing the arrangement back into the ordered sequence.Let me try to formalize this. Suppose I have an arrangement where the numbers are not in order. Then there must be some point where a number is followed by a number that's more than 2 away. But wait, the condition says that the difference can't exceed 2, so that can't happen. Therefore, every number must be followed by a number that's either one or two more or less.But if I start at 1, the next number has to be 2 or 3. If I choose 3, then the next number has to be 1, 2, 4, or 5. But 1 is already used, so it can't be 1 again. If I choose 2, then the next number has to be 1, 3, or 4. But 1 and 3 are already used, so it has to be 4. This seems like it's leading me back to the ordered sequence.Alternatively, if I start at 1 and choose 2 as the next number, then the next number can be 3 or 4. If I choose 3, then the next number can be 1, 2, 4, or 5. But 1 and 2 are already used, so it has to be 4 or 5. If I choose 4, then the next number can be 2, 3, 5, or 6. But 2 and 3 are already used, so it has to be 5 or 6. This seems like it's also leading me back to the ordered sequence.Wait, maybe I can think about this in terms of graph theory. Each number is a node, and edges connect numbers that differ by at most 2. Then, arranging the numbers in a circle with the given condition is equivalent to finding a Hamiltonian cycle in this graph.So, if I can show that the only Hamiltonian cycle in this graph is the ordered sequence, then that would prove the uniqueness. But I'm not sure about that. Maybe I need a different approach.Another idea is to use induction. Let's assume that for n-1, the only arrangement is the ordered sequence. Then, when adding the nth number, it has to be placed next to n-1 or n-2 because the difference can't exceed 2. But in the ordered sequence, n is already next to n-1, so adding it there would just extend the sequence. But I'm not sure if this covers all cases.Wait, maybe I can think about the arrangement as a permutation of the numbers 1 to n. The condition is that adjacent numbers in the permutation differ by at most 2. So, I need to show that the only such permutation, up to rotation and reflection, is the identity permutation.I think the key is to realize that once you fix the position of 1, the rest of the numbers are forced into their positions because each number can only be adjacent to numbers that are one or two away. So, starting from 1, the next number has to be 2 or 3. If it's 2, then the next has to be 3 or 4, and so on. If it's 3, then the next has to be 2 or 4, but 2 is already used, so it has to be 4, and so on. This seems to force the arrangement into the ordered sequence.Alternatively, if I try to place 3 next to 1, then the next number has to be 2 or 4. If it's 2, then the next has to be 3 or 4, but 3 is already used, so it has to be 4. Then the next has to be 5 or 6, and so on. But this also leads to the ordered sequence.Wait, but what if I try to place 4 next to 1? Then the difference would be 3, which exceeds 2, so that's not allowed. So, 1 can only be adjacent to 2 or 3. Similarly, 2 can only be adjacent to 1, 3, or 4, and so on.This seems to create a chain where each number is forced to be next to the next number in the sequence. Therefore, the only possible arrangement is the ordered sequence, and any other arrangement would either violate the difference condition or would be equivalent to the ordered sequence under rotation or reflection.I think I'm getting closer. Maybe I can formalize this by considering the possible neighbors for each number and showing that the only way to arrange them without violating the difference condition is in the ordered sequence.Let me try to write this out step by step.1. Start with number 1. Its neighbors must be 2 or 3.2. Suppose the next number is 2. Then, the next number after 2 must be 3 or 4.3. If it's 3, then the next number must be 4 or 5.4. Continuing this way, we get the sequence 1, 2, 3, 4, ..., n.5. If at any point we choose a number that's two more instead of one more, we still end up forcing the next number to continue the sequence.Alternatively, if we start with 1 and choose 3 as the next number, then the next number must be 2 or 4. If it's 2, then the next number must be 3 or 4, but 3 is already used, so it has to be 4, and so on, leading back to the ordered sequence.Therefore, regardless of whether we choose 2 or 3 after 1, we end up with the ordered sequence. This suggests that the only possible arrangement is the ordered sequence, up to rotation and reflection.I think this makes sense. Each choice after 1 forces the next number to continue the sequence, and there's no way to deviate without violating the difference condition. Therefore, the arrangement is unique.So, in conclusion, the only way to arrange the numbers 1, 2, ..., n around a circle such that the difference between any two adjacent numbers does not exceed 2 is in the ordered sequence, and this arrangement is unique up to rotation and reflection.

💡Okay, so I have this problem about a right-angled triangle ABC with legs AC = a and CB = b. I need to find two things: a) The side length of the square with one vertex at C that has the largest area and lies entirely within the triangle ABC.b) The dimensions of the rectangle with one vertex at C that has the largest area and lies entirely within the triangle ABC.Hmm, let's start with part a). I remember that in a right-angled triangle, if you want to fit a square inside it with one vertex at the right angle, the square will touch both legs and the hypotenuse. So, the square will have one corner at point C, and the other two corners touching the legs AC and CB, and the fourth corner touching the hypotenuse AB.I think the key here is to relate the side length of the square to the sides of the triangle. Let me denote the side length of the square as s. So, the square will divide the triangle into smaller similar triangles and the square itself.Wait, similar triangles? That might be useful. If I draw the square inside the triangle, the remaining part of the triangle above the square should be similar to the original triangle ABC. That makes sense because all the angles will be the same.So, if the original triangle has legs a and b, the smaller triangle above the square will have legs (a - s) and (b - s). Since they are similar, the ratios of corresponding sides should be equal. So, the ratio of the legs of the smaller triangle should be equal to the ratio of the legs of the original triangle.Mathematically, that would be:(a - s) / a = (b - s) / bWait, is that correct? Or should it be (a - s) / (b - s) = a / b? Hmm, maybe I need to think about it differently.Let me consider the similar triangles. The original triangle ABC has legs a and b, and the smaller triangle above the square has legs (a - s) and (b - s). Since they are similar, the ratio of their corresponding sides should be equal. So, (a - s)/a = (b - s)/b.Yes, that seems right. So, setting up the equation:(a - s)/a = (b - s)/bCross-multiplying:b(a - s) = a(b - s)Expanding both sides:ab - bs = ab - asSubtracting ab from both sides:-bs = -asMultiplying both sides by -1:bs = asThen, s(b - a) = 0Wait, that would imply s = 0 or b = a. But s can't be zero because we have a square, and b doesn't necessarily equal a. Hmm, I must have made a mistake here.Maybe my assumption about the similar triangles is incorrect. Let me try a different approach.Let me consider the coordinates. Let's place point C at the origin (0,0), point A at (0,a), and point B at (b,0). Then, the hypotenuse AB can be represented by the line connecting (0,a) and (b,0). The equation of this line is y = (-a/b)x + a.Now, if I place a square with side length s inside the triangle, one corner at C (0,0), and the square extending along the x and y axes. The top corner of the square will be at (s,s). This point (s,s) must lie on the hypotenuse AB. So, substituting x = s and y = s into the equation of AB:s = (-a/b)s + aLet me solve for s:s + (a/b)s = as(1 + a/b) = as = a / (1 + a/b)Simplify the denominator:s = a / ((b + a)/b) = (a * b) / (a + b)So, s = (ab)/(a + b)Okay, that makes sense. So, the side length of the square is ab/(a + b). That seems reasonable because if a = b, then s = a/2, which is correct.Alright, so that solves part a). Now, moving on to part b), which is about finding the dimensions of the rectangle with one vertex at C that has the largest area within the triangle.This seems similar to the square problem, but now it's a rectangle, so the sides can be different. Let me denote the sides of the rectangle as x and y, with x along the x-axis and y along the y-axis. So, the rectangle will have one corner at (0,0), extend to (x,0) along the x-axis, to (0,y) along the y-axis, and the opposite corner at (x,y) must lie on the hypotenuse AB.So, similar to the square, the point (x,y) must satisfy the equation of the hypotenuse AB, which is y = (-a/b)x + a.Therefore, y = (-a/b)x + a.So, the area of the rectangle is A = x * y.Substituting y from the equation of AB:A = x * [(-a/b)x + a] = (-a/b)x^2 + a xSo, A(x) = (-a/b)x^2 + a xTo find the maximum area, we can take the derivative of A with respect to x and set it to zero.dA/dx = (-2a/b)x + aSetting dA/dx = 0:(-2a/b)x + a = 0(-2a/b)x = -ax = (-a) / (-2a/b) = (a) / (2a/b) = b/2So, x = b/2Then, y = (-a/b)(b/2) + a = (-a/2) + a = a/2So, the dimensions of the rectangle are x = b/2 and y = a/2.Therefore, the rectangle with the largest area has sides equal to half of the legs of the triangle.Wait, that's interesting. So, the maximum area rectangle is actually a square only if a = b. Otherwise, it's a rectangle with sides half of each leg.Let me verify this result. If I plug x = b/2 into the area formula:A = (-a/b)(b/2)^2 + a*(b/2) = (-a/b)(b^2/4) + (a b)/2 = (-a b /4) + (a b)/2 = (a b)/4So, the maximum area is (a b)/4, which is indeed the maximum because the second derivative of A(x) is negative, confirming a maximum.Alternatively, I could have used the method of completing the square or using the vertex formula for a quadratic function, but taking the derivative seems straightforward.So, summarizing:a) The side length of the square is s = ab/(a + b)b) The dimensions of the rectangle are x = b/2 and y = a/2I think that's it. Let me just double-check if these results make sense.For part a), if a = b, then s = a^2/(2a) = a/2, which is correct because the square would fit perfectly in the middle, each side being half the leg length.For part b), if a = b, then the rectangle would also be a square with sides a/2, which is consistent with part a). However, in the case where a ≠ b, the rectangle is not a square, but its area is maximized when each side is half the length of the respective leg.Yes, that seems consistent. I don't see any errors in the reasoning.