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Difference between revisions of "2023 WSMO Team Round Problems/Problem 10"

(Created page with "==Problem== Let square <math>ABCD</math> be a square with side length <math>4</math>. Define ellipse <math>\omega</math> as the ellipse that is able to be inscribed inside <m...")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
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<asy>
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import math;
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unitsize(2cm);
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pair A=(-2,2),B=(2,2),C=(2,-2),D=(-2,-2);
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draw(A--B--C--D--cycle);
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label("$A$",A,NW);
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label("$B$",B,NE);
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label("$C$",C,SE);
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label("$D$",D,SW);
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transform T = rotate(45);
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draw(T*ellipse((0,0),sqrt(6),sqrt(2)),red);
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pair F1 = (sqrt(2),sqrt(2));
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pair F2 = (-sqrt(2),-sqrt(2));
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pair F2_ = (-sqrt(2),-4+sqrt(2));
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pair X = (-1,-2);
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dot(F1);label("$F_1$",F1,NE);
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dot(F2);label("$F_2$",F2,W);
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dot(F2_);label("$F_2'$",F2_,W);
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dot((0,0));label("$O$",(0,0),SE);
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dot(X);label("$X$",(-1,-2),S);
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draw(F1--X--F2,blue);
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draw(X--F2_,blue+dotted);
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</asy>
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 +
We will use analytic geometry. Let <math>O = (0,0)</math> be the center of the square. We have <math>A = (-2,2),B = (2,0),C = (2,-2),</math> and <math>D = (-2,-2).</math> Now, let the length of the equilaterial triangle be <math>2t</math>. So, the minor and major axes have length <math>2t</math> and <math>2t\sqrt{3},</math> respectively, meaning the focal distance is <math>2t\sqrt{2}</math>. Define points <math>F_1</math> and <math>F_2</math> to be the two focus points of the ellipse. Note that the major axis coincides with the diagonal of the square, meaning the angle of the ellipse with respect to the horizontal must be <math>45^{\circ}</math>. This means <math>F_1 = (t,t)</math> and <math>F_2 = (-t,-t)</math>. By the definition of an ellipse, all points <math>P</math> located on the ellipse must have <math>F_1P+F2P = 2t\sqrt{3}</math>. Let <math>X</math> be the tangency point of the ellipse with <math>CD</math> and <math>F_2'</math> be the reflection of <math>F_2</math> across <math>CD</math>. Due to tangency, <math>F_1P+F_2P</math> is minimized at <math>P=X</math> over all points <math>P</math> on <math>CD</math>. By properties of a reflection, we have <math>F_1X+F_2X = F_1X+XF_2'</math>. From the triangle inequality, <math>X</math> must be located on  <math>F_1F_2'</math>. So, <math>F_1X+XF_2'=F_1F_2'</math>. Since <math>F_2'</math> is the reflection of <math>F_2=(-t,-t)</math> across <math>CD:y=-2,</math> we have <math>F_2'=(-t,-4+t)</math>. So, <math>F_1F_2' = (2t)^2+(-4)^2 = 4t^2+16</math>. Setting this equal to <math>2t\sqrt{3},</math> we have <math>4t^2+16 = 12t^2\implies t^2=2.</math> This means the major and minor axes have length <math>2\sqrt{6},2\sqrt{2},</math> respectively, meaning our answer is <math>(\sqrt{6})(\sqrt{2})\pi = 2\pi\sqrt{3}\implies2+3 = \boxed{5}</math>.
 +
 +
~pinkpig

Latest revision as of 16:16, 13 September 2025

Problem

Let square $ABCD$ be a square with side length $4$. Define ellipse $\omega$ as the ellipse that is able to be inscribed inside $ABCD$ such that 2 of its vertices on its minor axis and 1 of its vertices on its major axis form an equilateral triangle. The largest possible area of $\omega$ is $m\pi\sqrt{n},$ for squarefree $n.$ Find $m+n.$

Solution

[asy] import math; unitsize(2cm); pair A=(-2,2),B=(2,2),C=(2,-2),D=(-2,-2); draw(A--B--C--D--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); transform T = rotate(45); draw(T*ellipse((0,0),sqrt(6),sqrt(2)),red); pair F1 = (sqrt(2),sqrt(2)); pair F2 = (-sqrt(2),-sqrt(2)); pair F2_ = (-sqrt(2),-4+sqrt(2)); pair X = (-1,-2); dot(F1);label("$F_1$",F1,NE); dot(F2);label("$F_2$",F2,W); dot(F2_);label("$F_2'$",F2_,W); dot((0,0));label("$O$",(0,0),SE); dot(X);label("$X$",(-1,-2),S); draw(F1--X--F2,blue); draw(X--F2_,blue+dotted); [/asy]

We will use analytic geometry. Let $O = (0,0)$ be the center of the square. We have $A = (-2,2),B = (2,0),C = (2,-2),$ and $D = (-2,-2).$ Now, let the length of the equilaterial triangle be $2t$. So, the minor and major axes have length $2t$ and $2t\sqrt{3},$ respectively, meaning the focal distance is $2t\sqrt{2}$. Define points $F_1$ and $F_2$ to be the two focus points of the ellipse. Note that the major axis coincides with the diagonal of the square, meaning the angle of the ellipse with respect to the horizontal must be $45^{\circ}$. This means $F_1 = (t,t)$ and $F_2 = (-t,-t)$. By the definition of an ellipse, all points $P$ located on the ellipse must have $F_1P+F2P = 2t\sqrt{3}$. Let $X$ be the tangency point of the ellipse with $CD$ and $F_2'$ be the reflection of $F_2$ across $CD$. Due to tangency, $F_1P+F_2P$ is minimized at $P=X$ over all points $P$ on $CD$. By properties of a reflection, we have $F_1X+F_2X = F_1X+XF_2'$. From the triangle inequality, $X$ must be located on $F_1F_2'$. So, $F_1X+XF_2'=F_1F_2'$. Since $F_2'$ is the reflection of $F_2=(-t,-t)$ across $CD:y=-2,$ we have $F_2'=(-t,-4+t)$. So, $F_1F_2' = (2t)^2+(-4)^2 = 4t^2+16$. Setting this equal to $2t\sqrt{3},$ we have $4t^2+16 = 12t^2\implies t^2=2.$ This means the major and minor axes have length $2\sqrt{6},2\sqrt{2},$ respectively, meaning our answer is $(\sqrt{6})(\sqrt{2})\pi = 2\pi\sqrt{3}\implies2+3 = \boxed{5}$.

~pinkpig

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