2023 WSMO Team Round Problems/Problem 10
Problem
Let square 
be a square with side length 
. Define ellipse 
as the ellipse that is able to be inscribed inside 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 is 
for squarefree 
Find 
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]](http://latex.artofproblemsolving.com/b/9/3/b9366b85dd92172d4120e0c4c227d8036dceed99.png)
We will use analytic geometry. Let 
be the center of the square. We have 
and 
Now, let the length of the equilaterial triangle be 
. So, the minor and major axes have length and 
respectively, meaning the focal distance is 
. Define points 
and 
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 
. This means 
and 
. By the definition of an ellipse, all points 
located on the ellipse must have 
. Let 
be the tangency point of the ellipse with 
and 
be the reflection of across . Due to tangency, 
is minimized at 
over all points on . By properties of a reflection, we have 
. From the triangle inequality, must be located on 
. So, 
. Since is the reflection of 
across 
we have 
. So, 
. Setting this equal to we have 
This means the major and minor axes have length 
respectively, meaning our answer is 
.
~pinkpig
