Problem:
Let a and b be positive real numbers with aβ₯b. Let Ο be the maximum possible value of baβ for which the system of equations
a2+y2=b2+x2=(aβx)2+(bβy)2
has a solution (x,y) satisfying 0β€x<a and 0β€y<b. Then Ο2 can be expressed as a fraction nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let ABCD be a rectangle such that AB=CD=a, and BC=DA=b. Let E and F be points on the sides AB and BC respectively, such that AE=x and CF=y. Solving the given system of equations is thus equivalent to requiring that β³DEF be equilateral. Let β ADE=Ξ±. Then β FDC=30ββΞ±,tanΞ±=bxβ, and β EDF=60β. Thus
Squaring and adding 1 yields (b3β+x)24(x2+b2)β=a2y2+a2β=a2x2+b2β. Thus 4a2=(b3β+x)2, and xβ₯0 implies that x=2aβb3β, which is a positive real number because aβ₯b. Equation (1) shows that yβ₯0 if and only if bβx3β=bβ(2aβb3β)3β=4bβ2a3ββ₯0. It follows that baββ€3β2β, and so Ο=3β2β. Hence Ο2=34β, and m+n=7β. This value of Ο is achieved when a=2,b=3β,x=1, and y=0.