Problem:
Let a and b be positive real numbers with a≥b. Let ρ be the maximum possible value of ba for which the system of equations
a2+y2=b2+x2=(a−x)2+(b−y)2
has a solution (x,y) satisfying 0≤x<a and 0≤y<b. Then ρ2 can be expressed as a fraction nm, 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 △DEF be equilateral. Let ∠ADE=α. Then ∠FDC=30∘−α,tanα=bx, and ∠EDF=60∘. Thus
ay=tan(30∘−α)=1+31⋅bx31−bx=b3+xb−x3.
Squaring and adding 1 yields (b3+x)24(x2+b2)=a2y2+a2=a2x2+b2. Thus 4a2=(b3+x)2, and x≥0 implies that x=2a−b3, which is a positive real number because a≥b. Equation (1) shows that y≥0 if and only if b−x3=b−(2a−b3)3=4b−2a3≥0. It follows that ba≤32, and so ρ=32. Hence ρ2=34, and m+n=7. This value of ρ is achieved when a=2,b=3,x=1, and y=0.