Problem:
Square ABCD is inscribed in a circle. Square EFGH has vertices E and F on CD and vertices G and H on the circle. The ratio of the area of square EFGH to the area of square ABCD can be expressed as m/n where m and n are relatively prime positive integers and m<n. Find 10n+m.
Solution:
Let O be the center of the circle, and represent the lengths of each side of the small square and the large square by x and s, respectively. Draw OL perpendicular to BC at L and FK perpendicular to OL at K. Then GK=GF+FK=GF+CL=x+2s​,OK=x/2, and the circle's radius is (1/2)s2​. Applying the Pythagorean Theorem to triangle OKG, we obtain (x+2s​)2+(2x​)2=(2s2​​)2. Expanding yields x2+sx+4s2​+4x2​=2s2​ which leads to 5x2+4sx−s2=0, or (5x−s)(x+s)=0, so x=s/5. The ratio of the squares' areas is thus 1/25, and 10n+m=251​.
