Question

Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ 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 $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers and $m \lt n.$ Find $10n + m.$

Answer

Answer: 251

Concepts:square (geometry)circle coordinate geometry quadratic

Difficulty rating: 2390

Solution:

Center the circle at the origin and let $ABCD$ have side $s,$ so the circle is $x^2 + y^2 = \frac{s^2}{2}$ and side $\overline{CD}$ lies on the line $y = \frac{s}{2}.$ The small square sits on $\overline{CD},$ outside $ABCD:$ if its side is $t,$ then by symmetry $G = \left(\frac{t}{2}, \frac{s}{2} + t\right),$ which must lie on the circle.

Substituting, $\frac{t^2}{4} + \left(\frac{s}{2} + t\right)^2 = \frac{s^2}{2},$ which expands to $5t^2 + 4st - s^2 = 0,$ or $(5t - s)(t + s) = 0.$ Since $t \gt 0,$ we get $t = \frac{s}{5}.$

The ratio of areas is $\frac{t^2}{s^2} = \frac{1}{25},$ so $m = 1,$ $n = 25,$ and $10n + m = 251.$

2001 AIME II Problem 6

Answer: 251

Difficulty rating: 2390

Solution:

Problem 6 in Other Years