HMMT November, 2013

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Chords $\overline{AB}$ and $\overline{CD}$ of circle $\omega$ intersect at $E$ such that $AE=8$ , $BE=2$ , $CD=10$ , and $\angle AEC=90^{\circ}$ . Let $R$ be a rectangle inside $\omega$ with sides parallel to $\overline{AB}$ and $\overline{CD}$ , such that no point in the interior of $R$ lies on $\overline{AB}$ , $\overline{CD}$ , or the boundary of $\omega$ . What is the maximum possible area of $R$ ?

Solution

Solution:

Answer: $26+6 \sqrt{17}$

By power of a point, $(CE)(ED) = (AE)(EB) = 16$ , and $CE + ED = CD = 10$ . Thus $CE$ , $ED$ are $2$ , $8$ . Without loss of generality, assume $CE = 8$ and $DE = 2$ .

Assume our circle is centered at the origin, with points $A = (-3,5)$ , $B = (-3,-5)$ , $C = (5,-3)$ , $D = (-5,-3)$ , and the equation of the circle is $x^{2} + y^{2} = 34$ . Clearly the largest possible rectangle must lie in the first quadrant, and if we let $(x, y)$ be the upper-right corner of the rectangle, then the area of the rectangle is $(x+3)(y+3) = 9 + 6(x+y) + x y \leq 9 + 12 \sqrt{\frac{x^{2} + y^{2}}{2}} + \frac{x^{2} + y^{2}}{2} = 26 + 6 \sqrt{17}$ , where equality holds if and only if $x = y = \sqrt{17}$ .

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