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Harvard-MIT November Tournament

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Let $ABCD$ be a rectangle with $AB = 3$ and $BC = 7$ . Let $W$ be a point on segment $AB$ such that $AW = 1$ . Let $X, Y, Z$ be points on segments $BC, CD, DA$ , respectively, so that quadrilateral $WXYZ$ is a rectangle, and $BX < XC$ . Determine the length of segment $BX$ .

Solution

Solution:

Answer: $\frac{7-\sqrt{41}}{2}$

We note that
$\angle YXC = 90^\circ - \angle WXB = \angle XWB = 90^\circ - \angle AWZ = \angle AZW$
gives us that $XYC \cong ZWA$ and $XYZ \sim WXB$ . Consequently, we get that $YC = AW = 1$ . From $XYZ \sim WXB$ , we get that
$\frac{BX}{BW} = \frac{CY}{CX} \Rightarrow \frac{BX}{2} = \frac{1}{7 - BX}$
from which we get
$BX^2 - 7BX + 2 = 0 \Rightarrow BX = \frac{7 - \sqrt{41}}{2}$
(since we have $BX < CX$ ).

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