Library / /539 of 1394

, 2015

Geometry Difficulty 5.2 AIME, harder Find the answer United States

Problem:

Let $ABCD$ be a cyclic quadrilateral with $AB=3$ , $BC=2$ , $CD=2$ , $DA=4$ . Let lines perpendicular to $\overline{BC}$ from $B$ and $C$ meet $\overline{AD}$ at $B'$ and $C'$ , respectively. Let lines perpendicular to $\overline{AD}$ from $A$ and $D$ meet $\overline{BC}$ at $A'$ and $D'$ , respectively. Compute the ratio $\frac{[BCC'B']}{[DAA'D']}$ , where $[\varpi]$ denotes the area of figure $\varpi$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

$\boxed{\dfrac{37}{76}}$

To get a handle on the heights $CB'$ , etc. perpendicular to $BC$ and $AD$ , let $X = BC \cap AD$ , which lies on ray $\overrightarrow{BC}$ and $\overrightarrow{AD}$ since $\widehat{AB} > \widehat{CD}$ (as chords $AB > CD$ ).

By similar triangles we have equality of ratios $XC : XD : 2 = (XD + 4) : (XC + 2) : 3$ , so we have a system of linear equations: $3XC = 2XD + 8$ and $3XD = 2XC + 4$ , so $9XC = 6XD + 24 = 4XC + 32$ gives $XC = \frac{32}{5}$ and $XD = \frac{84/5}{3} = \frac{28}{5}$ .

It's easy to compute the trapezoid area ratio
$\frac{[BC'C'B']}{[AA'D'D]} = \frac{BC(CB' + BC')}{AD(AD' + DA')} = \frac{BC}{AD} \cdot \frac{XC + XB}{XA + XD}$
(where we have similar right triangles due to the common angle at $X$ ). This is just
$\frac{BC}{AD} \cdot \frac{XC + BC/2}{XD + AD/2} = \frac{2}{4} \cdot \frac{32/5 + 1}{28/5 + 2} = \frac{37}{76}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.