# 1965 AHSME Problems/Problem 12

## Problem

A rhombus is inscribed in $\triangle ABC$ in such a way that one of its vertices is $A$ and two of its sides lie along $AB$ and $AC$. If $\overline{AC} = 6$ inches, $\overline{AB} = 12$ inches, and $\overline{BC} = 8$ inches, the side of the rhombus, in inches, is:

$\textbf{(A)}\ 2 \qquad \textbf{(B) }\ 3 \qquad \textbf{(C) }\ 3 \frac {1}{2} \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$

## Solution

$[asy] draw((0,0)--(6,0)); dot((0,0)); label("A", (-0.75,-0.5)); dot((6,0)); label("C", (6.75,-0.5)); label("6", (3,-1.75)); draw((0,0)--(12*29/36, 12*sqrt(455)/36)--(6,0)); dot((12*29/36, 12*sqrt(455)/36)); label("B", (12*29/36+0.4, 12*sqrt(455)/36+0.75)); label("12", (6*29/36-1.5, 6*sqrt(455)/36+1.5)); label("8", (6*29/36+4.5, 6*sqrt(455)/36-0.5)); dot((4*29/36, 4*sqrt(455)/36)); label("F", (4*29/36-0.75, 4*sqrt(455)/36)); dot((4*29/36+4, 4*sqrt(455)/36)); label("E", (4*29/36+4+0.65, 4*sqrt(455)/36-0.25)); dot((4,0)); label("D", (4,-0.75)); draw((0,0)--(4,0)--(4*29/36+4, 4*sqrt(455)/36)--(4*29/36, 4*sqrt(455)/36)); [/asy]$

As in the diagram, suppose the rhombus $ADEF$ is inscribed in $\triangle ABC$ with $D$ on $\overline{AC}$, $E$ on $\overline{BC}$, and $F$ on $\overline{AB}$. Let the side length of the rhombus be $x$. Because a rhombus is a parallelogram, its opposite sides are parallel. Thus, by AA similarity, $\triangle BFE \sim \triangle EDC$, and so $\frac{FE}{FB}=\frac{DC}{DE}$. Because $\overline{FE}$ and $\overline{DE}$ are sides of the rhombus, they both have length $x$. Furthermore, $FB=12-x$, and $DC=6-x$, because, combined with sides of the rhombus, they form sides of the triangle. Thus, by substituting into the proportion derived above, we see that: \begin{align*} \frac{FE}{FB}&=\frac{DC}{DE} \\ \frac{x}{12-x}&=\frac{6-x}{x} \\ x^2&=72-12x-6x+x^2 \\ 18x&=72 \\ x&=4 \\ \end{align*}

Thus, the side of the rhombus has length $\boxed{\text{(D) }4}$.