# 1950 AHSME Problems/Problem 47

## Problem

A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then: $\textbf{(A)}\ x=\dfrac{1}{2}h \qquad \textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad \textbf{(C)}\ x=\dfrac{bh}{2h+b} \qquad \textbf{(D)}\ x=\sqrt{\dfrac{hb}{2}} \qquad \textbf{(E)}\ x=\dfrac{1}{2}b$

## Solution 1

Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get $\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h + b)x = bh \implies x = \dfrac{bh}{2h + b}$. The answer is $\boxed{\textbf{(C)}}$.

## Solution 2

You can also use area to solve this problem. Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle. Therefore the smaller triangle (similar to the big one) has an area of $\dfrac{2x(h-x)}{2}=x(h-x)$. Now note that because the rectangle is right, and the two other pieces are complementary triangles, we can add them together to create one triangle. Therefore the area of these two triangles is $\dfrac{x(b-2x)}{2}$. This is due to the fact that we subtract $2x$ from the base of the triangle. Lastly the area of the rectangle is $2x^2$. These areas together sum to the area of the big rectangle, which is $\dfrac{bh}{2}$. Solving we get that $x = \dfrac{bh}{2h + b}$. The answer is $\boxed{\textbf{(C)}}$.

## Video Solution

~MathProblemSolvingSkills.com

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 