Art of Problem Solving

1950 AHSME Problems/Problem 47: Difference between revisions

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[[Category:Introductory Geometry Problems]]
[[Category:Introductory Geometry Problems]]
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Revision as of 11:30, 5 July 2013

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

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 $\boxed{\textbf{(C)}}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 46
Followed by
Problem 48
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