1958 AHSME Problems/Problem 19

Problem

The sides of a right triangle are $a$ and $b$ and the hypotenuse is $c$ . A perpendicular from the vertex divides $c$ into segments $r$ and $s$ , adjacent respectively to $a$ and $b$ . If $a : b = 1 : 3$ , then the ratio of $r$ to $s$ is:

$\textbf{(A)}\ 1 : 3\qquad \textbf{(B)}\ 1 : 9\qquad \textbf{(C)}\ 1 : 10\qquad \textbf{(D)}\ 3 : 10\qquad \textbf{(E)}\ 1 : \sqrt{10}$

Solution

Let the triangle be triangle $ABC$ with $A$ opposite to side $a$ , and define $B$ and $C$ similarly. Call the base of the perpendicular $D$ and say it has length $x$ .

We know the area of triangle $ABC$ , denoted as $[ABC],$ is $\frac{ab}{2} = \frac{cx}{2} = \frac{(r+s)x}{2}.$ So, $x = \frac{ab}{r+s}.$

Now, by simple angle chasing that $\triangle BCD \sim \triangle CDA \sim ACB.$ So, $\frac{AC}{CB} = \frac{CD}{BD}.$ Plugging in our variables for the side lenghts: $\frac{b}{a} = \frac{x}{r} = \frac{\frac{ab}{r+s}}{r}.$

We now get that $br = \frac{a^2b}{r+s},$ so $r = \frac{a^2}{r+s}.$ Similarly, we get that $s = \frac{b^2}{r+s}.$ So, $\frac{r}{s} = \frac{\frac{a^2}{r+s}}{\frac{b^2}{r+s}} = \frac{a^2}{b^2} = \boxed{\textbf{(B)}\ 1 : 9}$

$\fbox{}$

~Makethan

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1958 AHSME Problems/Problem 19

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