1967 AHSME Problems/Problem 5

Problem

A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:

$\text{(A)}\text{ independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2}$

Solution

The area $K$ of the triangle can be expressed in terms of its inradius $r$ and its semiperimeter $s$ as: $K = r \times s = r \times \frac{P}{2}$

So, $\frac{P}{K} = \boxed{\textbf{(C) } \frac{2}{r}}$ .

~ proloto

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1967 AHSME Problems/Problem 5

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