1960 AHSME Problems/Problem 15

Problem

Triangle $I$ is equilateral with side $A$ , perimeter $P$ , area $K$ , and circumradius $R$ (radius of the circumscribed circle). Triangle $II$ is equilateral with side $a$ , perimeter $p$ , area $k$ , and circumradius $r$ . If $A$ is different from $a$ , then:

$\textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad \\ \textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad \\ \textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad \\ \textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad \\ \textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes}$

Solution

$[asy] pair A=(0,50),O=(0,0),B=(43.301,-25),C=(-43.301,-25); draw(A--B--C--A); draw(circle(O,50)); draw(B--O--C); draw(anglemark(C,O,B,200)); draw((0,0)--(0,-25)); label("$60^{\circ}$",(-7,-12)); [/asy]$

First, find $P$ , $K$ , and $R$ in terms of $A$ . Since all sides of equilateral triangle are the same, $P=3A$ . From the area formula, $K=\frac{A^2\sqrt{3}}{4}$ . By using 30-60-90 triangles, $R=\frac{A\sqrt{3}}{3}$ .

Using the same steps, $p=3a$ , $k=\frac{a^2\sqrt{3}}{4}$ , and $r=\frac{a\sqrt{3}}{3}$ .

Note that $P/p = 3A/3a = A/a$ and $R/r = \frac{A\sqrt{3}}{3} \div \frac{a\sqrt{3}}{3} = A/a$ . That means $P/p = R/r$ , so the answer is $\boxed{\textbf{(B)}}$

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1960 AHSME Problems/Problem 15

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