Difference between revisions of "1960 AHSME Problems/Problem 15"

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Triangle <math>II</math> is equilateral with side <math>a</math>, perimeter <math>p</math>, area <math>k</math>, and circumradius <math>r</math>. If <math>A</math> is different from <math>a</math>, then:  
 
Triangle <math>II</math> is equilateral with side <math>a</math>, perimeter <math>p</math>, area <math>k</math>, and circumradius <math>r</math>. If <math>A</math> is different from <math>a</math>, then:  
  
<math>\textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad
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<math>\textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad \\
 
\textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad \\
 
\textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad \\
\textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad
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\textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad \\
\textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad
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\textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad \\
 
\textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes}    </math>
 
\textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes}    </math>
  

Revision as of 01:20, 13 May 2018

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 $\boxed{\textbf{(B)}}$

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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