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1960 AHSME Problems/Problem 15: Difference between revisions

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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
<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
\textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad \\
\textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad
\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>


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</asy>
</asy>


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


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


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


==See Also==
==See Also==
{{AHSME 40p box|year=1960|num-b=14|num-a=16}}
{{AHSME 40p box|year=1960|num-b=14|num-a=16}}
[[Category:Introductory Geometry Problems]]

Latest revision as of 18:01, 17 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 an 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 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AHSME Problems and Solutions
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