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 | + | <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> | ||
Revision as of 01:20, 13 May 2018
Problem
Triangle is equilateral with side , perimeter , area , and circumradius (radius of the circumscribed circle). Triangle is equilateral with side , perimeter , area , and circumradius . If is different from , then:
Solution
First, find , , and in terms of . Since all sides of equilateral triangle are the same, . From the area formula, . By using 30-60-90 triangles, .
Using the same steps, , , and .
Note that and . That means , so the answer is
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |