Difference between revisions of "1960 AHSME Problems/Problem 15"
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− | 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>. |
Revision as of 11:34, 14 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 an 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 | |
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All AHSME Problems and Solutions |