Difference between revisions of "1960 AHSME Problems/Problem 15"
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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 19:01, 17 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 | |
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 |