Difference between revisions of "1967 AHSME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | The radius of our circle is <math>\frac{\sqrt{3}}{6}s</math>. This means the square has side length <math>\frac{\sqrt{6}}{6}s</math> which has area of <math>\frac{s^2}{6}=\fbox{B}</math>. |
== See also == | == See also == | ||
| − | {{AHSME box|year=1967|num-b=2|num-a=4}} | + | {{AHSME 40p box|year=1967|num-b=2|num-a=4}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:34, 16 August 2023
Problem
The side of an equilateral triangle is 
. A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:
Solution
The radius of our circle is 
. This means the square has side length 
which has area of 
.
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
