Difference between revisions of "1968 AHSME Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | The triangle that goes through all the vertices of the triangle is the circumcircle of the triangle. |
| + | <math>(7\frac{1}{2})^{2}+10^{2}=(12\frac{1}{2})^{2}</math>, so the triangle is a right triangle.The radius of a circumcircle | ||
| + | of a right triangle is half the hypotenuse. <math>\frac{1}{2}\cdot \frac{25}{2}=\frac{25}{4}\implies</math> <math>\fbox{C}</math> | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=11|num-a=13}} | + | {{AHSME 35p box|year=1968|num-b=11|num-a=13}} |
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 01:52, 16 August 2023
Problem
A circle passes through the vertices of a triangle with side-lengths 
The radius of the circle is:
Solution
The triangle that goes through all the vertices of the triangle is the circumcircle of the triangle.
, so the triangle is a right triangle.The radius of a circumcircle
of a right triangle is half the hypotenuse. 
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
