Difference between revisions of "1974 AHSME Problems/Problem 16"
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Revision as of 11:43, 5 July 2013
Problem
A circle of radius 
is inscribed in a right isosceles triangle, and a circle of radius 
is circumscribed about the triangle. Then 
equals
Solution
![[asy] draw((0,0)--(1,0)--(0,1)--cycle); draw(circle((0.5,0.5),sqrt(2)/2)); draw(circle((1-sqrt(2)/2,1-sqrt(2)/2),1-sqrt(2)/2)); label("$A$",(0,0),SW); label("$B$",(0,1),NNW); label("$C$",(1,0),SSE); label("$D$",(0.5,0.5),NE); draw((0,0)--(0.5,0.5)); draw((1-sqrt(2)/2,1-sqrt(2)/2)--(0,1-sqrt(2)/2)); draw((1-sqrt(2)/2,1-sqrt(2)/2)--(1-sqrt(2)/2,0)); label("$E$",(1-sqrt(2)/2,0),S); label("$F$",(0,1-sqrt(2)/2),W); label("$I$",(1-sqrt(2)/2,1-sqrt(2)/2),NNW); [/asy]](http://latex.artofproblemsolving.com/1/8/8/1882135a1a49e9a11aad02c00fdb6cce17582b4d.png)
Label the points as in the figure above. Let the side length 
. Therefore, 
. Since the circumradius of a right triangle is equal to half of the length of the hypotenuse, we have 
.
Now to find the inradius. Notice that 
is a square with side length , and also 
. Therefore, 
, and so 
.
Finally, 
.
See Also
| 1974 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
