Difference between revisions of "1974 AHSME Problems/Problem 16"
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− | Label the points as in the figure above. Let the side length <math> AB=AC=s </math>. Therefore, <math> | + | Label the points as in the figure above. Let the side length <math> AB=AC=s </math>. Therefore, <math> BC=s\sqrt{2} </math>. Since the circumradius of a right triangle is equal to half of the length of the hypotenuse, we have <math> R=\frac{s\sqrt{2}}{2} </math>. |
Now to find the inradius. Notice that <math> IFAE </math> is a square with side length <math> r </math>, and also <math> AD=R</math>. Therefore, <math> s=AD=AI+ID=r\sqrt{2}+r </math>, and so <math> r=\frac{R}{\sqrt{2}+1} </math>. | Now to find the inradius. Notice that <math> IFAE </math> is a square with side length <math> r </math>, and also <math> AD=R</math>. Therefore, <math> s=AD=AI+ID=r\sqrt{2}+r </math>, and so <math> r=\frac{R}{\sqrt{2}+1} </math>. | ||
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{{AHSME box|year=1974|num-b=15|num-a=17}} | {{AHSME box|year=1974|num-b=15|num-a=17}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:11, 3 March 2018
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
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.