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Difference between revisions of "1974 AHSME Problems/Problem 16"

(Created page with "==Problem== A circle of radius <math> r </math> is inscribed in a right isosceles triangle, and a circle of radius <math> R </math> is circumscribed about the triangle. Then <mat...")
 
(corrected solution)
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Label the points as in the figure above. Let the side length <math> AB=AC=s </math>. Therefore, <math> AC=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>.
 
Label the points as in the figure above. Let the side length <math> AB=AC=s </math>. Therefore, <math> AC=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>.
  
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Now to find the inradius. Notice that <math> IFAE </math> is a square with side length <math> r </math>, and also <math> AD=AC=s </math>. Therefore, <math> s=AD=AI+ID=r\sqrt{2}+r </math>, and so <math> r=\frac{s}{\sqrt{2}+1} </math>.
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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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Finally, <math> \frac{R}{r}=\frac{\frac{s\sqrt{2}}{2}}{\frac{s}{\sqrt{2}+1}}=\frac{\sqrt{2}+1}{\sqrt{2}}=\frac{2+\sqrt{2}}{2}, \boxed{\text{B}} </math>.
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Finally, <math> \frac{R}{r}=1+\sqrt{2}, \boxed{\text{A}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=15|num-a=17}}
 
{{AHSME box|year=1974|num-b=15|num-a=17}}

Revision as of 09:15, 30 May 2012

Problem

A circle of radius $r$ is inscribed in a right isosceles triangle, and a circle of radius $R$ is circumscribed about the triangle. Then $R/r$ equals

$\mathrm{(A)\ } 1+\sqrt{2} \qquad \mathrm{(B) \ }\frac{2+\sqrt{2}}{2} \qquad \mathrm{(C) \ } \frac{\sqrt{2}-1}{2} \qquad \mathrm{(D) \ } \frac{1+\sqrt{2}}{2} \qquad \mathrm{(E) \ }2(2-\sqrt{2})$

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]

Label the points as in the figure above. Let the side length $AB=AC=s$. Therefore, $AC=s\sqrt{2}$. Since the circumradius of a right triangle is equal to half of the length of the hypotenuse, we have $R=\frac{s\sqrt{2}}{2}$.

Now to find the inradius. Notice that $IFAE$ is a square with side length $r$, and also $AD=R$. Therefore, $s=AD=AI+ID=r\sqrt{2}+r$, and so $r=\frac{R}{\sqrt{2}+1}$.

Finally, $\frac{R}{r}=1+\sqrt{2}, \boxed{\text{A}}$.

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
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