Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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We can see that Circle <math>O</math> is the incircle of <math>AB'C.</math> We can use the formula for finding the radius of the incircle to solve this problem. The area of <math>AB'C</math> is <math>12\times5 = 60.</math> The semiperimeter is <math>5+13 = 18.</math> Simplifying <math>\dfrac{60}{18} = \dfrac{10}{3}.</math> Our answer is therefore <math>\boxed{\textbf{(D)}\ \frac{10}{3}}.</math> | We can see that Circle <math>O</math> is the incircle of <math>AB'C.</math> We can use the formula for finding the radius of the incircle to solve this problem. The area of <math>AB'C</math> is <math>12\times5 = 60.</math> The semiperimeter is <math>5+13 = 18.</math> Simplifying <math>\dfrac{60}{18} = \dfrac{10}{3}.</math> Our answer is therefore <math>\boxed{\textbf{(D)}\ \frac{10}{3}}.</math> | ||
| − | ==Solution== | + | ==Similar Triangles Solution== |
We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</math>. Drawing radius <math>OD</math> with length <math>x</math> such that <math>OD</math> is tangent to <math>AB</math>, we immediately see that <math>ODB\cong OCB</math> because of HL congruence, so <math>BD=5</math> and <math>DA=8</math>. By similar triangles <math>ODA</math> and <math>BCA</math>, we see that <math>\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{D}}</math>. | We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</math>. Drawing radius <math>OD</math> with length <math>x</math> such that <math>OD</math> is tangent to <math>AB</math>, we immediately see that <math>ODB\cong OCB</math> because of HL congruence, so <math>BD=5</math> and <math>DA=8</math>. By similar triangles <math>ODA</math> and <math>BCA</math>, we see that <math>\frac{8}{12}=\frac{x}{5}\implies 12x=40\implies x=\frac{10}{3}\implies\boxed{\textbf{D}}</math>. | ||
Revision as of 15:56, 22 November 2017
Problem 22
In the right triangle 
, 
, 
, and angle 
is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?
![[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("$A$", (0,0), W); label("$C$", (12,0), E); label("$B$", (12,5), NE); label("$12$", (6, 0), S); label("$5$", (12, 2.5), E);[/asy]](http://latex.artofproblemsolving.com/1/7/f/17f66d4cef8b689eb626ec8c226380e1e72d8719.png)
Solution
We can reflect triangle on line 
This forms the triangle 
and a circle out of the semicircle. Let us call the center of the circle 
We can see that Circle 
is the incircle of 
We can use the formula for finding the radius of the incircle to solve this problem. The area of is 
The semiperimeter is 
Simplifying 
Our answer is therefore 
Similar Triangles Solution
We immediately see that 
, and we label the center of the semicircle . Drawing radius 
with length 
such that is tangent to 
, we immediately see that 
because of HL congruence, so 
and 
. By similar triangles 
and 
, we see that 
.
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
