https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ronaldo+777&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T19:36:49ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_22&diff=994312017 AMC 8 Problems/Problem 222018-12-13T17:28:45Z<p>Ronaldo 777: I have made this change in order to make sure that the readers get a clear view of the formula being used as I myself couldn't just understand. For the ease of understanding, I have made this edit. Thanks.</p>
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<div>==Problem 22==<br />
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In the right triangle <math>ABC</math>, <math>AC=12</math>, <math>BC=5</math>, and angle <math>C</math> is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br />
<asy><br />
draw((0,0)--(12,0)--(12,5)--(0,0));<br />
draw(arc((8.67,0),(12,0),(5.33,0)));<br />
label("$A$", (0,0), W);<br />
label("$C$", (12,0), E);<br />
label("$B$", (12,5), NE);<br />
label("$12$", (6, 0), S);<br />
label("$5$", (12, 2.5), E);</asy><br />
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<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}</math><br />
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==Solution 1==<br />
We can reflect triangle <math>ABC</math> over line <math>AC.</math> This forms the triangle <math>AB'C</math> and a circle out of the semicircle. Let us call the center of the circle <math>O.</math><br />
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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: Area of a triangle = Semi-perimeter <math> \times </math> inradius . 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><br />
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==Solution 2==<br />
We immediately see that <math>AB=13</math>, and we label the center of the semicircle <math>O</math> And the point where the circle is tangent to the triangle <math>D</math>. Drawing radius <math>OD</math> with length <math>x</math> such that <math>OD</math> is perpendicular to <math>AB</math>, we immediately see that <math>ODB\cong OCB</math> because of <math>\operatorname{HL}</math> 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)}\ \frac{10}{3}}</math>.<br />
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==Solution 3==<br />
Let the center of the semicircle be <math>O</math>. Let the point of tangency between line <math>AB</math> and the semicircle be <math>F</math>. Angle <math>BAC</math> is common to triangles <math>ABC</math> and <math>AFO</math>. By tangent properties, angle <math>AFO</math> must be <math>90</math> degrees. Since both triangles <math>ABC</math> and <math>AFO</math> are right and share an angle, <math>AFO</math> is similar to <math>ABC</math>. The hypotenuse of <math>AFO</math> is <math>12 - r</math>, where <math>r</math> is the radius of the circle. (See for yourself) The short leg of <math>AFO</math> is <math>r</math>. Because <math>AFO</math> ~ <math>ABC</math>, we have <math>r/(12 - r) = 5/13</math> and solving gives <math>r = \boxed{\textbf{(D)}\ \frac{10}{3}}</math><br />
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==Solution 4==<br />
Let the tangency point on <math>AB</math> be <math>D</math>. Note <cmath>AD = AB-BD = AB-BC = 8</cmath> By Power of a Point, <cmath>12(12-2r) = 8^2</cmath><br />
Solving for <math>r</math> gives <br />
<cmath>r = \boxed{\textbf{(D) }\frac{10}{3}}</cmath><br />
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==See Also==<br />
{{AMC8 box|year=2017|num-b=21|num-a=23}}<br />
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{{MAA Notice}}</div>Ronaldo 777