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Difference between revisions of "2017 AMC 10B Problems/Problem 21"

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<math>\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3</math>
 
<math>\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3</math>
 
==Solution==
 
==Solution==
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We can use the formula that states that the area of a triangle is equal to the inradius times the semiperimeter. We know that <math>AD=BD=CD=5</math>, and that <math>[ABD]=[ACD]=12</math>, where <math>[P]</math> is the area of polygon <math>P</math>. We can determine the semiperimeters of <math>ABD</math> and <math>ACD</math> as <math>\frac{5+5+6}{2}=8</math> and <math>\frac{5+5+8}{2}=9</math>, respectively. Thus, the sum of the inradii is <math>\frac{12}{8}+\frac{12}{9}=\boxed{\frac{17}{6}(\text{D})}</math>.
+
We note that by the converse of the Pythagorean Theorem, <math>\triangle ABC</math> is a right triangle with a right angle at <math>A</math>. Therefore, <math>AD = BD = CD = 5</math>, and <math>[ADB] = [ADC] = 12</math>. Since <math>A = rs</math>, the inradius of <math>\triangle ADB</math> is <math>\frac{12}{(5+5+6)/2} = \frac 32</math>, and the inradius of <math>\triangle ADC</math>  is <math>\frac{12}{(5+5+8)/2} = \frac 43</math>. Adding the two together, we have <math>\boxed{\textbf{(D) } \frac{17]6}</math>.
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~willwin4sure
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:35, 16 February 2017

Problem

In $\triangle ABC$, $AB=6$, $AC=8$, $BC=10$, and $D$ is the midpoint of $\overline{BC}$. What is the sum of the radii of the circles inscibed in $\triangle ADB$ and $\triangle ADC$?

$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3$

Solution

We note that by the converse of the Pythagorean Theorem, $\triangle ABC$ is a right triangle with a right angle at $A$. Therefore, $AD = BD = CD = 5$, and $[ADB] = [ADC] = 12$. Since $A = rs$, the inradius of $\triangle ADB$ is $\frac{12}{(5+5+6)/2} = \frac 32$, and the inradius of $\triangle ADC$ is $\frac{12}{(5+5+8)/2} = \frac 43$. Adding the two together, we have $\boxed{\textbf{(D) } \frac{17]6}$ (Error compiling LaTeX. Unknown error_msg).

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 AMC 10 Problems and Solutions

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