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

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edit by Lej: Area of Incircle = (inradius)(Semiperimeter)
 
edit by Lej: Area of Incircle = (inradius)(Semiperimeter)
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==Solution(More explanation)==
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We have <math>\triangle ABC</math> a right triangle by dividing each side lengths by <math>2</math> to create a well known <math>3-4-5</math> triangle. We also can know that the median of a right triangle must be equal to half the hypotenuse. Using this property, we have <math>BD=5</math>, <math>CD=5</math>, and <math>AD=5</math>. Now, we can use the Heron's formula to get the area of <math>\triangle ABD</math> as <math>\sqrt{8(2)(3)(3)}=\sqrt{144}=12</math>. Afterward, we can apply this formula again on <math>\triangle ADC</math> to get the area as <math>\sqrt{9(4)(4)(1)}=\sqrt{144}=12</math>. Notice we want the inradius. We can use another property, which is <math>A=rs</math>. This states that <math>\text{Area}=\text{Radius}(\text{Semiperimeter})</math>. (This can be proved by connecting the center of the inscribed circles to the vertices and we can notice the inradius is just the heights of each of the three triangles divided) Finally, we can derive the radii of each inscribed circle. Plugging the semiperimeter and area into the formula, we have <math>12=8r</math> and <math>12=9r</math> for <math>\triangle ABD</math> and <math>\triangle ADC</math>, respectively. Simplifying, we have the radii lengths as <math>r=\frac{3}{2}</math> and <math>\frac{4}{3}</math>. We want the sum, so we have <math>\frac{17}{6}</math>, or <math>\textbf{(D)</math> ~Solution by twinbrian
  
 
==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 17:42, 20 January 2019

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 inscribed 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 $\boxed{\textbf{(D) } \frac{17}6}$.


edit by asdf334: what do you mean by A = rs?

edit by Lej: Area of Incircle = (inradius)(Semiperimeter)

Solution(More explanation)

We have $\triangle ABC$ a right triangle by dividing each side lengths by $2$ to create a well known $3-4-5$ triangle. We also can know that the median of a right triangle must be equal to half the hypotenuse. Using this property, we have $BD=5$, $CD=5$, and $AD=5$. Now, we can use the Heron's formula to get the area of $\triangle ABD$ as $\sqrt{8(2)(3)(3)}=\sqrt{144}=12$. Afterward, we can apply this formula again on $\triangle ADC$ to get the area as $\sqrt{9(4)(4)(1)}=\sqrt{144}=12$. Notice we want the inradius. We can use another property, which is $A=rs$. This states that $\text{Area}=\text{Radius}(\text{Semiperimeter})$. (This can be proved by connecting the center of the inscribed circles to the vertices and we can notice the inradius is just the heights of each of the three triangles divided) Finally, we can derive the radii of each inscribed circle. Plugging the semiperimeter and area into the formula, we have $12=8r$ and $12=9r$ for $\triangle ABD$ and $\triangle ADC$, respectively. Simplifying, we have the radii lengths as $r=\frac{3}{2}$ and $\frac{4}{3}$. We want the sum, so we have $\frac{17}{6}$, or $\textbf{(D)$ (Error compiling LaTeX. Unknown error_msg) ~Solution by twinbrian

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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