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2017 AMC 10B Problems/Problem 21: Difference between revisions

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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 1==
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> we have <math>r = \frac As</math>, so 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>.
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> we have <math>r = \frac As</math>, so 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>.


==Solution 2==
We have
<asy>
draw((0,0)--(8,0));
draw((0,0)--(0,6));
draw((8,0)--(0,6));
draw((0,0)--(4,3));
label("A",(0,0),W);
label("B",(0,6),N);
label("C",(8,0),E);
label("D",(4,3),NE);
label("H",(2.3,4.2),NE);
label("K",(2.3,1.8),S);
draw(circle((1.54,3),1.49));
draw(circle((4,1.35),1.33));
dot((4,1.35));
dot((1.54,3));
label("F",(1.54,3),S);
label("J",(4,1.35),SW);
label("G",(0,3),W);
label("$x$",(1,3),S);
label("$y$",(4,1),E);
draw((1.54,3)--(0,3));
draw((1.54,3)--(2.3,1.8));
draw((1.54,3)--(2.3,4.2));
draw((4,1.35)--(4,0));
draw((4,1.35)--(3.12,2.4));
draw((4,1.35)--(4.8,2.3));
label("L",(4.9,2.4),NE);
label("E",(3.11,2.3),S);
label("I",(4,0),S);
</asy>
Let <math>x</math> be the radius of circle <math>F</math>, and let <math>y</math> be the radius of circle <math>J</math>. We want to find <math>x+y</math>.
We form 6 kites: <math>GAKF</math>, <math>HFKD</math>, <math>GFHB</math>, <math>EJIA</math>, <math>LJIC</math>, and <math>JEDL</math>
Since <math>G</math> and <math>I</math> are the midpoints of <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, this means that <math>BG = AG = \frac{6}{2} = 3</math>, and <math>AI = IC = \frac{8}{2} = 4</math>.
Since <math>AGFK</math> is a kite, <math>GF = FK = x</math>, and <math>AG = AK = 3</math>. The same applies to all kites in the diagram.
Now, we see that <math>AK = 3</math>, and <math>KD = 2</math>, thus <math>AD</math> is <math>5</math>, making <math>\triangle ADC</math> and <math>\triangle ABD</math> isosceles. So, <math>DI=3</math> using the Pythagorean Theorem, and <math>GD=4</math> also using the Theorem. Hence, we know that <math>[ADC] = [ABD] = 12</math>.
Notice that the area of the kite (if the <math>2</math> opposite angles are right) is <math>\frac{s_1 \cdot s_2}{2} \cdot 2</math>, where <math>s_1</math> and <math>s_2</math> denoting each of the 2 congruent sides. This just simplifies to <math>s_1 \cdot s_2</math>.
Hence, we have
<cmath>4b+4b+b = 12</cmath>
and
<cmath>3a+3a+2a = 12</cmath>
Solving for <math>a</math> and <math>b</math>, we find that <math>a = \frac{3}{2}</math> and <math>b = \frac{4}{3}</math>, so <math>a+b = \frac{3}{2} + \frac {4}{3} = \boxed{\textbf{(D)} ~\frac{17}6}</math>.
~MrThinker
==Video Solution==
==Video Solution==


Revision as of 13:25, 5 September 2022

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 1

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 $A = rs,$ we have $r = \frac As$, so 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 $\boxed{\textbf{(D) } \frac{17}6}$.

Solution 2

We have [asy] draw((0,0)--(8,0)); draw((0,0)--(0,6)); draw((8,0)--(0,6)); draw((0,0)--(4,3)); label("A",(0,0),W); label("B",(0,6),N); label("C",(8,0),E); label("D",(4,3),NE); label("H",(2.3,4.2),NE); label("K",(2.3,1.8),S); draw(circle((1.54,3),1.49)); draw(circle((4,1.35),1.33)); dot((4,1.35)); dot((1.54,3)); label("F",(1.54,3),S); label("J",(4,1.35),SW); label("G",(0,3),W); label("$x$",(1,3),S); label("$y$",(4,1),E); draw((1.54,3)--(0,3)); draw((1.54,3)--(2.3,1.8)); draw((1.54,3)--(2.3,4.2)); draw((4,1.35)--(4,0)); draw((4,1.35)--(3.12,2.4)); draw((4,1.35)--(4.8,2.3)); label("L",(4.9,2.4),NE); label("E",(3.11,2.3),S); label("I",(4,0),S); [/asy] Let $x$ be the radius of circle $F$, and let $y$ be the radius of circle $J$. We want to find $x+y$.

We form 6 kites: $GAKF$, $HFKD$, $GFHB$, $EJIA$, $LJIC$, and $JEDL$ Since $G$ and $I$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, this means that $BG = AG = \frac{6}{2} = 3$, and $AI = IC = \frac{8}{2} = 4$.

Since $AGFK$ is a kite, $GF = FK = x$, and $AG = AK = 3$. The same applies to all kites in the diagram.

Now, we see that $AK = 3$, and $KD = 2$, thus $AD$ is $5$, making $\triangle ADC$ and $\triangle ABD$ isosceles. So, $DI=3$ using the Pythagorean Theorem, and $GD=4$ also using the Theorem. Hence, we know that $[ADC] = [ABD] = 12$.

Notice that the area of the kite (if the $2$ opposite angles are right) is $\frac{s_1 \cdot s_2}{2} \cdot 2$, where $s_1$ and $s_2$ denoting each of the 2 congruent sides. This just simplifies to $s_1 \cdot s_2$. Hence, we have

\[4b+4b+b = 12\]

and

\[3a+3a+2a = 12\]

Solving for $a$ and $b$, we find that $a = \frac{3}{2}$ and $b = \frac{4}{3}$, so $a+b = \frac{3}{2} + \frac {4}{3} = \boxed{\textbf{(D)} ~\frac{17}6}$.

~MrThinker

Video Solution

https://youtu.be/EfKFDwTDRjs

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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