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

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(→‎Solution 4 (using Shoelace and general inradius))
 
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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>, 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>. Also, the median to the hypotenuse will be half of the hypotenuse. 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
 +
 
 +
==Solution 3 (Stewart's)==
 +
Applying [https://artofproblemsolving.com/wiki/index.php/Stewart%27s_theorem] gives us the length of <math>\overline{AD}.</math> Using that length, we can find the areas of triangles <math>\triangle ABD</math> and <math>\triangle ACD</math> by using Heron’s formula. We can use that area to find the inradius of the circles by the inradius formula <math>A=sr.</math> Therefore, we get <math>\boxed{\textbf{(D) }\frac{17}{6}}.</math> Although this solution works perfectly fine, it takes time and has room for error so apply Stewart’s and Heron’s with caution.
 +
 
 +
~peelybonehead
 +
 
 +
Edited by ~Jadon_Jung
 +
 
 +
==Solution 4 (using Shoelace and general inradius)==
 +
 
 +
First, start by plotting <math>\triangle{ABC}</math> on a grid; with B at (0,0), D at (5,0), and C at (10,0).
 +
 
 +
We can now solve for the position of point A. Let point A be (a,b). We can then set up the equation a^2 + b^2 = 36 and the equation (10-a)^2 + b^2 = 64. Solving for a and b, we get that a = <math>\frac{18}{5}</math> and b = <math>\frac{24}{5}</math>. We can then go ahead and use the Shoelace method to get the area of <math>\triangle{ABD}</math>, which ends up being 12. Since D is the midpoint of BC, <math>\triangle{ABD}</math> and <math>\triangle{ACD}</math> have the same area.
 +
 
 +
Using the distance formula once again, AD has length 5, and now we can get the semiperimeter of <math>\triangle{ABD}</math> and <math>\triangle{ACD}</math>, which turns out to be 8 for <math>\triangle{ABD}</math> and 9 for <math>\triangle{ACD}</math>. Dividing Area by Semiperimeter and adding them, we get <math>\frac{3}{2}</math> + <math>\frac{4}{3}</math> = <math>\boxed{\textbf{(D) }\frac{17}{6}}.</math>
 +
 
 +
~BanSpeedrun
  
 
==Video Solution==
 
==Video Solution==
  
−
https://youtu.be/EfKFDwTDRjs - Happytwin
+
https://youtu.be/EfKFDwTDRjs
  
 
==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}}

Latest revision as of 17:39, 17 October 2024

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 $A$. Also, the median to the hypotenuse will be half of the hypotenuse. 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

Solution 3 (Stewart's)

Applying [1] gives us the length of $\overline{AD}.$ Using that length, we can find the areas of triangles $\triangle ABD$ and $\triangle ACD$ by using Heron’s formula. We can use that area to find the inradius of the circles by the inradius formula $A=sr.$ Therefore, we get $\boxed{\textbf{(D) }\frac{17}{6}}.$ Although this solution works perfectly fine, it takes time and has room for error so apply Stewart’s and Heron’s with caution.

~peelybonehead

Edited by ~Jadon_Jung

Solution 4 (using Shoelace and general inradius)

First, start by plotting $\triangle{ABC}$ on a grid; with B at (0,0), D at (5,0), and C at (10,0).

We can now solve for the position of point A. Let point A be (a,b). We can then set up the equation a^2 + b^2 = 36 and the equation (10-a)^2 + b^2 = 64. Solving for a and b, we get that a = $\frac{18}{5}$ and b = $\frac{24}{5}$. We can then go ahead and use the Shoelace method to get the area of $\triangle{ABD}$, which ends up being 12. Since D is the midpoint of BC, $\triangle{ABD}$ and $\triangle{ACD}$ have the same area.

Using the distance formula once again, AD has length 5, and now we can get the semiperimeter of $\triangle{ABD}$ and $\triangle{ACD}$, which turns out to be 8 for $\triangle{ABD}$ and 9 for $\triangle{ACD}$. Dividing Area by Semiperimeter and adding them, we get $\frac{3}{2}$ + $\frac{4}{3}$ = $\boxed{\textbf{(D) }\frac{17}{6}}.$

~BanSpeedrun

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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