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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 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(More explanation)==
+
==Solution 2==
−
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>\boxed{\text{(D)}}</math> ~Solution by twinbrian
+
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==
 +
 
 +
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 16:39, 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)
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Problem 20 		Followed by
Problem 22
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