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>. 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 [[Stewart’s 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 | ||
==Video Solution== | ==Video Solution== | ||
− | https://youtu.be/EfKFDwTDRjs | + | 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 00:34, 8 August 2023
Problem
In ,
,
,
, and
is the midpoint of
. What is the sum of the radii of the circles inscribed in
and
?
Solution 1
We note that by the converse of the Pythagorean Theorem, is a right triangle with a right angle at
. Therefore,
, and
. Since
we have
, so the inradius of
is
, and the inradius of
is
. Adding the two together, we have
.
Solution 2
We have
Let
be the radius of circle
, and let
be the radius of circle
. We want to find
.
We form 6 kites: ,
,
,
,
, and
.
Since
and
are the midpoints of
and
, respectively, this means that
, and
.
Since is a kite,
, and
. The same applies to all kites in the diagram.
Now, we see that , and
, thus
is
, making
and
isosceles. So,
using the Pythagorean Theorem, and
also using the Theorem. Hence, we know that
.
Notice that the area of the kite (if the opposite angles are right) is
, where
and
denoting each of the 2 congruent sides. This just simplifies to
.
Hence, we have
and
Solving for and
, we find that
and
, so
.
~MrThinker
Solution 3 (Stewart's)
Applying Stewart’s theorem gives us the length of Using that length, we can find the areas of triangles
and
by using Heron’s formula. We can use that area to find the inradius of the circles by the inradius formula
Therefore, we get
Although this solution works perfectly fine, it takes time and has room for error so apply Stewart’s and Heron’s with caution.
~peelybonehead
Video Solution
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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