Difference between revisions of "2018 AMC 10A Problems/Problem 15"

m (Solution 2)
(Video Solution)
(14 intermediate revisions by 10 users not shown)
Line 1: Line 1:
Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points  <math>A</math> and <math>B</math>, as shown in the diagram. The distance <math>AB</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
+
== Problem ==
 +
 
 +
Two circles of radius <math>5</math> are externally tangent to each other and are internally tangent to a circle of radius <math>13</math> at points  <math>A</math> and <math>B</math>, as shown in the diagram. The distance <math>AB</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
  
 
<asy>
 
<asy>
Line 26: Line 28:
 
label("$Z$", (5,-6.25),NE);
 
label("$Z$", (5,-6.25),NE);
 
</asy>
 
</asy>
Call center of the largest circle <math>X</math>. The circle that is tangent at point <math>A</math> will have point <math>Y</math> as the center. Similarly, the circle that is tangent at point <math>B</math> will have point <math>Z</math> as the center. Connect <math>AB</math>, <math>YZ</math>, <math>XA</math>, and <math>XB</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math>. Writing out the ratios, we get
+
Let the center of the surrounding circle be <math>X</math>. The circle that is tangent at point <math>A</math> will have point <math>Y</math> as the center. Similarly, the circle that is tangent at point <math>B</math> will have point <math>Z</math> as the center. Connect <math>AB</math>, <math>YZ</math>, <math>XA</math>, and <math>XB</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math>. Writing out the ratios, we get
 
<cmath>\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.</cmath>
 
<cmath>\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.</cmath>
 
Therefore, our answer is <math>65+4= \boxed{\textbf{D) } 69}</math>.
 
Therefore, our answer is <math>65+4= \boxed{\textbf{D) } 69}</math>.
Line 46: Line 48:
 
label("$O$", (0,0), N);
 
label("$O$", (0,0), N);
 
</asy>
 
</asy>
Let the center of the large circle be <math>O</math>. Let the common tangent of the two smaller circles be <math>C</math>. Draw the two radii of the large circle, <math>\overline{OA}</math> and <math>\overline{OB}</math> and the two radii of the smaller circles to point <math>C</math>. Draw ray <math>\overrightarrow{OC}</math>. Draw <math>\overline{AB}</math>. This sets us up with similar triangles, which we can solve.
+
Let the center of the large circle be <math>O</math>. Let the common tangent of the two smaller circles be <math>C</math>. Draw the two radii of the large circle, <math>\overline{OA}</math> and <math>\overline{OB}</math> and the two radii of the smaller circles to point <math>C</math>. Draw ray <math>\overrightarrow{OC}</math> and <math>\overline{AB}</math>. This sets us up with similar triangles, which we can solve.
The length of  <math>\overline{OC}</math> is equal to <math>\sqrt{39}</math> by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, <math>OB</math> is 13, and therefore half of <math>AB</math> is <math>\frac{65}{8}</math>. Doubling gives <math>\frac{65}{4}</math> which results in <math>65+4=\boxed{\textbf{D) }69}</math>.
+
The length of  <math>\overline{OC}</math> is equal to <math>\sqrt{39}</math> by Pythagorean Theorem, the length of the hypotenuse is <math>8</math>, and the other leg is <math>5</math>. Using similar triangles, <math>OB</math> is <math>13</math>, and therefore half of <math>AB</math> is <math>\frac{65}{8}</math>. Doubling gives <math>\frac{65}{4}</math>, which results in <math>65+4=\boxed{\textbf{D) }69}</math>.
<math>QED \blacksquare</math>
+
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah
 +
 
 +
== Video Solution ==
 +
https://youtu.be/NsQbhYfGh1Q?t=1328
 +
 
 +
~ pi_is_3.14
 +
 
 +
==See Also==
 
{{AMC10 box|year=2018|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2018|ab=A|num-b=14|num-a=16}}
 +
{{MAA Notice}}
 +
 +
[[Category:Introductory Geometry Problems]]

Revision as of 04:32, 2 February 2021

Problem

Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B$, as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

[asy] draw(circle((0,0),13)); draw(circle((5,-6.2),5)); draw(circle((-5,-6.2),5)); label("$B$", (9.5,-9.5), S); label("$A$", (-9.5,-9.5), S); [/asy]

$\textbf{(A) }   21   \qquad    \textbf{(B) }  29   \qquad    \textbf{(C) }  58   \qquad   \textbf{(D) } 69 \qquad  \textbf{(E) }   93$

Solution 1

[asy] draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$X$", (0,0), N); label("$Y$", (-5,-6.25),NW); label("$Z$", (5,-6.25),NE); [/asy] Let the center of the surrounding circle be $X$. The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$, $YZ$, $XA$, and $XB$. Now observe that $\triangle XYZ$ is similar to $\triangle XAB$. Writing out the ratios, we get \[\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.\] Therefore, our answer is $65+4= \boxed{\textbf{D) } 69}$.

Solution 2

[asy] draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S); label("$C$", (0,-6.25), NE); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((0,0)--(0,-13)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$O$", (0,0), N); [/asy] Let the center of the large circle be $O$. Let the common tangent of the two smaller circles be $C$. Draw the two radii of the large circle, $\overline{OA}$ and $\overline{OB}$ and the two radii of the smaller circles to point $C$. Draw ray $\overrightarrow{OC}$ and $\overline{AB}$. This sets us up with similar triangles, which we can solve. The length of $\overline{OC}$ is equal to $\sqrt{39}$ by Pythagorean Theorem, the length of the hypotenuse is $8$, and the other leg is $5$. Using similar triangles, $OB$ is $13$, and therefore half of $AB$ is $\frac{65}{8}$. Doubling gives $\frac{65}{4}$, which results in $65+4=\boxed{\textbf{D) }69}$.

Video Solution

https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah

Video Solution

https://youtu.be/NsQbhYfGh1Q?t=1328

~ pi_is_3.14

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png