Difference between revisions of "2018 AMC 10A Problems/Problem 15"
(→Solution 1 (SOMEONE HELP ME ADD A PICTURE PLEASE)) |
(→Video Solution (HOW TO THINK CREATIVELY!)) |
||
(30 intermediate revisions by 19 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 11: | Line 13: | ||
<math>\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93 </math> | <math>\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93 </math> | ||
− | ==Solution | + | ==Solution== |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
<asy> | <asy> | ||
Line 25: | Line 21: | ||
label("$A$", (-8.125,-10.15), S); | label("$A$", (-8.125,-10.15), S); | ||
label("$B$", (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((0,0)--(8.125,-10.15)); | draw((0,0)--(8.125,-10.15)); | ||
draw((-5,-6.25)--(5,-6.25)); | draw((-5,-6.25)--(5,-6.25)); | ||
− | |||
draw((-8.125,-10.15)--(8.125,-10.15)); | draw((-8.125,-10.15)--(8.125,-10.15)); | ||
− | label("$ | + | label("$X$", (0,0), N); |
+ | label("$Y$", (-5,-6.25),NW); | ||
+ | label("$Z$", (5,-6.25),NE); | ||
</asy> | </asy> | ||
− | Let the center of the | + | |
− | + | 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> by SAS. | |
− | + | ||
+ | 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> | ||
+ | Therefore, our answer is <math>65+4= \boxed{\textbf{(D) } 69}</math>. | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!)== | ||
+ | https://youtu.be/xFnLbr-qt6I | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | |||
+ | https://youtu.be/HJALwsbHZXc | ||
+ | |||
+ | - Whiz | ||
+ | |||
+ | https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah | ||
+ | |||
+ | == Video Solution 2 by OmegaLearn == | ||
+ | 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]] |
Latest revision as of 17:10, 15 October 2023
Contents
Problem
Two circles of radius are externally tangent to each other and are internally tangent to a circle of radius
at points
and
, as shown in the diagram. The distance
can be written in the form
, where
and
are relatively prime positive integers. What is
?
Solution
Let the center of the surrounding circle be . The circle that is tangent at point
will have point
as the center. Similarly, the circle that is tangent at point
will have point
as the center. Connect
,
,
, and
. Now observe that
is similar to
by SAS.
Writing out the ratios, we get
Therefore, our answer is
.
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solution 1
- Whiz
https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah
Video Solution 2 by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=1328
~ pi_is_3.14
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
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.