Difference between revisions of "2017 AMC 12A Problems/Problem 16"
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==Solution== | ==Solution== | ||
+ | Connect the centers of the tangent circles! (call the center of the large circle <math>C</math>) | ||
+ | |||
+ | <asy> | ||
+ | size(5cm); | ||
+ | draw(arc((0,0),3,0,180)); | ||
+ | draw(arc((2,0),1,0,180)); | ||
+ | draw(arc((-1,0),2,0,180)); | ||
+ | draw((-3,0)--(3,0)); | ||
+ | pair P = (9/7,12/7); | ||
+ | draw(circle(P,6/7)); | ||
+ | dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | ||
+ | draw((-1,0)--P); | ||
+ | draw((2,0)--P); | ||
+ | draw((0,0)--(9/5,12/5)); | ||
+ | </asy> | ||
+ | |||
+ | Notice that we don't even need the circles anymore; thus, draw triangle <math>\Delta ABP</math> with cevian <math>PC</math>: | ||
+ | |||
+ | <asy> | ||
+ | size(5cm); | ||
+ | draw((-1,0)--(2,0)); | ||
+ | pair P = (9/7,12/7); | ||
+ | dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | ||
+ | draw((-1,0)--P); | ||
+ | draw((2,0)--P); | ||
+ | draw((0,0)--P); | ||
+ | </asy> | ||
+ | |||
+ | and use [[Stewart's Theorem]]: | ||
+ | |||
+ | <cmath>AB \cdot AC \cdot BC + AB \cdot {CP}^2 = AC \cdot {BP}^2 + BC \cdot {AP}^2</cmath> | ||
+ | |||
+ | From what we learned from the tangent circles, we have <math>AB = 3</math>, <math>AC = 1</math>, <math>BC = 2</math>, <math>AP = 2 + r</math>, <math>BP = 1 + r</math>, and <math>CP = 3 - r</math>, where <math>r</math> is the radius of the circle centered at <math>P</math> that we seek. | ||
+ | |||
+ | Thus: | ||
+ | |||
+ | <cmath>3 \cdot 1 \cdot 2 + 3 {\left(3-r\right)}^2 = 1 {\left(1+r\right)}^2 + 2 {\left(2+r\right)}^2</cmath> | ||
+ | <cmath>6 + 3\left(9 - 6r + r^2\right) = \left(1 + 2r + r^2\right) + 2\left(4 + 4r + r^2\right)</cmath> | ||
+ | <cmath>33 - 18r + 3r^2 = 9 + 10r + 3r^2</cmath> | ||
+ | <cmath>28r = 24</cmath> | ||
+ | <cmath>r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2017|ab=A|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:19, 8 February 2017
Problem
In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ?
Solution
Connect the centers of the tangent circles! (call the center of the large circle )
Notice that we don't even need the circles anymore; thus, draw triangle with cevian :
and use Stewart's Theorem:
From what we learned from the tangent circles, we have , , , , , and , where is the radius of the circle centered at that we seek.
Thus:
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 12 Problems and Solutions |
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