Difference between revisions of "2017 AMC 12A Problems/Problem 16"
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<math>8r + 2 = -20r + 26</math> | <math>8r + 2 = -20r + 26</math> | ||
<math>28r = 24</math>, so <math>r</math> = <math>6/7</math> <math>\boxed{(B)}</math> | <math>28r = 24</math>, so <math>r</math> = <math>6/7</math> <math>\boxed{(B)}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | <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> | ||
+ | |||
+ | Let <math>C</math> be the center of the largest semicircle. Notice that <math>\Delta ACP</math> and <math>\Delta CBP</math> are bounded by the same two parallel lines and therefore have the same heights. Since the bases of these two triangles differ by a factor of 2, the area of <math>\Delta CBP</math> must be twice that of <math>\Delta ACP</math>. | ||
+ | We know that <math>AC</math> <math>=</math> 1, <math>CB</math> <math>=</math> 2, <math>AP</math> <math>=</math> <math>r</math> <math>+</math> 2, <math>BP</math> <math>=</math> <math>r</math> <math>+</math> 1, <math>CP</math> <math>=</math> <math>3</math> <math>-</math> <math>r</math>. | ||
+ | Again, we don't even need the circles and semicircles anymore; just focus on the triangles. | ||
+ | |||
+ | <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> | ||
+ | |||
+ | The semiperimeter <math>s_1</math> of <math>\Delta ACP</math> is <cmath>[(r + 2) + (3 - r) + 1] / 2 = 6/2 = 3</cmath> | ||
+ | |||
+ | [[Heron's Formula]] states that the area of an triangle with sides <math>a</math> <math>b</math> and <math>c</math> is | ||
+ | <cmath>\sqrt{s(s-a)(s-b)(s-c)}</cmath> | ||
+ | where <math>s</math>, or the semiperimeter, is half of the triangle's perimeter. | ||
+ | |||
+ | Use Heron's Formula to obtain | ||
+ | |||
+ | <cmath>A_1 = \sqrt{3(2)(3-2-r)(3-3+r)} = \sqrt{6r(1-r)} = \sqrt{6r-6r^2}</cmath> | ||
+ | |||
+ | Using the same formula again, find the area of the second, larger triangle (<math>\Delta CBP</math>) with sides <math>r+1</math>, <math>2</math>, and <math>3-r</math>. | ||
+ | |||
+ | <cmath>s_2 = ((r + 1) + 2 + (3 - r)) / 2 = 3</cmath> | ||
+ | <cmath>A_2 = \sqrt{3(3-2)(3-1-r)(3-3+r)} = \sqrt{6(2r-r^2)} = \sqrt{6r-3r^2}</cmath> | ||
+ | |||
+ | |||
+ | Now, | ||
+ | <cmath>A_2 = 2 \cdot A_1</cmath> | ||
+ | <cmath>2\sqrt{6r-6r^2} = \sqrt{6r-3r^2}</cmath> | ||
+ | <cmath>4(6r-6r^2) = 6r-3r^2</cmath> | ||
+ | <cmath>24r-24r^2 = 6r-3r^2</cmath> | ||
+ | <cmath>18r = 21r^2</cmath> | ||
+ | <cmath>r = \frac{18}{21} = \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 21:48, 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 1
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:
NOTICE to proficient editors: please label the points on the diagrams, thanks!
Solution 2
Like the solution above, connecting the centers of the circles results in triangle with cevian . The two triangles and share angle , which means we can use Law of Cosines to set up a system of 2 equations that solve for respectively:
(notice that the diameter of the largest semicircle is 6, so its radius is 3 and is 3 - r)
We can eliminate the extra variable of angle by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find :
, so =
Solution 3
Let be the center of the largest semicircle. Notice that and are bounded by the same two parallel lines and therefore have the same heights. Since the bases of these two triangles differ by a factor of 2, the area of must be twice that of . We know that 1, 2, 2, 1, . Again, we don't even need the circles and semicircles anymore; just focus on the triangles.
The semiperimeter of is
Heron's Formula states that the area of an triangle with sides and is where , or the semiperimeter, is half of the triangle's perimeter.
Use Heron's Formula to obtain
Using the same formula again, find the area of the second, larger triangle () with sides , , and .
Now,
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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