Difference between revisions of "2017 AMC 12A Problems/Problem 16"
(→Solution 3) |
(→Solution 3) |
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<asy> | <asy> | ||
− | size(5cm); | + | size(7.5cm); |
draw(arc((0,0),3,0,180)); | draw(arc((0,0),3,0,180)); | ||
draw(arc((2,0),1,0,180)); | draw(arc((2,0),1,0,180)); | ||
Line 108: | Line 108: | ||
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | ||
draw((-1,0)--P); | draw((-1,0)--P); | ||
+ | draw((-3,12/7)--P); | ||
+ | draw((3,12/7)--P); | ||
draw((2,0)--P); | draw((2,0)--P); | ||
draw((0,0)--(9/5,12/5)); | draw((0,0)--(9/5,12/5)); | ||
Line 116: | Line 118: | ||
</asy> | </asy> | ||
− | Let <math>C</math> be the center of the largest semicircle. | + | Let <math>C</math> be the center of the largest semicircle. We know that <math>AC = 1</math>, <math>CB = 2</math>, <math>AP = r + 2</math>, <math>BP = r + 1</math>, and <math>CP = 3 - r</math>. Notice that <math>\Delta ACP</math> and <math>\Delta CBP</math> are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of <math>\Delta CBP</math> must be twice that of <math>\Delta ACP</math>, since the area of a triangle is <math>\frac{1}{2} Base \cdot Height</math>. |
− | We know that <math>AC | + | |
− | Again, we don't | + | Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. |
<asy> | <asy> | ||
Line 137: | Line 139: | ||
</asy> | </asy> | ||
− | + | Let <math>A_1</math> equal to the area of <math>\Delta ACP</math> and <math>A_2</math> equal to the area of <math>\Delta CBP</math>. | |
− | |||
[[Heron's Formula]] states that the area of an triangle with sides <math>a</math> <math>b</math> and <math>c</math> is | [[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> | <cmath>\sqrt{s(s-a)(s-b)(s-c)}</cmath> | ||
where <math>s</math>, or the semiperimeter, is half of the triangle's perimeter. | where <math>s</math>, or the semiperimeter, is half of the triangle's perimeter. | ||
+ | The semiperimeter <math>s_1</math> of <math>\Delta ACP</math> is <cmath>\frac{[(r + 2) + (3 - r) + 1]}{2} = \frac{6}{2} = 3</cmath> | ||
Use Heron's Formula to obtain | 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> | <cmath>A_1 = \sqrt{3(2)(3-2-r)(3-3+r)} = \sqrt{6r(1-r)} = \sqrt{6r-6r^2}</cmath> | ||
− | Using | + | Using Heron's Formula again, find the area of <math>\Delta CBP</math> with sides <math>r+1</math>, <math>2</math>, and <math>3-r</math>. |
− | <cmath>s_2 = | + | <cmath>s_2 = \frac{(r + 1) + 2 + (3 - r)}{2} = 3</cmath> |
− | <cmath>A_2 = \sqrt{3(3-2)(3-1-r)(3-3+r)} = \sqrt{ | + | <cmath>A_2 = \sqrt{3(3-2)(3-1-r)(3-3+r)} = \sqrt{3(2r-r^2)} = \sqrt{6r-3r^2}</cmath> |
Now, | Now, | ||
− | <cmath> | + | <cmath>2 \cdot A_1 = A_2</cmath> |
<cmath>2\sqrt{6r-6r^2} = \sqrt{6r-3r^2}</cmath> | <cmath>2\sqrt{6r-6r^2} = \sqrt{6r-3r^2}</cmath> | ||
<cmath>4(6r-6r^2) = 6r-3r^2</cmath> | <cmath>4(6r-6r^2) = 6r-3r^2</cmath> |
Revision as of 22:18, 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. We know that , , , , and . Notice that and are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of must be twice that of , since the area of a triangle is .
Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles.
Let equal to the area of and equal to the area of . 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.
The semiperimeter of is Use Heron's Formula to obtain
Using Heron's Formula again, find the area of 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.