Difference between revisions of "2018 AMC 12B Problems/Problem 25"
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<math>\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554</math> | <math>\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554</math> | ||
− | == Solution == | + | == Solution 1 == |
Let <math>O_i</math> be the center of circle <math>\omega_i</math> for <math>i=1,2,3</math>, and let <math>K</math> be the intersectoin of lines <math>O_1P_1</math> and <math>O_2P_2</math>. Because <math>\angle P_1P_2P_3 = 60^\circ</math>, it follows that <math>\triangle P_2KP_1</math> is a <math>30-60-90^\circ</math> triangle. Let <math>d=P_1K</math>; then <math>P_2K = 2d</math> and <math>P_1P_2 = \sqrt 3d</math>. The Law of Cosines in <math>\triangle O_1KO_2</math> gives <cmath>8^2 = (d+4)^2 + (2d-4)^2 - 2(d+4)(2d-4)\cos 60^\circ,</cmath>which simplifies to <math>3d^2 - 12d - 16 = 0</math>. The positive solution is <math>d = 2 + \tfrac23\sqrt{21}</math>. Then <math>P_1P_2 = \sqrt 3 d = 2\sqrt 3 + 2\sqrt 7</math>, and the required area is <cmath>\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.</cmath>The requested sum is <math>300 + 252 = \boxed{552}</math>. | Let <math>O_i</math> be the center of circle <math>\omega_i</math> for <math>i=1,2,3</math>, and let <math>K</math> be the intersectoin of lines <math>O_1P_1</math> and <math>O_2P_2</math>. Because <math>\angle P_1P_2P_3 = 60^\circ</math>, it follows that <math>\triangle P_2KP_1</math> is a <math>30-60-90^\circ</math> triangle. Let <math>d=P_1K</math>; then <math>P_2K = 2d</math> and <math>P_1P_2 = \sqrt 3d</math>. The Law of Cosines in <math>\triangle O_1KO_2</math> gives <cmath>8^2 = (d+4)^2 + (2d-4)^2 - 2(d+4)(2d-4)\cos 60^\circ,</cmath>which simplifies to <math>3d^2 - 12d - 16 = 0</math>. The positive solution is <math>d = 2 + \tfrac23\sqrt{21}</math>. Then <math>P_1P_2 = \sqrt 3 d = 2\sqrt 3 + 2\sqrt 7</math>, and the required area is <cmath>\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.</cmath>The requested sum is <math>300 + 252 = \boxed{552}</math>. | ||
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+ | == Solution 2 == | ||
+ | |||
+ | Let <math>O_1</math> and <math>O_2</math> be the centers of <math>\omega_1</math> and <math>\omega_2</math> respectively and draw <math>O_1O_2</math>, <math>O_1P_1</math>, and <math>O_2P_2</math>. Note than <math>\angle{OP_1P_2}</math> and <math>\angle{O_2P_2P_3}</math> are both right. Furthermore, since <math>\triangle{P_1P_2P_3}</math> is equilateral, <math>m\angle{P_1P_2P_3} = 60^\circ</math> and <math>m\angle{O_2P_2P_1} = 30^\circ</math>. Mark <math>M</math> as the base of the altitude from <math>O_2</math> to <math>P_1P_2</math>. By special right triangles, <math>O_2M = 2</math> and <math>P_2M = 2\sqrt{3}</math>. since <math>O_1O_2 = 8</math> and <math>O_1P_1 = 4</math>, we can find find <math>P_1M = \sqrt{8^2 - (4 + 2)^2} = 2\sqrt{7}</math>. Thus, <math>P_1P_2 = P_1M + P_2M = 2\sqrt{3} + 2\sqrt{7}</math>. This makes <math>\left[P_1P_2P_3\right] = \frac{\left(2\sqrt{3} + 2\sqrt{7}\right)^2\sqrt{3}}{4} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{252} + \sqrt{300}</math>. This makes the answer <math>252 + 300 = 552</math>. <math>\boxed{\textbf{D}.}</math> | ||
==See Also== | ==See Also== |
Revision as of 17:05, 16 February 2018
Contents
Problem
Circles , , and each have radius and are placed in the plane so that each circle is externally tangent to the other two. Points , , and lie on , , and respectively such that and line is tangent to for each , where . See the figure below. The area of can be written in the form for positive integers and . What is ?
Solution 1
Let be the center of circle for , and let be the intersectoin of lines and . Because , it follows that is a triangle. Let ; then and . The Law of Cosines in gives which simplifies to . The positive solution is . Then , and the required area is The requested sum is .
Solution 2
Let and be the centers of and respectively and draw , , and . Note than and are both right. Furthermore, since is equilateral, and . Mark as the base of the altitude from to . By special right triangles, and . since and , we can find find . Thus, . This makes . This makes the answer .
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
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