Difference between revisions of "2018 AMC 12B Problems/Problem 25"
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== Solution 2 == | == 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> Since <math>\triangle P_2O_2M</math> is a 30-60-90 triangle, <math>O_2M = 2</math> and <math>P_2M = 2\sqrt{3}</math>. Also, 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 <cmath>\left[P_1P_2P_3\right] = \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> So, our answer is <math>252 + 300 = \boxed{\textbf{D)}552}</math>. | + | 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> Since <math>\triangle P_2O_2M</math> is a 30-60-90 triangle, <math>O_2M = 2</math> and <math>P_2M = 2\sqrt{3}</math>. Also, 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 <cmath>\left[P_1P_2P_3\right] = \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> So, our answer is <math>252 + 300 = \boxed{\textbf{D) }552}</math>. |
== Solution 3 == | == Solution 3 == |
Revision as of 19:19, 14 February 2021
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 intersection 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 so the area of 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 Since is a 30-60-90 triangle, and . Also, since and , we can find find . Thus, . This makes So, our answer is .
Solution 3
Let be the center of circle for . Let be the centroid of , which also happens to be the centroid of . Because and , . is the height of , thus is .
Applying cosine law on , one finds that . Multiplying by to solve for the height of , one gets . Simply multiplying by and then calculating the equilateral triangle's area, one would get the final result of .
This makes the answer .
~AlbeePach~
Solution 4
First, note that because the , the arcs inside the shaded equilateral triangle are each . Also, the distances between the centers of any two of the given circles are each . Draw the circle concentric with with radius . Because the arc of inside said triangle is , touches , say at a point . Thus, is a common tangent of and , and it can be seen from inspection of the given diagram that the line is an common internal tangent. The length of the common internal tangent segment of and is then , and it is easily seen that . Because , the area of the shaded equilateral triangle is . We get
~crazyeyemoody907
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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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