2018 AMC 12B Problems/Problem 25
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 ?
Let and be the centers of and respectively and draw , , and . Note that 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 . Thus, . This makes So, our answer is .
Note by diyarv:
A way to see that is through similar triangles. Call the intersection between and . Since angles and are both right angles, and and are congruent, triangles and are similar by AA similarity. And, since is and is 2, the common ratio is . Using the fact that , we see that these lengths are and respectively. Using the Pythagorean theorem, we see that and are and respectively, giving us a sum of .
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 .
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 .
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
Solution 5 (Pure Coordinate Bash)
This seems like a coordinate bashable problem. To do this, we notice that it is easier to graph the equilateral triangle first, then the circles, rather than the other way around. Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to 's side length.
WLOG let . Place triangle on the coordinate plane with , , and . Let be the radius of the circles.
Now, we find the coordinates of the centers of circles and . Since is tangent to the x-axis at , the center of is . Draw a right triangle with legs parallel to the x and y axes, and with hypotenuse as the segment from the center of to . Since the slope of is , the slope of the hypotenuse is , so the right triangle is . It's easy to see that the center of is .
Since and are tangent, the distance between the centers is , so we have By the Quadratic Formula, . We take the positive value to get the radius of the circle is .
Therefore, the ratio of the radius to the side length of the equilateral triangle is .
The side length of the equilateral triangle is , so its area is .
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