2012 AIME I Problems/Problem 13
Three concentric circles have radii and An equilateral triangle with one vertex on each circle has side length The largest possible area of the triangle can be written as where and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find
Reinterpret the problem in the following manner. Equilateral triangle has a point on the interior such that and A counter-clockwise rotation about vertex maps to and to
Note that angle is and which tells us that triangle is equilateral and that We now notice that and which tells us that angle is because there is a -- Pythagorean triple. Now note that and so and Applying the law of cosines on triangle yields
and thus the area of equals
so our final answer is
Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties.
We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have two cases to consider; either the center of the circles lies in the interior of triangle or it does not (and we shall show that both can happen). To see that the first case can occur proceed as follows. Using the notation from Solution 1, let so that . Let . The law of cosines on and yields Solving this system will yield the value of . Since we have that Substituting these into the equation we obtain After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain so that by the quadratic formula . Under the hypothesis that lies in the interior of triangle , must be . To see this, note that the other value for is roughly so that , but since and we have a contradiction. We then obtain the area as in Solution 1.
Now, suppose does not lie in the interior of triangle . We then obtain convex quadrilateral with diagonals and intersecting at . Here . We may let denote the measure of angle so that angle measures . Note that the law of cosines on triangles and yield the same equations as in the first case with replaced with . Thus we obtain again . If then , but this is impossible since but the shortest possible distance from to is the height of equilateral triangle which is ; a contradiction. Hence in this case . But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).
Let's call the circle center . It has a distance of 3, 4, 5 to an equilateral triangle . Consider ’s pedal triangle . Since ’s antipedal triangle is equilateral, must be the one of the isogonic centers of . We’ll take the one inside , i.e., the Fermat point, because it leads to larger . Now we construct the three equilateral triangles , , and , the same way the Fermat point is constructed. Then we have . Since is concyclic with =4 as diameter, we have . Similarly, , and . So is a 3-4-5 right triangle with . With some more angle chasing we get By Law of Cosines, we have And the area follows. By Mathdummy
Let be the equilateral triangle with Assume the coordinates of the vertices are and Let be such that and Then Subtraction and addition of the first two equations yield The third equation gives Then We can then solve for in terms of and have a substitution. We have Simplify it we have a quadratic equation for So The larger one leads to the solution. The smaller one relates to another equilateral triangle, as indicated in Solution 2.
We have , , and . Because is the median of , by Stewart's Theorem we have Because is the altitude of equilateral triangle , we have . Then in , we have , so , and the Law of Cosines givesTo calculate we apply the Law of Cosines to to get Finally, we get and and thus the area of equals so our final answer is
Video Solution by Richard Rusczyk
2020 AMC 12A Problems/Problem 24
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