Triangle in a Triangle
An equilateral triangle of length 2 is drawn. A square is inscribed in the equilateral triangle. In the three smaller triangles formed by the square and the equilateral triangle, the incircle is drawn. The area of the triangle formed by connecting the centers of the three incircles has area , where , , and are integers and has no perfect square factors other than 1. Find .
The answer is 0.
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1 solution
Let the equilateral triangle be denoted as A at the apex, B the lower left vertex, and C the lower right vertex. Let S be the side of the square, and P,Q the points on BC where the square intersects the base of the triangle. Let t = BP = CQ, so that S = 2 - 2t.(Recall the side of the equilateral triangle is given as 2.) Then tan(60) = S/t, or sqrt(3) = (2 - 2t)/t. Solving for t, and simplifying, t =4 - 2 sqrt(3). Then S = 2 - 2t = 4 sqrt(3) - 6. Defining d as the hypotenuse of the right triangle with S and t as legs, we have cos(60) = t/d = 1/2, so d = 2t, or d = 8 - 4 sqrt(3).We can now compute the inradius, r, of the triangle with sides d,S,t from the equation r = Area/(semi-perimeter).The Area = (1/2)(t)(S) =14 sqrt(3) - 24. The semi-perimeter = (1/2)(d + S + t) = 3 - sqrt(3). Then r = 3 sqrt(3) - 5.We now need to compute the inradius, R, of the larger circle in the triangle whose sides are: 2 - d, 2 - d, and S.We note that 2 - d = 2 - (8 - 4 sqrt(3) = 4 sqrt(3) - 6. This makes life a little easier, for we see that 2 - d = S, and our triangle is equilateral with sides = 4 sqrt(3) - 6. Then the Area is ([sqrt(3)]/4) S^2, and the semi perimeter = 3S/2 =6 sqrt(3) - 9. The Area is (21/2) sqrt(3) - 18. The ratio, R = 2 - sqrt(3). Finally, we need to compute the area of the triangle whose vertices are the centers of the 3 circles. The base of this triangle is b = S + 2r, and the altitude, h, is h = S - r +R. Calculating (1/2)(b)(h), we have A = (1/2)(10 sqrt(3) - 16) (1) = A = 5 sqrt(3) - 8, so a + b + c = 5 + 3 +(-8) = 0. Ed Gray