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1998 AIME Problems/Problem 12

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Problem

Let ABC be equilateral, and D, E, and F be the midpoints of \overline{BC}, \overline{CA}, and \overline{AB}, respectively. There exist points P, Q, and R on \overline{DE}, \overline{EF}, and \overline{FD}, respectively, with the property that P is on \overline{CQ}, Q is on \overline{AR}, and R is on \overline{BP}. The ratio of the area of triangle ABC to the area of triangle PQR is a + b\sqrt {c}, where a, b and c are integers, and c is not divisible by the square of any prime. What is a^{2} + b^{2} + c^{2}?

Image:1998_AIME-12.png

Solution

We let x = EP = FQ, y = EQ, k = PQ. Since AE = \frac {1}{2}AB and AD = \frac {1}{2}AC, \triangle AED \sim \triangle ABC and ED \parallel BC.

By alternate interior angles, we have \angle PEQ = \angle BFQ and \angle EPQ = \angle FBQ. By vertical angles, \angle EQP = \angle FQB.

Thus \triangle EQP \sim \triangle FQB, so \frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y.

Since \triangle EDF is equilateral, EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1. Solving for x and y using x^{2} = y and x + y = 1 gives x = \frac {\sqrt {5} - 1}{2} and y = \frac {3 - \sqrt {5}}{2}.

Using the Law of Cosines, we get

k^{2}  =  x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}
=  \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\ri...
= 7 - 3\sqrt {5}

We want the ratio of the squares of the sides, so \frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5} so a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}.

See also

1998 AIME (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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