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 $a^{2} + b^{2} + c^{2}$ ?

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

$k^{2} = x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}$

$= \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\right)\left(\frac {3 - \sqrt {5}}{2}\right)\cos{\frac {\pi}{3}}$

$= 7 - 3\sqrt {5}$

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 $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}$ .

1998 AIME (Problems • Resources)
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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1998 AIME Problems/Problem 12

From AoPSWiki

Problem

Solution

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