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2007 AIME I Problems/Problem 15

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Problem

Let ABC be an equilateral triangle, and let D and F be points on sides BC and AB, respectively, with FA = 5 and CD = 2. Point E lies on side CA such that angle DEF = 60^{\circ}. The area of triangle DEF is 14\sqrt{3}. The two possible values of the length of side AB are p \pm q \sqrt{r}, where p and q are rational, and r is an integer not divisible by the square of a prime. Find r.

Solution

Image:AIME I 2007-15.png

Denote the length of a side of the triangle x, and of \overline{AE} as y. The area of the entire equilateral triangle is \frac{x^2\sqrt{3}}{4}. Add up the areas of the triangles using the \frac{1}{2}ab\sin C formula (notice that for the three outside triangles, \sin 60 = \frac{\sqrt{3}}{2}): \frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5 \cdot y + (x - 2)(x - 5) + 2(x - y)) + 14\sqrt{3}. This simplifies to \frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56). Some terms will cancel out, leaving y = \frac{5}{3}x - 22.

\angle FEC is an external angle to \triangle AEF, from which we find that \displaystyle 60 + \angle CED = 60 + \angle AFE, so \displaystyle \angle CED = \angle AFE. Similarly, we find that \angle EDC = \angle AEF. Thus, \triangle AEF \sim \triangle CDE. Setting up a ratio of sides, we get that \frac{5}{x-y} = \frac{y}{2}. Using the previous relationship between x and y, we can solve for x.

\displaystyle xy - y^2 = 10

\frac{5}{3}x^2 - 22x - \left(\frac{5}{3}x - 22\right)^2 - 10 = 0

\frac{5}{3}x^2 - \frac{25}{9}x^2 - 22x + 2 \cdot \frac{5 \cdot 22}{3}x - 22^2 - 10= 0

10x^2 - 462x + 66^2 + 90 = 0

Use the quadratic formula, though we only need the root of the discriminant. This is \sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}= \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}. The answer is 989.

See also

2007 AIME I (ProblemsResources)
Preceded by
Problem 14
Followed by
Last Question
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