Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it Ireland

$ABC$ is an equilateral triangle. $D$ is the midpoint of $BC$ , $E$ is a point on $AC$ between $A$ and $C$ and $F$ is a point on $AB$ between $A$ and $B$ . The area of $\triangle EDC$ is $24\sqrt{3}$ , the area of $\triangle AFE$ is $24\sqrt{3}$ and the area of $\triangle FBD$ is $54\sqrt{3}$ . Find the length of the sides of the triangle $ABC$ .

Solution

Let $x = |BD| = |DC|$ . Because $\triangle ABC$ is equilateral, considering the areas of $\triangle BDF$ and $\triangle DCE$ gives
$54\sqrt{3} = \frac{1}{2}x \cdot |BF| \cdot \sin(60^\circ) = \frac{1}{4}\sqrt{3} \cdot x \cdot |BF| \quad \Rightarrow \quad x \cdot |BF| = 216$
$24\sqrt{3} = \frac{1}{2}x \cdot |CE| \cdot \sin(60^\circ) = \frac{1}{4}\sqrt{3} \cdot x \cdot |CE| \quad \Rightarrow \quad x \cdot |CE| = 96.$

Because $|AF| = 2x - |BF|$ and $|AE| = 2x - |CE|$ we obtain for the area of $\triangle AFE$ the equation $24\sqrt{3} = \frac{1}{2}(2x - |BF|)(2x - |CE|) \sin(60^\circ)$ from which we get $96x^2 = (2x^2 - x \cdot |BF|)(2x^2 - x \cdot |CE|) = (2x^2 - 216)(2x^2 - 96)$ . This easily translates into the equation $0 = x^4 - 108x^2 + 5184 = (x^2 - 144)(x^2 - 36)$ . Because $0 \le |AF| = 2x - \frac{216}{x}$ , we need to have $2x^2 \ge 216$ , hence $x^2 \ne 36$ . As $x > 0$ we see that the only solution is $x = 12$ . Therefore, $|AB| = |BC| = |CA| = 24$ .

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