Q27

Q27 - In $\triangle ABC$, $AB = 5$, $BC = 7$, and $CA = 8$. Let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively, and let $M$ be the midpoint of $BC$. The area of triangle $MEF$ can be expressed as $\frac{a\sqrt{b}}{c}$ for positive integers $a$, $b$, and $c$ such that the greatest common divisor of $a$ and $c$ is 1 and $b$ is not divisible by the square of any prime.
Compute $a + b + c$.

Solution:

We first observe that quadrilateral EFBC is cyclic with circumcenter M since
$\angle BEC = \angle CFB = 90^\circ$. Thus, $MB = MF = ME = MC = BC/2 = 7/2$ as these segments are radii of
the circumscribed circle of EFBC, so triangles $\triangle MBF$, $\triangle MEC$ and $\triangle MEF$ are isosceles.
From these observations, we reduce that
$\angle BFM = \angle B$ and $\angle CEM = \angle C$, so $\angle BMF = 180^\circ - 2\angle B$ and $\angle CME = 180^\circ - 2\angle C$.
Therefore,

$$\angle EMF = 180^\circ - (180^\circ - 2\angle B + 180^\circ - 2\angle C) = 2(\angle B + \angle C) - 180^\circ = 2(180^\circ - \angle A) - 180^\circ = 180^\circ - 2\angle A$$

Now, by the Law of Cosines, we calculate

$$\cos A = \frac{AB^2 + AC^2 - BC^2}{2(AB)(AC)} = \frac{5^2 + 7^2 - 8^2}{2(5)(8)} = \frac{25 + 49 - 64}{80} = \frac{40}{80} = \frac{1}{2}$$

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So A = 60 degree.
∠EMF = 180 - 120 = 60.
Area of △MEF = 1/2.sin(60).ME.MF = 49.sqrt(3)/16. a + b+ c = 49 + 3 + 16 =68.