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Belarusian Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Belarus

The tangents to the circumcircle of the acute triangle $ABC$ , passing through the vertices $B$ and $C$ , meet at point $F$ . The points $M$ , $L$ and $N$ are the feet of the perpendiculars from the vertex $A$ to the lines $FB$ , $FC$ and $BC$ respectively.
Prove the inequality $AM + AL \ge 2AN$ .

Solution

Note that $\angle ABM = \angle ACB$ by the property of the angle between the tangent to the circle at point $B$ and the chord $AB$ . The right triangles $AMB$ and $ANC$ are similar since they have equal acute angles, therefore $\frac{AM}{AN} = \frac{AB}{AC}$ , whence $AN = \frac{AM \cdot AC}{AB}$ .

By analogy, the triangles $ANB$ and $ALC$ are similar and $\frac{AL}{AN} = \frac{AC}{AB}$ , whence $AN = \frac{AL \cdot AB}{AC}$ . Therefore
$AN^2 = \frac{AM \cdot AC}{AB} \cdot \frac{AL \cdot AB}{AC} = AM \cdot AL.$

$\text{Finally, from the AM GM inequality } AM + AL \ge 2\sqrt{AM \cdot AL} = 2\sqrt{AN^2} = 2AN.$

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