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Belarus — Selection and Training Session

Geometry Difficulty 5.2 AIME, harder Prove it Belarus

Given the isosceles triangle $ABC$ ( $CA = CB$ ). The bisector of the angle $\angle B$ intersects the side $AC$ at point $L$ and the circumcircle of the triangle $ABC$ at point $D$ . It is known that $\angle C > 60^\circ$ .
Prove that $DC + DL \le BC$ .

Solution

We prove the statement for $\angle C \ge 36^\circ$ . Let $\angle ABC = \angle BAC = 2x$ . We have $DC + DL \le BC \Leftrightarrow \frac{BC}{DC} \ge 1 + \frac{DL}{DC}$ . Now, by the law of sines,
$\frac{BC}{DC} = \frac{\sin \angle BDC}{\sin \angle DBC} = \frac{\sin 2x}{\sin x} = 2 \cos x, \\ \frac{DL}{DC} = \frac{\sin 3x}{\sin x} = \frac{1}{3 - 4 \sin^2 x} = \frac{1}{4 \cos^2 x - 1},$
so we need to prove that $2 \cos x \ge 1 + \frac{1}{4 \cos^2 x - 1} = \frac{4 \cos^2 x}{4 \cos^2 x - 1}$ . Note that $x \le 36^\circ$ since $\angle C \ge 36^\circ$ , so $\cos x > \frac{1}{2}$ and $4 \cos^2 x - 1 > 0$ . Therefore the inequality is equivalent to
$4 \cos^2 x - 1 \ge 2 \cos x \Leftrightarrow 4 \cos^2 x - 2 \cos x - 1 \ge 0 \Leftrightarrow \cos x \ge \frac{-1 + \sqrt{5}}{4},$
which is true for $x \le 36^\circ$ .

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