50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010)

Question

Let be a circumscribed circle of ABC. Chord AD is a bisector of triangle ABC and meets BC at L, chord DK is perpendicular to AC and meets it at M. Find the ratio AMMC, if BLLC = 12. FIGURE_0 Fig.15 FIGURE_1

Accepted Answer

Using the property of our bisector, we get

\frac{B L}{A B} = \frac{L C}{A C} \Rightarrow \frac{A C}{A B} = \frac{L C}{B L}

, or

\frac{A C}{A B} = 2

, hence,

A C = 2 A B

(Fig.15).

∠ B A D = ∠ D A C (A D is a bisector of ∠ B A C), thus B D = D C . Consider S - midpoint of A C . Then A S = S C = A B . Thus, △ B A D = △ S A D (they have two equal sides and angle between them). This implies, that B D = D S and D C = B D = D S .

Hence, triangle

△ S D C

is isosceles with base

S C

. Then, its altitude

D M

is also median. Thus,

M C = \frac{1}{2} S C = \frac{1}{2} \cdot \frac{1}{2} A C = \frac{1}{4} A C

. We have

A M = A C - M C = A C - \frac{1}{4} A C = \frac{3}{4} A C

. Hence,

\frac{A M}{M C} = \frac{\frac{3}{4} A C}{\frac{1}{4} A C} = 3

.

Note: In acute-angled triangle

A B C

,

A C = 2 A B

, thus

∠ A C B

does not exceed

3 0^{\circ}

(Fig.16). Last observation implies that perpendicular from point

D

to the line

A C

meets segment

A C

.

Fig.16