Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it Bulgaria

Problem:

Let $M$ be an interior point of $\triangle ABC$ . The lines $AM$ , $BM$ and $CM$ meet the lines $BC$ , $CA$ and $AB$ at points $A_1$ , $B_1$ and $C_1$ , respectively, such that $S_{CB_1M} = 2 S_{AC_1M}$ . Prove that $A_1$ is the midpoint of the segment $BC$ if and only if $S_{BA_1M} = 3 S_{AC_1M}$ .

Solution

Solution:

Let $A_1$ be the midpoint of the segment $BC$ . Then Ceva's theorem implies that
$\frac{AC_1}{C_1B} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} = 1$
i.e., $\frac{AC_1}{C_1B} = \frac{B_1A}{B_1C}$ . Hence we have $B_1C_1 \parallel BC$ , i.e. $S_{BC_1M} = S_{CB_1M} = 2 S_{AC_1M}$ and we get that $S_{AB_1M} = S_{AC_1M}$ . Then

$\frac{1}{3} = \frac{S_{AC_1M}}{S_{AMC}} = \frac{C_1M}{MC} = \frac{S_{BC_1M}}{S_{BMC}} = \frac{2 S_{AC_1M}}{2 S_{BA_1M}}$
and therefore $S_{BA_1M} = 3 S_{AC_1M}$ .

Conversely, let $S_{AC_1M} = 1$ , $S_{CB_1M} = 2$ , $S_{BA_1M} = 3$ , $S_{BC_1M} = x$ , $S_{CA_1M} = 3y$ and $S_{AB_1M} = 2z$ . We have to show that $y = 1$ . Note that
$\frac{1}{2(z+1)} = \frac{S_{AC_1M}}{S_{AMC}} = \frac{C_1M}{CM} = \frac{S_{C_1MB}}{SCMB} = \frac{x}{3(y+1)}$

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