2006 Indonesia MO Problems/Problem 5

Revision as of 18:48, 14 December 2019 by Rockmanex3 (talk | contribs) (Solution to Problem 5 -- Menelaus' Theorem)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

In triangle $ABC$, $M$ is the midpoint of side $BC$ and $G$ is the centroid of triangle $ABC$. A line $l$ passes through $G$, intersecting line $AB$ at $P$ and line $AC$ at $Q$, where $P\ne B$ and $Q\ne C$. If $[XYZ]$ denotes the area of triangle $XYZ$, show that $\frac{[BGM]}{[PAG]}+\frac{[CMG]}{[QGA]}=\frac32$.

Solution (credit to Moonmathpi496)

First, since $G$ is the centroid of the triangle and $BM = MC$, $[BMG] = [CMG] = \frac16 [ABC]$. Also, note that $[BGA] = \frac13 [ABC]$, so $[APG] = \frac{AP}{AB} \cdot \frac13 [ABC]$. Similarly, $[AQG] = \frac{AQ}{AC} \cdot \frac13 [ABC]$. Substituting the areas results in \begin{align*} \frac{[BGM]}{[PAG]}+\frac{[CMG]}{[QGA]} &= \frac{\frac16 [ABC]}{\frac{AP}{AB} \cdot \frac13 [ABC]} + \frac{\frac16 [ABC]}{\frac{AQ}{AC} \cdot \frac13 [ABC]} \\ &= \frac12 \cdot \left( \frac{AB}{AP} + \frac{AC}{AQ} \right). \end{align*} Draw $B'C'$ such that $B'C' \parallel BC$, $B'C'$ passes through $G$, $B'$ is on $AB$, and $C'$ and $AC$. By AA Similarity, $\triangle AB'G \sim \triangle ABM$ and $\triangle AGC' \sim \triangle AMC$. Since $AG = \tfrac23 \cdot AM$, lengths on $\triangle AB'G$ and $\triangle AGC'$ are $\tfrac23$ the lengths of $\triangle ABM$ and $\triangle AMC$, respectively. [asy] pair B=(0,0),A=(20,90),C=(100,0),M=(50,0),G=(40,30); pair b=(20/3,30),c=(220/3,30); draw(A--B--C--A); draw(A--M); draw(b--c); dot(G); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$M$",M,S); label("$B'$",b,W); label("$C'$",c,E); label("$G$",G,NW); draw((4,18)--(65.714,38.571)); label("$P$",(4,18),W); label("$Q$",(65.714,38.571),E); [/asy] By applying Menelaus' Theorem, $\frac{C'Q}{QA} \cdot \frac{PA}{PB'} \cdot \frac{B'G}{GC'} = 1$. Note that $B'G = GC'$, so $\frac{C'Q}{QA} \cdot \frac{PA}{PB'} = 1$ and $\frac{C'Q}{QA} = \frac{PB'}{PA}$.


Adding $\frac{QA}{QA} = 1$ to both sides results in $\frac{C'A}{QA} = \frac{PB'}{PA} + 1$, and rearranging both sides results in $\frac{C'A}{QA} - \frac{PB'}{PA} = 1$. Adding $\frac{PA}{PA} = 1$ to both sides results in $\frac{C'A}{QA} + \frac{AB'}{PA} = 2$.


Note that $C'A = \frac{2 CA}{3}$ and $B'A = \frac{2 BA}{3}$, so substituting those values results in $\frac{2 CA}{3 QA} + \frac{2 BA}{3 PA} = 2$. Thus, $\frac{AB}{AP} + \frac{AC}{AQ} = 3$, so $\frac{[BGM]}{[PAG]}+\frac{[CMG]}{[QGA]} = \frac32$.

See Also

2006 Indonesia MO (Problems)
Preceded by
Problem 4
1 2 3 4 5 6 7 8 Followed by
Problem 6
All Indonesia MO Problems and Solutions
Invalid username
Login to AoPS