2009 Indonesia MO Problems/Problem 3

Solution (credit to Moonmathpi496)

Since $AG = \tfrac23 \cdot AM$ , lengths on $\triangle AB'G$ and $\triangle AGC'$ are $\tfrac23$ the lengths of $\triangle ABM$ and $\triangle AMC$ , respectively.

Draw $GH$ such that $GH \parallel BC$ , $GH$ passes through $P$ , $G$ is on $AB$ , and $H$ on $AC$ . By AA Similarity, $\triangle AGP \sim \triangle ABD$ and $\triangle APH \sim \triangle ADC$ . Thus, $\frac{GP}{BD}=\frac{AP}{AD}=\frac{PH}{DC}$ . This also means $\frac{GP}{PH}=\frac{BD}{DC}$

$[asy] pair B=(0,0),A=(80,90),C=(100,0),D=(50,0),P=(60,30); pair g=(80/3,30),h=(280/3,30); draw(A--B--C--A); draw(A--D); draw(g--h); dot(P); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$G$",g,W); label("$H$",h,E); label("$P$",P,NW); draw((8,9)--(90.58824,42.35294)); label("$F$",(8,9),W); label("$E$",(90.58824,42.35294),E); [/asy]$

Using Menelaus's Theorem, $\frac{HE}{EA}\cdot\frac{AF}{FG}\cdot\frac{GP}{PH}=\frac{HE}{EA}\cdot\frac{AF}{FG}\cdot\frac{BD}{DC}=1$

Next, $\frac{AE+EH}{AC}=\frac{AP}{AD}$ , and $\frac{AF-GF}{AB}=\frac{AP}{AD}$

Solving $EH$ and $GF$ yields $EH=\frac{AP\cdot AC-AE\cdot AD}{AD}$ , $GF=\frac{AF\cdot AD-AP\cdot AB}{AD}$ Plugging these into the Menelaus's equation above yields

$\frac{AP\cdot AC-AE\cdot AD}{AD\cdot EA}\cdot \frac{AF\cdot AD}{AF\cdot AD-AP\cdot AB}\cdot \frac{BD}{DC}=1$

$AP\cdot AC\cdot AF\cdot BD - AE\cdot AD\cdot AF\cdot BD = EA\cdot AF\cdot AD\cdot DC -EA\cdot AP\cdot AB\cdot DC$

$AB\cdot AE\cdot AP\cdot DC+AC\cdot DB\cdot AF\cdot AP=AE\cdot AD\cdot AF\cdot BD +AE\cdot AD\cdot AF\cdot DC$ $AB\cdot AE\cdot AP\cdot DC+AC\cdot DB\cdot AF\cdot AP=AE\cdot AD\cdot AF\cdot (BD+DC)$ $AB\cdot AE\cdot AP\cdot DC+AC\cdot DB\cdot AF\cdot AP=AE\cdot AD\cdot AF\cdot (BC)$ dividing both sides by $AF\cdot AE\cdot AP$ yields the result $\frac{AB}{AF}\cdot DC+\frac{AC}{AE}\cdot DB=\frac{AD}{AP}\cdot BC$

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2009 Indonesia MO Problems/Problem 3

Solution (credit to Moonmathpi496)