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2009 Indonesia MO Problems/Problem 3: Difference between revisions

Victorzwkao (talk | contribs)
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Created page with "==Solution (credit to Moonmathpi496)== Since <math>AG = \tfrac23 \cdot AM</math>, lengths on <math>\triangle AB'G</math> and <math>\triangle AGC'</math> are <math>\tfrac23</m..."
 
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==Solution (credit to Moonmathpi496)==
==Solution (credit to Moonmathpi496)==
Since <math>AG = \tfrac23 \cdot AM</math>, lengths on <math>\triangle AB'G</math> and <math>\triangle AGC'</math> are <math>\tfrac23</math> the lengths of <math>\triangle ABM</math> and <math>\triangle AMC</math>, respectively.


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


Using Menelaus's Theorem, <math>\frac{HE}{EA}\cdot\frac{AF}{FG}\cdot\frac{GP}{PH}=\frac{HE}{EA}\cdot\frac{AF}{FG}\cdot\frac{BD}{DC}=1</math>
Using [[Menelaus' Theorem]], <math>\frac{HE}{EA}\cdot\frac{AF}{FG}\cdot\frac{GP}{PH}=\frac{HE}{EA}\cdot\frac{AF}{FG}\cdot\frac{BD}{DC}=1</math>


Next, <math>\frac{AE+EH}{AC}=\frac{AP}{AD}</math>, and <math>\frac{AF-GF}{AB}=\frac{AP}{AD}</math>
Next, <math>\frac{AE+EH}{AC}=\frac{AP}{AD}</math>, and <math>\frac{AF-GF}{AB}=\frac{AP}{AD}</math>

Latest revision as of 21:59, 17 September 2024

Solution (credit to Moonmathpi496)

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' 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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