Let ABC be an acute triangle...

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Let ABC be an acute triangle...

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Source: Bulgarian IMO TST 2005, Day 1, Problem 1

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#1 h Jul 7, 2013, 9:04 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABC$ be an acute triangle. Find the locus of the points $M$ , in the interior of $\bigtriangleup ABC$ , such that $AB-FG= \frac{MF.AG+MG.BF}{CM}$ , where $F$ and $G$ are the feet of the perpendiculars from $M$ to the lines $BC$ and $AC$ , respectively.

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#2 h Jul 7, 2013, 3:25 PM • 1 Y

Y by Adventure10

The expression is $(AB)(CM)-(FG)(CM)=(MF)(AG)+(MG)(BF)$ .
Use Ptolemy's Theorem in cyclic quad $MGCF$ :
It becomes: $(AB)(CM)-(MG)(FC)-(MF)(GC)=(MF)(AG)+(MG)(BF)$ .
$(AB)(CM)=MF(AG+GC)+MG(BF+FC)$
i.e. $(AB)(CM)=(MF)(AC)+(MG)(BC)$
i.e $(AB)(CM)=(MC\sin\angle MCB)(AC)+ (MC\sin\angle MCA)(BC)$
i.e $AB=AC\sin\angle MCB +BC\sin\angle MCA$
Let $CD'$ be the isogonal of $CM$ , where $D'$ lies on $AB$ .
So $AB=AC\sin\angle ACD' +BC\sin\angle BCD'$
$\frac{1}{2}(AB)(CD')=\frac{1}{2}(CD')(AC)\sin\angle ACD' +\frac{1}{2}(CD')(BC)\sin\angle BCD'$
$\frac{1}{2}(AB)(CD')=[ACD']+[BCD']$
$\frac{1}{2}(AB)(CD')=[ABC]$
So $AB\perp CD'$ .
So the locus of $M$ is all points in the interior of $\triangle ABC$ such that $M$ lies on $AO$ , where $O$ is the circumcentre of the triangle. (because circumcentre, orthocentre are isogonal conjugates)

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