Orthogonal projections and areas

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Source: Polish MO

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#1 h Feb 25, 2017, 7:29 PM • 1 Y

Y by Adventure10

In an acute triangle $ABC$ the bisector of $\angle BAC$ crosses $BC$ at $D$ . Points $P$ and $Q$ are orthogonal projections of $D$ on lines $AB$ and $AC$ . Prove that $[APQ]=[BCQP]$ if and only if the circumcenter of $ABC$ lies on $PQ$ .

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#2 h Feb 25, 2017, 8:10 PM • 2 Y

Y by Adventure10, Mango247

Nice problem!

This post has been edited 3 times. Last edited by math1181, Feb 25, 2017, 8:11 PM

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#3 h Feb 25, 2017, 10:52 PM • 1 Y

Y by Adventure10

I only found not-quite short bash solution,here is sketch:
$\frac{[APC]}{[APQ]}=\frac{b}{AQ}$ and $\frac{[CPB]}{ACP}=\frac{PB}{AP}$ so we have $[CPB]=\frac{b\cdot PB}{AQ\cdot AP}\cdot [APQ]$ so from $[APQ]=[BCQP]$ we have $AP\cdot CQ+b\cdot PB=AQ\cdot AP$ and now from from $P-O-B$ applying in bary formula for colinearity we have:
$\frac{QC(AP\cdot \sin(2C))}{\sin(2A)}+AQ\cdot PB\cdot \frac{\sin(2B)}{\sin(2A)}=AQ\cdot AP$ so we have to prove these to are equivalent,ie. $LHS$ of these to are equal.

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#4 h Feb 26, 2017, 9:54 PM • 4 Y

Y by Elnino2k, Saramahmoodabadi, Adventure10, Mango247

This problem was proposed by Burii.

The assumption that $D$ is the feet of angle bisector is redundant. It can be any point on the side $BC$ as can be clearly seen in the following solution.

Let $AX$ be a diameter of the circumcircle of $ABC$ . Observe that $BX \parallel DP$ , so $[BDP]=[XDP]$ . Analogously, $[CDQ]=[XDQ]$ . Thus $[BCQP]=[BDP]+[PDQ]+[CDQ]=[XDP]+[BDP]+[XDQ]=[PXQ].$ Therefore $[APQ]=[BCQP] \iff [APQ]=[PQX] \iff O \in PQ$ , where $O$ is the midpoint of $AX$ , i.e. the circumcenter of $ABC$ .

This post has been edited 1 time. Last edited by timon92, Feb 26, 2017, 9:56 PM

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#5 h Mar 30, 2017, 9:58 PM • 2 Y

Y by Adventure10, Mango247

Notice that $[APQ]=[BCQP] \iff [APQ]=\frac{[ABC]}{2} \iff \frac{AP \cdot AQ \cdot \sin \angle BAC}{2}=\frac{AB \cdot AC \cdot \sin \angle BAC}{4} \iff AP \cdot AQ=\frac{AB \cdot AC}{2}$ . Consider transformation $\phi$ - composition of symetry wtr bisector of $\angle BAC$ and inversion in $A$ with radius $\sqrt{\frac{1}{2} \cdot AB \cdot AC}=AP=AQ$ . Denote $M, N$ as midpoints of $AB, AC$ and let $H$ be a feet of $A$ -altitude. It is sufficient to show that $A, \ \phi(O), \ \phi(P)=Q, \phi(Q)=P$ are concyclic. Notice that $\phi(B)=N, \ \phi(C)=M$ .
Since $B, H, C$ are collinear, $A, M, \phi(H), N$ are concyclic. However, as $AH$ and $AO$ are isogonals wtr $\angle BAC$ and $\angle CHA=\angle BHA=\frac{\pi}{2}=\angle AN\phi(H)=\angle AM\phi(H)$ , hence $\phi(H)=O$ and $\phi(O)=H$ . Points $A, P, H, D, Q$ are concyclic, so we are done.

This post has been edited 1 time. Last edited by Pinionrzek, Mar 30, 2017, 10:00 PM

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