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Source: Korea National 2013 #6

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#1 h Nov 10, 2013, 8:46 AM • 2 Y

Y by ImSh95, Adventure10

Let $O$ be circumcenter of triangle $ABC$ . For a point $P$ on segmet $BC$ , the circle passing through $P, B$ and tangent to line $AB$ and the circle passing through $P, C$ and tangent to line $AC$ meet at point $Q ( \ne P )$ . Let $D, E$ be foot of perpendicular from $Q$ to $AB, AC$ . ( $D \ne B, E \ne C$ ) Two lines $DE$ and $BC$ meet at point $R$ . Prove that $O, P, Q$ are collinear if and only if $A, R, Q$ are collinear.

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mathuz

#2 h Nov 10, 2013, 12:52 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

we have that $Q$ lies on the circle $(O)$ , because triangles $QO_1B$ and $QO_2C$ are similar.
So it's Simson's line and $QR\perp BC$ ...

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Sardor

#3 h Jul 22, 2014, 7:04 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Nice and easy problem.
Here my solution:
We have by tangent circles $\angle BQC=\angle BQP + \angle PQC=\angle B + \angle C=180- \angle A$ ,hence $Q$ lies on the circle $(ABC)$ .From the Simson's line theorem $QR \perp BC$ .
$(i)$ Let $A,R,Q$ are collinear, i.e. $AR \perp BC \implies \angle BPQ=\angle CPO=90- \angle B + \angle C \implies O,P,Q$ are also collinear.
$(ii)$ Let $O,P,Q$ are collinear $\implies \angle OPC = \angle BPQ \implies 90- \angle B + \angle C= \angle C + \angle BCQ= \angle C+ \angle BAQ \implies \angle BAQ =90- \angle B \implies AR \perp BC \implies A,R,Q$ are collinear, so we are done !

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#4 h May 14, 2022, 9:34 PM • 2 Y

Y by ImSh95, centslordm

when you solve the last part of a simple angle chasing problem with inversion...

Since $(ADQE)$ is cyclic, $\measuredangle ACQ=\measuredangle CPQ=\measuredangle BPQ=\measuredangle ABQ$ and $\measuredangle QBR=\measuredangle QAC=\measuredangle QAE=\measuredangle QDE$ so $Q$ lies on $(ABC),(BDR),$ and similarly $(CER).$ Notice $O$ lying on $\overline{PQ}$ is equivalent to $\measuredangle CQP+\measuredangle ACQ=90.$

Invert about $Q$ with arbitrary radius. In the inverted problem, we have and $A',B',C'$ collinear, $P$ the intersection of the tangents to $(A'B'Q)$ and $(A'C'Q)$ at $B'$ and $C',$ respectively. Note $P$ lies on $(QB'C').$ Also, $D'$ and $E'$ are the intersections of the line perpendicular to $\overline{A'Q'}$ at $A'$ with $(A'B'Q)$ and $(A'C'Q),$ respectively. Finally, we have $R'=(QB'C')\cap\overline{D'B'}.$ If $\measuredangle C'QP'+\measuredangle A'C'Q=90,$ then $\measuredangle D'R'A'=90-\measuredangle A'D'R=90-\measuredangle C'B'P'=90-\measuredangle C'QP'=\measuredangle B'C'Q=\measuredangle B'R'Q.$ Conversely, $A',Q,R'$ collinear implies $\measuredangle A'C'Q=\measuredangle B'R'Q=90-\measuredangle A'D'R=90-\measuredangle C'QP'=90-\measuredangle C'QP'.$ $\square$

This post has been edited 4 times. Last edited by Mogmog8, May 14, 2022, 10:55 PM

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