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Source: BrMO

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#1 h Jan 25, 2017, 7:32 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with $P$ a point in it s interior. Let $AP$ meet the circumcircle of $ABC$ at $A'$ . The points $B'$ and $C'$ defined similiarly. Let $O_A$ be the circumcenter of BCP. The circumcenters $O_B$ , $O_C$ are defined similiarly. Let $O_A'$ be the circumcenter of $B'C'P$ , $O_B'$ and $O_C$ define similiarly. Prove that lines $O_AO_A'$ , $O_B'O_B$ and $O_C'O_C$ concur.

This post has been edited 6 times. Last edited by problem333, Jan 25, 2017, 8:21 PM

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#3 h Jan 25, 2017, 7:38 PM • 1 Y

Y by Adventure10

How are $O_A$ , $O_B$ , $O_C$ defined?

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#4 h Jan 25, 2017, 8:22 PM • 2 Y

Y by Adventure10, Mango247

My bad fixed now

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#5 h Jan 26, 2017, 2:38 AM • 3 Y

Y by Phie11, problem333, Adventure10

Let $Q_A$ be the image of $O_A$ under a homothety centered at $P$ with factor $2$ and define $Q_B,Q_C,Q_A', Q_B', Q_C'$ similarly. It suffices to show $Q_AQ_A', Q_BQ_B', Q_CQ_C'$ concur. Invert at $P$ with power $PA \cdot PA'$ , switching $A,B,C\iff A',B',C'$ . Since we originally had $\angle PBQ_A=\angle PCQ_A=90^{\circ}$ , it's clear then that $Q_A$ maps to the foot of the altitude from $P$ to the new segment $BC$ and similarly for the other points.

Now define $D=BC\cap B'C', E=CA\cap C'A', F=AB\cap A'B'$ . We want to show that $(PQ_AQ_A'), (PQ_BQ_B'), (PQ_CQ_C')$ concur at a point other than $P$ . It's clear that these circles are just $(PD), (PE), (PF)$ , and since $D,E,F$ are collinear by Desargues on $ABC, A'B'C'$ , we know that the three circles share a second common point which is the foot of the altitude from $P$ on line $DEF$ .

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oolite

#6 h Nov 19, 2022, 5:32 PM

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You're not expected to know about inversion or Desargues at BMO2.

Here's an elementary approach. Let $O$ be the centre of $(ABC)$ . It suffices to prove that $PO_AOO_A^\prime$ is a parallelogram, since then $O_AO_A^\prime$ goes through its centre (the midpoint of $OP$ ) and so by analogy do $O_BO_B^\prime$ and $O_CO_C^\prime$ .

Let $\beta=\angle PBC$ and $\gamma=\angle BCP$ . Noting that $OO_A\perp BC$ (line joining centres $\perp$ common chord), we angle-chase in $\triangle BCP$ to get $\begin{align*}\angle O_APB&=90^\circ-\gamma,\\ {\color{blue}\angle OO_AP}&{\color{blue}=|\gamma-\beta|}.\end{align*}$ Since $\triangle BCP\sim\triangle C^\prime B^\prime P$ (cyclic quad), and $O_A,O_A^\prime$ are the respective circumcentres, and $OO_A^\prime\perp B^\prime C^\prime$ for similar reasons to before, we also get $\begin{align*}\angle C^\prime P O_A^\prime&=90^\circ-\gamma,\\ {\color{blue}\angle PO_A\prime O}&{\color{blue}=|\gamma-\beta}|.\end{align*}$ But it's also obvious that $\angle BP C^\prime=\beta+\gamma$ (external angle), whence ${\color{blue}\angle O_APO_A^\prime=180^\circ-|\gamma-\beta|}.$ The angles in blue make a parallelogram as required, so we are done.

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This post has been edited 1 time. Last edited by oolite, Nov 19, 2022, 5:33 PM
Reason: midpoint clarification

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#7 h May 20, 2023, 3:41 PM • 1 Y

Y by frontlinerbd

If the solution below is correct, this problem is way too easy for a P4 at BMO2.

Because $AP$ is the radical axis of the circles $(APB)$ and $(APC)$ it follows that $O_BO_C\perp AP$ . But because $A'P$ is the radical axis of $(A'PB')$ and $(A'PC')$ it follows that $O_B'O_C'\perp A'P$ , so $O_B'O_C'\perp AP$ , so because $O_BO_C\perp AP$ it follows that $O_B'O_C'\parallel O_BO_C$ . Similarly we get $O_A'O_C'\parallel O_AO_C$ and $O_A'O_B'\parallel O_AO_B$ , so $\triangle O_AO_BO_C$ and $\triangle O_A'O_B'O_C'$ are homothetic, so the lines $O_AO_A'$ , $O_B'O_B$ and $O_C'O_C$ concur in their centre of homothety.

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