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Consider $\triangle ABC$ and its incenter $I$. Let $M$ be the midpoint of arc $BC$ (not containing $A$) and $P$ be the second intersection point of the circumcircle of $\triangle ABC$ and the circle with center $I$ and radius $IA$. Prove that if $I'$ is the reflection of $I$ with respect to $BC$, then $I', P, M$ are collinear (lie on the same line).

Please help, I am lost, I don't have any idea! Thanks a lot in advance! Anyway, this problem is based on a problem connected to the reflection of the incenter, which appeared in Mathematical Excalibur.

For help, here is a figure: enter image description here

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5 Answers 5

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@Alex Zhao's proof is incorrect, but I think I've managed to make the idea work. As in the original, this is just a big angle chase.

enter image description here

Let $K$ be the intersection of $(I)$ and $BC$ closest to $C$.

The triangle $ABK$ is symmetric about $BI$.$\qquad(*)$

Therefore $$ \begin{aligned} \measuredangle IKB =\measuredangle BAI=\measuredangle IAC&\implies\measuredangle IAC+\measuredangle CKI=\pi\\ &\implies CKIA \text{ is cyclic}\\ &\implies \measuredangle KIA=\pi-\measuredangle ACB.\qquad(**) \end{aligned} $$

We claim $\measuredangle KPB+\measuredangle KI'B=\pi$, because $$\begin{aligned} \measuredangle KPB &\overset{}=\measuredangle APB-\measuredangle APK\\ &= \pi-\measuredangle ACB-\measuredangle KIA/2\\ &=(\pi-\measuredangle ACB)/2 \qquad\text{ (by $(**)$)}\\ &=(\measuredangle CAB+\measuredangle CBA)/2\\ &=\measuredangle IAB+\measuredangle IBA\\ &=\pi-\measuredangle AIB\\ &=\pi-\measuredangle BIK\qquad\text{ (by $(*)$)}\\ &=\pi-\measuredangle KI'B. \end{aligned} $$ Thus $PBI'K$ is cyclic.

Therefore $$ \begin{aligned} \measuredangle BPI' &=\measuredangle BKI'\\ &=\measuredangle IKB\\ &=\measuredangle BAI\\ &=\measuredangle BAM\\ &=\measuredangle BPM. \end{aligned} $$ Thus $P,I',M$ are collinear and we're done.

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Denote circumcircle of $\triangle ABC$ as $\Omega$ and circle with center $M$ and radius $MI$ as $\Gamma$. It's well-known that $\Gamma$ is the circumcircle of $\triangle BIC$.

Let $\Phi$ be the inversion with respect to the circle $\Gamma$. Then, for any point $X\neq M$ of the plane denote its image under the $\Phi$ as $X^*=\Phi(X)$.

Since $I$ and $J$ (here $J=I'$) are symmetric with respect to the line $BC$ their images $I^*$ and $J^*$ are symmetric with respect to $\Phi(BC)=\Omega$ (because inversion preserves symmetry eith respect to generalized circles; $\Phi(\{I,B,C\})=\{A,B,C\}$). However, $\Phi(I)=I$, so $J^*$ is the image of $I$ under the inversion with respect to the circle $\Omega$. In particular, $J^*$ lies on the line $OI$.

It's clear that $A$ and $P$ are symmetric with respect to $OI$. Hence, if $D$ is the point symmetric to $M$ with respect to $OI$, then $D\in\Omega$, $D, I, P$ are collinear and lines $MP$, $AD$ and $OI$ are concurrent. Denote the intersection of $MP$, $AD$ and $OI$ as $K$.

Now note that the collinearity of $P$, $M$ and $J$ is equivalent to the collinearity of $P$, $M$ and $J^*$ (since $M$ is the center of $\Gamma$). Thus, we need to prove that $J^*=K$.

Indeed, we have cyclic quadrilateral $ADMP$ with $AP\parallel MD$, where $K=AD\cap PM$, $I=AM\cap PD$. It means that $I$ and $K$ are inverse images of each other with respect to $\Omega$. Therefore, $J^*=K$, as desired.

enter image description here

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  • $\begingroup$ +1. This is really elegant. $\endgroup$
    – brainjam
    Commented Nov 13, 2020 at 17:36
  • $\begingroup$ This a solution from other world. Very nice. But now it seem to me very artificial problem. Someone just compose two inversion and get the problem. $\endgroup$
    – nonuser
    Commented Nov 13, 2020 at 19:37
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After a suggestion by @brainjam I found an offical solution:

enter image description here

This was posted initial and it is not a solution, just partial result if someone find it usefull. enter image description here Let $PM$ meet smaller circle at $D$. Then $AD\bot BC$.

Proof: It is easy to see that angle between altitude from $A$ and angle bisector from is $|\beta-\gamma|$. Now we see that $\angle MBC = \angle MAC = \alpha $ so $$\angle DPA = \angle MPA = \angle MBA = \alpha+2\beta$$ From here we get (notice: $\alpha +\beta+\gamma = 90^{\circ}$ and that $ADI$ is isosceles) $$\angle DJA = 360^{\circ}-2(\alpha+2\beta) \implies \angle DAI = \beta - \gamma$$ and we are done. Let perpendicular from $I$ to $BC$ meet $PM$ at $J$. If we prove that $BC$ bisect $IJ$ we are done, i.e. $J=I'$. But no idea how to finish.

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    $\begingroup$ Another observation (possibly irrelevant): if you repeat the construction for the other two sides (e.g. midpoint of other two arcs, two other circles centered at $I$, etc), you get two more lines. All three lines are concurrent, presumably at an obscure triangle center. $\endgroup$
    – brainjam
    Commented Nov 12, 2020 at 17:02
  • $\begingroup$ Are you working on this problem? @brainjam $\endgroup$
    – nonuser
    Commented Nov 13, 2020 at 16:11
  • $\begingroup$ Yes, but happy to stop if the new answer is correct. The concurrency point referred to in my earlier comment is X(36), by the way. It shows up in @richrow's answer as $K$. $\endgroup$
    – brainjam
    Commented Nov 13, 2020 at 17:13
  • $\begingroup$ I think we can all stop now, new answer looks really good. FWIW, there is another (less elegant IMO) answer at AoPS $\endgroup$
    – brainjam
    Commented Nov 13, 2020 at 17:40
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    $\begingroup$ OP mentioned Math Excalibur, so I found mention of problem G2 of 2018 Croatian Math Olympiad here. (That issue was after OP posted, so there must have been an earlier reference). AoPS has a lot of olympiad solutions, so wasn't too hard to track down. $\endgroup$
    – brainjam
    Commented Nov 13, 2020 at 17:57
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Hint: The line through $P$ and $I$ must intersect the perpendicular bisectors of the triangle; in particular it must intersect the bisector of the side $BC$.

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Let the circle centered at I intersect BC again at K. I claim that PBI'K is cyclic. Note that \begin{align*} \measuredangle KPB &=\measuredangle APB-\measuredangle APK\\ & = \measuredangle ACB-\frac{\measuredangle AIK}2\\ &=\measuredangle ACB-\measuredangle AIB\\ &=\measuredangle ACB-(\measuredangle ACB+\measuredangle IAC+\measuredangle CBI)\\ &=\measuredangle CAI+\measuredangle IBC\\ &=\measuredangle IAB+\measuredangle ABI\\ &=\measuredangle AIB\\ &=\measuredangle BIK\\ &=\measuredangle KI'B. \end{align*} Therefore \begin{align*} \measuredangle BPI' &=\measuredangle BKI'\\ &=\measuredangle IKB\\ &=\measuredangle BAI\\ &=\measuredangle BAM\\ &=\measuredangle BPM. \end{align*} Thus BPI'M are collinear and we're done.

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