# A geometry problem with the reflection of the incenter

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:

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.

• +1. This is really elegant. Nov 13 '20 at 17:36
• 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.
– Aqua
Nov 13 '20 at 19:37

@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.

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.

After a suggestion by @brainjam I found an offical solution:

This was posted initial and it is not a solution, just partial result if someone find it usefull. 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.

• 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. Nov 12 '20 at 17:02
• Are you working on this problem? @brainjam
– Aqua
Nov 13 '20 at 16:11
• 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 $K$. Nov 13 '20 at 17:13
• I think we can all stop now, new answer looks really good. FWIW, there is another (less elegant IMO) answer at AoPS Nov 13 '20 at 17:40
• 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. Nov 13 '20 at 17:57

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 $BC$.

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.