Let there be a scalene triangle $ABC$, and denote $M$ by the midpoint of $BC$. The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at point $P$, on the same side with $A$ with respect to $BC$. Let the incenters of $ABM$ and $AMC$ be $I,J$, respectively. Let $\angle BAC=\alpha$, $\angle ABC=\beta$, $\angle BCA=\gamma$. Find $\angle IPJ$.
Problem
Source: KJMO 2018 p3
Tags: geometry, incenter
electrovector
01.05.2020 20:31
Use angle-chasing and do trigonometric ceva theorem in $PMC$ and $PMB$ triangles according to $J$ and $I$, respectively.Divide side by side and use Masmis Theorem.
$\frac{\angle BAC}{2}$
I will post the solution after if wanted.
mathroyal
11.10.2020 14:29
electrovector wrote:
Use angle-chasing and do trigonometric ceva theorem in $PMC$ and $PMB$ triangles according to $J$ and $I$, respectively.Divide side by side and use Masmis Theorem.
$\frac{\angle BAC}{2}$
I will post the solution after if wanted. I heard it uses iso-conj
sarjinius
11.10.2020 17:50
Let $I'$ be the reflection of $I$ with respect to $PM$.
Claim: $I'$ and $J$ are isogonal conjugates with respect to $\triangle PMC$.
Proof: $\angle PMI' = \angle PMI = 90^{\circ} - \angle IMB = 90^{\circ} - \frac{\angle AMB}{2} = \frac{180^{\circ} - \angle AMB}{2} = \frac{AMC}{2} = \angle JMC$ and
$\angle PCJ = \angle PCM - \angle JCM = 90^{\circ} - \frac{\angle BPC}{2} - \frac{\angle ACM}{2} = 90^{\circ} - \frac{\alpha}{2} - \frac{\gamma}{2} = \frac{\beta}{2} = \angle IBM = \angle I'CM$.
Thus, $I'$ and $J$ are isogonal conjugates with respect to $\triangle PMC$. $\square$
Now, by the claim, $\angle MPI = \angle MPI' = \angle JPC$. Thus, $\angle IPJ = \angle IPM + \angle MPJ = \angle JPC + \angle MPJ = \angle MPC = \frac{\angle BPC}{2} = \frac{\angle BAC}{2} = \boxed{\frac{\alpha}{2}}.$
Olympiadium
20.07.2021 19:21
Elegant Proof Without Isogonal Conjugate [asy][asy] import graph; size(9cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -10.655684210393872, xmax = 11.231090215956694, ymin = -7.676992027190916, ymax = 4.809327064604368; pen qqffff = rgb(0,1,1); pen wwwwww = rgb(0.4,0.4,0.4); pen zzttqq = rgb(0.6,0.2,0); pen xfqqff = rgb(0.4980392156862745,0,1); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((-3.4063636363636305,2.5872727272727305)--(-5.060909090909085,-3.194545454545449)--(3.7209090909090965,-3.0672727272727216)--cycle, linewidth(0.8) + qqffff); draw((-3.4063636363636305,2.5872727272727305)--(-5.060909090909085,-3.194545454545449), linewidth(0.8) + qqffff); draw((-5.060909090909085,-3.194545454545449)--(3.7209090909090965,-3.0672727272727216), linewidth(0.8) + qqffff); draw((3.7209090909090965,-3.0672727272727216)--(-3.4063636363636305,2.5872727272727305), linewidth(0.8) + qqffff); draw(circle((-0.6963046001639144,-1.3158916795986), 4.7517483484245), linewidth(0.8) + wwwwww); draw((-3.4063636363636305,2.5872727272727305)--(-0.67,-3.1309090909090855), linewidth(0.8) + zzttqq); draw((-3.0499630395341804,-1.6553606927240363)--(1.7099630395341918,-4.606457489094135), linewidth(0.8) + zzttqq); draw((0.12776263521186038,-1.8441701479771047)--(-0.67,-3.1309090909090855), linewidth(0.8) + zzttqq); draw((-3.0499630395341804,-1.6553606927240363)--(0.12776263521186038,-1.8441701479771047), linewidth(0.8) + blue); draw((0.12776263521186038,-1.8441701479771047)--(1.7099630395341918,-4.606457489094135), linewidth(0.8) + blue); draw((-3.0499630395341804,-1.6553606927240363)--(-5.060909090909085,-3.194545454545449), linewidth(0.8) + xfqqff); draw((-0.7651632871005909,3.435357719032072)--(3.7209090909090965,-3.0672727272727216), linewidth(0.8) + fuqqzz); draw((-0.7651632871005909,3.435357719032072)--(-3.0499630395341804,-1.6553606927240363), linewidth(0.8) + fuqqzz); draw((-0.7651632871005909,3.435357719032072)--(0.12776263521186038,-1.8441701479771047), linewidth(0.8) + red); draw((-5.060909090909085,-3.194545454545449)--(-0.7651632871005909,3.435357719032072), linewidth(0.8) + fuqqzz); draw((1.7099630395341918,-4.606457489094135)--(3.7209090909090965,-3.0672727272727216), linewidth(0.8) + xfqqff); draw((3.7209090909090965,-3.0672727272727216)--(0.12776263521186038,-1.8441701479771047), linewidth(0.8) + green); draw((-0.7651632871005909,3.435357719032072)--(3.0666562815922997,-0.6208575920054793), linewidth(0.8) + fuqqzz); draw((3.0666562815922997,-0.6208575920054793)--(3.7209090909090965,-3.0672727272727216), linewidth(0.8) + xfqqff); draw((0.12776263521186038,-1.8441701479771047)--(3.0666562815922997,-0.6208575920054793), linewidth(0.8) + blue); dot((-3.4063636363636305,2.5872727272727305),dotstyle); label("$A$", (-3.698204350320291,2.852089403762315), NE * labelscalefactor); dot((-5.060909090909085,-3.194545454545449),dotstyle); label("$B$", (-5.5268643546106775,-3.476788579836438), NE * labelscalefactor); dot((3.7209090909090965,-3.0672727272727216),dotstyle); label("$C$", (3.930736605078666,-3.3482109232847703), NE * labelscalefactor); dot((-0.67,-3.1309090909090855),linewidth(4pt) + dotstyle); label("$M$", (-1.126651219286935,-3.5339342049705125), NE * labelscalefactor); dot((-0.7651632871005909,3.435357719032072),linewidth(4pt) + dotstyle); label("$P$", (-0.883782312467118,3.709273780773433), NE * labelscalefactor); dot((-3.0499630395341804,-1.6553606927240363),linewidth(4pt) + dotstyle); label("$I$", (-3.055316067561952,-2.1052935766186494), NE * labelscalefactor); dot((0.12776263521186038,-1.8441701479771047),linewidth(4pt) + dotstyle); label("$J$", (0.24484378393085504,-1.548123731561423), NE * labelscalefactor); dot((1.7099630395341918,-4.606457489094135),linewidth(4pt) + dotstyle); label("$E$", (1.6306251934321636,-5.005434052172931), NE * labelscalefactor); dot((3.0666562815922997,-0.6208575920054793),linewidth(4pt) + dotstyle); label("$D$", (3.2878483223203268,-0.5195024791480815), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $D$ be the point outside the triangle $PBC$ such that $\triangle PBI \equiv \triangle PCD$. Let $E$ be the reflection of $I$ with respect to $M$. Then, $\triangle MIJ \equiv \triangle MEJ$ and $\triangle MIB \equiv \triangle MEC(SAS)$. which leads to $CD=BI=CE$ and \begin{align*} \angle DCJ &= \angle DCP + \angle PCA + \angle ACJ \\ &= \angle IBP + \angle PBA + \angle BCJ \\ &= \angle ABI + \angle BCJ \\ &= \angle MBI + \angle BCJ \\ &= \angle MCE + \angle MCJ \\ &= \angle ECJ \end{align*}Therefore, $\triangle DCJ \equiv \triangle ECJ(SAS)$ and $\triangle PIJ \equiv \triangle PDJ(SSS)$. \begin{align*} \angle IPJ &= \frac{1}{2}\angle IPD = \frac{1}{2}\angle BPC \\ &= \frac{1}{2}\angle BAC = \boxed{\frac{\alpha}{2}} \end{align*}