segments of midpoints of incenters and excenters perpendicular to side

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Source: 2018 Oral Moscow Geometry Olympiad grades 10-11 p4

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#1 h Jul 25, 2019, 2:33 AM • 3 Y

Y by kiyoras_2001, Adventure10, Mango247

On the side $AB$ of the triangle $ABC$ , point $M$ is selected. In triangle $ACM$ point $I_1$ is the center of the inscribed circle, $J_1$ is the center of excircle wrt side $CM$ . In the triangle $BCM$ point $I_2$ is the center of the inscribed circle, $J_2$ is the center of excircle wrt side $CM$ . Prove that the line passing through the midpoints of the segments $I_1I_2$ and $J_1J_2$ is perpendicular to $AB$ .

This post has been edited 2 times. Last edited by parmenides51, Apr 11, 2020, 12:13 PM
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#2 h Feb 3, 2020, 12:08 PM • 5 Y

Y by mueller.25, amar_04, AmirKhusrau, MamaChandu, Adventure10

Nice problem. Heres a solution found with amar_04 and mueller.25. Alter the labelling a bit i.e. make $A$ as the apex.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.131864028953908, xmax = 20.253735857649307, ymin = -9.714561632147108, ymax = 13.653720890369517; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); /* draw figures */ draw((4.619335093416707,7.054662233638681)--(-6.64,-0.51), linewidth(0.4) + rvwvcq); draw((-6.64,-0.51)--(7.16,-0.39), linewidth(0.4) + rvwvcq); draw((7.16,-0.39)--(4.619335093416707,7.054662233638681), linewidth(0.4) + rvwvcq); draw((4.619335093416707,7.054662233638681)--(-0.19913787669058447,-0.45399250327557034), linewidth(0.4)); draw((-4.989046817214253,8.22130074168339)--(7.785421206986891,3.9545398495955895), linewidth(0.4) + dtsfsf); draw((-4.989046817214253,8.22130074168339)--(-1.1127166413234106,1.2006451493330466), linewidth(0.4) + dtsfsf); draw((7.785421206986891,3.9545398495955895)--(3.988370419640953,1.8580657592817953), linewidth(0.4) + dbwrru); draw((1.3981871948863192,6.087920295639489)--(1.437826889158771,1.5293554543074208), linewidth(0.4) + wvvxds); draw((-1.1127166413234106,1.2006451493330466)--(-0.19913787669058447,-0.45399250327557034), linewidth(0.4) + dtsfsf); draw((-0.19913787669058447,-0.45399250327557034)--(3.988370419640953,1.8580657592817953), linewidth(0.4) + linetype("4 4")); draw((-4.989046817214253,8.22130074168339)--(7.16,-0.39), linewidth(0.4) + wvvxds); draw((-6.64,-0.51)--(7.785421206986891,3.9545398495955895), linewidth(0.4) + wvvxds); draw(circle((8.053985450375514,-1.018644062623383), 4.980430200051047), linewidth(0.4) + linetype("4 4")); draw(circle((1.437826889158771,1.5293554543074206), 2.5716381482395336), linewidth(0.4) + linetype("4 4")); draw((3.083603259288729,2.4993689196212827)--(3.1090345176054646,-0.42522578680343076), linewidth(0.4) + dtsfsf); draw(circle((6.13011804457036,3.849612753256479), 5.2346146200537245), linewidth(0.4) + linetype("4 4")); draw((-1.1127166413234106,1.2006451493330466)--(3.1090345176054646,-0.42522578680343076), linewidth(0.4) + linetype("4 4") + dbwrru); draw((3.988370419640953,1.8580657592817953)--(3.1090345176054646,-0.42522578680343076), linewidth(0.4) + dtsfsf); draw((-0.19913787669058447,-0.45399250327557034)--(3.083603259288729,2.4993689196212827), linewidth(0.4) + linetype("4 4")); draw((7.785421206986891,3.9545398495955895)--(3.1090345176054646,-0.42522578680343076), linewidth(0.4) + dtsfsf); draw((7.785421206986891,3.9545398495955895)--(11.257372438088742,2.7948840655710425), linewidth(0.4) + dtsfsf); draw((-1.1127166413234106,1.2006451493330466)--(11.257372438088742,2.7948840655710425), linewidth(0.4) + wvvxds); /* dots and labels */ dot((4.619335093416707,7.054662233638681),dotstyle); label("$A$", (4.720836167166338,7.28890844323372), NE * labelscalefactor); dot((-6.64,-0.51),dotstyle); label("$B$", (-6.538218342207413,-0.2707424416315403), NE * labelscalefactor); dot((7.16,-0.39),dotstyle); label("$C$", (7.248379016209425,-0.15585413031139958), NE * labelscalefactor); dot((-0.19913787669058447,-0.45399250327557034),dotstyle); label("$M$", (-0.10447290827955526,-0.22478711710348404), NE * labelscalefactor); dot((-4.989046817214253,8.22130074168339),linewidth(4pt) + dotstyle); label("$J_2$", (-4.90680432146142,8.414813894171099), NE * labelscalefactor); dot((-1.1127166413234106,1.2006451493330466),linewidth(4pt) + dotstyle); label("$I_1$", (-1.0235793988406778,1.3836492413784862), NE * labelscalefactor); dot((3.988370419640953,1.8580657592817953),linewidth(4pt) + dotstyle); label("$I_2$", (4.077461623773552,2.0500014470353025), NE * labelscalefactor); dot((7.785421206986891,3.9545398495955895),linewidth(4pt) + dotstyle); label("$J_1$", (7.868775897338183,4.140968713061864), NE * labelscalefactor); dot((1.3981871948863192,6.087920295639489),linewidth(4pt) + dotstyle); label("$M_1$", (1.480985787938381,6.277891303616482), NE * labelscalefactor); dot((1.437826889158771,1.5293554543074208),linewidth(4pt) + dotstyle); label("$M_2$", (1.5269411124664372,1.7053365130748803), NE * labelscalefactor); dot((3.083603259288729,2.4993689196212827),linewidth(4pt) + dotstyle); label("$I$", (3.181332795476458,2.6933759904280907), NE * labelscalefactor); dot((11.257372438088742,2.7948840655710425),linewidth(4pt) + dotstyle); label("$L$", (11.33840289920642,2.969107937596428), NE * labelscalefactor); dot((3.1090345176054646,-0.42522578680343076),linewidth(4pt) + dotstyle); label("$D$", (3.204310457740486,-0.24776477936751218), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Proof: (with amar_04 and mueller.25) Firstly we start of with a claim that is a kinda bazooka for this problem. We name it the incentric bazooka.
Claim: $\angle I_1DB=\angle I_2DI$ .
Proof : (given by mueller.25) Let $I$ be the incentre of $ABC$ . $\odot(I) \cap BC=D$ and let the incircles of $ABM$ and $ACM$ touch $BC$ at $X$ and $Y$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.86, xmax = 9.86, ymin = -6.25, ymax = 6.25; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((-2.64,4.17)--(-5.46,-3.29), linewidth(0.4)); draw((-5.46,-3.29)--(4.42,-3.19), linewidth(0.4)); draw((4.42,-3.19)--(-2.64,4.17), linewidth(0.4)); draw((-2.64,4.17)--(-0.2003694050991509,-3.2367648725212463), linewidth(0.4)); draw((-2.760492710585694,-1.4042574548855635)--(-2.7416847224016037,-3.2624866874737015), linewidth(0.4) + sexdts); draw((-2.760492710585694,-1.4042574548855635)--(-1.6316803560008093,-3.2512518254655958), linewidth(0.4) + sexdts); draw((-1.6316803560008093,-3.2512518254655958)--(0.8942701578595993,-1.707487430439186), linewidth(0.4) + sexdts); draw((-0.2003694050991509,-3.2367648725212463)--(-6.937663844801672,1.5857145795784886), linewidth(0.6) + linetype("4 4") + wrwrwr); draw((-0.2003694050991509,-3.2367648725212463)--(4.983357551510355,4.005214102940234), linewidth(0.6) + linetype("4 4") + wrwrwr); draw((0.8942701578595993,-1.707487430439186)--(0.9096349613016435,-3.225530010513142), linewidth(0.4) + sexdts); draw((-5.46,-3.29)--(-1.6581684500135365,-0.6342281370081773), linewidth(0.4) + linetype("4 4") + wrwrwr); draw((-1.6581684500135365,-0.6342281370081773)--(4.42,-3.19), linewidth(0.4) + linetype("4 4") + wrwrwr); draw(circle((-0.933111276363048,-1.5558724426623756), 1.8336602767852548), linewidth(0.4) + dtsfsf); /* dots and labels */ dot((-2.64,4.17),linewidth(3pt) + dotstyle); label("$A$", (-2.56,4.29), NE * labelscalefactor); dot((-5.46,-3.29),linewidth(3pt) + dotstyle); label("$B$", (-5.8,-3.53), NE * labelscalefactor); dot((4.42,-3.19),linewidth(3pt) + dotstyle); label("$C$", (4.5,-3.07), NE * labelscalefactor); dot((-0.2003694050991509,-3.2367648725212463),linewidth(3pt) + dotstyle); label("$M$", (-0.22,-3.63), NE * labelscalefactor); dot((-1.6581684500135365,-0.6342281370081773),linewidth(3pt) + dotstyle); label("$I$", (-1.58,-0.51), NE * labelscalefactor); dot((-1.6316803560008093,-3.2512518254655958),linewidth(3pt) + dotstyle); label("$D$", (-1.8,-3.59), NE * labelscalefactor); dot((-2.760492710585694,-1.4042574548855635),linewidth(3pt) + dotstyle); label("$I_1$", (-2.98,-1.03), NE * labelscalefactor); dot((0.8942701578595993,-1.707487430439186),linewidth(3pt) + dotstyle); label("$I_2$", (0.98,-2.15), NE * labelscalefactor); dot((4.983357551510355,4.005214102940234),linewidth(3pt) + dotstyle); label("$J_1$", (5.06,3.69), NE * labelscalefactor); dot((-6.937663844801672,1.5857145795784886),linewidth(3pt) + dotstyle); label("$J_2$", (-7.38,1.33), NE * labelscalefactor); dot((-2.7416847224016037,-3.2624866874737015),linewidth(3pt) + dotstyle); label("$X$", (-3.02,-3.65), NE * labelscalefactor); dot((0.9096349613016435,-3.225530010513142),linewidth(3pt) + dotstyle); label("$Y$", (1,-3.57), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
This is the notation we use for lengths:
$BC=a \;\;\; AC=b \;\;\; AB=c \;\;\; AM=d \;\;\; BM=m \;\;\; CM=n \;\;\; s=\frac{a+b+c}{2} \;\;\; I_1X=r_1 \;\;\; I_2Y=r_2$
We claim that $\triangle I_1XD \sim DYI_2$ .
Note that $\angle I_1XD=90^{\circ}=\angle I_2YD$ .
Now, we prove $r_1 \cdot r_2 = DX \cdot DY$

$r_1 \cdot r_2= ((\frac{m+d-c}{2}) \tan{\frac{\angle AMB}{2}}) \cdot ((\frac{n+d-b}{2}) \tan{\frac{\angle AMC}{2}})=\frac{m+d-c}{2} \cdot \frac{n+d-b}{2}=MX \cdot MY$ [The last step follows from the identity $\tan \theta=\tan{90^{\circ} -\theta}$ ]
$DX=BD-BX=\frac{a+c-b}{2}-\frac{m+c-d}{2}=\frac{n+d-b}{2}=MY$ $DY=DM+MY=DM+DX=MX$ This implies $r_1 \cdot r_2 = DX \cdot DY$ which gives us $\triangle I_1DX \sim \triangle DI_2Y$
$\implies \angle I_1DB=\angle DI_2Y=\angle IDI_2$ $\square$ .

Now back to the main problem. Clearly by the incentric bazooka we have that $I_1I_2DM$ is cyclic. By a similiar arguement as the incentric bazooka we can also show that $J_1J_2DM$ is cyclic $\implies D$ is the miquel point of $I_1I_2J_1J_2$ . Now let $M_1,M_2$ be the midpoints of $I_1I_2$ and $J_1J_2$ . Notice that $M_1,M_2$ are the centres of $\odot(I_1I_2MD),\odot(J_2J_1MD)$ (since $I_1I_2$ , $J_1J_2$ are diameters) $\implies M_1M_2\perp DM$ since $DM$ is the radical axis. Done $\blacksquare$ .

This post has been edited 3 times. Last edited by GeoMetrix, Feb 3, 2020, 12:48 PM

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