Collinear points

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Collinear points

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Source: circumcircle,incircle and excircle

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439 posts

#1 h Sep 2, 2009, 3:12 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Assume that $O$ , $I$ and $I_a$ are the circumcenter,incenter and the excenter corresponding to the edge $BC$ of the triangle $ABC$ ,respectively.Let $II_a$ intersect the segment $BC$ and $(O)$ at $A_1$ and $M$ respectively.( $M \in II_a$ ).Let $N$ be the midpoint of the arc $MCA$ and $S,T$ be the intersections of lines $NI,NI_a$ with $(O)$ ,respectively.
Prove that $S,T,A_1$ are collinear.

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843 posts

shoki

#2 h Sep 2, 2009, 8:56 PM • 2 Y

Y by Adventure10, Mango247

use menelaus for the line $TA_1S$ and the triangle $NII_{a}$ :
$T,A_1,S$ r collinear iff: $IS.NT.I_{a}A_1 = SN.TI_{a}.A_1I$
there r SOME computations to do

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859 posts

#3 h Sep 3, 2009, 1:13 AM • 2 Y

Y by Adventure10, Mango247

Use the concept of radical center.
http://www.mathlinks.ro/viewtopic.php?t=291967

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#4 h Jan 4, 2021, 11:36 PM • 1 Y

Y by mijail

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -99.81870241417839, xmax = 70.43943654885625, ymin = -41.841466281108254, ymax = 64.5420506134408; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw(circle((-2.709274226113634,7.219890998869171), 20.33026651545786), linewidth(0.8) + red); draw(circle((-15.055211549657772,-8.932432926580905), 13.87497728453352), linewidth(0.8) + red); draw((2.1603549766224512,26.95834200236504)--(-22.541924725659296,2.7493529157724113), linewidth(0.8) + blue); draw((-22.541924725659296,2.7493529157724113)--(-1.8180397157717145,-13.090831234402664), linewidth(0.8) + blue); draw((-1.8180397157717145,-13.090831234402664)--(2.1603549766224512,26.95834200236504), linewidth(0.8) + blue); draw((2.1603549766224512,26.95834200236504)--(-21.055943997668578,-21.442676918354664), linewidth(0.8) + blue); draw((-23.02658296781961,6.494148562639172)--(-5.356249818188496,-12.937322518988577), linewidth(0.8) + blue); draw((15.62132115508931,-1.572663212517206)--(-23.02658296781961,6.494148562639172), linewidth(0.8) + blue); draw((15.62132115508931,-1.572663212517206)--(-21.055943997668578,-21.442676918354664), linewidth(0.8) + blue); draw((15.62132115508931,-1.572663212517206)--(-15.055211549657772,-8.932432926580905), linewidth(0.8) + blue); draw((15.62132115508931,-1.572663212517206)--(2.1603549766224512,26.95834200236504), linewidth(0.8) + blue); draw(circle((-18.60108996809332,-7.231597379632428), 14.421549499357262), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((2.1603549766224512,26.95834200236504)--(-5.356249818188496,-12.937322518988577), linewidth(0.8) + blue); draw(circle((-29.126640394061177,12.19709600564352), 34.594370390294806), linewidth(0.8) + linetype("4 4") + qqwuqq); /* dots and labels */ dot((2.1603549766224512,26.95834200236504),dotstyle); label("$A$", (2.5587406224176044,28.042266574306808), NE * labelscalefactor); dot((-22.541924725659296,2.7493529157724113),dotstyle); label("$B$", (-22.14538150163056,3.894543597196827), NE * labelscalefactor); dot((-1.8180397157717145,-13.090831234402664),dotstyle); label("$C$", (-1.3360534061485478,-12.018472005230487), NE * labelscalefactor); dot((-15.055211549657772,-8.932432926580905),linewidth(4pt) + dotstyle); label("$M$", (-14.578353103273466,-8.012398147276757), NE * labelscalefactor); dot((-9.054479101646967,3.577811065192854),linewidth(4pt) + dotstyle); label("$I$", (-8.56924231634283,4.450942744134846), NE * labelscalefactor); dot((-21.055943997668578,-21.442676918354664),linewidth(4pt) + dotstyle); label("$I_{a}$", (-20.5874638902041,-20.587018868075965), NE * labelscalefactor); dot((-12.963570251786978,-4.57180808514041),linewidth(4pt) + dotstyle); label("$A_{1}$", (-12.464036344908983,-3.6724848011602176), NE * labelscalefactor); dot((15.62132115508931,-1.572663212517206),linewidth(4pt) + dotstyle); label("$N$", (16.02359997831773,-0.6679294076949203), NE * labelscalefactor); dot((-23.02658296781961,6.494148562639172),linewidth(4pt) + dotstyle); label("$S$", (-22.590500819180978,7.344218308212539), NE * labelscalefactor); dot((-5.356249818188496,-12.937322518988577),linewidth(4pt) + dotstyle); label("$T$", (-4.897007946551887,-12.018472005230487), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Nice one

$\color{black}\rule{25cm}{1pt}$
Since we have that:
$\angle I_aMN = 180 - \angle AMN = 180-\angle MAN = \angle MTN$ and in conjunction with $\angle MNT = \angle MNI_a$ we have that $NMT \sim NI_aM$ .
From that relation we have that:
$\frac{NM}{NT}=\frac{NI_a}{NM}$ this implies that $NT.NI_a=NM^2=NA^2$ , thus we have that $AN$ is tangent to the circumcircle of $AI_aT$ .
Thus we have that:
$\angle TI_aI \angle TI_aA = \angle TAN = TSN = \angle TSI$ thus we must have that $TI_aSI$ is a cylic quad.

Now let's apply the radical center on the circles around $TI_aSI$ , $ABC$ and $IBC$ .
The radical axis of $(ABC)$ and $(TI_aSI)$ is the line $ST$ , where $BC$ is the radical axis of $(ABC)$ and $(IBC)$ , while $II_a$ is the radical axis of $(TI_aSI)$ and $(IBC)$ .
Since $A_1$ is the intersection of $BC$ and $II_a$ we have that $S,A_1$ and $T$ are colinear points.

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