mop 1997 [two orthogonal circles in a triangle]

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

mop 1997 [two orthogonal circles in a triangle]

G H J

Reveal topic tags

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: kiran kedlaya geometry unbound

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

paul_mathematics

58 posts

paul_mathematics

#1 h Jan 25, 2005, 2:49 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle, and let $O$ be the circumcenter of the triangle $ABC$ . The lines $AB$ and $AC$ meet the circumcircle of triangle $BOC$ at two points $B_{1}$ and $C_{1}$ (apart from $B$ and $C$ ), respectively. Let $D$ be the point of intersection of the lines $BC$ and $B_{1}C_{1}$ . Show that the circle tangent to the line $AD$ at $A$ and having its center on the line $B_{1}C_{1}$ is orthogonal to the circle with diameter $OD$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

7849 posts

#2 h Jan 25, 2005, 3:51 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I'll let you prove that the conclusion is equivalent to $AO\perp B_1C_1$ . This last assertion follows from the fact that $O$ is the orthocenter of $AB_1C_1$ . We can show $OB_1\perp AC_1,\ OC_1\perp AB_1$ by a simple angle chase.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

199 posts

mathpk

#3 h Jun 12, 2007, 2:59 PM • 2 Y

Y by Adventure10, Mango247

Does someone have a solution with Inversion?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

143 posts

cuenca

#4 h Aug 19, 2007, 5:06 PM • 1 Y

Y by Adventure10

I finish the solution:
Click to reveal hidden text
Can somebody provide another solution?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

786 posts

Amir.S

#5 h Aug 20, 2007, 9:01 AM • 2 Y

Y by Adventure10, Mango247

we know that $AO\perp B_{1}C_{1}$ , let $AO\cap B_{1}C_{1}\equiv H$ and $AO$ intersect circumcircles of $ABC$ and $BOC$ at $A'$ and $O'$ in order.and let the circle with center on $B_{1}C_{1}$ and tangent to AD at A be $\omega$
cause $\angle H =\angle A'CA = 90^{\circ}$ hence $AH.AA' = 2R.AH = AC_{1}.AC$ also we have $AO.AO' = R.AO' = AC_{1}.AC$ hence $2AH = AO'$ therefore $H$ is midpoint of AO' hence $DO'$ is tangent to the $\omega$ .
let $DO$ intersect $\omega$ at $X,Y$ we have $(DO,XY) =-1$ hence $RO^{2}= RX.RY$

Another solution :Use Inversion

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

904 posts

#6 h Nov 19, 2023, 1:24 PM • 3 Y

Y by Om245, ehuseyinyigit, rama1728

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(40.43578cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.9984, xmax = 30.43738, ymin = -11.55818, ymax = 7.60822; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-0.7666754448139819,-1.052078394243146)--(-1.3016609822709255,-1.2328308623090223)--(-1.1209085142050494,-1.7678163997659662)--(-0.5859229767481056,-1.5870639317000899)--cycle, linewidth(1.6) + rvwvcq); draw((-0.7791294488206318,-4.633454919515893)--(-0.8745100121529052,-4.076872914651903)--(-1.4310920170168955,-4.172253477984177)--(-1.335711453684622,-4.728835482848167)--cycle, linewidth(1.6) + rvwvcq); /* draw figures */ draw(circle((-1.4117814564201705,0.8572863299500568), 5.586639737152943), linewidth(2.) + rvwvcq); draw(circle((-0.975345025853369,-4.6670800674876185), 5.541579274093659), linewidth(2.) + wvvxds); draw((xmin, 3.028571428571429*xmin + 15.841428571428576)--(xmax, 3.028571428571429*xmax + 15.841428571428576), linewidth(2.) + sexdts); /* line */ draw((xmin, -1.1319648093841643*xmin + 2.5277126099706746)--(xmax, -1.1319648093841643*xmax + 2.5277126099706746), linewidth(2.) + sexdts); /* line */ draw((-6.,-2.33)--(3.62,-1.57), linewidth(2.) + sexdts); draw((2.664809357855145,-0.48875780683896225)--(-6.403717953141712,-3.5526886580863235), linewidth(2.) + rvwvcq); draw((xmin, -5.835381815947613*xmin-12.523221811032363)--(xmax, -5.835381815947613*xmax-12.523221811032363), linewidth(2.) + dtsfsf); /* line */ draw((xmin, -2.959768921450996*xmin-3.3212605486431874)--(xmax, -2.959768921450996*xmax-3.3212605486431874), linewidth(2.) + wrwrwr); /* line */ draw((-1.335711453684622,-4.728835482848167)--(0.3764370871596592,-4.435427340099887), linewidth(2.) + wrwrwr); draw((xmin, -5.835381815947613*xmin-7.380997508936245)--(xmax, -5.835381815947613*xmax-7.380997508936245), linewidth(2.) + dtsfsf); /* line */ /* dots and labels */ dot((-3.2,6.15),dotstyle); label("$A$", (-4.11538,6.14412), NE * labelscalefactor); dot((-6.,-2.33),dotstyle); label("$B$", (-6.83062,-2.32104), NE * labelscalefactor); dot((3.62,-1.57),dotstyle); label("$C$", (4.08358,-1.57568), NE * labelscalefactor); dot((-1.4117814564201705,0.8572863299500568),linewidth(4.pt) + dotstyle); label("$O$", (-1.18718,1.24604), NE * labelscalefactor); dot((-6.403717953141712,-3.5526886580863235),linewidth(4.pt) + dotstyle); label("$B_1$", (-7.17668,-3.65204), NE * labelscalefactor); dot((2.664809357855145,-0.48875780683896225),linewidth(4.pt) + dotstyle); label("$C_1$", (2.7792,-0.2713), NE * labelscalefactor); dot((-1.8036087062514856,-1.998476363487643),linewidth(4.pt) + dotstyle); label("$D$", (-2.57142,-1.76202), NE * labelscalefactor); dot((-0.5859229767481056,-1.5870639317000899),linewidth(4.pt) + dotstyle); label("$E$", (-0.36196,-1.14976), NE * labelscalefactor); dot((0.3764370871596592,-4.435427340099887),linewidth(4.pt) + dotstyle); label("$E'$", (0.80932,-4.90318), NE * labelscalefactor); dot((-1.335711453684622,-4.728835482848167),linewidth(4.pt) + dotstyle); label("$D'$", (-1.93254,-5.51544), NE * labelscalefactor); dot((2.028154046503791,-9.324127863400186),linewidth(4.pt) + dotstyle); label("$O'$", (2.43314,-9.56168), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let the circumcircle of $\triangle ABC$ be $\omega$ and circumcircle of $\triangle BOC$ be $\gamma$ .

Consider the inversion centred at $A$ which fixes $\gamma$ . Then $B \mapsto B_1$ , $C \mapsto C_1$ , and vice versa. Therefore, $\omega \mapsto B_1C_1$ , and $B_1C_1 \mapsto \omega$ .

Now, the line $AO$ is orthogonal to $\omega$ , therefore their images under inversion, i.e. $AO$ and $B_1C_1$ are orthogonal. Let $AO$ intersect $B_1C_1$ at $E$ . So $E$ lies on the circle with diameter $OD$ . Also, note that if $E \mapsto E'$ , then $E'$ lies on $AO$ as well as $\omega$ , therefore $E'$ is the antipode of $A$ in $\omega$ .

Now, suppose $D \mapsto D'$ . We have $D$ lies on $B_1C_1$ , therefore $D'$ lies on $\omega$ . Therefore, $AD' \perp D'E'$ since $AE'$ is a diameter.

Finally, the circle with centre on $B_1C_1$ and tangent to $AD$ at $A$ maps to a line $\ell$ parallel to $AD$ and orthogonal to $\omega$ . Therefore $\ell$ is the line through $O$ parallel to $AD$ . Since $O$ is the midpoint of $AE'$ , $\ell$ bisects $E'D'$ . Thus, $\ell$ is the perpendicular bisector of $D'E'$ .

So if $O \mapsto O'$ , then we are reduced to showing that $\ell$ is orthogonal to circumcircle of $O'D'E'$ , which is now obvious since $\ell$ is the perpendicular bisector of $D'E'$ .

This post has been edited 1 time. Last edited by Rijul saini, Nov 30, 2023, 10:35 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

21 posts

Lxrdux

#7 h Nov 19, 2023, 2:50 PM

Y by

The statement is equivalent to the fact that $O$ lies on the polar of $D$ w.r.t. the other circle. but that polar goes through $A$ and is perpendicular to $B_1C_1$ , so it remains to prove $AO\perp B_1C_1$ .

But this follows immediately from the fact that $B_1C_1$ is antiparallel to $BC$ and $AO$ is isogonal to the $A$ -altitude of $BC$ , which are orthogonal.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a