the centers of the circles are on opposite sides of AB

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the centers of the circles are on opposite sides of AB

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Source: 6th JBMO 2002, Problem 2

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165 posts

#1 h Jun 19, 2004, 3:14 PM • 5 Y

Y by Adventure10, Adventure10, jhu08, samrocksnature, Mango247

Two circles with centers $O_{1}$ and $O_{2}$ meet at two points $A$ and $B$ such that the centers of the circles are on opposite sides of the line $AB$ . The lines $BO_{1}$ and $BO_{2}$ meet their respective circles again at $B_{1}$ and $B_{2}$ . Let $M$ be the midpoint of $B_{1}B_{2}$ . Let $M_{1}$ , $M_{2}$ be points on the circles of centers $O_{1}$ and $O_{2}$ respectively, such that $\angle AO_{1}M_{1}= \angle AO_{2}M_{2}$ , and $B_{1}$ lies on the minor arc $AM_{1}$ while $B$ lies on the minor arc $AM_{2}$ . Show that $\angle MM_{1}B = \angle MM_{2}B$ .

Ciprus

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3660 posts

Arne

#2 h Jun 19, 2004, 4:30 PM • 4 Y

Y by AlastorMoody, Adventure10, jhu08, Mango247

Iris Aliaj wrote:

I really need the solution of this problem.

I've posted it here before but nobody replied. It was getting old and the number of views kept being the same so I posted it again.

You shouldn't do that

Be patient please, and remember, you cannot command us to solve a problem.

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Maja

#3 h Jun 19, 2004, 4:35 PM • 3 Y

Y by Adventure10, jhu08, Mango247

Maybe this could work:

It is easy to show that B1, A and B2 are colinear (angle B1AB=angle B2AB=90).
Let us denote angle AO1M1=angle AO2M1=2x, then we have angle ABM1=x and angle ABM2=180-x, so M1,B,M2 are colinear.
It is also obvious that in quadrilateral B1M1M2B2 there is angle B1M1B=angle BM2B2=90, so we can conclude that B1M1 and B2M2 are parallel and that B1M1M2B2 is trapezium (I'm not sure whether this is correct expression or not, but the point is that it has 2 parallel sides), so its midline MN (N is midponint of M1M2) is parallel to B1M1 and B2M2 and therefore it is perpendicular to M1M2. Now, triangle MM1M2 has 2 equal sides, MM1 and MM2, and that's the end.
Am I right ?

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#4 h Jun 19, 2004, 5:40 PM • 4 Y

Y by Adventure10, jhu08, ehuseyinyigit, Mango247

Did that sound like I was trying to command you?

I didn't mean that at all.

Can somebody who says please when asking sth try to command the others?
.In fact i'm grateful to all the members of this club who spend their time to help me with my problems.

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Arne

#5 h Jun 19, 2004, 6:22 PM • 3 Y

Y by Adventure10, jhu08, Mango247

No

I was rather joking, you should just show a little bit patience
(and please make no double postings

)

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1144 posts

Tiks

#6 h Oct 30, 2005, 6:18 PM • 3 Y

Y by Adventure10, jhu08, Mango247

$\angle B_1AB=\angle B_2AB=90^{\circ}$ => $A$ lie in the $B_1B_2. B_1M=MB_2$ and $BO_2=B_2O_2$ => $MO_2$ paralel to $BB_1$

$B_1M=MB_2$ and $B_1O_1=BO_1$ => $O_1M$ paralel to $BO_2$ => $\angle O_1MO_2=\angle O_1BO_2$ ,but

$\bigtriangleup{O_1AO_2}=\bigtriangleup{O_1BO_2}$ => $\angle O_1AO_2=\angle O_1BO_2$ => $\angle O_1MO_2=\angle O_1AO_2$ => $O_1,A,M,O_2$ lie in the same circle => $\angle AO_1M=\angle AO_2M$ => $\angle M_1O_1M=\angle M_2O_2M$ .

$MO_2$ paralel to $BB_1$ and $O_1M$ paralel to $BO_2$ => $O_1M=BO_2=O_2M_2$ and

$O_2M=BO_1=O_1M_1$ => $O_1M=O_2M_2$ , $O_1M_1=MO_2$ and $\angle M_1O_1M=\angle M_2O_2M$ .=> $\bigtriangleup{M_1O_1M}=\bigtriangleup{M_2O_2M}$ => $MM_1=MM_2$

$\angle M_1BA=\angle M_1O_1A/2$ and $M_2BA=180^{\circ}-\angle AO_2M/2$ =>

$B$ lie in the $M_2M_1$ => $\angle MM_1B=\angle MM_2B$ .

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KRIS17

#7 h Dec 9, 2018, 8:40 PM • 2 Y

Y by jhu08, Adventure10

It's easy to see that $B_{1}AB_{2}$ forms a straight line and is parallel to line $O_{1}O_{2}$ .

Let us define $\angle AO_{1}M_{1}= \angle AO_{2}M_{2} = \theta$ .

Let circumradii of the 2 circles be $R_{1}$ and $R_{2}$ respectively.

Now $\angle ABM_{2} = 180^{\circ} - \theta/2$ and $\angle ABM_{1} = \theta/2$ , this implies that:

$\angle ABM_{2} + \angle ABM_{1} = 180^{\circ}$ . So $M_{1}BM_{2}$ forms a straight line.

Now since $M$ is the midpoint of $B_{1}B_{2}$ , $MO_{2}$ is parallel to $BB_{1}$ and its length is equal to $R_{1}$ .

Similarly, we see that, $MO_{1}$ is parallel to $BB_{2}$ and it's length is equal to $R_{2}$ .

So $\angle AMO_{1} = \angle AB_{2}B = \angle AO_{2}O_{1}$ (since $O_{2}$ is the circumcenter).

So $AMO_{2}O_{1}$ forms a cyclic quadrilateral.

Thus, we have $\angle MO_{2}A = \angle MO_{1}A$ .

Adding $\theta$ to both sides we have:
$\angle MO_{2}A + \theta = \angle MO_{1}A + \theta$
or, $\angle MO_{2}M_{2} = \angle MO_{1}M_{1}$

Thus by SAS, $\triangle MO_{2}M_{2}$ is congruent to $\triangle MO_{1}M_{1}$

So, we have $MM_{1} = MM_{2}$ , hence $\triangle MM_{1}M_{2}$ is an isoceles triangle.

So, we get $\angle MM_{1}M_{2} = MM_{2}M_{1}$ -- (1)

Now $\angle B_{1}M_{1}B = 90^{\circ} = B_{2}M_{2}B$
So, $\angle B_{1}M_{1}B - \angle MM_{1}M_{2} = \angle B_{2}M_{2}B - \angle MM_{2}M_{1}$

giving $\angle B_{1}M_{1}M = \angle B_{2}M_{2}M$ .

Attachments:

This post has been edited 1 time. Last edited by KRIS17, Dec 9, 2018, 8:43 PM
Reason: Added attachment

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gvole

#8 h Aug 4, 2020, 3:12 AM • 1 Y

Y by jhu08

Maja wrote:

Maybe this could work:

It is easy to show that B1, A and B2 are colinear (angle B1AB=angle B2AB=90).
Let us denote angle AO1M1=angle AO2M1=2x, then we have angle ABM1=x and angle ABM2=180-x, so M1,B,M2 are colinear.
It is also obvious that in quadrilateral B1M1M2B2 there is angle B1M1B=angle BM2B2=90, so we can conclude that B1M1 and B2M2 are parallel and that B1M1M2B2 is trapezium (I'm not sure whether this is correct expression or not, but the point is that it has 2 parallel sides), so its midline MN (N is midponint of M1M2) is parallel to B1M1 and B2M2 and therefore it is perpendicular to M1M2. Now, triangle MM1M2 has 2 equal sides, MM1 and MM2, and that's the end.
Am I right ?

trapezoid i think

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#9 h Aug 3, 2021, 4:41 PM • 2 Y

Y by jhu08, BVKRB-

$[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 = 0.3681090964262815, xmax = 26.250661674223267, ymin = -7.713842348766126, ymax = 4.868142351179191; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqffff = rgb(0.,1.,1.); pair O_1 = (8.597107655042617,-1.416175259182274), B = (10.003237142682393,-3.6518608831291033), O_2 = (15.,-1.), A = (9.701998499398552,0.9827216888317629), B_1 = (7.1909781674028395,0.8195103647645573), B_2 = (19.9967628573176,1.651860883129098), M = (13.593870512360219,1.2356856239468277), M_1 = (6.264181000120974,-2.6542962419172973), M_2 = (18.01170874000472,-5.7884878261428465); /* draw figures */ draw(circle(O_1, 2.641115360057683), linewidth(0.8) + blue); draw(circle(O_2, 5.656854708735154), linewidth(0.8) + blue); draw(B--A, linewidth(0.8) + qqwuqq); draw((xmin, 0.5307157771647164*xmin-8.960736657470747)--(xmax, 0.5307157771647164*xmax-8.960736657470747), linewidth(0.8) + yqqqyq); /* line */ draw((xmin, -1.5899571437759157*xmin + 12.252857472763345)--(xmax, -1.5899571437759157*xmax + 12.252857472763345), linewidth(0.8) + yqqqyq); /* line */ draw(B_1--B_2, linewidth(0.8) + red); draw((xmin, 0.5307157771647166*xmin-5.978795929696952)--(xmax, 0.5307157771647166*xmax-5.978795929696952), linewidth(0.8)); /* line */ draw((xmin, -1.5899571437759084*xmin + 22.849357156638625)--(xmax, -1.5899571437759084*xmax + 22.849357156638625), linewidth(0.8)); /* line */ draw(M_1--M_2, linewidth(0.8) + red); draw(A--O_1, linewidth(0.8) + ffxfqq); draw(A--M_1, linewidth(0.8) + qqffff); draw(A--O_2, linewidth(0.8) + ffxfqq); draw(A--M_2, linewidth(0.8) + qqffff); draw(O_1--O_2, linewidth(0.8) + ffxfqq); /* dots and labels */ dot(O_1,dotstyle); label("$O_1$", (8.096267802211656,-1.3957336524573782), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (9.913063006729622,-3.985344727553882), NE * labelscalefactor); dot(O_2,dotstyle); label("$O_2$", (15.336332273947429,-1.1245701890964879), NE * labelscalefactor); dot(A,linewidth(4.pt) + dotstyle); label("$A$", (9.51987598485633,1.2481101153113037), NE * labelscalefactor); dot(B_1,linewidth(4.pt) + dotstyle); label("$B_1$", (7.120079334112451,1.2209937689752146), NE * labelscalefactor); dot(B_2,linewidth(4.pt) + dotstyle); label("$B_2$", (20.02746019009083,1.885344254209396), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (13.57376976210164,1.5057154055041495), NE * labelscalefactor); dot(M_1,linewidth(4.pt) + dotstyle); label("$M_1$", (5.8862855758204,-2.5210620254050737), NE * labelscalefactor); dot(M_2,linewidth(4.pt) + dotstyle); label("$M_2$", (18.020850561220243,-5.436069256534646), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
My Solution:
In order to do that we can see here that $BO_2=M_2O_2=B_2O_2=AO_2=R_2$ here lets suppose $R_2$ is radius of the circle $O_2$
similarly, $M_1O_1=B_1O_1=AO_1=BO_1=R_1$ here $R_1$ is the radius of the circle $O_1$ .
In triangle $B_1BB_2$ we have
By Mid-Point theorem
$O_1M$ is parallel to $BB_2$ and since $BB_2$ has a midpoint $O_2$ so we can say that $BO_2=O_1M$
Similarly,
$O_2M$ is parallel to $BB_1$ so we same as above we can say by mid-point theorem that
$BO_1=O_2M$
hence we can say that quadrilateral $BO_2MO_1$ is a parallelogram.
now observe that $MM_1=R_1+R_2$ and $MM_2=R_1+R_2$ hence $MM_1=MM_2$ so we can say that $\triangle MM_1M_2$ is isosceles triangle so $\angle MM_1M_2=\angle MM_2M_1$ and we are done $\blacksquare$

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