[P1] JBMO 2013-2 (Three collinear midpoints)
by MSTang, Jan 3, 2016, 1:59 AM
JBMO 2013-2. Let 
be an acute-angled triangle with 
and let 
be the centre of its circumcircle 
. Let 
be a point on the line segment 
such that 
. Let 
be the second point of intersection of and the line 
. If 
, 
and 
are the midpoints of the line segments 
, 
and 
, respectively, show that the points , and are collinear.
be an acute-angled triangle with 
and let 
be the centre of its circumcircle 
. Let 
be a point on the line segment 
such that 
. Let 
be the second point of intersection of and the line 
. If 
, 
and 
are the midpoints of the line segments 
, 
and 
, respectively, show that the points , and are collinear.![[asy]
size(8cm);
pair A=dir(100),B=dir(212),C=dir(-32),D=foot(A,B,C),E=dir(-100),O=(0,0),M=(B+E)/2,P=(A+C)/2,N=(O+D)/2;
draw(A--B--C--cycle^^B--E--A^^D--O);
draw(M--P,dashed+gray);
draw(unitcircle);
dot("$A$",A,dir(90));
dot("$B$",B,SW);
dot("$C$",C,.7*SE);
dot("$D$",D,SE);
dot("$E$",E,S);
dot("$M$",M,.5*SW);
dot("$O$",O,dir(90));
dot("$N$",N,SE);
dot("$P$",P,NE);
markscalefactor /= 6;
draw(rightanglemark(A,D,B));
add(pathticks(B--M,2));
add(pathticks(M--E,2));
add(pathticks(A--P,3));
add(pathticks(P--C,3));
add(pathticks(O--N,1));
add(pathticks(N--D,1));
[/asy]](http://latex.artofproblemsolving.com/1/3/8/138e10c6d467a6aa71c22689191f6e46f539b44e.png)
The condition 
)
by considering triangle 
.
,
by the inscribed angle theorem. Therefore, 
must both equal 
.
.
,
.
to 
and 
,
and 
,
.
or 
,![[asy]
size(8cm);
pair A=dir(100),B=dir(212),C=dir(-32),D=foot(A,B,C),E=dir(-100),O=(0,0),M=(B+E)/2,P=(A+C)/2,N=(O+D)/2;
draw(A--B--C--cycle^^B--E--A^^D--O--P^^M--D);
draw(M--P,dashed+gray);
draw(unitcircle);
dot("$A$",A,dir(90));
dot("$B$",B,SW);
dot("$C$",C,.7*SE);
dot("$D$",D,SE);
dot("$E$",E,S);
dot("$M$",M,.5*SW);
dot("$O$",O,dir(90));
dot("$N$",N,SE);
dot("$P$",P,NE);
markscalefactor /= 6;
draw(rightanglemark(A,D,B));
draw(rightanglemark(O,P,C));
add(pathticks(B--M,2));
add(pathticks(M--E,2));
add(pathticks(A--P,3));
add(pathticks(P--C,3));
add(pathticks(O--N,1));
add(pathticks(N--D,1));
[/asy]](http://latex.artofproblemsolving.com/3/2/b/32bfe0f232b39f7ae369607be5ccb9342b39dd4a.png)
Wait - are 
and 
are congruent? If so, that would immediately imply that 
- this gives us a way to use point 
.
into 
,
,
,
,![\[\angle MDN = (90^\circ - C) + 90^\circ + \angle ADO = (180^\circ - C) + \angle ADO.\]](http://latex.artofproblemsolving.com/5/4/e/54ee83646d91366ae5cd4bd8c9b0002b849e4522.png)
We can tease out 
by considering quadrilateral 
:![\[\angle NOP = 360^\circ - 90^\circ - C - \angle ODC = 270^\circ - C - \angle ODC.\]](http://latex.artofproblemsolving.com/6/4/b/64b64aa0ffb9cb12306685e93280c404f0bb6c81.png)
This works! Angles 
and 
sum to 
,
.
,
,
(since 
)
.
.
:![\[BE = 2R \sin\angle BAE = 2R \sin (90^\circ - B) = 2R \cos B.\]](http://latex.artofproblemsolving.com/9/b/f/9bfb740165cf6816602ee06eb754f2f7e395c512.png)
So indeed, 
,
by SAS, so 
)
![\[\angle BAD = \angle CAO = \dfrac{\pi - \angle AOC}{2} = \dfrac{\pi - 2 \cdot \angle B}{2} = \dfrac{\pi}{2} - \angle B.\]](http://latex.artofproblemsolving.com/d/3/7/d37c3ccee5187e56d254bfe074dc1e17e6c4d866.png)
Therefore, 
,
and 
are congruent. For one, 
.
is right, ![\[\begin{aligned} \angle NOP &= \dfrac{3\pi}{2} - \angle C - \angle ODC \\ &= \dfrac{3\pi}{2} - \angle C - \left(\dfrac{\pi}{2} - \angle ADO \right) \\ &= \pi - \angle C + \angle ADO \\ &= \pi - \angle C + \left(\angle MDO - \angle MDB - \dfrac{\pi}{2}\right) \\ &= \dfrac{\pi}{2} - \angle C + \angle MDO - \angle MBD \\ &= \dfrac{\pi}{2} - \angle C + \angle MDO - \angle DAC \\ &= \dfrac{\pi}{2} - \angle C + \angle MDO - \left(\dfrac{\pi}{2} - \angle C\right) \\ &= \angle MDO. \end{aligned}\]](http://latex.artofproblemsolving.com/e/d/5/ed51c3dd9db8079ef1185ba597bd4eea48a9c4a4.png)
Finally, considering 
,![\[OP = AO \cos \angle AOP = R \cos\left(\dfrac{1}{2} \angle AOC\right) = R \cos \left(\dfrac{1}{2} \cdot 2\angle B\right) = R \cos \angle B\]](http://latex.artofproblemsolving.com/5/e/5/5e58e240b3dd2cbca1dc602ac735fc724ffd8373.png)
where 
is the radius of 
,![\[MD = \dfrac12 BE = \dfrac12 (2R \sin \angle BAE) = R \sin \left(\dfrac{\pi}{2} - \angle B\right) = R \cos \angle B\]](http://latex.artofproblemsolving.com/a/b/2/ab24c5fb42a969dfcec48e5a366e36d7e517798d.png)
by the extended law of sines, so 
.
,
are collinear. This post has been edited 2 times. Last edited by MSTang, Jan 4, 2016, 7:31 AM
Reason: removed "[P1]"
Reason: removed "[P1]"
,
where 
is the midpoint of 
to hit the circumcircle again at 
,
is a parallelogram. It follows that 
.
That's the proof of 
,
on the unit circle so 
Then 
and 
,
.
.![\[\frac{p-n}{p-m}=\frac{\tfrac{1}{4}\left(a+c+\tfrac{bc}{a}-b\right)}{\tfrac{1}{2a}\left(a^2+ca+bc-ab\right)}=\frac{1}{2}\in \mathbb{R},\]](http://latex.artofproblemsolving.com/7/5/f/75f6762439b56d7d15699d4037614658b218bcc9.png)
so 
are collinear with 
.
,
is a parallelogram but would the argument above also work?
are already collinear. All of this stuff (parallelogram, equal lengths, equal angles) are basically the same, so yes. All of that should work. --MS
March 2019
February 2019