Bary Bash (oops)
by agbdmrbirdyface, Apr 11, 2017, 3:39 AM
I channeled my sadness from not making JMO into this problem.
USA TSTST 2012/7 wrote:
Triangle 
is inscribed in circle 
. The interior angle bisector of angle 
intersects side 
and at 
and 
(other than ), respectively. Let 
be the midpoint of side . The circumcircle of triangle 
intersects sides 
and 
again at 
and 
(other than ), respectively. Let 
be the midpoint of segment 
, and let 
be the foot of the perpendicular from to line 
. Prove that line 
is tangent to the circumcircle of triangle 
.
is inscribed in circle 
. The interior angle bisector of angle 
intersects side 
and at 
and 
(other than ), respectively. Let 
be the midpoint of side . The circumcircle of triangle 
intersects sides 
and 
again at 
and 
(other than ), respectively. Let 
be the midpoint of segment 
, and let 
be the foot of the perpendicular from to line 
. Prove that line 
is tangent to the circumcircle of triangle 
.![[asy]
import graph;
size(250);
pair A, B, C, D, M, H, L, Q, P, N, d, m;
A = (0,0);
B = dir(250);
C = 1.5*dir(320);
draw(A--B--C--A, red);
path circum = circumcircle(A, B, C);
draw(circum, blue);
M = midpoint(B--C);
d = bisectorpoint(B, A, C);
D = extension(A, d, B, C);
m = rotate(90, M) * C;
L = extension(A, D, M, m);
draw(M--L, orange);
path circ = circumcircle(A, D, M);
draw(circ, magenta);
pair[] abinters = intersectionpoints(circ, A--B);
Q = abinters[1];
pair[] acinters = intersectionpoints(circ, A--C);
P = acinters[1];
N = midpoint(Q--P);
draw(Q--P);
H = foot(L, D, N);
draw(N--H, orange);
draw(A--L);
draw(circumcircle(N, M, H), magenta);
draw(M--H--L, orange);
draw(N--M);
dot(M);
dot(D);
dot(L);
dot(Q);
dot(P);
dot(N);
dot(H);
label("$A$", A, dir(100));
label("$B$", B, dir(215));
label("$C$", C, dir(345));
label("$D$", D, dir(220));
label("$M$", M, dir(315));
label("$P$", P, dir(20));
label("$Q$", Q, dir(170));
label("$N$", N, dir(65));
label("$L$", L, dir(270));
label("$H$", H, dir(290));
[/asy]](http://latex.artofproblemsolving.com/c/7/f/c7f12ad4de9fe474d68f91883839448a8b3e08d7.png)
and 
.
is cyclic yields 
,
and 
are parallel. This looks like a job for barycentric coordinates!
and 
.
into 
and we get 
.
We solve for 
and 
and we find that 
and 
.
to find 
to find 
We have to find the midpoint of this segment 
We have our point at infinity along 
.
thus implying 
are collinear and so 
and we are done.
May 2017
April 2017