is described on a circle 
,
be 
respectively. A circle with a center 
,
and lines 
.
,
.
,
and lines 
,
and lines 
are concurrent.
such that 
is a divisor of 
.
,
and 
.
is prime, 
with 
and 
.
Y by Davi-8191, raknum007, cosmicgenius, samrocksnature, megarnie, HWenslawski, Adventure10, DroneChaudhary, ItsBesi, ehuseyinyigit, Rounak_iitr, Yiyj1, and 1 other user
In acute triangle 
angle 
is greater than
. Let 
is midpoint of 
. 
and 
are the feet of the altitude from and respectively. 
and 
are midpoint of 
and 
respectively. If 
intersect the line through 
parallel to in 
, prove that 
.
angle 
is greater than
. Let 
is midpoint of 
. 
and 
are the feet of the altitude from and respectively. 
and 
are midpoint of 
and 
respectively. If 
intersect the line through 
parallel to in 
, prove that 
.
=
, then it`s easy to see 
and the line through 
.
and 
.
the circumcenter of 
,
, 
the midpoint of 
,
, 
and 
.
is the incircle of 
thereafter, deduce that 
,
which can be reduced to 
.
and 
.
is the midpoint of 
and that 
,
.
, 
and 
be midpoints of 
, 
and 
respectively. By an angle chasing, we get that 
and then the quadrilateral 
is cyclic. Furthermore, triangles 
and 
are homothetic with respect to 
hence, 
,
and 
because 
.
,
and 
,
is cyclic with center 
,
,
.
and hence that 
meet 
and meet 
at 
,
at 
at 
;
meet 
.
on 
similar 
(
then
is 
that here 
is 
.
. by Cayce theorem we get tha
,
is the length of the perpendicular from 
is the midpoint of 
.
.
.
that 
is the feet of the perpendicular from 
that 
.
)
that 
is the circumcenter's Radius. so 
,
a
.
.
onto 
and projection of 
be 
.
,
.
.
where 
also we've 
and from thiose two equations we get 
and so that implies 
and we had 
thus finally we get 
and so done.
be a midpoint of 
be a center of circumcircle of 
.
.
,
and 
,
.
and 
,
are inscribed in a circle of diameter 
Hence, 
is isoceles. Now, let 
Then since 
Now here is the key to the whole solution: The points 
are really annoying, and we would love to get them out of the picture. We can do this by proving that the circumcenter of 
lies on the line through 
This is because then said circumcenter must be the intersection of the perpendicular bisector of 
and the line through 
be the circumcenter of 
Then since 
,![\[\angle T'AM = 90^{\circ} - \frac{1}{2}\angle AT'M = 90^{\circ} - \left(180^{\circ} - \angle AFM\right) = \angle AFM - 90^{\circ} = \angle AFM - \angle EFM = \angle AFE.\]](http://latex.artofproblemsolving.com/c/c/d/ccd555eb573240fde31935f865fdf004b65bd423.png)
Finally, notice that since 
is a cyclic quadrilateral, we have 
Hence, because 
Therefore 
,
,
(call it 
)
;
,
,
and 
.
;
;
.
is given by 
and displacement vector 
is given by 
.
which is a straightforward computation.
are all tangent to 
.
cyclic 
.
,
.
.
and similarly for 
and 
.
,
;![[asy]
pair A = dir(130); pair B = dir(210); pair C = dir(330); pair M = midpoint(B--C); pair D = foot(C, A, B); pair E = foot(B, A, C); pair K = midpoint(E--M); pair L = midpoint(D--M); filldraw(A--B--C--cycle, invisible, blue); pair T = extension(K, L, A, A+C-B);
draw(T--A, lightred); draw(L--T, lightred); draw(M--T, lightred); filldraw(circumcircle(A, D, E), invisible, orange); draw(D--M--E, orange);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$D$", D, dir(220)); dot("$E$", E, dir(45)); dot("$K$", K, dir(315)); dot("$L$", L, dir(L)); dot("$T$", T, dir(T));
/* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A = dir 130 B = dir 210 C = dir 330 M = midpoint B--C D 220 = foot C A B E 45 = foot B A C K 315 = midpoint E--M L = midpoint D--M A--B--C--cycle / 0.1 yellow / blue T = extension K L A A+C-B T--A / lightred L--T / lightred M--T / lightred circumcircle A D E / 0.1 red / orange D--M--E orange */
[/asy]](http://latex.artofproblemsolving.com/f/a/7/fa75f8985ca3776c0001eb1da0764aaf0713dd2e.png)
and the circle centered at ![\[ TM^2 = \operatorname{Pow}_{(AED)}(T) = TA^2. \]](http://latex.artofproblemsolving.com/d/b/c/dbc1336a9951a4d24d9778b3add0417df9e33769.png)
.
.
is the perpendicular bisector of 
,
.
.
,
,
.
,
,
.
lies on 
;
;
is perpendicular to 
are reflections in 
is the 
and 
are tangent to
at 
and 
,![\[
TA = \text{Pow}_{(AEF)}(T) = \text{Pow}_{\omega}(T) = TM
\]](http://latex.artofproblemsolving.com/1/b/1/1b15569742a72edcf30d4e24d286a3a13682478e.png)
![[asy]
size(200);
defaultpen(linewidth(0.4)+fontsize(10));
pair A, B, C, E, F, H, M, K, L, T;
A = dir(120); B = dir(200); C = dir(340); H = orthocenter(A, B, C); E = foot(H, C, A); F = foot(H, A, B); M = .5B + .5C; K = .5M + .5E; L = .5M + .5F; T = extension(K, L, A-.75, A+.75);
draw(A--B--C--cycle); draw(circumcircle(A, E, F), blue); draw(E--M--F); draw(B--E); draw(C--F); ; draw(L--T); draw(A--T--M, heavyred);
label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$E$", E, dir(75)); label("$F$", F, W); label("$M$", M, S); label("$L$", L, S); label("$K$", K, NE); label("$T$", T, NE);
dot(A); dot(B); dot(C); dot(E); dot(F); dot(M, blue); dot(K); dot(L);
[/asy]](http://latex.artofproblemsolving.com/e/3/e/e3e779c498531ea912f35ca86379f2825b433df7.png)
![\[pow(K, (AEF)) = \frac{KE^2}{4} = \frac{KM^2}{4} = pow(K, \odot M)\]](http://latex.artofproblemsolving.com/6/1/d/61da6726ce9c5507d17fbf68c65386eb71dbcb55.png)
and analogously for ![\[pow(T, (AEF)) = pow(T, \odot M) \implies TA = TM. \text{ } \blacksquare\]](http://latex.artofproblemsolving.com/a/d/5/ad52b8e7709e113bbf37c70f56b49876d6d88f5b.png)
denote the degenerate circle centered at 
—
—
.
,
as desired. 
,
are tangent to 
as a circle with radius ![\[TA^2 = \operatorname{Pow}_{(AEF)} (T) = \operatorname{Pow}_{\omega} (T) = TM^2-R^2 = TM^2. \ \square\]](http://latex.artofproblemsolving.com/c/5/b/c5bcb21e838260ef5d828ae7000406f8539081ad.png)
being the power of 
so 
finishes.
the circle 
the point circle at 
are tangent to 
,
.
implies 
.
thus, 
.
is the radical axis of these two circles and every point on it including 
