such that :
denotes the number of distinct prime factors dividing 
,
be the incenter and 
.
be the point on 
with 
and 
be the orthocenter of 
,
on 
also satisfies 
.
,
be nonnegative real numbers such that 
.
Y by Math-Ninja, jhu08, mathematicsy, mathlearner2357, Adventure10, Mango247, Rounak_iitr, and 4 other users
Let 
be three diameters of the circumcircle of an acute triangle 
. Let 
be an arbitrary point in the interior of 
, and let 
be the orthogonal projection of on 
, respectively. Let 
be the point such that 
is the midpoint of 
, let 
be the point such that 
is the midpoint of 
, and similarly let 
be the point such that 
is the midpoint of 
. Prove that triangle 
is similar to triangle .
be three diameters of the circumcircle of an acute triangle 
. Let 
be an arbitrary point in the interior of 
, and let 
be the orthogonal projection of on 
, respectively. Let 
be the point such that 
is the midpoint of 
, let 
be the point such that 
is the midpoint of 
, and similarly let 
be the point such that 
is the midpoint of 
. Prove that triangle 
is similar to triangle .
be the reflection of 
.
be the reflection of 
the nine-point center. We prove that 
and the same for the other sides.
and we'll do the same thing for the other sides(HY,HZ) and we'll be done.
be the reflection of 
.
pass thru 
.
wrt 
is the perpendicular bisector of 
.
is 
and 
then we have 
a
therefor it is on the perpendicular bisector of 
and so of 
.
,
,
be the midpoints of 
,
,
,
,
,
are the midpoints opposite 
,
,
,
.
and 
.
and 
,
is a parallelogram. Similarly, we get 
and 
are parallelograms.
,
to intersect 
,
,
at 
,
,
is homothetic to 
,
.
,
cyclic, so 
,
,
,![\[ d = \frac{1}{2} \left( p + \frac{(b-c)\bar p + \bar b c - b \bar c}{\bar b - \bar c} \right) \]](http://latex.artofproblemsolving.com/f/0/b/f0be5134ac912c14494cfaa6180f91e84ab58639.png)
Therefore, ![\[ \frac{x-y}{a-b} = \bar p c \implies \frac{\lvert XY \rvert}{\lvert AB \rvert} = \lvert \bar p c \rvert = \lvert \bar p \rvert. \]](http://latex.artofproblemsolving.com/f/b/d/fbd820a53ecc2c4642756f2c694de77b71f03327.png)
Hence we get the conclusion.
,
and 
lie on the unit circle, corresponding to points 
,
,
are equal to 
,
and 
respectively. We can calculate that:
.
as desired. 
,
,
.
,
,
,![\[x=a+b+c+p-bc\bar{p}\]](http://latex.artofproblemsolving.com/b/8/b/b8b82c22bc927bd504989391a32f906c3d65a90e.png)
![\[y=a+b+c+p-ca\bar{p}\]](http://latex.artofproblemsolving.com/c/c/3/cc3e4b592d286f8c637194949e4df8dd159ad43e.png)
![\[x=a+b+c+p-ab\bar{p}\]](http://latex.artofproblemsolving.com/d/5/f/d5f69f08aa0e140518cb044dd7da3ddb47fe3177.png)
Negating all three of these and adding 
and then dividing by 
(
)
,
,
.
reflected across the 
,
so 
Similarly, 
Since similarly does not care about shifting or scaling by -1, define 
.
Substituting 
makes this 
If we take the conjugate, this becomes 
after multiplying top and bottom by 
,
,
(with egmo hints)
denote 
resp. Take 
is collinear, we get 
as 
we must have 
.
and 
.
are collinear 
Now the condition 
gives us 
and 
so that 
.
and 
.
It is now clear that 
.
(![\[d = \frac{1}{2}(b + c + p - bc\overline{p})\]](http://latex.artofproblemsolving.com/2/0/4/2040890bb158812634b8bb0f7e04342ee3c57b47.png)
![\[e = \frac{1}{2} (a + c + p - ac\overline{p})\]](http://latex.artofproblemsolving.com/f/e/8/fe8076d6adff8c73c6faee0e0ebc488fb33487ec.png)
![\[f = \frac{1}{2} (a + b + p - ab\overline{p})\]](http://latex.artofproblemsolving.com/2/7/b/27b3e21edbb3355c96613b31458a9769c42df35d.png)
![\[x = \frac{1}{2} (a + b + c + p - bc\overline{p})\]](http://latex.artofproblemsolving.com/9/0/4/904b6e07534e39948e30a31ec4c810c094d2b2d0.png)
![\[y = \frac{1}{2} (a + b + c + p - ac\overline{p})\]](http://latex.artofproblemsolving.com/f/7/7/f77283259d99593dcd492ce864b137ebfe17153d.png)
![\[z = \frac{1}{2} (a + b + c + p - ab\overline{p})\]](http://latex.artofproblemsolving.com/2/8/f/28f118464bb1bf993a764806118a216c4a7ed822.png)
from all of these equations, since 
does not depend on that.
,
is a real number (and cyclic variants). Note that the bottom fraction is flipped because 
are in opposite orientation.
.
,
,
Similarly, 
and 
Now, we see that
and analogously for 
and 
So,
Now,
and
by symmetry.
and by symmetry, we are done.
move at constant velocity along lines through the orthocenter 
This is easy to see, since 
we have 
Now we may assume 
be the 
antipodes and let 
be the feet from 
respectively. Then the projections of 
and 
onto 
have the same directed length.
onto 
We want to show 
with directed lengths, or 
To do this we show 
and 
share a midpoint. We consider the Newton-Gauss line of complete quadrilateral 
By a well-known lemma (for example, 10.9 in EGMO) the Newton-Gauss line is perpendicular to the Simson line 
and 
Now the Newton-Gauss line is the 
so the midpoint of 
of 
and let 
be the reflection of 
over this Simson line. By homothety, the directed length 
is twice the directed length of the projection of 
which is twice the directed length of the projection of 
By symmetry these are the same as the directed length 
and these are all parallel to the Simson line, so 
which is a reflection of 
This finishes.
be the center of the unit circle, which we also allow to be 
)
,
,
are (respectively) reflections of 
,
,
,
.
by the well known formula for the foot of an altitude onto a segment connected by two vertices on the unit circle. It follows that 
Thus, we have 
which directly implies the result. 
are 
respectively, then 
which is constant thus 
.
and 
.
so 
,
,
,
.
and similarly for 
.
and similarly for 
and we are done.
is parallel to the imaginary axis. Thus 
Then we find the locations of 
similarity.
and 
and 
respectively. Then, by the Complex Foot formula, note that 
Thus, since 
it follows that
Analogously, we have 
Now, let us perform the transformation 
to get 
Obviously, this preserves similarity. Now, I claim that 
and 
Expanding by minors, we see that this determinant is equal to
concluding the proof. 
.
,
,
.
,
,
![$$
\begin{vmatrix}
1&x&\overline{a}\\
1&y&\overline{b}\\
1&z&\overline{c}\\
\end{vmatrix}=-\overline{p}\begin{vmatrix}
1&bc&\overline{a}\\
1&ca&\overline{b}\\
1&ab&\overline{c}\\\end{vmatrix}=-\overline{p}\begin{vmatrix}
ca-bc&\overline{b}-\overline{a}\\
ab-bc&\overline{c}-\overline{a}\\
\end{vmatrix}=-\overline{p}[(ca-bc)(\overline{c}-\overline{a})-(ab-bc)(\overline{b}-\overline{a})]=0.
$$](http://latex.artofproblemsolving.com/f/1/2/f12716f572695c6e55a4f495f6b083d224fa2ec1.png)
