lie on a circle 
and point 
and 
are tangent to 
are collinear, and (iii) 
.
bisects 
.
of positive integers has the property that, for each 
and each positive integer divisor 
of 
,
satisfying 
.
could be equal.)
Y by ChimkinGang, lpieleanu, zhenghua, Rounak_iitr
Let 
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
be the reflection of 
be the foot from 
to 
be the circumcenter of 
.
is right, by homothety by a factor of 
at 
is right. If we apply a homothety by a factor of 
at 
and 
map to the perpendicular bisectors of 
and 
,
maps to 
,
.
be midpoint of 
be center of 
.
be the point so that 
is a rectangle; it suffices to show 
.
Now checking 
we get 
.
.
be the foot from 
so 
.
with 
.![\[
-a^2S_CS_B-b^2S_Bp-c^2pS_C=0.\tag{1}
\]](http://latex.artofproblemsolving.com/1/1/6/1161ffe16824cb3b00142d28a4efa40a707dace5.png)
Note that 
.
so 
.
gives 
.
.
to intersect with 
and![\[
-a^2yz+S_By-S_Cz=0
\]](http://latex.artofproblemsolving.com/9/6/4/9642d2d3f74e0b48f987465ca82ddea330aa1cc1.png)
so![\[
-a^2y(1-y)+S_By-S_C(1-y)=0.
\]](http://latex.artofproblemsolving.com/1/0/f/10f9d162db73b06ae53a918cc9dbdf0456660aab.png)
Solving yields 
so 
and 
.
,
.
note that 
Let the line through 
at point 
Then 
by the parallel lines, so 
is a diameter in 
Thus if 
then 
Since 
by midlines we get 
and this implies, once again by midlines, that 
Hence 
and together with 
this implies 
as desired.
.
.
is an isosceles trapezoid, so the perpendicular bisector of 
,
,
.
,
,
and we are done.
?
Let 
.
.
in terms of 
gives: 
Simplifying 
gives ![\[a(d-b) = b^2+bc.\]](http://latex.artofproblemsolving.com/d/7/8/d788c34f9d0515a0f15efdc7d847e72fe28259ce.png)
Solving 
for 
gives 
.
,![\[d^2 - bd - bc-c^2=(d-b-c)(d+c) = 0.\]](http://latex.artofproblemsolving.com/d/c/2/dc25cd031e8c1a1dd008b9a4e268010d380574b6.png)
Since 
,
as desired.
,
,
.
intersect on the line 
,
.
and 
,
is the line 
,
has slope 
.
.
,
,
.
,
.
,![\[\begin{cases}
y=\frac{bx}{a} \\
y=-\frac{ax}{b}+\frac{ac}{b}
\end{cases} \implies \frac{x(a^2+b^2)}{ab}=\frac{ac}{b} \implies x=\frac{a^2c}{a^2+b^2},\]](http://latex.artofproblemsolving.com/7/f/c/7fc201fc320294bbf76988ed102f0365ef813bba.png)
and![\[y=\frac{bx}{a} \implies y=\frac{abc}{a^2+b^2}.\]](http://latex.artofproblemsolving.com/d/b/1/db177eeb8d3cf0c4c7a85ea3972887ef4e9d63d2.png)
Thus,![\[F=\left(\frac{a^2c}{a^2+b^2},\frac{abc}{a^2+b^2}\right).\]](http://latex.artofproblemsolving.com/8/5/d/85d5b65be4dc7ab740364ecfe55a38f3337f1f38.png)
The midpoint of ![\[\left(\frac{1}{2}\left(a+\frac{a^2c}{a^2+b^2}\right),\frac{1}{2}\left(b+\frac{abc}{a^2+b^2}\right)\right)=\left(\frac{a^3+a^2c+ab^2}{2(a^2+b^2)},\frac{a^2b+abc+b^3}{2(a^2+b^2)}\right).\]](http://latex.artofproblemsolving.com/2/9/4/29480f23081b3d4f152fbdba1d72ca0ca86d32d3.png)
Since 
,
is 
,![\[y=-\frac{a}{b}\left(x-\frac{a^3+a^2c+ab^2}{2(a^2+b^2)}\right)+\frac{a^2b+abc+b^3}{2(a^2+b^2)}.\]](http://latex.artofproblemsolving.com/9/8/3/98361d93bd1ab6da7e65c6b825be56b719e894d8.png)
At 
-
The midpoint of ![\[\left(\frac{1}{2}\left(a+\frac{a^2c}{a^2+b^2}\right),\frac{1}{2}\left(\frac{a^2-ac}{b}+\frac{abc}{a^2+b^2}\right)\right)=\left(\frac{a^3+a^2c+ab^2}{2(a^2+b^2)},\frac{a^4+a^2b^2-a^3c}{2b(a^2+b^2)}\right).\]](http://latex.artofproblemsolving.com/1/6/d/16d8b38820d8e74de949540dba40ead29beb3d4a.png)
The slope of ![\[\frac{\frac{abc}{a^2+b^2}-\frac{a^2-ac}{b}}{\frac{a^2c}{a^2+b^2}-a}=\frac{\frac{2ab^2c+a^3c-a^4-a^2b^2}{b(a^2+b^2)}}{\frac{a^2c-a^3-ab^2}{a^2+b^2}}=\frac{2b^2c+a^2c-a^3-ab^2}{abc-a^2b-b^3}.\]](http://latex.artofproblemsolving.com/c/c/3/cc3bfbaef81e1ce30aefb09ee42ad1b06b09837d.png)
The slope of the perpendicular bisector of ![\[y=\frac{b^3+a^2b-abc}{2b^2c+a^2c-a^3-ab^2}\left(x-\frac{a^3+a^2c+ab^2}{2(a^2+b^2)}\right)+\frac{a^4+a^2b^2-a^3c}{2b(a^2+b^2)}.\]](http://latex.artofproblemsolving.com/e/a/8/ea8752d7b641762523f4f5b32447db58b0e619ce.png)
At 
Hence, the perpendicular bisectors of 
is a parallelogram and hence a rectangle. Since 
,
.
we would be done, as we would have 
by parallelogram and 
with 
.
simple terms are equivalent.
be the two intersection of line 
.
to be the reflection of ![[asy]
import geometry;
size(9cm);
defaultpen(fontsize(11pt));
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair H = orthocenter(A,B,C);
pair F = extension(C,H,B,A);
pair D = extension(A,H,B,C);
pair P = 2*D - H;
pair Q = 2*C - H;
pair X = intersectionpoints(line(D,C), circumcircle(A,F,P))[0];
pair Y = intersectionpoints(line(D,C), circumcircle(A,F,P))[1];
draw(A--B--C--A, fuchsia);
draw(circumcircle(A,B,C), red);
draw(A--P, fuchsia);
draw(F--Q, fuchsia);
draw(P--Q, fuchsia);
draw(C--Y, fuchsia);
draw(C--P, fuchsia);
draw(circumcircle(F,A,P), red);
markscalefactor = 0.01;
draw(rightanglemark(A,D,C), deepgreen);
draw(rightanglemark(A,P,Q), deepgreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(273)*1.5);
dot("$H$", H, dir(210));
dot("$F$", F, dir(F));
dot("$D$", D, dir(D));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
[/asy]](http://latex.artofproblemsolving.com/1/b/5/1b5a1ba4cbefc17a0251dea99deb4a38aebdc22a.png)
.![\[AD \cdot HD = FH \cdot HC \implies AD \cdot 2HD = FH \cdot 2CH \implies AH \cdot HP = FH \cdot HQ\]](http://latex.artofproblemsolving.com/7/a/8/7a80afd263cf98dd67bdf2903cf52d1354aefecd.png)
which proves the claim. Since ![\[\overline{CP} = \overline{CH} = \overline{CQ}\]](http://latex.artofproblemsolving.com/9/2/5/92574cd31cb489d626cab182192c9b2bffef0d1c.png)
which means 
in 
at 
,now notice two spiral similar triangles
,now we have here: 
and 
so so we would have also 
and we are done:
be the feet of the 
-
.
and 
,
.
with 
;
and 
respectively. Then it suffices to show that 
is a harmonic quadrilateral. Projecting from 
is harmonic. But this is easy as 
and 
is a cyclic isosceles trapezoid it follows that the midpoint of 
lies on the perpendicular bisectors of 
,
,
.
:
:
:
.
:
:
Now clearly 
since 
,
,
,
Q
with 
with the way you labeled your coordinates, so idt this is right
,
,
and that whether 
be the center of 
,
,
& 
makes 
.
be the point on 
such that 
.
Since 
& 
,
,
.
,
.
,
,
,
is a rectangle, so 
,
.
axis. Then 
has the same 
