and 
be positive integers with 
.
.
such that 
is an integer for every positive integer 
Y by ChimkinGang, lpieleanu, zhenghua
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 reflection of 
is a chord in 
;
and 
lie on 
and 
are tangent to ![[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,H,D,EE,F,P,Q,X; A=dir(130); B=dir(210); C=dir(330); H=A+B+C; D=foot(A,B,C); EE=foot(B,C,A); F=foot(C,A,B); P=B+unit(C-B)*sqrt(abs(A-B)*abs(F-B)); Q=2B-P; X=extension(Q,F,P,EE);
draw(circumcircle(A,F,Q),gray+Dotted); draw(B--EE,gray); draw(C--F,gray); draw(circumcircle(A,P,Q),dashed); draw(circumcircle(A,EE,F),gray); draw(B--C--A--B--Q--X); draw(P--EE);
dot("\(A\)",A,N); dot("\(C\)",B,S); dot("\(B\)",C,SE); dot("\(H\)",H,S); dot("\(D\)",D,S); dot("\(F\)",EE,dir(30)); dot("\(E\)",F,dir(110)); dot("\(X'\)",P,SE); dot("\(Y'\)",Q,SW); dot(X); dot("\(P\)",2D-H,S);[/asy]](http://latex.artofproblemsolving.com/3/5/5/355d5c98fa2b262274923140965ddaefd640ed49.png)
.
,
.
.
,
.
,
and 
intersect on the circumcircle of 
,
and 
,
.
(as everyone else has pointed out), and 
and the circle with diameter 
and 
is on the radical axes of 
,
,
.
,
is a symmedian of 
.
are collinear. Similarly, by orthocenter reflections, 
are collinear. Thus, 
are known to be collinear, and thus 
.
with 
,
is the midpoint of 
,![[asy]
size(300);
defaultpen(linewidth(0.6)+fontsize(11));
pair A = (0,7), B = (-3,0), C = (6,0), H = orthocenter(A,B,C), D = foot(A,B,C), P = 2*D - H, F = foot(C,A,B), Xx = 3*C-2*B;
path circ = circumcircle(A,F,P);
pair[] oops = intersectionpoints(circ, Xx--B);
pair X = oops[0], Y = oops[1];
draw(C--A--B--Y, rgb(0.8,0.1,0.1));
draw(A--P--C--F, rgb(0.9,0.4,0.1));
draw(circ,rgb(0.1,0.5,0.1));
draw(circumcircle(A,B,C),rgb(0.1,0.1,0.7));
draw(F--X--P--Y--F,rgb(0.8,0.1,0.8)+linetype("3 3"));
dot("$A$",A,N,linewidth(3.3));
dot("$B$",B,SW,linewidth(3.3));
dot("$C$",C,SE,linewidth(3.3));
dot("$D$",D,NE,linewidth(3.3));
dot("$F$",F,NW,linewidth(3.3));
dot("$X$",X,SW,linewidth(3.3));
dot("$Y$",Y,SE,linewidth(3.3));
dot("$P$",P,S,linewidth(3.3));
dot("$H$",H,SW,linewidth(3.3));
[/asy]](http://latex.artofproblemsolving.com/f/a/c/facf88a5a8e1915fa22b64036aa479f08c6c9785.png)
.
and
It follows that 
and
It follows that 
.
,
be the reflection of 
and 
It is well-known that 
Then 
But 
so 
is the diameter of 
thus 
so 
is an isosceles trapezoid. But as 
we have that 
so 
QED
be the midpoint of 
be the center of the nine point circle of 
.
be the projection of 
be the projection of 
the intersection of 
with the circumcircle of 
,
,
and 
.
,
,
is a parallelogram, and since diagonals in a parallelogram bisect each other, 
intersect 
.
is isosceles, so 
where 
too, so we are done.
cut 
implying 
.
and triangle 
is isosceles. Then it is easy to observe
Thus, 
.
implies 
as desired.
but i kept getting complicated expression that doesnt equal that. Here's what I did:
is parallel to 
c
.
conj
then why when i try to calculate it doesn't work?
