[/quote]
,
and 
are externally tangent at a point 
.
be a line tangent to 
and 
.
and 
be diameters in 
lies on 
and 
are tangent to 
and 
lies on 
and 
are tangent to 
.
,
times the radius of 
,
,
outside 
.
and 
intersect at 
and define 
similarly. Prove that line 
bisects 
.
and 
Prove that
Where 
be a square inscribed in an equilateral triangle 
.
inside the square. Line 
meets 
at 
.
is the isosceles right triangle.
,
,
,
is equilateral. Show that the center of 
such that there exists exactly 
different functions 
such that ![$f:[1,5]\rightarrow[1,5]$](http://latex.artofproblemsolving.com/6/2/0/620e1621905a4670627d46fcb367e1a8c461bb49.png)
satisfying 
.
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
and 
be a 
grid of unit squares.
or 
rectangle. A subset 
edges of the grid squares.
times as many students as Lara’s high school. The two high schools have a total of
students. How many students does Portia’s high school have?
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 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 
.
.
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 
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.
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 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?
