[/quote]
,
real numbers 
,
and transform the sequence into 
.
where 
are lengths of parallel sides and 
are other two sides.
?
be a cyclic quadrilateral. Let 
,
,
be the feet of the perpendiculars from 
to the lines 
,
,
if and only if the bisectors of 
and 
are concurrent with 
be reals such that 
. Prove that
.
to 
to 
and 
as the reflections of 
be the circumcircle of 
.
with 
with 
is the circumcenter of 
can be inscribed within a circle.
and 
Prove that 
be a triangle. A circle passing through 
intersect at 
and 
.
if and only if 
.
and 
.
be reals such that 
Prove that
Prove that
such that 
.
of positive integers satisfying the following two conditions:
,
is in 
,
and 
;
such that 
is not divisible by 
.
Y by budu, tworigami, kevinmathz, Mango247
Let 
be an acute, scalene triangle with orthocenter 
, and let 
meet the circumcircle of at a point 
. Points 
and 
are chosen on 
and 
such that 
and 
. Show that 
, 
, and the line through parallel to 
concur.
be an acute, scalene triangle with orthocenter 
, and let 
meet the circumcircle of at a point 
. Points 
and 
are chosen on 
and 
such that 
and 
. Show that 
, 
, and the line through parallel to 
concur.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/f/1/3/f134468be5304baf255d345a69ae02fd6d95ee5c.png)
and let 
be the antipode of 
.
and 
are cyclic, so
meaning that 
is cyclic. Let 
be the circumcenter of 
.
,
.
.
since 
is a diameter of 
collinear. Also, by Brokard's on 
so 
.
collinear.
is the midpoint of 
.
,
.
,
.
,
.
,
by easy angle chase, we conclude that 
.
be the intersection of 
and 
,
,
.
.
has slope 
.
,
.
:
:
and Point 
The 
-
.
.
The equation of the line through 
and 
is 
and 
into 
and solve for 
.
and 
So 
and similarly 
\textbf{Claim}: The intersection of 
and 
is 
where the 
in the 
and the 
and 
,
so 
be ![\[A = (0,a), B = (-b, 0), C = (c, 0)\]](http://latex.artofproblemsolving.com/f/c/b/fcb5aca7a1ceea34a6009d3c85c590091b773df5.png)
and 
are all positive. \bigskip \bigskip
is given by ![\[\frac{c}{a} (x+b) = y,\]](http://latex.artofproblemsolving.com/6/0/5/6050fa886629af04e273c7eaec1abb7ccb1b2825.png)
while the altitude from 
.
.
\bigskip \bigskip![\[-\frac{a}{c} (x-c) = y,\]](http://latex.artofproblemsolving.com/e/d/b/edbf1bad601e72614d9add8fc8e9cafc2eef8c93.png)
while the perpendicular from ![\[\frac{c}{a} x = y + \frac{bc}{a}.\]](http://latex.artofproblemsolving.com/a/c/3/ac3ccf1012ab802f569d1d6b08d6094f3e21ead4.png)
Solving these two equations, we get ![\[E = \left(\frac{c(a^2+bc)}{a^2+c^2}, \frac{ac(c-b)}{a^2+c^2}\right)\]](http://latex.artofproblemsolving.com/9/b/b/9bba30be5c2170906172262b58e0bdb1a1d550cf.png)
Similarly, 
is given by ![\[\frac{a}{b} (x+b) = y,\]](http://latex.artofproblemsolving.com/f/4/d/f4da765083e94e04b50700bc4833c1ae369f2e4e.png)
and the perpendicular from ![\[-\frac{b}{a} x = y + \frac{bc}{a}.\]](http://latex.artofproblemsolving.com/8/a/9/8a9233be64e5e1043f8d00451b5dc4f325025e72.png)
We can solve these two equations, and we get ![\[F = \left(-\frac{b(bc+a^2)}{a^2+b^2}, \frac{ab(b-c)}{a^2+b^2}\right)\]](http://latex.artofproblemsolving.com/9/e/0/9e07a4d953fe73e074e7de18efa306f128a6d9e4.png)
\bigskip \bigskip
.
.
.![\[\frac{ac(c-b)}{c(a^2+bc)} = \frac{a(c-b)}{a^2+bc}.\]](http://latex.artofproblemsolving.com/4/5/e/45e703c4cb00fc5dbb7f4e76f6b1c290a9b1da95.png)
Thus, the equation of the line passing through ![\[\frac{a(c-b)}{a^2+bc} x = y - \frac{bc}{a}.\]](http://latex.artofproblemsolving.com/4/9/b/49b4d64b62500e5abf16f6aa94c80af0435e8691.png)
\bigskip \bigskip
and 
.![\[\frac{\frac{ac(c-b)}{a^2+c^2}}{\frac{c(a^2+bc)}{a^2+c^2} + b} = \frac{ac^2 - abc}{a^2b + 2bc^2 + a^2c}.\]](http://latex.artofproblemsolving.com/2/7/e/27ef33979b4e701f1487d8b3858beb563394556d.png)
Thus, the equation of ![\[\frac{ac^2 - abc}{a^2b + 2bc^2 + a^2c} (x+b) = y.\]](http://latex.artofproblemsolving.com/b/7/7/b7714b3b9dca001fac445146b620f21d80f763d5.png)
The slope of ![\[\frac{\frac{ab(b-c)}{a^2+b^2}}{-\frac{b(bc+a^2)}{a^2+b^2}-c} = \frac{abc-ab^2}{2b^2c+a^2b+a^2c}.\]](http://latex.artofproblemsolving.com/b/c/1/bc1604acf792c2ed32b974cc5405e8215638913c.png)
The equation of ![\[\frac{abc-ab^2}{2b^2c+a^2b+a^2c}(x-c) = y.\]](http://latex.artofproblemsolving.com/f/d/c/fdc53d12d3cf7cef093724d2a1d1094bce58c3a2.png)
We need to show that these two lines and the line through ![\[\left(\frac{2bc(a^2+bc)}{a^2(b-c)}, -\frac{bc}{a}\right).\]](http://latex.artofproblemsolving.com/b/3/6/b36ef5557830145c319cc8885d289539069fa541.png)
We simply prove that this point lies on all three lines, and we're done. We first show it lies on ![\[\frac{a(c-b)}{a^2+bc} \cdot \frac{2bc(a^2+bc)}{a^2(b-c)} = -\frac{2bc}{a} = -\frac{bc}{a}-\frac{bc}{a}.\]](http://latex.artofproblemsolving.com/b/5/2/b52fe00b87dc739dc8aa4b2f77d293dd540a13e8.png)
We now show it lies on 
Lastly, we show the point lies on 
Thus, we have shown that the point lies on all three lines, meaning they are concurrent.
