[/quote]
,
.
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
is the incenter 
of 
.
sides 
.
consecutive vertices is constant and its value is 
.
is marked with 
and 
is marked with 
,
.
intersects circle 
at 
,
intersects circle 
at 
.
and 
are isogonal conjugates in triangle 
,
.
.
are such that 
. Prove that 
.
denote the set of positive real numbers. Find all functions 
such that![\[x+f(yf(x)+1)=xf(x+y)+yf(yf(x))\]](http://latex.artofproblemsolving.com/1/6/c/16c10ac3b45b6b3f864f9314c3aeae654db514db.png)
for all 
is the midpoint of the hypotenuse 
.
is chosen on the leg 
so that the line segment 
meets 
again at 
)
be the reflection of 
meet again at 
)
.
that satisfy the functional equation:![\[
f(f(x)(1 + y)) = f(x) + f(xy), \quad \forall x, y > 0.
\]](http://latex.artofproblemsolving.com/9/f/c/9fcb6b114b87842f23f31a97ddae6c722fe71611.png)
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
of positive integers such that 
is an integer.
and 
be positive integers. Prove that there exists a positive integer 
,
are all greater than 
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Points 
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
and Z is outside 
Though it's not competition-oriented, I believe it's around the same audience.
be the midpoint of 
is cyclic. Let 
Then, 
so 
colinear.
.
,
,
By Power of a Point and cyclic quadrilaterals, 
So 
are on the same circle, which implies our desired conclusion.
is similar to 
,
is similar to 
.
.
is a median
be the point at infinity along lines 
and 
,
.
is harmonic,![\[-1 = (A,C ;B , D) \stackrel{E}{=} (A, C; M, P_{\infty}) \]](http://latex.artofproblemsolving.com/3/b/a/3bac455ec07839e6b78f06a8fd19884b103e48b3.png)
so 
,
.
cyclic. Now let 
.
and clearly 
where 
is the center of 
,
and thus 
.
is perpendicular to 
is cyclic.
and the tangent criterion, we have ![\[\angle BMP = \angle BMD = \angle DBP = \angle BOP\]](http://latex.artofproblemsolving.com/a/3/6/a363b3afbc14fcf16302abdb30b3f845fadecc28.png)
as desired.
,
and 
let the center of the circle be 
and let 
so we have 
we have 
from which it follows that 
so we are done. 
and 
and 
respectively. Then, by the formula for the intersection of tangents, we get 
Then, since 
and 
must have the same measures, so it follows that 
Then, since 
lie on the unit circle while 
we have
Therefore, the midpoint of 
It now suffices to show that 
and 
are collinear. By the complex number collinearity condition, this is equivalent to
concluding the proof. 
,
.
are all concyclic.![\[ \angle{BMP} = \angle{MED} = \angle{BDP}, \]](http://latex.artofproblemsolving.com/2/d/d/2dddfc10aac77de9281ee88e4f3cf7a47c46cf03.png)
where the last part comes from the so-called Tangent Lemma. In particular, 
is a cyclic quadrilateral, so 
.
is clearly cyclic 
)
tells us that 
,
is a symmedian. Also, 
is cyclic with 
,
implies 
,
.
,
is cyclic. But since 
has diameter 
,
and hence 
.
;
is isosceles.
is cyclic. We can do so by proving 
:![\[\angle{BXO} = \angle{BED} = \frac{\overarc{BD}}{2} = \frac{\angle{BOD}}{2} = \angle{BOP}\]](http://latex.artofproblemsolving.com/f/a/7/fa7332189b642ee5ad9f46b558a4b8e5aae1b7d4.png)
,
(this can be shown by proving 
,
,
(as 
by the tangent condition). Since 
and then project through 
onto 
is cyclic and by the 
criterion, 
.
is cyclic. This means 
,
and we're done. 
,
