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be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
, 
be the incenters of triangles 
, 
, respectively. Show that the lines 
, 
, 
meet at a point.
+1 w
and 
, 
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that 
for all 
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist.
, a line 
to tangent , and another circle 
disjoint from such that and lie on opposite sides of . The tangents to from a variable point 
on meet at 
and 
. Prove that, as varies over , the circumcircle of 
is tangent to two fixed circles.
be a triangle and let 
be its incircle. Denote by 
and 
the points where is tangent to sides 
and 
, respectively. Denote by 
and 
the points on sides and , respectively, such that 
and 
, and denote by 
the point of intersection of segments 
and 
. Circle intersects segment at two points, the closer of which to the vertex 
is denoted by 
. Prove that 
.
![\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}](http://latex.artofproblemsolving.com/0/c/d/0cd22d36acedbcb10d76d1b884aa35418273ed52.png)
be positive integers. Prove that the expression![\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]](http://latex.artofproblemsolving.com/1/b/7/1b78a2027c3b7156fbb566901d2cfd088cc09f00.png)
is always a positive integer, and determine all possible values it can take.
be a convex pentagon such that 
, 
, and 
. Prove that the perpendicular line from 
to and the line segments and 
are concurrent.
midpoints of segments 
.
, 
.![\[ MA^2 - MD^2 + BD^2 - BC^2 + NC^2 - NA^2 = 0 \]](http://latex.artofproblemsolving.com/c/4/1/c415699bc1b9dbd4254853ea0729e54e1e67e357.png)
Notice that, by power of a point and Pythagorean's: 
.
. Thus, the above sum translates to![\[ -DB\cdot DC + CB\cdot CD + BD^2 - BC^2 = (DB - DC)\cdot CD - (DB - DC)\cdot CD = 0 \]](http://latex.artofproblemsolving.com/6/f/c/6fcaff49af054b5ce1f5d4295c570d35d629cbd4.png)
And we are done.
and 
circles, 
and 
their centers and and 
their intersections and we will move point 
along .
are concyclic.
is the bisector of segment 
and 
is the bisector of segment 
.
.
.
be the intersection of the perpendicular in on the 
line with the 
circle (other than ).
and 
and then we will just have to prove that 
for 
points 
.
is projective ( is fixed), 
is projective (
is the perpendicular from to ) and 
is projective (the circle is fixed and 
is the intersection of with it), we have - projective.
is projective ( is fixed), 
is projective (
is the perpendicular in on line ) and 
is projective (the circle is fixed and is the intersection of with it), we have - projective. for points .
becomes the tangent to in 
.
. is the antipode of in 
.
.
and 
, we can conclude that is the orthocenter of 
.
.
becomes the tangent in at 
and we similarly get 
is a parallelogram. Since is the reflection of across 
, is well-known that is the reflection of across the bisector of segment .
is an isoscelles trapezoid, so is also the reflection of across the bisector of segment , so 
. and are equal in different points , so for all points .
.
and be the centers of and respectively.
is cyclic. Note 
and 
by radax. This implies 
.![\[\measuredangle OBC = \measuredangle OBO_1 + \measuredangle O_1BC = \measuredangle FAB + 90 - \measuredangle CAB = 90\]](http://latex.artofproblemsolving.com/f/e/c/fecf6647b00f5878563150d3b07cc330b7b56e0e.png)
which lies on 
by triangle Miquel. Then 
, so 
, so 
cyclic. Now we can access the argument of 
, so the problem dies:![\[\angle OBC=\angle OBE+\angle EBC=\angle OFE+\angle EAC=90^\circ.\]](http://latex.artofproblemsolving.com/4/2/6/426b83aeadb1d67b1294fd7c6178fb76a01087e6.png)
,
intersects 
,
.
,
.