AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
with no parallel sides is inscribed in a circle 
. Circles 
are inscribed in triangles 
, respectively. Common external tangents are drawn between 
and 
, and 
, and 
, and and , not containing any sides of quadrilateral . A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle 
. Prove that the lines joining the centers of and , and , and the centers of and all intersect at one point.
be a quadrilateral inscribed in a circle 
such that 
. Let 
and 
be the midpoints of 
and 
respectively. The line 
meets again at 
. Prove that the tangent at to , the line and the line 
are concurrent.
be a continuous function. A chord is defined as a segment of integer length, parallel to the x-axis, whose endpoints lie on the graph of 
. It is known that the graph of contains exactly 
chords, one of which has length 2025. Find the minimum possible value of .
has been partitioned into disjoint pairs 
(
) so that for all 
, 
equals 
or 
. Prove that the sum ![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
with 
and 
. Lines 
and intersect at 
, and lines 
and intersect at 
. Let 
be the midpoints of sides 
, respectively. Let 
and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
, the extensions of sides and meet at point 
, and the extensions of sides and meet at point 
. Prove that the distance between the orthocenters of triangles 
and 
is equal to the distance between the orthocenters of triangles 
and 
.
. Let 
and be the feet of the altitudes from 
to and from 
to respectively.
and 
as the reflections of across lines and , respectively. Let be the circumcircle of 
. Denote by the second intersection of line 
with , and by the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
be a cyclic quadrilateral, and let , 
, 
, and 
be the midpoints of , , , and 
respectively. Let 
, , and 
be the orthocenters of triangles 
, 
, 
and 
, respectively. Prove that the quadrilaterals and 
have the same area.
be a quadrilateral inscribed in a circle 
Let the tangent to 
at meet rays 
and at and 
respectively. A point is chosen inside 
so that 
and 
Let 
be a point on segment 
satisfying 
Prove that lines 
and 
are concurrent.
be a cyclic quadrilateral. Let , , 
be the feet of the perpendiculars from to the lines , 
, , respectively. Show that 
if and only if the bisectors of 
and 
are concurrent with .
be a triangle with 
. The points and lie on the side such that is perpendicular to and 
. The line that bisects 
meets 
and 
at and , respectively. Prove that the quadrilateral 
is cyclic.
be a cyclic quadrilateral. A circle centered at passes through and and meets lines and again at points and (distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar., find the locus of the points lying inside the rhombus and satisfying 
.
=
and 
=
.
= and =.
.. Since 
= 
= 
, and 
= 
(analogously), then 
= 
. So, by the reflection: 
= . We can conclude that D' lies on 
. Again, the reflection also gives us that 
= 
= . Then, E is the center of 
. By this, we have:
= 
= 
- . Analogously: 
= - . So: 
= and D' must lie on 
. (In fact, D' is the Miquel point of the quadrilateral ). Noting that AD' is the radical axis of and 
, we conclude that 
is perpendicular to 
. Analogously, 
is perpendicular to 
. Then: 
= 
- 
= - = - . As we want to show: O lies on .
be the intersection of and , notice that belongs to 
, and 
iff 
. So notice we just need to prove 
, and since we already have a common side 
and a common angle (
, which is true). Then, we'll prove 
to conclude by SAA congruence.
, but since 
, I spent very much time on angle chase, but I couldn't do anything, then I thought: what if I tried the SAS similarity to prove it? Here it is:
, which is true because is the miquel point of 
, and there's a spiral similarity taking 
.
be the unit circle.
lies on the perpendicular from to 
. Also 
is cyclyc 
