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.
and 
Show that
Where 
and 
Show that
Where
be a quadrilateral inscribed in a circle 
. The lines 
and 
meet at 
, the lines 
and 
meet at 
, and the diagonals 
and 
meet at 
. Let 
be the midpoint of the segment 
, and let 
be the common point of the segment 
and the circle . Prove that the circumcircle of the triangle 
and are tangent to one another.
be an acute triangle with orthocenter 
, and let 
be a point on the side , lying strictly between 
and 
. The points and 
are the feet of the altitudes from and , respectively. Denote by 
is the circumcircle of 
, and let 
be the point on such that 
is a diameter of . Analogously, denote by 
the circumcircle of triangle 
, and let 
be the point such that 
is a diameter of . Prove that 
and are collinear.
be an acute triangle. Define 
as its incenter. Let 
and 
be the incircle's tangent points to 
and 
, respectively. Let 
be the midpoint of 
. Let 
be the intersection point of a perpendicular line passing through to 
. Line 
intersects the circumcircle of 
at 
. The circumcircle of 
intersects line 
at 
. Prove that quadrilateral 
is cyclic., the diagonal bisects neither the angle nor the angle 
. The point lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be a cyclic quadrilateral in which the lines and meet at a point . Let be the point of the line 
, different from , such that 
. We construct the parallelograms 
and 
. Prove that the points 
lie on the same circle.
point 
was chosen such that 
and 
. Prove that if perpendicular from to passes from the intersection of diagonals of , then 
.
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
be an acute triangle. Let be the altitude on , and let be any interior point on . Lines 
, when extended, intersect 
at 
respectively. Prove that 
.
be a point on 
such that 
is an isosceles trapezoid, we have 
and 
. be the second intersection of 
with .
. So, 
.
also lies on 
.
and 
. So 
.
intersects again at 
, note that 
.
, i.e. is the midpoint of .
.
and over to get 
and 
, respectively. Also, reflect over to get 
.
and 
.
. It follows that 
is the reflection of 
over 
.
is the reflection of 
over 
.
.
. Since is the midpoint of 
, is also the midpoint of , done.
. Note that this preserve all the properties of the configuration.
. Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral lie on the unit circle.
are collinear we have,![\[\frac{d-c}{q-c}=\frac{\bar{d}-\bar{c}}{\bar{q}-\bar{c}}\implies \bar{q}=\frac{c-q+d}{cd}\]](http://latex.artofproblemsolving.com/5/f/6/5f67db63556f789eeed8ff27bc40a9a71bee7907.png)
Again since 
,![\[|b-c|=|c-q|\implies (c-b)(\frac{1}{c}-\frac{1}{b})=(\frac{1}{c}-\frac{c-q+d}{cd})(c-q)\]](http://latex.artofproblemsolving.com/8/0/d/80dfe807f60e0b8766813c45bbcf1135c2fc0345.png)
![\[\implies q=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+c\]](http://latex.artofproblemsolving.com/2/6/6/26694a785e0a97a46d61d5f5d8aa5adfbb4910f6.png)
Similarly,![\[p=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+a\]](http://latex.artofproblemsolving.com/a/f/6/af651b78c203dde9d541284ef4a818deaec8b70a.png)
We can rotate the quadrilateral, so WLOG we can assume 
.
and the other has 
sign. WLOG assume 
has sign.![\[m=\frac{p+q}{2}=\frac{\sqrt{d}(a-c)+c+a}{2}\]](http://latex.artofproblemsolving.com/9/8/6/9865f8e2dde2a80af2ec2ae72b3e6a90156797a9.png)
Let 
, we wish to show 
. Now,![\[\delta=\frac{2a+\sqrt{d}(c-a)-c-a}{2c+\sqrt{d}(c-a)-c-a}=\frac{(c-a)(\sqrt{d}-1)}{(c-a)(\sqrt{d}+1)}=\frac{\sqrt{d}-1}{\sqrt{d}+1}\]](http://latex.artofproblemsolving.com/2/d/1/2d1f5feeda0bf1fec98970da2de594d2923039b8.png)
Hence,![\[\bar{\delta}=\frac{\frac{1}{\sqrt{d}}-1}{\frac{1}{\sqrt{d}}+1}=\frac{1-\sqrt{d}}{1+\sqrt{d}}\]](http://latex.artofproblemsolving.com/0/3/8/038f3e9a08b91b2fcc5a1bfed126a7ca40300187.png)
Thus, . And we are done. 
. Note that this preserve all the properties of the configuration.. Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral lie on the unit circle. are collinear we have,Again since ,Similarly,We can rotate the quadrilateral, so WLOG we can assume . and the other has sign. WLOG assume has sign.Let , we wish to show . Now,Hence,Thus, . And we are done. 
." in the very first line of my proof, so I think it should be fine.
." in the very first line of my proof, so I think it should be fine.
be the reflection of point through point 
is a parallelogram.
, it follows that that 
. 
.
and 
be the spiral center. Then, since 
, 
, which means that 
passes through the midpoint of . Seeing this, we observe that 
or
or 
as desired.
and 
to 
and 
.
.
and 
such that 
and 
.
then 
wich gives 
so 
and similarly 
.
but 
and replacing in 
gives 
so 
and so 
but we proved 
and 
thus 
bisects 
and 
bisects 
so 
as desired
be the midpoints of 
,
with ratio 
,
.
be the midpoints of 
respectively. So 
hence 
are cyclic. The fact that 
and 
finishes.