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such that:
with all 
is real number
be a triangle, and let 
, 
, and 
be collinear points such that 
, 
, and 
. Points 
, 
, and 
are the reflections of , , and over 
, 
, and 
, respectively. Prove that if 
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of .
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that is a multiple of 
and has exactly positive factors.
.
.
, 
. Similarly we have 
.
be the intersection of 
,
.
. Now extend its own length through 
to 
to form parallelogram 
. Since 
. But note that this is also 
. Doing the similar thing on the other side of the diagram (extending through 
) we get 
. Thus 
.
are fixed, then the whole figure is fixed.
on tangent of such that 
, where 
is the intersection point of 
and the circle. (Just move closer and further from and observe length of 
to see this fact)(Which is a diameter) as the axis of reflection, and are sent to each other, also and 
are sent to each other. Conclusion follows.
and 
by power of a point, 
. This implies that 
by Law of Sines. Thus 
, implying 
, or 
is a trapezoid.
.
at 
at 
.
and 
and similarly 
.
and 
:![\[\frac{\sin \angle KBA}{\sin \angle KAB} = \frac{KA}{KB} = \sqrt{2} = \frac{MB}{MA} = \frac{\sin \angle MAB}{\sin \angle MBA}.\]](http://latex.artofproblemsolving.com/9/4/4/94494277fb5697711229ddeaff39ec098442ebdf.png)
,
.
gives that 
,
are concurrent at some point 
or parallel. Suppose they aren't parallel. Let 
,
be the infinite points on 
,
,
and 
