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with 
intersect at 
, and 
and 
are the circumcenter and incenter of 
, respectively. Prove that if 
then 
and 
are perpendicular
be real numbers such that 
and 
, for all 
. Prove that 
for all 
for which the number 
is a perfect square.
such that the centroid of the triangle lies on the circle inscribed in the triangle. Prove that 
we can find positive integers 
such that 
. Show that in any such equation there must be two of the 
numbers with a common divisor (
).
is a triangle. Show that 
be the common chord of two circles 
and 
. A line through 
intersects at 
and at 
. Let 
be a point on the segment , different from 
and . The line 
intersect at and 
. The line 
intersects at 
and 
.
lie on the circumference of a circle 
. is , prove that 
is perpendicular to 
.
inhabitants of an island, where is even, every two are either friends or enemies. Some day, the chief of the island orders that each inhabitant (including himself) makes and wears a necklace consisting of marbles, in such a way that two necklaces have a marble of the same type if and only if their owners are friends.
different types of stones. types.
for which the equation 
has positive integer solutions.
, consider the numbers 
is even. for which 
and 
are primes.
with intersect at , and and are the circumcenter and incenter of , respectively. Prove that if then and are perpendicular
be a triangle with circumcentre and Incentre , and suppose 
are points on rays 
respectively, such that 
. Prove that 
.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/b/c/7/bc7fe01f13f832751a0ff585efa44b77f7fc9595.png)
and circumcircle be denoted as 
. For any point in the plane, let 
. So, we know that 
is a linear function on any line.
, since in that case 
will be parallel to the radical axis of and , and hence perpendicular to .
. We have 
Interchanging 
will give us the expression for 
by symmetry, therefore 
and we are done. 
be a triangle with circumcentre and Incentre , and suppose are points on rays respectively, such that . Prove that .
. Now, 
. Thus, 
For any line 
, we have the infinite point as 
. Thus, we have 
. Thus, 
Note that 
, therefore, we have 
Also, 
, thus 
Finally, for two infinite points, the directions corresponding to them are perpendicular if and only if their isogonal conjugates are antipodes on the circumcircle.
So we need to show that 
We have 
, thus we are reduced to showing that
Now, 
Thus, it reduces to proving that 
which is readily verified to be true.
, , and on the unit circle with 
, 
, and 
so that the incenter is at 
.
.
can be checked to be 0 after some relatively painless calculations.
