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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 
for all 
. If
, show that the number 
is a square. and , we denote 
.
for all 
without loss of generality. Note that![\[
\{a_1+b_1\} + \{a_2,b_2\} = \{a_1+a_2,a_1+b_2,b_1+a_2,b_1+b_2\},
\]](http://latex.artofproblemsolving.com/b/3/8/b389639c051fc35b8f912973f8af6268c28f3558.png)
and![\[
a_1+a_2 > \max\{a_1+b_2,b_1+a_2\} > \min\{a_1+b_2,b_1+a_2\} > b_1+b_2.
\]](http://latex.artofproblemsolving.com/5/2/9/5299bfbe7ff6f19cecca8e1fa7f57042a31d2a05.png)
We this immediately deduce that![\[
a_1 + a_2 = a_3 + a_4 \quad\text{and}\quad b_1+b_2 = b_3 + b_4.
\]](http://latex.artofproblemsolving.com/f/e/1/fe1e8b9fdb61c171eae8547d84bd4daa120b9304.png)
There are now two cases to consider.
and 
.
and 
,![\[
\prod_{1\le i\le 4}|a_i-b_i| = |a_1-b_1|^2 \cdot |a_2-b_2|^2,
\]](http://latex.artofproblemsolving.com/b/1/c/b1c723f6ebcfa7fb07f9d5eaf7a90cbb91be2b34.png)
which is perfect square.
and 
.