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.
is not a prime number.
and 
be two points in the coordinate plane. A circle is drawn such that 
is a diameter.
be non-nagative real numbers such that 
.
.
be an isosceles triangle with 
, and let 
be its incenter. Incircle touches sides 
at 
, respectively. Foot of altitudes from 
to 
are 
, respectively. Rays 
intersect 
at 
, respectively. Prove that 
touches incircle at 
.
with two altitudes 
and 
.Let be the point on the line such that 
.The lines 
and 
intersect at point 
, and 
is the point on segment 
such that 
.Suppose that bisects .Prove that triangle 
is isosceles.
-plane given points 
on the planes 
respectively. Let 
be the intersection point of the line 
and the 
-plane.
. When the point 
moves on the perimeter of the circle with center 
, radius 1 on the plane 
,.
and 
on the plane 
. When the point 
moves on the perimeter of the circle with center 
, radius 1 on the plane and the point moves on the perimeter of the square 
, draw the domain swept by the point on the -plane, then find the area.
with 
, let its incircle be tangent to sides , 
, and 
at , 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at and , respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
. is tangent to the circumcircle of 
. is tangent to the incircle of .
and 
be positive integers with 
such that 
divides 
. Prove that
and 
; or
and 
. with their incenter . Point lies on on the same side as 
wrt 
. Point lies on the shorter arc of the circumcircle . It is given that 
Prove that 
is the angle bisector of 
.
. We call a distribution the act of distributing the integers from 
represented by tokens to players 
to 
so that they all have the same number of tokens in their urns.
beats 
when, when and each draw a token from their urn, has a strictly greater chance of drawing a larger number than . We then denote 
. A distribution is said to be chicken-fox-viper when 
What is 
such that the following equations are both true:
heads and 
tails, his score will be: 
. We also know 
. Let's list out possible values of and and see which pairs give to be a perfect square(we will be using complementary counting). If 
, this is impossible. If 
, this is also impossible. If 
, this will give us that 
, which is a perfect square. If 
, this will give us that 
, which is a perfect square. If 
, this is impossible. If 
, this is impossible. So, Mike will only get a perfect square if he gets either 
heads and 
tails or heads and tails. If Mike gets heads and tails, there are 
possible ways to do this(
factorial ways to arrange the flips, then divide by the factorial ways to arrange the heads and the factorial ways to arrange the tails). Similarly, there are possible ways to do the second case. So we can have a total of 
total possible occurrences for which Mike WILL get a perfect square. The probability is simply 
. Taking the complementary of this, we get our final probability of 
.
.
:
:
,
.
,
for this case
to get 
