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 called stylish if the set 
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
. For example, 
is stylish because the set 
can be partitioned as 
, with sums , 
, and respectively. Determine the number of stylish numbers less than 
.
that verify the following equation for all 
:
is a prime number
to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
, but am confused on how to prove this statement for general vectors 
and 
although its supposed to be a quick comparison.
is a function for which the expression 
for all 
is always a perfect square. Prove that 
for all 
.
such that 
.
be a quadrilateral inscribed in a circle 
, and let 
be a point on the extension of 
such that 
and 
are tangent to . The tangent at 
intersects at 
and the line 
at 
. Let 
be the second point of intersection between 
and . Prove that 
, , are collinear.
be a triangle with incenter 
, and suppose the incircle is tangent to 
and 
at and 
. Denote by 
and 
the reflections of and over . Let be the intersection of 
with 
, and let 
be the midpoint of . Prove that 
and 
are perpendicular.
chance of landing heads when flipped. Krithik is playing a game where he starts with 
point. Every turn, he flips the coin, and if it lands heads, he gains point, and if it lands tails, he loses point. However, after the turn, if he has a negative number of points, his point counter resets to . Krithik wins when he earns 
points. Find the expected number of turns until Krithik wins.
be an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of . Let 
be the circle with diameter . Let 
be the intersection of 
and . Let be the orthocenter of . Let 
be the intersection of and . Let 
be the lines through tangent to 
respectively. Let be the intersection of 
and 
. Let be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter , centroid and orthocenter . Prove that 
is constant.
opposite the vertex 
be tangent to the side at the point 
. Define the points 
on and 
on analogously, using the excircles opposite and , respectively. Suppose that the circumcentre of triangle 
lies on the circumcircle of triangle . Prove that triangle is right-angled. and , Jaqueline has a ruler that allows tracing the line 
. Jaqueline also has a special object that allows the construction of a circle of diameter . Also, always when two circles (or a circle and a line, or two lines) intersect, she can mark the points of the intersection with a pencil and trace more lines and circles using these dispositives by the points marked. Initially, she has an acute scalene triangle . Show that Jaqueline can construct the incenter of .
be an isosceles triangle with 
Let be a point on the circumcircle of on minor arc 
Let 
intersect the extension of 
at 
Let be the midpoint of segment 
and let be the intersection of 
and 
Let the extension of 
intersect 
and the circumcircle of at and 
respectively. Given that 
and 
compute the length of 
we have 
and 
. Furthermore is an interior pont in the triangle so that 
and 
. Determine 
. to get a new triangle 
with an interior point 
(clearly 
and 
). Due to this rotation we know 
. Since 
, triangle 
is an isosceles right triangle, so 
. Now 
and 
. Thus triangle 
is also an isosceles right triangle.
. Then 
, and 
, as desired.
so 
and let 
Applying the law of cosines to 
gives 
and doing so on 
gives 
since 
Rearrange these to isolate the sines and cosines, giving us 
and 
Squaring and adding, we have 
which rearranges and simplifies to 
a quadratic in 
Solving gives 
but we discard the solution 
since the resulting positive value for 
would result in an impossible configuration. Thus, 
and the answer is 