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.
+1 w
be a triangle. Distinct points 
, 
, 
lie on sides 
, 
, and 
, respectively, such that quadrilaterals 
and 
are cyclic. Line 
meets the circumcircle of 
again at 
. Let 
denote the reflection of across . Show that lies on the circumcircle of 
.
lie on a circle 
and point lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects .
be an acute triangle with circumcenter 
such that 
, let be the intersection of the external bisector of 
with , and let be a point in the interior of such that 
is similar to 
. Show that 
.
be the set of real numbers. Find all functions 
that satisfy the following condition. Here, 
is the function obtained by composing 
times, that is, 
, 
be an odd prime number and 
be the set of positive integers. Suppose that a function 
satisfies the following properties:
.
for any pair of relatively prime positive integers 
not both equal to 1;
for any pair of relatively prime positive integers .
with circumcenter 
. Let 
be a point on 
such that 
. Denote 
the point on satisfying 
. If 
is the projection of onto , show that 
passes through the orthocenter of .
be a triangle with 
. Let the incenter and incircle of triangle be 
and , respectively. Let 
be the point on line different from 
such that the line through parallel to is tangent to . Similarly, let 
be the point on line different from 
such that the line through parallel to is tangent to . Let 
intersect the circumcircle of triangle at 
. Let 
and be the midpoints of and , respectively.
. be an acute triangle with orthocenter . Points 
, 
, 
are chosen in the interiors of sides , 
, , respectively, such that 
has orthocenter . Define 
, 
, and 
.
has orthocenter .
be a positive integer. Initially, a bishop is placed in each square of the top row of a 
to 
from left to right. A jump is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.
of the numbers 
with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order 
, from left to right.
is a rectangle with 
and 
. Point 
lies 
, point 
lies on 
, and 
is a right angle. The areas of 
and 
are equal. What is the area of 
?![[asy]
pair X = (0, 0);
pair W = (0, 4);
pair Y = (8, 0);
pair Z = (8, 4);
label("$X$", X, dir(180));
label("$W$", W, dir(180));
label("$Y$", Y, dir(0));
label("$Z$", Z, dir(0));
draw(W--X--Y--Z--cycle);
dot(X);
dot(Y);
dot(W);
dot(Z);
pair M = (2, 0);
pair A = (8, 3);
label("$A$", A, dir(0));
dot(M);
dot(A);
draw(W--M--A--cycle);
markscalefactor = 0.05;
draw(rightanglemark(W, M, A));
label("$M$", M, dir(-90));
[/asy]](http://latex.artofproblemsolving.com/6/6/8/668d1b9909d619b769900471cd1fd94a582d030b.png)
is a rectangle with and . Point lies , point lies on , and is a right angle. The areas of and are equal. What is the area of ?
Then, 
Since 
by 
similarity, we get 
so 
Now, 
so 
Thus, we have that 
and 
for an answer of 
is a pair of real numbers such that
.
,
(and 
)
,
,
,![$[AEF]=\tfrac{2\sqrt{5}\cdot3\sqrt{5}}{2}=\boxed{\textbf{(C) }15}$](http://latex.artofproblemsolving.com/4/b/8/4b81e12b8bf9d1e36762858c13ea4d5eaaf493a6.png)
.
are 
triangles. 
and 
,
and 
Thus, 
and 
This gives us ![$[WXM]=2x$](http://latex.artofproblemsolving.com/8/1/1/811dcbea963aa58243cd7ff5c299a9c892bf2dbe.png)
and ![$[WZA]=4y.$](http://latex.artofproblemsolving.com/1/9/e/19e8c5fbf0de0494cea878a05ffb663222543c08.png)
We are given that ![$[WXM]=[WZA],$](http://latex.artofproblemsolving.com/7/8/c/78ce3c981bfaebf23b66b9f3d1f9fb7f3ada0338.png)
so 
Thus, ![$[WXM]=4y.$](http://latex.artofproblemsolving.com/f/3/5/f3593db3cec4f8cbb28967d631edc9b6887e492e.png)
![$[MYA]=\frac{(8-x)(4-y)}{2},$](http://latex.artofproblemsolving.com/9/6/b/96b72947f4067d376cefc6ef79ca8e4b409fdc6a.png)
or 
That expression can be simplified to 
![$[WXM]=[WZA]=4y,$](http://latex.artofproblemsolving.com/b/1/6/b16f182abf3b74c3e08de025b8d08ed29b79e9e3.png)
so the total area inside the rectangle but outside ![$[WMA]$](http://latex.artofproblemsolving.com/a/1/1/a11400bdda6f35bc7d6b9e1f096f7592d23a637f.png)
is 
must be a positive integer (because of the answer choices), so the only way that 
can equal something that is 32 minus one of the answer choices is if 
So, the area inside the rectangle but outside 
