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.
be a cyclic pentagon inscribed in a circle of centre 
which has angles 
. Show that the diagonals 
and 
meet at a point belonging to the diameter 
.
and 
are the incenter and the 
-excenter of triangle 
, respectively.
be the midpoint of arc 
on the circumcircle, and 
be the foot of the to 
. The line 
intersects the circumcircle again at 
. For
on the circumcircle of triangle , let 
intersect at 
. Prove
are concyclic.
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
. 
are the midpoints of 
respectively. is the midpoint of arc , is symmetric to through . Let 
be the incenter and tangent of angles of triangle 
respectively. Prove that 
have a common point
be the circumcircle of triangle , is the midpoint of and 
be the second intersection of 
and . Tangent line of at intersects at , let 
be a transversal of and 
be intersections of 
with . Let 
be a tangent line of . Prove that 
is cyclic
be integers such that there are consecutive integers 
satisfy 
.
is a perfect square.
on one side and a on the other. Harry has 
of these coins arranged in a line from left to right. He repeatedly performs the following operation: if there are exactly 
coins showing , then he turns over the 
th coin from the left; otherwise, all coins show and he stops. For example, if 
the process starting with the configuration 
would be 
, which stops after three operations., let 
be the number of operations before Harry stops. For example, 
and 
. Determine the average value of over all 
possible initial configurations . be a positive integer. Find the number of permutations 
, 
, 
of the
, 
, 
, satisfying
.
such that
cm by 
cm is folded along a straight line, making two opposite vertices coincide. In this way a pentagon is formed. Calculate it's area.
cm by cm is folded along a straight line, making two opposite vertices coincide. In this way a pentagon is formed. Calculate your area.
.
a
Prove that
![[asy]
size(7cm);
pair A=(0,0), B=(9,0), C=(9,3), D=(0,3), E=(5,3), F=(4,0), G=(7.2,-2.4);
draw(A--B--E--D--cycle);
draw(B--C--E--F--G--cycle, blue);
label("A", A, dir(225));
label("B", B, dir(315));
label("C", C, dir(45));
label("D", D, dir(135));
label("E", E, dir(90));
label("F", F, dir(240));
label("G", G, dir(210));
[/asy]](http://latex.artofproblemsolving.com/d/0/c/d0ce2880e9ba765131898f23ae17836edc25a566.png)
is reflected to 
.
,
so ![$$[BCE] = \dfrac{1}{2} \cdot 3 \cdot 4 = 6.$$](http://latex.artofproblemsolving.com/b/9/1/b917138b507c0751d28b136c4b0e2db67692b866.png)
We can also see that ![$$[BEFG] = [ADEF] = \dfrac{1}{2}[ABCD] = \dfrac{27}{2},$$](http://latex.artofproblemsolving.com/e/8/4/e8474e2b843db34a77f67c3014283495bb29885d.png)
so ![$$[BCEFG] = 6 + \dfrac{27}{2} = \boxed{\dfrac{39}{2}}.$$](http://latex.artofproblemsolving.com/9/5/9/9590fb887b86033c519b4de8f2cd03e798c3a90c.png)
(again, refer you to solution above).
.