AoPS Academy
at 
and 
meet at 
. The circumcircle of 
meets sides 
and 
again at 
and 
respectively. Let 
be the circumcentre of . Show that 
is perpendicular to 
.
be an integer greater than zero. Then, the value of the sum of the reciprocals of the cubes of the roots of the equation![\[
mx^4 + 8x^3 - 139x^2 - 18x + 9 = 0
\]](http://latex.artofproblemsolving.com/a/3/a/a3a9bca412e927632e4921c51f2a6593409031d9.png)
is equal to:
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
+1 w
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
be triangle, inscribed in parabola. Tangents in points 
forms triangle . Prove that 
. (
is area of triangle 
).
such that the equality 
holds for all pairs of real numbers 
.
, in which 
: 
, but the proof is important.
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
be an acute triangle with circumcircle 
. Let 
be the midpoint of and let 
be the midpoint of . Let be the foot of the altitude from 
and let 
be the centroid of the triangle . Let 
be a circle through and that is tangent to the circle at a point 
. Prove that the points 
and 
are collinear.
, we have 
, 
, and 
centimeter. Find the length of diagonal .
, we have , , and 
centimeter. Find the length of diagonal .
, we have , , and centimeter. Find the length of diagonal .
, the point can be any point on some circle.
, we have , 
, and centimeter. Find the length of diagonal ., the point can be any point on some circle.
; any point onto the bigger arc 
has 
, that is, 
is cyclic, wherefrom 
.
so
But the LHS has the same sign as that of
so that 
.
such that 
is isosceles with angle 
=50° Then notice that 
is cyclic and 
=90° and then get that is the center of the circumscribed circle. From here it is obvious that is the radius 
(the case 
is analogous). Let the perpendicular bisector of 
intersect at 
(this is due to in ). Since 
is isosceles, the perpendicular bisector will pass through . We now have 
and 
, so 
. Also, 
and if we let 
, we will get 
, but 
too, so 
is cyclic. Now we have 
, so is isosceles and 
. Now in the case 
we directly get that is a deltoid, so 
, while 
and so .
is an obvious solution. Now, let 
.
or 
.
.
, we have 
hence 
are cyclic but 
li e on the 
implies 