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.
of each imon 
. During this procedure, the two copies and 
become entangled if and only if the original imons and 
are entangled, and each copy becomes entangled with its original imon ; no other entanglements occur or disappear at this moment.
members from party 
, 
members from party 
, 
members from party 
, and 
members from party 
, keeping them isolated in individual rooms in his castle, where he forces them to work.
or more people involved).
that if they were to be considered the number of members from the four parties, it is possible that, after some time, all of the kidnapped people belong to the same party, under the same rules imposed by the wizard.
at and meet at . The circumcircle of 
meets sides 
and 
again at 
and 
respectively. Let 
be the circumcentre of . Show that 
is perpendicular to 
.
for which 
is the square of an integer
of natural numbers with elements with odd number. A nonempty subset 
of is called 
if the product of the elements of divisible by the sum of the elements of , but not by its square. It is known that the set itself is good. Determine the maximum number of subsets (including itself). make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
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.
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 
.
satisfy 
; in other words, let be an isosceles triangle with base . Let be a point inside the triangle such that 
. Denote by the midpoint of the segment . Show that 
.
moves on a circle tangent in 
to 
respectively. Let be the second intersection of 
with this circle. 
is a harmonic cyclic quadrilateral, so, if 
is the intersection between the tangent at to the circle and , we find 
, meaning that 
is the symmedian of 
, which is precisely what we want.
satisfy ; in other words, let be an isosceles triangle with base . Let be a point inside the triangle such that . Denote by the midpoint of the segment . Show that .
, 
, 
, 
, 
,
, 
.
The theorem of Sinus in the triangle 
. The trigonometrical form of the Menelaus' theorem in for the inner point 
.
, 
, 
results: 
, i.e. the ray 
is in the triangle 
, meaning
be a triangle and be the midpoint of . Suppose is the intersection of the tangents of the circumcircle at and . Then 
.
and 
are the tangents of 
at and .
). 
is isosceles 
(U) is also tangent to CA at A, AB is polar of C WRT the circle (U). CP meets (U) at P and again at Q. If P' is a reflection of P in 
and CP' meets (U) again at Q'. PP'Q'Q is isosceles trapezod symmetrical WRT CM, the diagonals PQ', P'Q meet on CM and on the polar AB of C, 
. Let 
be on 
such 
is parallogram. Then 
. Hence 
are isogonal points wrt 
. Hence 
is isogonal line . This imply 
.
be a triangle and 
its circumcircle. Let the tangent to at and intersect at . Then 
coincides with the - symmedian of . The proof can be found here. and are tangents to the circumcircle of 
at points and respectively. The tangents and intersect at . So 
coincides with the - symmedian of . Since 
is the median of , 
. or 
. So 
.
is the symmedian.
. Then the problem is equivalent to 
, aka. 
is a symmedian in 
. Notice that and are the tangents to 
so we are done.
is the humpty point, and so 
is the midpoint of . Furthermore, we have that 
and 
. Hence, we get that 
where this follows from similar triangles 
and 
.
meet circumcircle of at S. Since then is tangent to and since 
then 
is symmedian of 
so 
and so 
.
. The angle condition implies that and are tangent to the circumcircle of , then we know that 
is the -symmedian of , which implies that 
thus we have that 
, which clearly gives us that 
be a symmedian in triangle 
, so it suffices to show that lies along the symmedian. But this is evident by the tangent intersection definition.
is an obvious solution. Now, let 
.
or 
.
.
or on the line 
.
be the point of intersection of the circumcircle of the triangle 
with line 
(Fig. 1 and 2). Since 
,
lie on a circle. Also![\[(1)\qquad\angle BPC=\angle BQC=\angle AQC.\]](http://latex.artofproblemsolving.com/a/9/8/a988a64bff438e45b57a05249ab94c426c3398f4.png)
![\[\angle APM+\angle BPC=\angle APM+\angle AQC=180^\circ.\]](http://latex.artofproblemsolving.com/2/5/9/259d348a3417b67c0faa0548f45e32447269a7a7.png)
![\[\angle APM+\angle BPC=\angle AQM+\angle AQC=180^\circ.\]](http://latex.artofproblemsolving.com/9/c/f/9cfac1d50178272a49af4608a3d80936c833cb06.png)
. Since 
, so by analogy with the above, we obtain the equality 
.![\[\angle APM+\angle BPC=180^\circ,\]](http://latex.artofproblemsolving.com/9/6/1/9618056d4f538976f67525c7e20375d9c00e9972.png)
.
.
.
,
and 
Hence, 
is the 
Notice that 
,
,
and 
.
So
Therefore 
,
,
is cyclic and 