[/quote]
,
with 
is called historic if 
.
,
are written on a blackboard. The players take turns making moves with Ingrid starting. A move consists of one of the players crossing out a number on the board that has not yet been crossed out. If the product of all currently crossed out numbers is 
after the move, the player whose move it was receives one point, otherwise, zero points are awarded. The game ends after all numbers have been crossed out.
be a positive integer. Given is a subset 
of 
with 
elements. Prove that there exist three elements 
from 
.
be an acute triangle and 
its orthocenter. Let 
.
as points on the parallels through 
to 
and through 
to 
similarly. If 
are positioned around the sides of 
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
and 
be real numbers. Let 
be the matrix with entries ![\[a_{ij} = \begin{cases}1,&\text{if }x_i + y_j\geq 0;\\0,&\text{if }x_i + y_j < 0.\end{cases}\]](http://latex.artofproblemsolving.com/c/0/8/c0856fe7b2b113180659952c4d216a01d8849f64.png)
Suppose that 
matrix with entries 
,
such that the sum of the elements in each row and each column of 
.
.
are collinear, and 
are arbitrary points
be the sum of the squares of the digits of 
.
such that 
.
and circumcircle 
centred at 
,
meets 
intersect at 
of the triangle 
lies on 
are collinear. Prove that 
.
and 
be points on sides 
is cyclic. Let lines 
and 
intersect at point 
and 
,
is the foot of the perpendicular from 
and 
intersect at second point 
different from 
and 
,
for all integers 
.
Y by codyj, Davi-8191, Abdollahpour, itslumi, jhu08, mathematicsy, megarnie, Lamboreghini, Adventure10, Mango247, Rounak_iitr, Shadow6885, NicoN9, and 7 other users
Let 
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 .
Proposed by Giorgi Arabidze, Georgia.
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 .Proposed by Giorgi Arabidze, Georgia.
and 
which are symmetric to 
,
and 
,
.
and 
are parallel.
be points on 
such that 
and 
and 
are similar, then 
and 
are similar. Then 
and 
are similar, then 
and 
are similar. Then 
is a parallelogram. Take 
is parallel to 
and 
.
and 
,
so 
and 
and 
so in the triangle 
we have ![\[\frac{\sin(x-A)}{\sin(x)}=\frac{AC}{AB}\]](http://latex.artofproblemsolving.com/e/0/a/e0a847df323278506b00883a5d252f2c3cc34890.png)
and by the same way ![\[\frac{\sin(y-A)}{\sin(y)}=\frac{AB}{AC}\]](http://latex.artofproblemsolving.com/5/5/3/5538c403fa6f3417655b9b5b8b06d74da013aada.png)
![\[\sin(x-A)\sin(y-A)=\sin(x)\sin(y)\]](http://latex.artofproblemsolving.com/e/8/1/e81c6ab921fae057c0cefcd3f0bd8f39ca0c120c.png)
so 
and therefore ![\[\cos(x+y-2A)=\cos(x+y)\]](http://latex.artofproblemsolving.com/b/7/2/b72b67a7257ec66db6ec4dc99d04a8725c93b053.png)
which gives ![\[x+y=A\]](http://latex.artofproblemsolving.com/3/9/0/3909c3b8e840d629a69eb7da870cb712bb5a74ed.png)
and if ![\[<BRC+<BCR=A\]](http://latex.artofproblemsolving.com/9/9/4/99463ee90671b4a0056d7745b25ba43294524b8b.png)
is cycle
,
,
respectively.
is the double of 
,
is the double of 
.
has the same area as 
and 
respectively.
,
.
be the circumcenter of 
.
,
.
,
.
be the midpoints of 
.
.
lies on circle 
,
with diameter 
.
lie on 
,
and 
are collinear, i.e 
is on 
.
,
and 
are collinear, so by converse of Pascal's Theorem, 
which is precisely circle 
and 
be the midpoints of 
be the intersection between 
and 
.
and 
.
,
and 
are similar, where we get 
(because both 
and finally:
.
is cyclic.
is cyclic as well. This ends the proof.
,
,
.
, 
and 
so the triangles 
and 
are similar hamotetic (with ratio 2) .On the other 
so we have 
lies on the circumcircle of the triangle 
.
be similar to 
.
is similar to 
and we are done!
.
.
.
.
so there exists a homothety mapping 
to 
,
we have 
,
.
.
is 
which clearly lies on the circumcircle.
From similar triangles 
and 
we have 
Combined with the fact that 
we deduce that 
implying 
Let 
We have that 
We then have 
Similarly, 
From sine rule in 
and from sine rule in 
Combining them, we see that![\[\sin^2(\theta - \angle BAC) = \sin^2 \theta \implies \cos (2(\theta - \angle BAC)) + 1 = \cos^2 \theta \\ \implies 2 \theta = \angle BAC \implies \theta = \dfrac{\angle BAC}{2}.\]](http://latex.artofproblemsolving.com/4/6/c/46c111dede65fac5a72358f01f29f3bce84b0360.png)
we have that 
so 
is cyclic, as desired. 
such that 
.
.
.
.
be the intersection of the tangents to ![\[\angle AQC=180-\angle CAQ-\angle C=180-\angle B-\angle C=\angle A=\angle BCT,\]](http://latex.artofproblemsolving.com/4/b/e/4beb5e5549419656924300328427b2dfaa8cb097.png)
so 
,![\[-1=(AX;BC)\overset{B}{=}(AX;TY)\overset{C}{=}(AN';P_{\infty, CT}Q),\]](http://latex.artofproblemsolving.com/7/6/b/76b998ba4911ceec7d11a3c6e7807600d9f9c393.png)
so 
,
,
.
and 
.
and 
,
Therefore, 
be the reflection of 
is a parallelogram. By the definitions of 
.![\[PABM \sim ACBK, \qquad QACN \sim ABCJ.\]](http://latex.artofproblemsolving.com/7/5/1/7516788eef95600fd87f2981c8bc3d5a2a2f2945.png)
Now, letting 
as desired.
,
and thus 
.
.
,
Consider the tangent at 
Hence, 
which finishes.
Observe that from our conditions 
But 
so by SAS 
so 
is cyclic. Hence, 
so 
the reflection of 
and 
are similar, so triangles 
and 
are similar, implying that 
.
is a parallelogram, we should have:![\[ \angle ABT + \angle ACT = \angle ABM + \angle ACN = \angle CAC’ + 180^{\circ} - \angle CAC’ = 180^{\circ} \]](http://latex.artofproblemsolving.com/7/f/5/7f55d990bd0c68210e762ac3473749724d50ab08.png)
Thus providing that 
so then its immediate that 
and similarly 
.
and 
.
by parameterizing 
whicih gives 
.
be 
.
,
,
.
and 
.
and 
.
and 
.
,
.
and 
.
.
which clearly is true so 
.
with the segment 
is the intersection of 
,
.
Therefore 
and we are done.
,
is the 
and 
are collinear, where 
which is the composition of a inversion centered at 
and a reflection over the internal angle bissector of 
.
is the intersection of the parallel through 
is the refletion of 
is the intersection of parallels through 
is cyclic, but it's immediate, since it's a isosceles trapezoid. The result follows simillarly to show the other collinearity. 
are tangent to 
.
goes through the intersection of the 
bisects 
is tangent to 
lies on ![\[ \measuredangle D_APB = \measuredangle D_APC = \measuredangle D_AAC = \measuredangle D_ABA \]](http://latex.artofproblemsolving.com/0/c/3/0c3485b72baef54123f135b58d6ccf3ba9f1c1d0.png)
as desired.
Let 
and we use the usual notation 
Then, we clearly have 
so it results that
Therefore, we have 
and, analogously, 
Now, reflecting 
Therefore, it results that the intersection of the cevians 
Now, we see that
which implies that the intersection of 
as desired. 
