[/quote]
,
,
its shortest side. Let 
be the circle with center 
and radius 
.
meets 
.
be the intersection point of the lines 
and 
is tangent to the circumcircle of the triangle 
.
and circumcircle 
centred at 
meets 
;
intersect at 
;
and 
intersect at 
.
of the triangle 
lies on 
are collinear. Prove that 
.
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
is composite. 
- all divisors of 
are all divisors for some 
(except 
)
be a positive integer, with divisors 
.
is always less than 
,
we consider the set of ten circles, each of which contains three of the vertices of the pentagon on its circumference. Is it possible that none of these circles contains the pentagon? Prove your answer.
is tangent to the circle 
at 
is the in-centre of 
,
.
and 
and 
intersect at 
lie on a circle.
to 
.
as points on the parallels through 
to 
are positioned around the sides of 
and 
and 
are the reflections of 
are the intersections of 
and 
respectively with the circumcircle of 
and 
lies on 
.
,
.
in 
respectively and the line through 
in 
respectively. Prove that point 
lie on a circle.
kids in a class. Is it true that for all positive integers 
it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
be an integer. In a configuration of an 
board, each of the 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
lamps 
in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp 
and its neighbours (only one neighbour for 
or 
,
)
Prove that there are infinitely many integers 
for which all the lamps will eventually be off.
Prove that there are infinitely many integers 
Y by LuoJi, Rounak_iitr
The triangle 
has 
as its circumcenter, and 
as its orthocenter. The line 
and 
intersect the circumcircle of for the second time at points 
and 
, respectively. Let 
and 
be the circumcenters of triangle 
and 
respectively. If 
are not collinear, prove that 
intersects the midpoint of segment 
.
has 
as its circumcenter, and 
as its orthocenter. The line 
and 
intersect the circumcircle of for the second time at points 
and 
, respectively. Let 
and 
be the circumcenters of triangle 
and 
respectively. If 
are not collinear, prove that 
intersects the midpoint of segment 
.
is a parallelogram. Spiral sim at 
to 
and spiral sim at 
to 
gives 
clearly the two expressions are equal hence 
similarly we get opposite pair of equal sides yay.
motivate to use complex 
an
dan 
(imply 
by 
property). Because 
we have 
and 
.
:![\[|a'-a| = |a'-h| = |a'-e|,\]](http://latex.artofproblemsolving.com/e/9/1/e91a86dc7cb188c46dd0518b5e4c055cd6d3ba5a.png)
we have
Analogously, 
Therefore, the midpoint of ![\[m = \frac{a'+b'}{2} = \frac{1}{2}\left (\frac{-a^2-ac +b^2+bc}{b-a} \right )=\frac{(b-a)(a+b+c)}{2(b-a)}=\frac{a+b+c}{2},\]](http://latex.artofproblemsolving.com/2/4/2/24294137082e8e162ca95a4664b0447a55df0379.png)
which is also the midpoint of 
is the radical axis of 
and 
,
.
So, 
as well. Therefore, 
.
,
be midpoint of 
,
and 
similarly. We are going to use Barycentric Coordinates.
.
and 
.
.
and similarly 
.
.
is point of infinity of line ![\[
\begin{vmatrix}
x & 0 & z \\
0 & 1 & -1\\
S_bS_c + T & S_cS_a & S_aS_b
\end{vmatrix}
= 0
\]](http://latex.artofproblemsolving.com/4/6/0/46048d18efde6b79020101b3ee0011ef9ff74c5f.png)
.
.
be the midpoint of 
.
are collinear (and in fact ![\[
\begin{vmatrix}
S_bS_c & S_cS_a & S_aS_b \\
b^2S_b + c^2S_c & c^2S_c + a^2S_a & a^2S_a + b^2S_b\\
a^2S_a & b^2S_b & c^2S_c
\end{vmatrix}
= \begin{vmatrix}
S_bS_c & S_cS_a & S_aS_b \\
a^2S_a + b^2S_b + c^2S_c & c^2S_c + a^2S_a + b^2S_b & a^2S_a + b^2S_b + c^2S_c\\
a^2S_a & b^2S_b & c^2S_c
\end{vmatrix}
= 2T \begin{vmatrix}
S_bS_c & S_cS_a & S_aS_b \\
1 & 1 & 1\\
S_aS_b + S_aS_c & S_aS_b + S_bS_c & S_aS_c + S_bS_c
\end{vmatrix}
= 2T \begin{vmatrix}
S_bS_c & S_cS_a & S_aS_b \\
1 & 1 & 1\\
T & T & T
\end{vmatrix}
= 0
\]](http://latex.artofproblemsolving.com/e/4/b/e4b5728b3b7fbc5be5fb5ba06a4d81392e6ce829.png)
which is evident from the last determinant.
be the point on 
,
.
is parallel to 
,
is the point on 
is parallel to 
points: 
is a parallelogram, so the midpoint of 
is equal to the midpoint of 
.
.
)
.
is a parallelogram. To prove that, we can simply prove that 
is congruent to 
.
,
and 
are isosceles)
is acute, the obtuse one can be proved similarly.
and 
are isosceles triangles and also 
and similarly 
and 
are cyclic quads.
and 
,
and 
,
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/3/7/d/37df5ad1685a0113f47206c9125a81a0e5171916.png)
is a parallelogram, since the diagonals of a parallelogram intersect
,
,
as the midpoint of 
and 
.![\[OB'^2 = ON^2 + {(AN - AB')}^2 = ON^2 + AN^2 - 2AN \cdot AB + AB'^2 = R^2 - 2AN \cdot AB' + AB'^2 \]](http://latex.artofproblemsolving.com/4/7/7/477fca7cb9a87e7c65465d3f30af00cdca077deb.png)
We shall calculate the value of 
,
,
is equal to that of 
,
![\[\frac{BH}{\sin(90 - \beta)} = 2R \Longrightarrow 2R\cos(\beta) = BH \]](http://latex.artofproblemsolving.com/9/7/c/97c549d7fbc1af54db3ee07e47261575b8d42d82.png)
We know that the circumradius of 
is equal to 
,![\[A'B = \frac{BH}{2\sin(\angle BDH)} = \frac{BH}{2\sin(\gamma)} = \frac{R\cos(\beta)}{\sin(\gamma)}\]](http://latex.artofproblemsolving.com/c/e/0/ce0c0585a8196f8db6c998ca8c8be951f8e3d13e.png)
Similarly we find that![\[AB' = \frac{R\cos(\alpha)}{\sin(\gamma)}\]](http://latex.artofproblemsolving.com/1/9/5/1957259d428a542e4468e4c33642af406d76b012.png)
We wish to show that 
,![\[A'B^2 = OB'^2 \iff \frac{R^2\cos^2(\beta)}{\sin^2(\gamma)} = R^2 - 2AN \cdot AB' + AB'^2\]](http://latex.artofproblemsolving.com/b/e/1/be1fedfb4d34781abc1d48df9cce5a3e567e1a72.png)
We know that![\[AN = \frac{AC}{2} = R\sin(\beta)\]](http://latex.artofproblemsolving.com/f/d/c/fdc9b1741c5f0ad5394e9a7568ab56f7b7daf299.png)
Thus we need to show![\[\frac{R^2\cos^2(\beta)}{\sin^2(\gamma)} = R^2 - 2(R\sin(\beta)) \cdot \left(\frac{R\cos(\alpha)}{\sin(\gamma)}\right) + {\left(\frac{R\cos(\alpha)}{\sin(\gamma)}\right)}^2\]](http://latex.artofproblemsolving.com/0/d/c/0dc387df99deec389c7f8ae28932b3d6935ac101.png)
which is equivalent to showing![\[\frac{\cos^2(\beta)}{\sin^2(\gamma)} = 1 - \frac{2\sin(\beta)\cos(\alpha)}{\sin(\gamma)} + \frac{\cos^2(\alpha)}{\sin^2(\gamma)} \iff \cos^2(\beta) = \sin^2(\gamma) - 2\sin(\beta)\sin(\gamma)\cos(\alpha) + \cos^2(\alpha)\]](http://latex.artofproblemsolving.com/f/9/a/f9ae8f987cda0feb6e6b5ba657a4f416d9682878.png)
We can sub 
to show that it suffices to prove![\[\cos^2(\beta) - \cos^2(\alpha) = [\sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)][\sin(\alpha)\cos(\beta) - \sin(\beta)\cos(\alpha)]\]](http://latex.artofproblemsolving.com/8/e/9/8e9f1376f8730e5a4d6c8ae1387568e6adbaf6e0.png)
But this is true since it is equivalent to![\[\cos^2(\beta) - \cos^2(\alpha) = \sin^2(\alpha)\cos^2(\beta) - \sin^2(\beta)\cos^2(\alpha)\]](http://latex.artofproblemsolving.com/1/3/e/13e6d50e82caeb5c94fc17f238ef2a18a2066af2.png)
Which is equivalent to![\[\cos^2(a)\cos^2(\beta) = \cos^2(\beta)\cos^2(\alpha)\]](http://latex.artofproblemsolving.com/6/1/8/61850f182ee796454f8098e49914edfb92c1db09.png)
