[/quote]
,
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 
be positive integers such that ![$\sqrt[n]{a}+\sqrt[n]{b}$](http://latex.artofproblemsolving.com/d/a/a/daaf95cd3381060559529aa4a318e97b727eac31.png)
is an integer. Prove that 
are both the 
t
positive integers.
are written on the board.In each step,two random numbers 
and 
are chosen and deleted.Then,the number 
is written instead.What will be the number remained on the board after the last step.
such that 
is a prime number and 
.
is irreducible over ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
such that![\[ \frac {2^n + 1}{n^2}
\]](http://latex.artofproblemsolving.com/9/2/2/922444e7669c326efcad5fde3ea00f8c2d9fa5d8.png)
is an integer.
called ‘nice’ if and only if 
and 
are both perfect squares. How many positive integers 
such that 
are composites for all nice numbers 
satisfies the conditions![\[\begin{cases}f(x+24)\le f(x)+24\\ f(x+77)\ge f(x)+77\end{cases}\quad\text{for all}\ x\in\mathbb{R}\]](http://latex.artofproblemsolving.com/5/5/6/556f545540d4941d058a1c4099b4b86eb35caf0d.png)
for all real 
.
such that
for all 
.
,
of rational numbers, where 
are relatively prime positive integers for each 
such that the positive integers 
are all distinct.
![\[A = \left\{1+\frac{1}{k} : k=1,2,3,4,\cdots \right\}.\]](http://latex.artofproblemsolving.com/3/d/f/3dfb1d52835627a2977f0074f8df7e7ed677f2a5.png)
can be written as the product of one or more elements of 
,
of integers with 
,
,![\[f(xy)<f(x)+f(y).\]](http://latex.artofproblemsolving.com/c/f/f/cff5554b8e72d99f8989448c2621794242d784d8.png)
(Pairs 
and 
are different if 
or 
)
denote the set of natural numbers, and let 
,
,
.
whose product is a perfect cube.
prove that the triangle must have a right angle.
,
and 
intersect at 
and 
intersect again at 
. If 
, show that 
.
be a positive integer. A graph on 
vertices is given such that the size of the largest clique in the graph is 
Y by anantmudgal09, Understandingmathematics, Adventure10, Mango247, and 1 other user
We are given an acute-angled triangle 
and a random point 
in its interior, different from the centre of the circumcircle 
of the triangle. The lines 
and 
intersect for a second time in the points 
and 
respectively. Let 
and 
be the points that are symmetric of and in respect to 
and 
respectively. Prove that the circumcircle of the triangle and passes through a constant point that does not depend on the choice of .
and a random point 
in its interior, different from the centre of the circumcircle 
of the triangle. The lines 
and 
intersect for a second time in the points 
and 
respectively. Let 
and 
be the points that are symmetric of and in respect to 
and 
respectively. Prove that the circumcircle of the triangle and passes through a constant point that does not depend on the choice of .
.
is the orthocenter of 
.
lies on 
and so. 
be the anticomplementary triangle of 
is the line through 
and parallel to 
,
are isogonl conjugates wrt 
.
concurr at some point 
.
lie on the circle with diameter 
.
this is called Hagge circle look it up.
'
is equivalent to trigonometric ceva.
(*).
is in the circumcircle of 
.
.
are collinear. We need to show that 
are collinear.
we are left to prove 
,
and vary 
notice that the maps 
and 
are projective, so 
is also projective. Note that 
and 
so for 
such that 
are concyclic, inversion at 
is also projective.
for three choices of 
work, so we are done! 
,
,
,
respectively. Now define 
,
,
are the midpoints of the sides of 
,
,
meet the circumcircle again at 
,
,
,
be a point lying on the circumcircle and 
its reflection through 
be the nine-point center of 
and line 
are symmetric with respect to 
of 
(where 
is the circumradius). Let 
be the image of 
maps 
.
,
and 
and the midpoint of 
belongs to the nine-point circle, it follows easily what was wanted. So we have 
,
.
is a rhombus and 
.
going through 
be the reflection of 
and 
.
,
.
passes through 
.
as the unit circle. I'm sure one can give a version of this preliminary argument without complex numbers with rotations, translations etc, but its a lot easier to encode this way.
.
by standard complex number formulas. Thus, 
,
.
are concylic. We have that 
,
are on the unit circle, we see that 
and 
,
all concur at the isogonal conjugate of 
concur at some point 
be the reflection of some point 
,
are concyclic.
that is the image of 
.
![[asy]
unitsize(2.5inches);
pair A=dir(115);
pair B=dir(220);
pair C=dir(-40);
pair H=orthocenter(A,B,C);
pair P=0.27*A+0.35*B+0.38*C;
pair D=2*foot(H,B,C)-H;
pair A1=2*foot(0,A,P)-A;
pair X=foot(A,B,C);
pair Y=foot(B,A,C);
pair Z=foot(C,A,B);
pair A2=2*foot(A1,B,C)-A1;
pair A3=extension(Y,Z,A2,H);
pair U=extension(A,H,Y,Z);
draw(circumcircle(A,B,C));
draw(A--D);
draw(A--B--C--cycle);
draw(C--Z);
draw(B--Y);
draw(A--A1);
draw(A1--A2);
draw(A2--A3);
draw(Y--A3);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$H$",H,dir(125));
dot("$P$",P,dir(0));
dot("$D$",D,dir(D));
dot("$A_1$",A1,dir(A1));
dot("$X$",X,dir(X));
dot("$Y$",Y,dir(Y));
dot("$Z$",Z,dir(140));
dot("$U$",U,dir(145));
dot("$A_2$",A2,dir(45));
dot("$A_3$",A3,dir(A3));
[/asy]](http://latex.artofproblemsolving.com/2/0/d/20d39d915b6bcc08049ad0853ea182a242270005.png)
.
,
,
,
,
similarly. Finally, let 
,
and 
similarly.
.
are reflections over 
,
.
is 
,
.
.
by Ceva, so it suffices to show that 
.![\[(YZ;UA_3)\stackrel{H}{=}(B,C;X,HA_2\cap BC)=(B,C;X,DA_1\cap BC)\stackrel{D}{=}(BC;A_1A).\]](http://latex.artofproblemsolving.com/f/b/9/fb949f38d2314dc202e64dbbce5762b988393514.png)
It's easy to see that the concurrence of 
,
is equivalent to 
(say using trig Ceva), so we're done.
and coefficient 
,
since if 
,
and 
.
.
;
the orthocenter of 
and animate 
Observe the maps 
and 
are projective, implying the map 
is projective. Note that 
is cyclic, say, by angle chasing, and analogously, 
and 
are also cyclic. Consequently, by second intersection of circles, it follows the map 
is projective. It suffices to check these maps concur for three positions of 
and 
being easy.
lies on 
.
are collinear, so we will use Menalaus.
where the last equality is by Ceva.
.
in homogenous coordinates. We will say "degree" to denote the (upper bound) of the degree of homogenous coordinates (in terms of 
,
,
)
,
,
is degree 
.
and 
denote the circle points, and redefine the circumcircle of 
as the circumconic 
.
)
and 
,
over 
)
)
)
,
)
,
is cyclic.
)
)
,
,
,
,
denote the circumcenter, incenter, and three excenters, as usual.
)
,
,
.
.
)
,
)
,
,
.
,
,
.![\[ a_2=y^2+yz+z^2, \quad b_2=z^2+zx+x^2, \quad c_2=x^2+xy+y^2. \]](http://latex.artofproblemsolving.com/d/e/8/de8dae39e53ccc7eb7455a3a2045fb519da8baf8.png)
Let 
denote the incenter and 
denote the orthocenter. We want to show 
,
,
,
are concyclic.
to those four points. We want to show that 
are concyclic. This follows since each point has magnitude 
.
)
,
)
.
)
.
in place of 
.
in terms of 
,
,
,
,
is 
.![\[ \det \begin{bmatrix} aa_1 & a+a_1 & 1 \\ bb_1 & b+b_1 & 1 \\ cc_1 & c+c_1 & 1 \end{bmatrix} \overset{?}= 0. \tag{1} \]](http://latex.artofproblemsolving.com/4/2/1/421535c2e95219ec4ade5fe7782e67b6c1f4ee69.png)
Now we tackle condition (2). Let 
denote the orthocenter. The goal is to show 
,
,
,![\[ \varphi \colon z \mapsto \overline{(h-z)/abc}, \]](http://latex.artofproblemsolving.com/1/c/3/1c393b5e4dbe43e97e508e0dedf792dc6c62bf04.png)
which preserves concyclicity. We have ![\[ \varphi(a_2) = bc + aa_1, \quad \varphi(b_2) = ca + bb_1, \quad \varphi(c_2) = ab + cc_1, \quad \varphi(h) = 0. \]](http://latex.artofproblemsolving.com/7/f/0/7f09d6a3a3d0d50ba7de8dffaf58a6f82378f36e.png)
Label the first three points above by 
,
,
if and only if 
,
,
are collinear (invert around the unit circle). As such, we want to show ![\[ 0 \overset{?}= \det \begin{bmatrix} 1/\alpha & 1/\overline{\alpha} & 1 \\ 1/\beta & 1/\overline{\beta} & 1 \\ 1/\gamma & 1/\overline{\gamma} & 1 \end{bmatrix} = - \det \begin{bmatrix} \alpha & \overline{\alpha} & \alpha\overline{\alpha} \\ \beta & \overline{\beta} & \beta \overline{\beta} \\ \gamma & \overline{\gamma} & \gamma\overline{\gamma} \end{bmatrix}. \]](http://latex.artofproblemsolving.com/b/0/b/b0b4acf423cce6726ff0554f35df7fb0c5e2da02.png)
Since 
and so on, the above is equivalent to ![\[ 0 \overset{?}= \frac{\alpha \beta \gamma}{a_1b_1c_1(abc)^2} \cdot \det \begin{bmatrix} \alpha & a_1 & 1 \\ \beta & b_1 & 1 \\ \gamma & c_1 & 1 \end{bmatrix}. \tag{2} \]](http://latex.artofproblemsolving.com/b/6/f/b6f9c13c6eb25fe1becc24cc18ff6e6e539dbaad.png)
We will show that the determinant on the right is actually equivalent to (1). This will finish the problem, as the denominator on the left is nonzero.![\begin{align*} \det \begin{bmatrix} bc+aa_1 & a_1 & 1 \\ ca+bb_1 & b_1 & 1 \\ ab+cc_1 & c_1 & 1 \end{bmatrix} &= \sum_\text{cyc} (aa_1b_1 + bb_1c - aa_1c - bb_1a_1) \\ &= \sum_\text{cyc} aa_1 \bigl[(b+b_1) - (c+c_1)\bigr] \\ &= \det \begin{bmatrix} aa_1 & a+a_1 & 1 \\ bb_1 & b+b_1 & 1 \\ cc_1 & c+c_1 & 1 \end{bmatrix}, \end{align*}](http://latex.artofproblemsolving.com/8/e/2/8e2904d0f95fa8bdb7860b05aad5435e9c193545.png)
done. 
