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.
. Prove that: 
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
be the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
be non-negative real numbers such that 
Prove that
are positive real numbers such that 
, Prove that ![\[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \]](http://latex.artofproblemsolving.com/0/c/e/0ce097b0b507403c959c2992cae1217ecdd7e581.png)
be a finite sequence of positive integers. Prove that there exist nonnegative integers 
and 
such that 
holds for all integers 
be a positive integer. Dominoes are placed on a 
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each , determine the largest number of dominoes that can be placed in this way.
or 
. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
be a scalene triangle with orthocenter 
and circumcenter 
. Denote by 
, the midpoints of 
, 
. Suppose the circle 
with diameter meets the circumcircle of at 
, and meets line 
at a point 
. The tangent to at 
meets line 
at 
. Show that the circumcircles of 
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
be a cyclic quadrilateral, and angle bisectors of 
and 
meet at point 
. Show that if 
, then is the incenter of triangle 
.
![[asy]
unitsize(3cm);
draw(circle((0,0),1));
draw(dir(160) -- dir(70));
draw(dir(20) -- dir(140));
draw(dir(270) -- dir(120));
draw(dir(90) -- dir(180));
draw(dir(180) -- dir(20) -- dir(120) -- dir(160) -- dir(180));
label("$A$",dir(120), N);
label("$B$",dir(160), W);
label("$C$",dir(180), W);
label("$D$",dir(20), E);
label("$M$",dir(140), W);
label("$N$",dir(70), N);
label("$P$",dir(90), N);
label("$Q$",dir(270), S);
draw(dir(160) -- dir(20));
label("$I$", dir(130)*0.5, N);
[/asy]](http://latex.artofproblemsolving.com/2/0/d/20da2cef0478c7954a6bbd51e86bc9e2714a15f6.png)
intersect the circle at 
respectively. Start by doing some angle chasing:
Then
Hence
On the other hand, by similarity, we have
Because 
bisects 
Multiply these equations:
If 
then 
Because 
bisects , 
Hence 
Similarly we can show that it is impossible that 
Thus 
and is the incenter.
![[asy]
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draw((-4.782,-1.24)--(6.338,1.04), linewidth(0.4) + red);
draw((6.338,1.04)--(2.284,6.278), linewidth(0.4) + red);
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draw((-0.5576361412228421,6.4141552150868435)--(5.492817515943216,-2.2250378477853023), linewidth(0.4));
draw((-4.782,-1.24)--(5.1414961079806005,4.301770374523359), linewidth(0.4));
draw((6.338,1.04)--(-3.5909877249415154,4.7201818654478265), linewidth(0.4));
draw((6.338,1.04)--(2.108144524261927,2.607797024884341), linewidth(0.4) + blue);
draw((2.108144524261927,2.607797024884341)--(5.492817515943216,-2.2250378477853023), linewidth(0.4) + blue);
draw((5.492817515943216,-2.2250378477853023)--(6.338,1.04), linewidth(0.4) + blue);
draw(circle((3.5539946019787774,0.018752974104690016), 2.965405797544633), linewidth(0.4) + blue);
/* dots and labels */
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label("$A$", (2.4273798447351713,6.626034878135939), NE * labelscalefactor);
dot((-4.782,-1.24),dotstyle);
label("$B$", (-4.645923878631826,-0.8781299857407152), NE * labelscalefactor);
dot((6.338,1.04),dotstyle);
label("$D$", (6.484655584534515,1.3838910019350228), NE * labelscalefactor);
dot((2.108144524261927,2.607797024884341),linewidth(4pt) + dotstyle);
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dot((-0.5576361412228421,6.4141552150868435),linewidth(4pt) + dotstyle);
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dot((5.492817515943216,-2.2250378477853023),linewidth(4pt) + dotstyle);
label("$C$", (5.622933303515185,-1.955282837014876), NE * labelscalefactor);
dot((5.1414961079806005,4.301770374523359),linewidth(4pt) + dotstyle);
label("$P$", (5.299787448132936,4.579444460715034), NE * labelscalefactor);
dot((-3.5909877249415154,4.7201818654478265),linewidth(4pt) + dotstyle);
label("$Q$", (-3.461055742230247,5.010305601224698), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/f/d/1/fd19737339dfd20ee8163d32d1148fd1f4a45887.png)
to be the intersection of the circle 
and line 
, by this way we have that is the midpoint of the arc 
.Also let 
be the intersection of 
with the circle and let be the intersection of 
with ., this implies that the circumcircle of 
is tangent to . Thus we have that:
this implies that 
, and since we have that is on the angle bisector of , this implies that is the incenter of 
.
and 
then prove 
and 
.
(maybe at infinity)
we get the desired result.