[/quote]
,
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
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 
.
and 
and 
then prove 
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
and 
.
,![\[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \]](http://latex.artofproblemsolving.com/0/c/e/0ce097b0b507403c959c2992cae1217ecdd7e581.png)
be a finite sequence of positive integers. Prove that there exist nonnegative integers 
and 
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 
or 
.
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 
.
and circumcenter 
.
,
.
with diameter 
,
at a point 
.
meets line 
at 
.
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
Y by samrocksnature, The_Turtle, BOBTHEGR8, mathematicsy, Jupiter_is_BIG, centslordm, megarnie, Mango247, Rounak_iitr
Let 
be a triangle with an obtuse angle at 
. Let 
and 
be the intersections of the external bisector of angle with the altitudes of through 
and 
respectively. Let 
and 
be the points on the segments 
and 
respectively such that 
and 
. Prove that the points 
lie on a circle.
be a triangle with an obtuse angle at 
. Let 
and 
be the intersections of the external bisector of angle with the altitudes of through 
and 
respectively. Let 
and 
be the points on the segments 
and 
respectively such that 
and 
. Prove that the points 
lie on a circle.
.
be the feet of altitudes from points 
to the lines 
respectively. Define point 
.
.
to the picture, which is the intersection of 
and 
.
But quadrilateral 
is cyclic, therefore we have 
.
is cyclic.
.
to segment 
,
to 
.
to 
.
.
and 
are cyclic.
On the other hand 
,
.
and 
are cyclic.
But 
.
are cyclic, we simply see that six points 
lie on one circle.
.
.
and 
.
:
.
.
so we must have 
Let 
be the reflection of 
.
so 
collinear. Notice the cyclic quadrilaterals: 
and 
.
concur) 
cyclic 
cyclic and we're done by angles.
since 
Let 
cut circle 
,
.
and 
.
cyclic 
and 
and let 
are isogonal lines and so are 
in 
.
are also isogonal but as 
are isogonal, we have 
are collinear. Now, observe that 
and
.
Thus, 
and we are done.
at 
,
at 
where 
where 
be line through 
,
,
and so 
.
and 
,
,
.
and so 
are concyclic. But we also have 
and so 
are concylic. Now radical axis theorem gives 
are concyclic. Similarly, 
are concylic as well, and so 
are concyclic.
concurrent at say 
.
and hence 
is cyclic.
is cyclic and hence 
are concyclic.
and 
,
be the midpoint of 
is concurrent with 
,
,
at a point ![[asy] import graph; size(9.98895cm); pen ffxfqq = rgb(1.,0.49803,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pair H = (0.2,3.2), B = (-1.,0.2), C = (2.2,0.2), A = (0.2,1.), Y = (-0.55862,1.30344), Z = (1.21538,1.67692), F = (1.57146,1.14279), Q = (-0.68,2.76), M = (0.87710,0.51931); draw(H--B--C--cycle, linewidth(0.6) + zzttqq); draw(circle((0.34465,1.81068), 1.39682), linewidth(0.6) + ffxfqq); draw(circle((0.2,2.1), 1.1), linewidth(0.6) + yqqqqq); draw(circle((0.6,1.3), 1.94164), linewidth(0.6) + yqqqqq); draw(H--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--H, linewidth(0.6) + zzttqq); draw(B--Z, linewidth(0.6)); draw(C--Y, linewidth(0.6)); draw(B--F, linewidth(0.6) + blue); draw(C--(-0.71824,0.90439), linewidth(0.6) + blue); draw((0.6,0.2)--Q, linewidth(0.6) + green); draw(circle((-1.,1.5), 1.3), linewidth(0.6) + qqwuqq); draw((-0.71824,0.90439)--F, linewidth(0.6) + yqqqqq); dot("$H$", H, dir((-5.899, 6.902))); dot("$B$", B, dir((-8.675, -9.665))); dot("$C$", C, dir((5.858, -4.103))); dot("$A$", A, dir((1.801, 3.594))); dot("$Y$", Y, dir((-13.028, -3.649))); dot("$Z$", Z, dir((6.784, -1.212))); dot("$E$", (-0.71824,0.90439), dir((-10.327, -2.673))); dot("$F$", F, dir((7.966, -3.412))); dot("$Q$", Q, dir((-8.162, 8.978))); dot("$N$", (-0.15969,0.50808), dir((1.837, -9.672))); dot("$W$", (0.6,0.2), dir((-1.409, -11.804))); dot("$M$", M, dir((-1.312, -9.939))); [/asy]](http://latex.artofproblemsolving.com/b/2/f/b2f7e79e9656b1981aada6b0dedbde46882f9450.png)
,
is tangent to 
.![\[ \measuredangle QNF = \measuredangle QNB = \measuredangle QAB = \measuredangle QAZ = \measuredangle QHZ = \measuredangle QHF \]](http://latex.artofproblemsolving.com/3/8/d/38d1c5eea5adc40e8246617dacfe5e4143bf8b72.png)
and hence 
.
be 
,
,
.
be the feet of the 
-
be the midpoint of ![[asy]
import olympiad;
unitsize(1.7inch);
pair A, B, C, H, S, X, M, N, E, F, P, Q, R, I, J, Y, Z, O, L, n, m;
A = dir(120); B = dir(210); C = dir(330); I = dir(270);
P = foot(A, B, C); Q = foot(B, C, A); R = foot(C, A, B); H = extension(A, P, B, Q); J = foot(H, A, I);
X = .5B + .5C; S = foot(A, H, X);
E = extension(H, J, A, B); F = extension(H, J, A, C);
Y = foot(C, E, F); Z = foot(B, E, F);
O = circumcenter(Z, H, Q); n = 2 * foot(O, X, Q) - Q; L = circumcenter(Y, H, R); m = 2 * foot(L, X, R) - R;
M = extension(m, H, C, E); N = extension(n, H, B, F);
draw(A--B--C--cycle); draw(circumcircle(A, B, C), dashed+lightblue); draw(circumcircle(A, E, F), dotted+lightblue);
draw(E--C, purple); draw(B--F, purple);
draw(C--R, magenta); draw(B--Q, magenta); draw(A--P, magenta);
draw(Y--Z); draw(B--Z); draw(C--Y);
draw(circumcircle(m, n, X), dotted+lightblue);
draw(X--n, dotted+lightred); draw(X--m, dotted+lightred); draw(X--S, dotted+lightred);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E'$", Y, dir(Y)); dot("$F'$", Z, dir(Z)); dot("$H$", H, dir(120)); dot("$A'$", P, dir(260)); dot("$B'$", Q, dir(50)); dot("$C'$", R, dir(110)); dot("$X$", X, dir(X)); dot("$E$", E, dir(135)); dot("$F$", F, dir(60)); dot("$S$", S, dir(S)); dot("$N'$", n, dir(n)); dot("$M'$", m, dir(m)); dot("$N$", N, dir(255)); dot("$M$", M, dir(M));
[/asy]](http://latex.artofproblemsolving.com/5/1/b/51bec1ff599ca9d1ada36634f0db69658c6495d0.png)
and 
,
Let 
be the second intersection of 
and 
,
to 
.
too. Also note that 
because 
.
and take a 
homothety. 
that is the second intersection of 
and 
,
that is the foot from 
while 
,
and 
cyclic. Then 
,
,
is collinear. Also 
.
is collinear and 
.
is cyclic, 
is cyclic and
Now 
so 
.
and 
,
.
.
,
intersects the circumscribed circle of 
.
be the reflection of 
with 
and 
are both cyclic. This indicates that 
.
,
be the reflexion of 
and 
,
are concyclic.
with 
.
.
,
.
be the points on 
such that 
is cyclic. Let 
be the intersection of 
,
is the internal bisector of 
be 
are collinear. We have angle 
,
.
and thus 
are parallel. Note that the reflection of 
in 
.
and 
be the intersection of 
be the intersection of 
and 
,
due to angle bisectors. We also have 
,
,
,
are collinear, so 
are collinear, as needed.
by the reflection of 
.
and 
are cyclic. Then, we shall be done if we show that 
,
and 
concur. Let 
implying that 
passes through 
,
,
are concurrent due to Jacobi's Theorem giving us the desired conclusion.
Gives me vibe of IMO 2019 P2 somehow. Here's my solution.
and 
.
,
,
and 
as the feet of altitude from 
,
and 
midpoint of 
is cyclic. Notice that 
.
are collinear. Furthermore, let the reflection of 
by the reflection of orthocenter lemma. Therefore, 
.
is tangent to 
.
.
.
.
.
and 
are cyclic, we obtain 
being cyclic as well.
lies on 
and 
respectively. We'll prove that 
and 
collinear. By symmetry, it suffices to prove that 
and we are done.
you must reflect 
are 
concurring in one point. By desargue's theorem in 
we just need to have 
,
concurring. For this, we'll prove that 
,
,
,
,
is a rhombus. 
be foots of 
altitudes in triangle 
and 
.
is midpoint of 
is cyclic. Since 
bisects 
we get 
is cyclic. So 
is spiral similarity 
that maps 
to 
.
is cyclic.
is cyclic, 
.
.
is cyclic, which gives 
in 
Q
.![[asy]
size(10cm);
import olympiad;
void my_dot(pair A) {
fill(circle(A, .01),black);
}
pair H = dir(110);
pair B = dir(200);
pair C = dir(340);
pair Q=foot(B,H,C);
pair P=foot(C,H,B);
pair A=extension(B,Q,C,P);
pair I=incenter(A,B,P);
pair E=extension(A,I,B,H);
pair F=extension(A,I,C,H);
pair S=(1B+1C)/(1+1);
pair L=extension(P,Q,B,C);
pair K=extension(L,H,S,A);
pair N = intersectionpoints(circumcircle(A,B,K), circumcircle(K,E,F))[1];
pair M= intersectionpoints(circumcircle(A,C,K), circumcircle(K,E,F))[1];
draw(H--B--C--H);
draw(B--A--P);
draw(C--A--Q);
draw(E--A--F);
draw(B--F);
draw(C--E);
draw(arc(circumcenter(A,C,K), circumradius(A,C,K), 200, 270),red);
draw(arc(circumcenter(A,B,K),circumradius(A,B,K),270,420),orange);
draw(S--A--K);
draw(A--N);
draw(A--M);
draw(circumcircle(H,B,C),blue);
draw(circumcircle(H,E,F),dotted+darkgreen);
label("$B$", B, dir(150));
label("$A$", A, dir(90));
label("$C$", C, dir(340));
label("$P$",P,dir(170));
label("$Q$",Q,dir(30));
label("$E$",E,dir(160));
label("$F$",F,dir(0));
label("$H$",H,dir(100));
label("$S$",S,dir(270));
label("$K$",K,dir(120));
label("$M$",M,dir(250));
label("$N$",N,dir(300));
my_dot(H);
my_dot(B);
my_dot(C);
my_dot(P);
my_dot(Q);
my_dot(A);
my_dot(E);
my_dot(F);
my_dot(S);
my_dot(K);
my_dot(N);
my_dot(M);
[/asy]](http://latex.artofproblemsolving.com/6/5/a/65abe203542d70e7455c040055e60cc1f5ada3c7.png)
,
concur on 
to 
.
,
Humpty point in 
Hence 
is tangent to 
is tangent to 
and 
are cyclic.
Finally,
Hence 
cyclic, and so 
is cyclic, as desired.
denotes orthocenter of 
Note that by Isogonal theorem 
concur. Furthermore 
implies that 
is cyclic, and analogously for 
Then by radical axis 
is cyclic too.
denote the 
be the midpoint of minor arc 
major arc 
lie on 
.
,
.
be the feet of the altitudes from 
respectively.
centred at 
.
.
be the projections of 
,
is cyclic. Now we have 
,
and analogously 
.
cyclic.
is cyclic.
collinear means 
cyclic. Similarly, 
collinear means 
cyclic. Thus, by PoP 
,
.
.
.
,
.
is the harmonic conjugate of 
,
.
.
and by converse of alternate segment, 
.
.
.
,
is tangent to 
at 
collinear, and 
.
.
.
means 
is cyclic by converse of angles in a cyclic quad. Thus, 
is cyclic. 
,
,
,
,
,
,
,
,
,
.
,
.
,
,
.
,
.
be a point on this line, note that 
,
,
.
,
,
,
,
,
,
.
.
,
.
then we would have 
,
is a parallelogram, so 
,
.
,
,
.
,
,
.
,
,
,
be the feet of the altitudes from 
and 
respectively such that 
externally. Let 
and let 
such that 
.
be the so-called 
and 
are tangent to 
.
,
which is exactly what we want. Similarly, 
.
to 
by angle bisector theorem/similar triangles, by the gliding principle 
and thus 
is cyclic.
hence 
is cyclic as well. Likewise, 
is cyclic, so 
all lie on a circle, as desired. 
be the circle tangent at 
,
,
,
,
and 
(all intersections with 
are collinear.![\[(B,H;N,X)\stackrel{B}{=}(C,P;F,A)\stackrel{\triangle HXB\sim\triangle HAC}{=}(B,Y;E,X)\stackrel{H}{=}(B,Z;Q,X)=(X,Q;Z,B)\stackrel{E}{=}(B,H;N',X).\]](http://latex.artofproblemsolving.com/3/4/7/347ae8d03732dc65f9ba3ecebb7667b9872a7b08.png)
Thus, 
which is symmetric in 
is cyclic. We will prove that 
is cyclic; the proof for 
is analogous.
and 
.
.
and 
are perpendicular, the triangles are similar isosceles triangles. Hence, 
and 
are cyclic. Since ![\[\angle AQE=\angle APE=\angle AHE=\angle ACB=\angle AME,\]](http://latex.artofproblemsolving.com/b/1/6/b16ca92f41e9a76b2bd0dd72403d48f3aafe15fa.png)
so 
is cyclic. Consequently, ![\[CH \cdot CF=CA \cdot CQ=CE \cdot CM,\]](http://latex.artofproblemsolving.com/c/a/3/ca30e01437b748d7d32cf5873d0aa1a2ff447810.png)
so 
is cyclic by symmetrical reasons.
,
by power of a point in 
.
we get 
and similarly 
and 
are cyclic.
are cyclic
are cyclic
lie on same circle
are cyclic 
be the circumcircle of triangle 
and 
denote the midpoints of arcs 
and 
of 
is a parallelogram, which implies that the angle bisectors of 
are parallel. But the latter is the line 
,
,
is a rectangle, so we're done. 
.
,
.
is cyclic since 
as triangle 
is 
is also cyclic; let the circumcircles of the two quadrilaterals be 
and 
,
is a rhombus, 
,
.
.
is parallel to the angle bisector of 
)
externally bisects angle 
is cyclic, and hence that 
.
too.
concur.
,
,
,
and 
,
obviously. Now 
.
,
multiply to 
.
,
.
.![\[\frac{1 + \frac{s^2 + a}{s^2 - 1}}{s - \frac{s(s^2 + a)}{s^2 - 1}} = -\frac{2s^2 + a - 1}{(a + 1)s}\]](http://latex.artofproblemsolving.com/f/8/7/f878f1ecddc1b542c666654c229da5be2ed911ee.png)
and so has equation ![\[y = -\frac{2s^2 + a - 1}{(a + 1)s}(x - s) + 1.\]](http://latex.artofproblemsolving.com/3/9/a/39a933d217f7c2d0e25976fd1c6c717654ee9e17.png)
Similarly, line ![\[y = -\frac{2s^2 - a - 1}{(a - 1)s}(x - s) - 1.\]](http://latex.artofproblemsolving.com/3/6/2/3622bb7b054134edad3e93181ccac99df7105fe6.png)
Hence, if we let 
be the 
,![\[\frac{1}{x - s} = \frac 12 \left(\frac{2s^2 + a - 1}{(a + 1)s} - \frac{2s^2 - a - 1}{(a - 1)s}\right) = \frac{a^2 - 2s^2 + 1}{(a^2 - 1)s}.\]](http://latex.artofproblemsolving.com/7/d/4/7d4a52c0b0fe7bb49c693e15c038bf107475c692.png)
Now line ![\[y = -\frac as (x - s) + a.\]](http://latex.artofproblemsolving.com/5/6/8/568343977de7083f7ff4b5dab1d8933825818558.png)
Therefore, if we let 
,![\[\frac{1}{x - s} = \frac{1}{a - 1} \left(\frac as - \frac{2s^2 + a - 1}{(a + 1)s}\right) = \frac{a^2 - 2s^2 + 1}{(a^2 - 1)s}.\]](http://latex.artofproblemsolving.com/7/8/f/78f3ac7ead7ff9d51e1c00e6c5b2bef7d9f5ab57.png)
Hence the two values of 
finishes by Power of a Point.
meet 
.
on 
such that 
and 
.
are concyclic.
.
are concyclic. Notice that 
thus, 
,![\[\frac{QB}{QC}=\frac{\cos B}{\cos C}=\frac{HB}{HC}=\frac{BF}{CE}\]](http://latex.artofproblemsolving.com/b/e/7/be72224803ea8e2822a960df1fd1bef365ad29ca.png)
Also 
hence 
which gives that 
are concyclic.
and 
are concyclic. We have 
thus, 
bisects 
.![\[180-\measuredangle QME=\measuredangle BMQ=\measuredangle BHQ=\frac{\measuredangle A}{2}+\measuredangle FEQ+\frac{\measuredangle A}{2}=\measuredangle A+\measuredangle BCQ=\measuredangle CAQ=\measuredangle EAQ\]](http://latex.artofproblemsolving.com/a/3/9/a394dac287e869fb3eff44ed06d86c60317d66e3.png)
So 
.![\[\measuredangle FNQ=180-\measuredangle QNC=180-\measuredangle QHC=180-(\frac{\measuredangle A}{2}+\measuredangle QBC+\frac{\measuredangle A}{2})=180-\measuredangle A-\measuredangle QBC=\measuredangle FAQ\]](http://latex.artofproblemsolving.com/5/c/d/5cd4ac19ff402d596254d03f72b14279a5c899af.png)
Hence 
so 
Let 
equivalently the 
is tangent to 
is cyclic.
so 
is cyclic.
is cyclic.
so there is spiral centered at 
to 
However, from the earlier ratio, 
is sent to 
which implies 
is cyclic. This finishes.
