[/quote]
,
for real numbers 
not multiple of 
.
of the triangle 
to 
.
is extended beyond 
to 
so that 
.
,
.
Let 
.![$$ \frac{1}{2}\geq \frac{a}{a^2+3 }+ \frac{b}{b^2+3} \geq \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$](http://latex.artofproblemsolving.com/f/9/1/f9119e071c12778037616c624c7b9ca75356e8e3.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+3 }+ \frac{b}{b^3+3} \geq \frac{1}{2\sqrt[3]{9}}$$](http://latex.artofproblemsolving.com/2/b/2/2b2d8a011229fe88a81cb75f9b018ad3a6245faa.png)
![$$ \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2} \geq \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$](http://latex.artofproblemsolving.com/5/f/6/5f673cd494b1909ca237db63e96f949c24b6f2ad.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2} \geq \frac{\sqrt[3]{3}}{5}$$](http://latex.artofproblemsolving.com/b/c/3/bc343a07c88dbd98ebffdfb78f311b7b82683d65.png)
and 
Prove that
Let 
Prove that
segments 
are parallel. Angle bisectors of 
and 
meet at 
.
and 
meet at 
.
is cyclic
= 2∠AMP.
| 
with 
are positive integers and 
is prime 
)
Let ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
denote the set of natural numbers less than or equal to 
![$f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$](http://latex.artofproblemsolving.com/f/d/f/fdf7b2fcfb960e93d59f3f84cef4372262c69499.png)
Y by v4913, CyclicISLscelesTrapezoid, CT17, OronSH, mathmax12, mahaler, ETS1331, Danielzh, HWenslawski, Amir Hossein, ItsBesi
We are given an acute triangle 
. Let 
be the point on its circumcircle such that 
is a diameter. Suppose that points 
and 
lie on segments 
and 
, respectively, and that 
and 
are tangent to circle 
.
Show that line
passes through the orthocenter of triangle .
. Let 
be the point on its circumcircle such that 
is a diameter. Suppose that points 
and 
lie on segments 
and 
, respectively, and that 
and 
are tangent to circle 
.Show that line
passes through the orthocenter of triangle .
be the orthocenter of 
be the midpoint of 
.![\[\angle DBK=\angle DMK=\angle DCL=\angle DML=90^\circ,\]](http://latex.artofproblemsolving.com/a/c/4/ac423c5060ddd0251e32d3ec477edb472ac1cb45.png)
so 
and 
are cyclic. We have ![\[\angle ABM=\angle KDM=90^\circ-\angle DKM=90^\circ-\angle BAC=\angle ABH,\]](http://latex.artofproblemsolving.com/1/4/7/1474377acad7760ecd959f7412bd15738f3ba505.png)
and similarly, 
.
,
now 
so 
and 
so 
and from here using the cyclics we get 
which tells that 
thus we are done.
bisects 
and 
.
which means 
.
is a parallelogram. Now, 
so 
.
and 
are cyclic with diameters 
.
are isogonal in 
as median and symmedian, we have that 
so 
and similarly 
.
which satisfy the problem hypothesis, since a priori the property of 
be the midpoints of 
respectively, and 
the circumcenter of 
and by homothety 
,
,
(which is equivalent to saying that the conjugate of 
)
is a parallelogram. Since these points coincide, 
is the 
.
.
be the perpendicular bisector of 
,
and 
.
So in fact the orthocenter of 
,
and hence 
is cyclic. Hence, 
.
.
.
at 
s
is the perpendicular bisector of 
which gives 
is cyclic. And similarly 
is also cyclic.
which upon combining with the fact that the orthocenter lied on 
be a point on 
such that 
& 
.
is cyclic
. So Claim 
is proved .
.
is cyclic
we get that 
and 
.
intersect 
and 
intersect 
is a parallelogram
and 
are concylic
which means that
of 
.
cyclic
.
inversion centred at 
be the altitudes of 
onto 
respectively. The inversion sends 
to the point on 
is right, and similar for 
.
with 
,
is the intersection of 
with 
,
is tangent to 
and 
at 
.
point of 
.
are collinear or that 
are concyclic.
is parallel to 
is parallel to 
.
is parallel to 
.
to take the circumcircle of 
,
in this dilation.
at a point 
or just 
.
by repeated Power of a Point, so we are done, as 
are collinear.
and 
.
is a parallelogram, which is true as![\[\angle CMB = \angle DKB + \angle CLD = \angle ALK + \angle AKL = 180-\angle A = \angle BDC\]](http://latex.artofproblemsolving.com/1/c/9/1c90a65c0695a0b897ac7c5a0a1578348f04fbfd.png)
![\[\angle MBD = \angle MKD = \angle DLM = \angle DCM. \quad \blacksquare\]](http://latex.artofproblemsolving.com/f/8/d/f8dc498ad70ff51af1331c6cdaa8bcfb31ea777e.png)
.
and 
so 
,
are cyclic. Now note that
Thus 
and 
are altitudes, implying 
will be the isogonal conjugate and by angle chasing we will have 
)
So we have
and therefore 
.
,
.
,
.
.
,
and 
and 
and 
,
and 
are the heights from B and C respectively 
![[asy]
import geometry;
unitsize(4cm);
pair A = dir(110); pair B = dir(220); pair C = dir(320);
pair H = A+B+C; pair G = dir(250);
pair M = (B+C)/2; pair O = 0; pair Y = reflect( perpendicular(O, line(G,M))) * G; pair D = -A; pair T = intersectionpoint(perpendicular(B,line(O,B)), perpendicular(C,line(O,C)));
pair K = intersectionpoint(perpendicular(H,line(H,D)), line(A,B)); pair L = intersectionpoint(perpendicular(H,line(H,D)), line(A,C));
draw(A--B--C--cycle); draw(circle(A,K,L), dashed); draw(D--K); draw(D--L); draw(circle(A,B,C)); draw(Y--T); draw(A--G); draw(Y--B); draw(Y--C); draw(Y--G); draw(A--D); draw(K--L); draw(T--B); draw(T--C);
draw(A--H--D--cycle, red+1.1); draw(Y--M--T--cycle, red+1.1);
filldraw(A--H--D--cycle, invisible, red+1.1);
filldraw(Y--M--T--cycle, invisible, red+1.1);
filldraw(A--K--L--cycle, invisible, blue+1.1);
filldraw(Y--B--C--cycle, invisible, blue+1.1);
dot("$A$", A, dir(110)); dot("$D$", D, dir(290)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$H$", H, dir(0));
dot("$G$", G, dir(250)); dot("$M$", M, dir(135)); dot("$Y$", Y, dir(60)); dot("$T$", T, dir(270)); dot("$O$", O, dir(120)); dot("$K$", K, dir(180)); dot("$L$", L, dir(30)); [/asy]](http://latex.artofproblemsolving.com/c/e/5/ce599af5d4d5def8a5d4fbf8a35a37754c914327.png)
![\[ \frac{AK}{AL} = \frac{\sin KAD}{\sin LAD} = \frac{\cos C}{\cos B} = \frac{HC}{HB} \]](http://latex.artofproblemsolving.com/7/4/7/7472370d21b1c288518ad57ef5dc2f5865beeaed.png)
where 
as the reflection of 
intersect 
again at 
.
.
and 
are isogonal wrt 
so 
is harmonic, implying ![\[ \frac{YB}{BC} = \frac{DB}{DC} = \frac{GC}{GB} = \frac{HC}{HB} \]](http://latex.artofproblemsolving.com/9/2/e/92e9a02ba3612b18b1114c9b6bab046070bcc5d0.png)
hence 
be the pole of 
to 
(both are the segments on the 
is collinear by the definition of the symmedian.
.
.
by the definition of pole. Thus 
,![\[ \frac{YM}{YT} = \frac{OM}{OY} = \frac{OM}{OB} = \cos A = \frac{AH}{AD} \]](http://latex.artofproblemsolving.com/b/c/8/bc825a52cd3ca2cc8bd54023b86bf24e8e843711.png)
thus 
to 
to 
so 
is cyclic. Analogously we get 
cyclic, which implies 
However, we also have 
so indeed 
so 
are both cyclic and
same for 
so 
and we're done.
and 
intersect at 
.
so 
are collinear. Now 
and 
This implies 
is a parallelogram so 
is orthocenter of 
now we wish to prove that 
is the symmedian of 
is perpendicular to 
are concyclic points
is perpendicular to 
is the orthocenter with respect to 
so 
and 
are cyclic. Let 
,![\[
\angle HBD=\angle HKD = \alpha, \angle HCD=\angle HLD=\alpha
\]](http://latex.artofproblemsolving.com/5/4/f/54f27fa0cf765c212d8f5de1cf204a0537255d32.png)
also 
and 
so 
is a parallelogram and 
