[/quote]
,
be the real polonomial such that all roots are real and distinct. Prove that there is a rational number 
that all roots of 
are real numbers
there exists positive integers 
such that
.
be any other point on the arc 
of the circumcircle of 
and let line 
intersect line 
at 
and let line 
intersect line 
at 
.
and 
)
intersect the circumcircle of 
again at 
)
concur.
such that:
for any positive integer 
,
are all distinct, and
,
?
have altitudes 
circumcenter 
and orthocenter 
be a point on line 
.
are on lines 
respectively such that 
and 
.
is perpendicular to 
.
on line 
such that 
are in that order and 
and 
.
on line 
on line 
and 
are equidistant from the circumcenter of 
be an acute triangle inscribed in its circumcircle 
,
be its incircle. Let 
be the point where the 
incircle of triangle 
.
intersects segment 
at 
,
at 
.
perpendicular to 
at 
.![\[AI \parallel PA'.\]](http://latex.artofproblemsolving.com/0/2/6/0264e64b5edff4aca1e35037c5b0f86763c3b4e7.png)
whose corners are vertices of the 100-gon, so that
has three corners of one color and one corner of the other color.
be positive integers and 
be prime. If 
is a perfect square, prove that 
is also a perfect square.
be real numbers satisfying 
Prove that 
and determine all cases of equality.
be the set of rational numbers. A function 
is called aquaesulian if the following property holds: for every 
,![\[ f(x+f(y)) = f(x) + y \quad \text{or} \quad f(f(x)+y) = x + f(y). \]](http://latex.artofproblemsolving.com/4/0/e/40e7d033daf1bd62c2bfd5d022eeaea3f860c5b3.png)
Show that there exists an integer 
such that for any aquaesulian function 
there are at most 
for some rational number 
,
lines
such that for all 
,
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
possible.
Y by Adventure10, Mango247, Zhaom
(a) Let given triangle 
with orthocenter 
and altitudes 
. Let 
be in-centers of triangles 
. Prove that lines 
intersects in same point
(b) Let given triangle with orthocenter and altitudes . Let lines intersects sides of triangle 
at points 
. Let 
be In-centers of triangles 
. Prove that triangle formed by lines 
is perspective to triangle
with orthocenter 
and altitudes 
. Let 
be in-centers of triangles 
. Prove that lines 
intersects in same point(b) Let given triangle
with orthocenter and altitudes . Let lines intersects sides of triangle 
at points 
. Let 
be In-centers of triangles 
. Prove that triangle formed by lines 
is perspective to triangle This post has been edited 1 time. Last edited by solver6, Sep 12, 2016, 4:05 PM
is acute since otherwise the problem is not true. We invert centered at 
.
and 
be 
and 
,
and 
are coaxial.
and 
through 
and 
at 
and 
,
and 
are coaxial, or that 
and 
be the feet of the altitudes from 
to 
,
to 
,
and 
be the midpoints of 
and 
on 
and 
not containing 
and 
intersect 
and 
other than 
-
,
at 
and 
go through 
we see that it suffices that there is an involution swapping 
and 
,
and 
,
and 
gives that it suffices that the inverted images of 
is cyclic.
.
and 
and 
intersect 
and 
at 
and 
other than 
,
in 
and 
intersect 
is cyclic.
and 
intersect 
and 
,
and 
,
and 
and the point at infinity on 
.
and 
.
in 
is also perspective with 
.
is an isoconjugation. This implies that there exists an involution 
on 
is always perspective with 
,
.
.
and 
,
and 
concur.
and 
and 
,
is the radical center of 
and 
and 
is the radical center of 
and 
and 
concur. Reflect 
and 
.
.
and 
be the points such that 
and 
are isosceles right triangles with hypotenuses being the sides of 
and 
be the reflections of 
and 
,
.
lies on 
,
Now, we see that
Similarly, we see that 
.
since 
bisects 
,
.
and 
are isogonal conjugates since
Similarly, we see that 
and 
are isogonal conjugates and that 
and 
are the perpendicular bisectors of 
.
and the Bevan point of 
such that 
is an isosceles right triangle with 
be the isogonal conjugate of 
.
be the reflection triangle of 
be the isogonal conjugate of 
passes through 
.
in part 
we see that the tangent to 
the rectangular hyperbola through 
and 
.
in part 
.
is tangent to 
and 
of 
and 
and 
,
,
since 
and 
bisect 
,
,
with 
and 
concur at a point 
on the isopivotal cubic of 
and similarly 
.
is on the cubic and is 
.
also goes through this point.
and 
intersect 
and 
and 
,
and 
and 
be the images of 
centered at 
and 
concur on the rectangular hyperbola through 
we see that the intersections are all 
we see that the intersections are all the orthocenter of 
,
be the point on 
such that 
.
be the cevian triangle of 
.
.
and 
.
be the isogonal conjugate of 
.
and 
are isogonal conjugates in 
since 
and 
.
goes through 
.
and 
with 
,
,
,
,
This implies that 
and 
are isogonal in 
,
we get that 
be incenters of 
,
be 
concur.
that are isogonal conjugates of 
.
gives that 
are isogonal, so 
are collinear and 
are perspective.
and define 
similarly. DDIT from 
implies 
lies on 
.
are collinear along their perspectrix by Desargues. By Desargues again, this implies 
and 
are also perspective, since they have the same perspectrix. Let 
concur at 
in 
.
intersect 
meets 
.
,
by Cayley-Bacharach. Similarly 
,
be a hyperbola with foci 
lies on the line through 
WLOG we may let 
meet 
.
.
,
bisects 
.
,
meet 
and 
.
,
.
be the 
and 
meet 
and 
respectively. Then ![\[ (A, A'; M, N)_{(ABC)} = -\frac{CD}{BD}.\]](http://latex.artofproblemsolving.com/1/3/f/13fb67d97d9c0678bcb58548f9d1b260ba97cc4a.png)
.
.![\[ \frac{AM}{AN} \div \frac{A'M}{A'N} = \frac{AM}{A'M} \div \frac{AN}{A'N} = \frac{ID}{BD} \div \frac{ID}{CD}. \]](http://latex.artofproblemsolving.com/9/d/c/9dc927dc415b408a6fd3e10bb2cf0e4ebc57c405.png)
Accounting for sign, we are done.
meets the circumcircle of 
.
.
touches ![\[ (B, C; T, T')_{\mathcal H} = - \frac{BX}{CX}. \]](http://latex.artofproblemsolving.com/c/9/4/c94ddcd64015b8d0642a6aa8a012341a950d6d0b.png)
meet 
.
meet ![\[ (B, C; T, T')_{\mathcal H} \overset{T'}{=} (\overline{T'B} \cap \ell, \overline{T'C} \cap \ell, P', AT \cap \ell). \]](http://latex.artofproblemsolving.com/c/d/e/cdea6b1f28f93355cfc47ce363a5407db0fb4c79.png)
Add in the arc midpoints of 
,
and ![\[ (M_B, M_C; Q, P) = -\frac{BX}{CX}. \]](http://latex.artofproblemsolving.com/7/5/6/756146bd9b0cbb96b5f360c0732dc2c88ce409c5.png)
The conclusion is immediate from Lemma 3 and a quick manipulation of cross ratios.
meets 
.
meets 
and 
is ![\[ \frac{CT \cdot CL}{CK} = \frac{BT \cdot BL}{BK}.\]](http://latex.artofproblemsolving.com/c/b/9/cb97945984de66c8571ca323046b388dfc16369e.png)
Lengths are not directed in this statement.
.
and 
the incenters of 
,
respectively. Recall from say, MMP, that 
.
.
.
is tangent to both 
and 
,
,
is tangent to ![\[ -\frac{CT}{BT} = (B, C; K', \overline{AK'} \cap \mathcal H)_{\mathcal H} \overset{K'}{=} (B, C; K, L) = -\frac{|BK| \cdot |CL|}{|CK| \cdot |BL|}, \]](http://latex.artofproblemsolving.com/9/9/8/998d02a5cdcbc0205271500296ecc630b1866db1.png)
which rearranges into our desired claim.
,
,
to 
meet 
meet ![\[ \frac{BT \cdot BA'}{BX} = \frac{CT \cdot CA'}{CX} \implies \frac{BX}{CX} = \frac{BT}{CT} \cdot \frac{BA'}{CA'} = \frac{BT}{CT} \cdot \frac{BA}{CA}. \]](http://latex.artofproblemsolving.com/d/a/3/da3a0275496411da3ed0d3709f314e20f8f82673.png)
be the incenter of 
.
analogously. If we denote by 
,
and 
as 
maps to 
under reflection over 
.
,
concur, where 
.
are collinear from well-known config. Hence, it suffices to show that 
,
concur.
,![\[\frac{S_aB}{S_aC} \cdot \frac{S_bC}{S_bA} \cdot \frac{S_cA}{S_cB} = 1\]](http://latex.artofproblemsolving.com/a/b/a/aba8f6e1a3a2b8df128fef39e62eae6c0f5fea47.png)
which means that the alternate sides of the cyclic hexagon 
have equal products, and thus its diagonals concur.
be the incentres of 
.
meets 
.
are defined similarly, we have 
.
through 
is the 
,![\[\frac{SB}{SC} = \frac{KB}{KC} \cdot \frac{AB}{AC}\]](http://latex.artofproblemsolving.com/8/b/e/8bebe160cf5d857c239a3346d7f628c027d3967f.png)
Thus it suffices to show that the cyclic product of 
is unity. In particular, this is the same as wanting the product of the cross-ratios 
to be equal to unity, as the cyclic product of 
is unity.
denotes the point where the 
invert: ![\[(A'K;BC) = (\infty, R_a; CB) = \frac{R_aB}{R_aC}\]](http://latex.artofproblemsolving.com/1/5/e/15eaf509b06854fe5ee2bc2c2e480431bd3a4ac0.png)
First, due to the existence of the isotomic conjugate, we can redefine 
,![\[\frac{S_aB}{S_aC} = \frac{R_aB}{R_aC}\]](http://latex.artofproblemsolving.com/6/e/e/6ee4b6c66612b52bee412243f8a83bf5b5b41207.png)
where 
as the intersection of 
,
have product unity. By Menelaus and Desargue's, this finishes.
.
be 
.
be points on 
such that 
is tangent to 
and 
are collinear.
and vary 
move with degree 
cases.
then the three points are collinear along 
then the three points are collinear along 
they are collinear on 
from 
is the tangency point of 
and 
,![\[\frac{AC+BC-AB}{BC}=\frac{BD}{BM}=\frac{AD}{AX}=\frac{AC}{AE}=\frac{2AC}{AB+AC+BC},\]](http://latex.artofproblemsolving.com/b/0/5/b054c1c84cbe6d378ef04b6b34ff74760835c0c6.png)
which rearranges to 
which is true. Similarly this works when 
.
.![\[\frac{2AE}{AE+EK+AK}=\frac{AD}{AT}=\frac{CD}{CM}=\frac{2EC}{BC}\]](http://latex.artofproblemsolving.com/0/8/0/08079cf02ddb14f560cb18cdad12c5f489fa0c5c.png)
which becomes 
which becomes 
which follows from angle bisector theorem. Similarly this works when 
cases, which is enough.
be the 
respectively. Then 
concur.
respectively. Construct the common external tangent to 
such that 
are on the same side of it. By 
homothetic to 
,
and thus 
and 
.
are perpendicular to 
it follows that the bisectors are parallel, and thus they are the same line, so 
bisects 
.
bisects 
.
and 
similarly, it suffices to show ![\[\frac{H_BX_A}{X_AH_C}\cdot\frac{H_CX_B}{X_BH_A}\cdot\frac{H_AX_C}{X_CH_B}=-1\]](http://latex.artofproblemsolving.com/e/c/d/ecd34ffe20fc0ad5992facb3e5503a2cc2245997.png)
as Menelaus would give 
collinear along the desired perspectrix. To do this, it would also suffice to show ![\[(H_B,H_C;A',X_A)\cdot(H_C,H_A;B',X_B)\cdot(H_A,H_B;C',X_C)=-1\]](http://latex.artofproblemsolving.com/4/1/7/417b7e7a254c038fc17cd256e4b3d7ef7d34e0e1.png)
as 
concurring at 
.
be minor arc midpoints of 
on this circle, and let 
.
then 
by angle bisectors bundle, so projecting through 
.
,
similarly. Prove that ![\[\frac{BK_A}{K_AC}\cdot\frac{CK_B}{K_BA}\cdot\frac{AK_C}{K_CB}=1.\]](http://latex.artofproblemsolving.com/5/6/0/560afe491e486654e3d180fd591a900562a48fc1.png)
concurrent by Ceva. However, notice 
,
.
