[/quote]
,
such that:
for any positive integer 
,
are all distinct, and
,
?
lies on a line segment 
and 
and circles are drawn having 
and 
as diameters. A common tangent to both circles touches the circle with 
and the circle with 
.
and the common tangent to both circles at 
and 
denote three distinct integers, and let 
denote a polynomial having integer coefficients. Show that it is impossible that 
and 
.
is a minuan pair if it satisfies the following two conditions:
is even.
are all the positive divisors of 
,
is precisely
mod 3 can be written as the sum of one square and 2 cubes.
,
be the intersection point of 
with the diagonal 
.
at point 
be a point on the segment 
such that 
.
be the intersection point of the diagonal 
.
and 
,
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
be a polynomial of degree 
with integer coefficients and let 
be a positive integer. Consider the polynomial 
,
such that 
.
,
and 
meet at the orthocenter 
from 
.
meets 
.
is perpendicular to the angle bisector of 
be one to one maps. Show that the function 
defined by 
,
,
be a surjective function. Show that there exist surjective functions 
such that 
,
prove that for each 
there exists 
such that either 
or 
is multiple of 
.
and 
at 
)
intersects the circumcircle of triangle 
intersects the perpendicular bisector of 
lies on circle 
,
and 
is a trapezoid 
)
and 
meet on 
Y by DrMath, anantmudgal09, Catoptrics, itslumi, 554183, pog, MehmetBurak, Newmaths, Lcz, Adventure10, Mango247, Rounak_iitr, Funcshun840, X.Luser, and 1 other user
In triangle 
, 
, 
and 
are the points of tangency of incircle with the center of 
to 
, 
and 
respectively. Let 
be the foot of the perpendicular from to 
. 
is on 
such that 
. If 
is the orthocenter of 
, prove that 
bisects 
.
, 
, 
and 
are the points of tangency of incircle with the center of 
to 
, 
and 
respectively. Let 
be the foot of the perpendicular from to 
. 
is on 
such that 
. If 
is the orthocenter of 
, prove that 
bisects 
.This post has been edited 1 time. Last edited by khashi70, May 16, 2009, 6:07 PM
, 
, 
and 
are the points of tangency of incircle with the center of 
to 
, 
and 
respectively . Let 
be the feet of perpendicular from 
such that : 
. If 
be the orthocenter of 
bisects 
cut 
at 
and 
cut 
.
.
is cyclic and
.
and 
.
be midpoint of 
and points 
it's enough to prove that
.
we get 
,
.
.
as
is harmonical.
is orthic triangle of 
.
is bisector of 
.
we get that 
is exterior bisector of 
,
.
and 
(this result can be proved easily by harmonic division). Hence 
.
and 
be the projection of 
,
,
.
,
.
.
,
is the midpoint of 
Here is the proof without calculation.
to BC
are collinear.
midpoint of 
radius of incircle of 
.
at 
so 
)
)
(1)
so 
(2)
(3).
,
so
(4).
)
the intersection point of 
and 
be the feet of the perpendicular from 
and 
,
.
respectively, so 
and therefore 
.
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be the midpoint of 
,
.
is cyclic, from which it follows that 
.
,
is the orthic triangle of 
.
and 
now so equation of circle 
be 
.
point be 
,
in terms of 
.
is 
, now basically we need to show now slope of the line is 
and that’s quite easy to show. Note here we’re saying , solution is easy by CG because no stuffs involving any power is coming, and that makes our calculation easier. (I got a trig solution too, but there calculation is damn hard to do.)
![[asy]
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and 
be the feet of perpendiculars from 
respectively. Since 
and 
and 
are collinear. Furthermore, since 
is cyclic and since 
is also cyclic. Now, note that 
and 
so 
implying 
are collinear. Similarly, 
are collinear. Combining them, we see that 
are collinear. Let this line intersect 
Using cyclic quadrilaterals 
and 
we deduce that![\[\angle J_cJ_bI = 90^{\circ} - \angle HJ_bJ_c = 90^{\circ} - \angle CJ_bE = \angle IJ_bC,\]](http://latex.artofproblemsolving.com/c/a/b/cabeb187952837e752997606337f505af928bb9e.png)
internally bisects 
Since 
externally bisects 
so by the angle bisector theorem we have 
so 
implying 
is harmonic, or 
![\[\dfrac{PM}{PN} \cdot \dfrac{DH}{DX} \cdot \dfrac{XN}{MN} = 1,\]](http://latex.artofproblemsolving.com/2/3/4/2342f29f772fb8bd1a3c6b768eb4183c5672d27b.png)
we get
and 
we get
.
.
,
are collinear as desired.
note that 
is cyclic so 
thus 
are collinear similarly 
are collinear now let the perpendicular lines from 
intersect 
note that 
(because both of them are perpendicular to 
)
thus 
so 
is perpendicular bisector of 
similarly 
is perpendicular bisector of 
so 
now using 
are collinear.
,
,
.
and 
the pentagon 
is cyclic with 
as diameter, so 
and hence 
.
the homothety centered at 
.
from which the assertion follows since 
since 
,
.
so 
which means the pentagon 
is cyclic since 
.
,
which yields 
,
so we are done.
and 
denote the feet from 
respectively. Note that 
is an orthocentric system, so 
are collinear as well as 
By that egmo lemma, 
and 
Thus, 
,
are 
in 
.
Since 
is just the midpoint of 
,
.
.
Thus, 
.
is cyclic and thus 
which means that indeed 
is the foot of the perpendicular from 
the feet of the perpendiculars from 
to the opposite sides. Let 
be the 
W
.
)
.
maps to 
and thus the 
maps to the 
under this inversion. Further, since 
remains fixed under this inversion 
)
be the reflection of 
the intersection of 
and 
(with 
)![\[GM \cdot GW = \frac{AG \cdot GW}{2} = \frac{A'G\cdot GY_A}{GD \cdot GY_A} = GD \cdot GY_A\]](http://latex.artofproblemsolving.com/3/e/8/3e87e56bf62f0e0252be7d04be612b480a9df243.png)
Thus, 
is cyclic. Now, let 
.![\[X_A'D\cdot X_A'M = X_A'Y_A\cdot X_A'W = X_A'B \cdot X_A'C\]](http://latex.artofproblemsolving.com/3/4/b/34b33627a06e3a7e2d4a2956bfa9371f4210b4e8.png)
Thus, 
and we have that 
.
be the 
and 
is cyclic. Now, note that from this it follows that,![\[\measuredangle X_AY_AD = \measuredangle WY_AD = \measuredangle AMD=\measuredangle H_AMD = \measuredangle H_AHA = \measuredangle X_AHD\]](http://latex.artofproblemsolving.com/5/f/6/5f68a355b87b448551eaeeefc680e9b315f2dfd1.png)
and thus, 
is cyclic and 
.
.
is cyclic. Further, since 
and 
and we also have,![\[\measuredangle H_AY_AD = \measuredangle H_AA' =90^\circ\]](http://latex.artofproblemsolving.com/7/9/b/79b887b58c94a718e5ad2b1ababae7a3ece2cc6a.png)
H
.
is the 
.
.
by Midpoint Theorem, which implies that 
must indeed be cyclic.![\[\measuredangle DY_AM = \measuredangle DNM = \measuredangle DX_AZ = \measuredangle DY_AZ\]](http://latex.artofproblemsolving.com/f/5/b/f5b26a89df4f18b11aab72088234f0adf489d2d0.png)
This clearly implies that 
and we put these two together and conclude that indeed 
are collinear as claimed.
,
and the foot of the perpendicular from 
is the perpendicular bisector of 
bisects 
![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
,
a
.
with incenter 
a
and 
.
be the feet from 
and from 
,
is the orthic triangle of 
satisfies 
,
respectively. By the Iran Lemma, we see that 
are collinear. The condition that 
Since 
is the orthic triangle of 
we see that 
so the foot from 
is the 
Additionally, it is clear that 
.
,
.
trivially and cyclic variants. Now we write 
. Now we find 
. The circumcenter of 
,
.
.
.
,
,
.
is real, which is equivalent to showing 
which is obviously self conjugating, so we are done.
,
and 
![\[-1=(EF\cap BC,D;C,B)\stackrel{K}{=}(RD;HI)\stackrel{\infty_{\perp EF}}{=}(RM;SQ)\stackrel{H}{=}(D,M;\infty_{DM},HQ\cap DM),\]](http://latex.artofproblemsolving.com/f/1/4/f14d21a14f82237ebfaf055a0f65ee7d071de314.png)
so 
bisects ![\[(MD;P\infty) \overset{N}{=} (MN \cap HD, D; HI) = -1. \quad \blacksquare\]](http://latex.artofproblemsolving.com/c/f/5/cf5626d5f1492aa78b2ca5077f82c8b8c94ca28d.png)
and 
be the feet of the altitudes from 
and 
are cyclic.![\[\measuredangle H_BFI=\measuredangle H_BBI=90^\circ-\measuredangle BIH_B=90^\circ-\measuredangle BIC=-\tfrac{\measuredangle BAC}{2}=\measuredangle EFI\]](http://latex.artofproblemsolving.com/4/7/2/4720a84e40c3b1a10adce3b1505b9eb086fd380d.png)
with 
with 
and 
and 
.
be 
.
is linear. It suffices to show that 
.
.
traversing 
,![\[\frac{MN}{TN}\cdot \frac{TH}{DH}\cdot \frac{DP}{MP}=1\implies \frac{MN}{TN}=\frac{DH}{TH}\implies \frac{NT}{MT}=\frac{TH}{DH+TH}\]](http://latex.artofproblemsolving.com/f/7/2/f724e3b447b2087c44d611b11fcbe96833160a09.png)
Therefore,![\[f(N) = \frac{TH}{DH+TH}f(M)\]](http://latex.artofproblemsolving.com/f/4/1/f41bf130da346d3ba2f6f045d53855ac7a05fd94.png)
Since 
and 
,
.![\[\frac{TH}{DH+TH}=\frac{d(H,H_BH_C)}{2d(H,H_BH_C)+d(D,EF)}=\frac{H_BH_C}{2H_BH_C+EF}\]](http://latex.artofproblemsolving.com/6/5/3/6533ba724e44840f342667d06a5b03c625c6c0aa.png)
Let 
,
,
,
.
.
Now, we relate 
with 
and 
with 
.
;
.
implying 
while 
so![\[y=\cot{\angle C}=\frac{1-\tan^2\left(\frac{\angle C}{2}\right)}{2\tan\left(\frac{\angle C}{2}\right)} = \frac{1-u^2}{2u}\]](http://latex.artofproblemsolving.com/3/b/1/3b15b3181ef6abbf8b3c261d6d562b468ebebb29.png)
and the result follows.
and we are done.
be the orthic triangle of 
.
be the medial triangle of 
.
.
and 
.
and 
;
and so 
and also 
.
.
which is trivial just because 
is a parallelogram and gliding principle or whatever.
Define 
the intersection of 
with 
is bisected by 
By apollonian circles configuration, it follows 
but uniqueness of harmonic bundles implies the result.
and 
,
and 
.
and 
.
and I is the incenter of 
.
excenter for 
if we show that 
we will be ready. We know that IE = IF and N is the midpoint of EF 
