[/quote]
,
,
,
be positive integers such that 
.
grid as follows:
and a constant 
,
and 
are also friendly polynomials. Prove that 
.
for which the sequence of real numbers with 
for all 
is periodic from the beginning.
be reals such that 
and 
Prove that
Let 
Prove that
Prove that
and circumscribed about the incircle 
,
.
,
,
,
,
,
,
,
,
,
(distinct from 
)
intersects 
(distinct from 
be the midpoint of the arc 
of 
and 
at 
at 
.
.
.
intersects 
at 
and 
.
is the midpoint of 
.
be the altitude in 
,
intersects 
.
be the midpoint of 
be the reflection of 
with respect to 
and 
are similar.
as well as all values of 
which attain it.
such that 
(where 
)
that satisfies 
for all 
is the function obtained by composing 
)
such that 
divides 
for any positive integers 
such that for all 
,
.
Y by Davi-8191, Wizard_32, nguyendangkhoa17112003, Adventure10, Mango247, ItsBesi, and 4 other users
A point 
is chosen on the side 
of a triangle 
with 
in such a way that 
. The incircle of is tangent to 
and at points 
and 
, respectively. Let 
be the incenter of triangle 
. Prove that the line 
intersects the line segment 
at its midpoint.
is chosen on the side 
of a triangle 
with 
in such a way that 
. The incircle of is tangent to 
and at points 
and 
, respectively. Let 
be the incenter of triangle 
. Prove that the line 
intersects the line segment 
at its midpoint.This post has been edited 1 time. Last edited by djmathman, Jun 27, 2015, 12:16 AM
Reason: changed wording to reflect english version of ISL2006
Reason: changed wording to reflect english version of ISL2006
.
,
is the bisector of 
.
,
.
,
.
.
)
.
,
,
,
.
is equivalent with:
,
such that 
.![$D\in [AC]$](http://latex.artofproblemsolving.com/c/5/a/c5acd77baad8d15d98170c2ac60f572eeeed5507.png)
such that 
of ![$K\in [AB]$](http://latex.artofproblemsolving.com/f/7/c/f7c55ea4f0642b6ee4395cbe8c000c6fec53e3ef.png)
and ![$L\in [AC]$](http://latex.artofproblemsolving.com/c/4/5/c45e50a5a8063a6607d6afd834b15cea829eb243.png)
.
be the incircle of 
.
.
.
touches 
,
and 
.
.
. Since 
obtain that 
.
.
is well-known (see the lower remark). Thus, 
is cyclically. Analogously 
,
.
.
in 
: 
.
the quad. 
is cyclically 
or 
.
,
.
.
,
,
.
.
and 
be the excentre of 
opposite to 
and 
.
,
and 
.
we conclude that 
,
.
that 
is cyclic, hence 
,
and 
are similar. Hence![\[\frac{QP}{AK}=\frac{QA}{AD}= \frac{1}{2}\cdot\frac{QA}{AM}=\frac{1}{2\sin\frac{\alpha}{2}},\ \ \ \ \ \ \ \ (1) \]](http://latex.artofproblemsolving.com/6/6/6/66678fe4620aeb1de0b54bcc59301c5b49d9c53a.png)
and 
,
and 
are similar, so![\[\frac{QJ}{AI}=\frac{BQ}{BA}=\frac{\sin(90^\circ-\frac{\alpha}{2})}{\sin\frac{\alpha}{2}}= \frac{\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}.\ \ \ \ \ \ \ \ (2) \]](http://latex.artofproblemsolving.com/e/1/3/e1313f2d4b997f882bea9f07f92770b021e05b2b.png)
![\[\frac{QJ}{QP}=2\cdot\sin\frac{\alpha}{2}\cdot\frac{\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}\cdot \frac{AI}{AK}= 2\cdot\cos\frac{\alpha}{2}\cdot\frac{AI}{AK}.\ \ \ \ \ \ \ \ (3) \]](http://latex.artofproblemsolving.com/9/f/7/9f70baf3255f5b73868522dda2c0014f5e006d4c.png)
,![\[\frac{AI}{AK}=\frac{1}{\cos\frac{\alpha}{2}}\ \ \ \ \ \ \ \ (4) \]](http://latex.artofproblemsolving.com/d/c/2/dc241036d9d2ec8f5f59cb445bc829a28c1c9f90.png)
![\[\frac{QJ}{QP}=2, \]](http://latex.artofproblemsolving.com/3/5/5/3557203cc33b820edc4c09037efc646e099b1447.png)
be the midpoint of 
and let 
be the intersection of lines 
and 
.
be the foot of the perpendicular of 
on 
.
,
and 
are similar. This follows from a short angle-chase. Hence 
.
.
is cyclic.
.
and 
,
.
yields:![\[ \frac {AS}{SJ} \cdot \frac {JT}{DT} \cdot \frac {DL}{LA} = 1 \\
\Leftrightarrow JT = LA \cdot \frac { DT}{DL} = \frac{LA^2}{LK}.\]](http://latex.artofproblemsolving.com/9/3/f/93f8ca81d1da4270d6b10767136dd5e708c74af2.png)
,
follows from the power of a point wrt quadrilateral 
,
Which is obviously true Q.E.D
and 
and 
.
.
with collinear points 
,
Note that by Law of sines in 
,
.
Hence 
as desired.
,
,
at 
denote the semiperimeter of triangle 
.
and 
.
so 
Thus, basic length chasing yields 
and 
.
follows from properties of external bisectors, we know 
.
is a rectangle, so 
which means 
is a parallelogram. Hence, we know 
bisects 
,
.
so we have ![\[ND=2AL-ND-AD=2AL-AN=AL-(AN-AL)=PL-NL=PN\]](http://latex.artofproblemsolving.com/c/f/7/cf7f32bdaceb5c3993067a0189b06571c546d5a0.png)
Since 
,![\[\angle JPA=\angle JDC=\frac12 (180^\circ-\angle BDA)=90^\circ-\frac12\angle BAC=\angle KLA\]](http://latex.artofproblemsolving.com/8/8/0/880919e9e3efcf0f0805f4b421ecc9969ca36c77.png)
so 
.
and 
are equal -- the best way to view this is by showing that the distances 
)
(where 
,
with 
after we compute the base 
and the leg 
.
we have 
.
and 
we obtain 
.
,
and 
give 
and 
(we used 
from the problem conditions). Since 
from the isosceles triangle 
(here ![\[ IJ = \frac{AD}{2\cos \frac{\gamma}{2}} = \frac{c\cos\alpha}{\cos \frac{\gamma}{2}}.\]](http://latex.artofproblemsolving.com/4/d/0/4d0f9a2af00d07e1bd3450a0c5e1b8f9c7d6189b.png)
Now focus on the trapezoid 
.![\[ JS = JZ + ZS = IJ\sin\frac{\beta}{2} + IN = \frac{c\cos\alpha\sin\frac{\beta}{2}}{\cos \frac{\gamma}{2}} + \frac{b+c-a}{2}\frac{\sin^2 \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \]](http://latex.artofproblemsolving.com/6/1/2/6125bb65fa21df44e4918ca78a7f57eeac316c93.png)
and now it suffices to verify that the latter equals 
.
and 
)![\[ 4R\sin\frac{\gamma}{2}\sin\frac{\beta}{2}\cos\alpha - \frac{b+c-a}{2\cos \frac{\alpha}{2}}\cos\alpha = R\cos\alpha\left(4\sin\frac{\beta}{2}\sin \frac{\gamma}{2} - \frac{\sin \beta + \sin \gamma - \sin\alpha}{\cos \frac{\alpha}{2}}\right) \]](http://latex.artofproblemsolving.com/2/3/c/23c9882b56a29d3d699f542cb5a61a9e556997f6.png)
and so it remains to show 
.![\[ \sin\beta + \sin \gamma - \sin \alpha = \sin \beta + \sin \gamma - \sin(\beta + \gamma) = \sin\beta + \sin \gamma - \sin\beta\cos\gamma - \cos\beta\sin \gamma \]](http://latex.artofproblemsolving.com/f/2/7/f27f6477205f5138b1856d6e431bf7ef05442014.png)
![\[ = \sin\beta(1-\cos \gamma) + \sin\gamma(1-\cos \beta) = 4\sin\frac{\beta}{2}\cos\frac{\beta}{2} \sin^2\frac{\gamma}{2} + 4\sin\frac{\gamma}{2}\cos\frac{\gamma}{2} \sin^2\frac{\beta}{2} \]](http://latex.artofproblemsolving.com/3/e/a/3eab14cc061f282c81d9f602440264f80f3b5c66.png)
![\[ = 4\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\left(\sin\frac{\gamma}{2}\cos\frac{\beta}{2} + \cos\frac{\gamma}{2}\sin\frac{\beta}{2}\right) = 4\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\sin \frac{\beta+\gamma}{2} = 4\cos \frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}. \]](http://latex.artofproblemsolving.com/a/2/d/a2d2a905cad3373ba2de9ff616e450ed4579b52c.png)
and angles of 
(whose proof in fact goes through an expression such as 
,
be the reflection of 
,
.
So we must prove 
,
is isosceles, or 
.
Thus we have 
so we are done.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/5/c/2/5c26ca66967578c2c76d3e49e919ee972c56792e.png)
meet 
,
.
and we now need to show that 
.
-
.
,
,
.
so![\[
db=AD=2c\cos A=\frac{b^2+c^2-a^2}{b}.
\]](http://latex.artofproblemsolving.com/c/d/c/cdc74de3018396fcc0f43301e32136b70ca431a8.png)
Also 
and 
.
is given by 
where 
varies. Since 
,
.
is given by 
where ![\[
J=(a(1-d):b(1-d):c+ad)=\left(\frac{a(1-d)}{a+b+c-db},\frac{b(1-d)}{a+b+c-db},\frac{c+ad}{a+b+c-db}\right).
\]](http://latex.artofproblemsolving.com/0/5/7/05708601b3c2d0df75ac9c230da14c20f890d224.png)
Line ![\[
\begin{vmatrix}
x & y & x\\
s-b & s-a & 0\\
s-c & 0 & s-a
\end{vmatrix}
=0
\]](http://latex.artofproblemsolving.com/e/5/1/e5144f1d7cc06a8903f6bc28a5b07b14338adb28.png)
which simplifies to 
.
is![\[
\frac{1}{2}\left(1+\frac{a(1-d)}{a+b+c-db},\frac{b(1-d)}{a+b+c-db},\frac{c+ad}{a+b+c-db}\right).
\]](http://latex.artofproblemsolving.com/f/e/5/fe5e07b662d4c27cdb5b8f9ddaec95e4c19aa47b.png)
Now we just need to check that it lies on line 
We are done. 
at 
.
cyclic. Now 
. As 
and 
intersected at point 
. Let 
is the incenter of triangle 
.
collinear. Let 
and 
. By lemma 255 ( or by Iran Lemma ), We know 
. By angle chasing, we can find easily 
. So By Menelaus theorem for triangle 
and points 
we can find 
. Proved
be the 
and 
is perpendicular to 
and our new RTP is to prove 
.
,
is isosceles and 
.
thus 
is cyclic and 
and 
thus the question is done.
and 
we get that the triangle 
is isosceles 
.
we have:
.
and 
we get by AA criter that triangles 
, 
are similar 
and 
we get by AA criter that triangles 
, 
are similar 
,
and 
.
.
and 
we get that 
is the midpoint of 
