,
be its incenter. Incircle touches sides 
at 
,
to 
are 
, respectively. Rays 
intersect 
at 
,
touches incircle at 
.
a grid of 9*9, we choose a little square in 
.
s
such that 
be the divisors of 
.
.
a
that satisfy 
by ![\[\displaystyle f(x) = \left\{ \begin{array}{lr} x+\frac 12 & \text{if}\ \ x < \frac 12\\ x^2 & \text{if}\ \ x \ge \frac 12 \end{array} \right.\]](http://latex.artofproblemsolving.com/c/b/d/cbd9b412893774283a438296cdeb05b80cc878bf.png)
Let 
and 
be two real numbers such that 
.
and 
by 
,
,
for 
.![\[(a_n - a_{n-1})(b_n-b_{n-1})<0.\]](http://latex.artofproblemsolving.com/a/5/c/a5c8844d595bb1186d4804beadd69a3bc1d40ea3.png)
is defined recursively by ![\[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]](http://latex.artofproblemsolving.com/f/4/b/f4b8f95e49de7274cccd8ee498e0b496f31ddfea.png)
Show that 
for all 
.
be positive integers satisfying the conditions 
and 
Show that there exists a real number 
with the property that all the three numbers 
have their fractional parts lying in the interval ![$ \left(\frac {1}{3}, \frac {2}{3} \right].$](http://latex.artofproblemsolving.com/4/2/b/42ba58ddd6032a3176122f1c9b2015cb6f4ca925.png)
Y by tenplusten, Abdollahpour, RedFlame2112, Adventure10, Mango247, and 6 other users
Circles 
intersect at 
. 
is common tangent of two circles which is nearer to 
and 
is on 
and 
is on 
. 
intersects for the second time in 
and 
intersects in 
. Let 
be intersection point of 
and 
. Prove that if 
is the second intersection point of circumcircles of 
and 
![\[\angle QXA=\angle QKP\]](//latex.artofproblemsolving.com/c/4/d/c4db3e48ee306cf574c3ac06758b919fb0ab887c.png)
intersect at 
. 
is common tangent of two circles which is nearer to 
and 
is on 
and 
is on 
. 
intersects for the second time in 
and 
intersects in 
. Let 
be intersection point of 
and 
. Prove that if 
is the second intersection point of circumcircles of 
and 
![\[\angle QXA=\angle QKP\]](http://latex.artofproblemsolving.com/c/4/d/c4db3e48ee306cf574c3ac06758b919fb0ab887c.png)
This post has been edited 1 time. Last edited by Omid Hatami, Jun 9, 2010, 9:10 PM
and 
are concyclic . is it useful ?
with 
and 
the second intersection of 
with 
For convenience, denote 
it follows that 
Since 
then 
are concyclic. Therefore, 
concur at the radical center 
of 
and 
which becomes the exsimilicenter of 
since 
into each other. Thus, the tangent of 
at 
passes through 
Further, 
are tangent lines of 
at points 
Then 
is the polar of 
such that 
is tangent to 
If 
then the cross ratio 
is harmonic.
(midpoint of XY), it follows that 
Since 
are collinear, being K,R double points, 
But because of 
we deduce that 
we make an inversion with center in 
on circumcircle of 
intersect sides 
in 
respectively. Show that 
are collinear and need to get that 
is tangent but this is same
the circumcircle of 
and let 
,
hence 
,
,
are concurent at a point say 
,
meaning 
.
so 
is isocel, therefore 
so 
is tangent to 
.
,
thus 
is cyclic. From another hand 
so 
is also cyclic, as a result 
is also cyclic. Now applying radical axis theorem on circles 
, 
and 
it happen that 
and so 
which end the proof.
.
.
.
,
.
.
,
be the midpoints of 
.
,
is a cyclic quadrilateral.
.
.
,
.
.
,
be the circumcircle of 
with 
meets 
touch with 
then
and 
conllinear.
are similar
is conllinear.
lie on circle.(*)
(**)
then 
lie on a circle
follow from that? Thanks.
be the image of 
and 
are cyclic.So
is also cyclic.Now some trivial steps yeild you 
which means 
and 
.
and 
,
to 
.
to 
,
as well. By angle chasing, we find that 
,
,
.
,
.
.
,
,
.
and 
,
to 
,
are similar. Since quadrilateral 
is cyclic, 
.
is cyclic
are spirally similar, then the triangle formed by the midpoints of 
and 
is similar to 
and 
are similar to 
,
X
.
and 
are similar to 
are similar, so 
.
.
.
.
,
,
,
,
by spiral similarity. Thus, 
,
,
(same direction)
(same direction)
which is a spiral similarity that maps:
is cyclic
(because 
be the second intersection point of 
.
q.e.d.
is cyclic.Indeed
.
.
is cyclic 
.
which gives that 
is also cyclic 
.
we conclude that 
all lie on the same circle.
and 
are cyclic. Now let 
and 
and 
point 
so we have 
so we need to prove that 
is cyclic so it is enough to prove that 
.
and 
.
and 
are tangents to the circumcircle of 
.
.
so we need to prove that 
which is equivalent to 
being the tangent to the circle 
so we need to prove that circumcircles 
are tangent to each other. Now we invert at 
.
is the point inside the isosceles triangle 
and the equivalent when we switch 
.
are cyclic we have that 
and 
and because 
and 
are cyclic we have 
.
is tangent to circumcircle 
.
and we need to prove that the same holds for 
it is enough to prove that intersections of tangents at 
and 
respcetively in any triangle 
and similarly 
,
,
is cyclic. This means that the circumcircle of 
is symmetric with respect to the angle bisector of 
.
be the reflection of 
be the reflection of 
is a rectangle that is symmetric with respect to 
inversion with respect to 
maps to 
maps to 
and 
.
and 
.
.
,
and 
are reflections of each other across 
is the reflection of 
across 
.
from which it follows that 
.
as the other intersection point of circles 
and 
.
.
followed by reflection in the bisector of angle 
and 
are interchanged under this map. As 
and 
the points 
and 
Thus, we have 
which holds as 
and 
proving the result. 
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[/asy]](http://latex.artofproblemsolving.com/5/7/1/571b65e22f01fbd514cf27fcb1bf1beb8454baae.png)
and 
.
to 
send 
.
so 
is cyclic. Moreover, ![\[\measuredangle BPL=\measuredangle BCL=\measuredangle XYK=\measuredangle YCK=\measuredangle BXK=\measuredangle BPK,\]](http://latex.artofproblemsolving.com/a/4/5/a456c0f2c23315a46d18dac62596db1c479dbfa4.png)
whence 
.
,
looks like we even ended up with the same letters!):
and 
.
Adding the two gives 
,
,
and 
and 
and 
respectively, and 
followed by a reflection over the angle bisector of 
.
and 
are tangent to ![\[\measuredangle AXY=\measuredangle BXY=\measuredangle BPX=\measuredangle YPC=\measuredangle XYC=\measuredangle XYA,\]](http://latex.artofproblemsolving.com/6/8/8/68805c7fd038d1319fd1ef962a7bcdce211c0a58.png)
so 
so 
is cyclic and 
,
bisects 
;
.![\[\frac{BX}{PX}=\frac{\sin\angle BPX}{\sin\angle PBX}=\frac{\sin\angle BPX}{\sin\angle PXY}\quad\text{and}\quad\frac{CY}{PY}=\frac{\sin\angle CPY}{\sin\angle PYX}.\]](http://latex.artofproblemsolving.com/9/c/a/9ca866ee09e7502b0527512bd6386d4eab357b06.png)
From this, ![\[\frac{BX}{PX}\Bigg/\frac{CY}{PY}=\frac{\sin\angle PYX}{\sin\angle PXY}=\frac{PX}{PY}\implies \frac{BX}{CY}=\left(\frac{PX}{PY}\right)^2.\]](http://latex.artofproblemsolving.com/a/3/7/a3777d895fcd6c5e7f24fd6fca417a19c8f133e3.png)
To finish, note that ![\[\frac{QX}{QY}=\frac{BX}{CY}=\left(\frac{PY}{PY}\right)^2=\left(\frac{KX}{KY}\right)^2=\frac{DX}{DY},\]](http://latex.artofproblemsolving.com/0/e/1/0e162b436cc408a867a733ef685c8b4c91757bfa.png)
where the third equality is since 
and 
imply ![\[\frac{PX}{KX}=\frac{MP}{MX}=\frac{MY}{MX}=\frac{PY}{KY}.\]](http://latex.artofproblemsolving.com/8/1/0/810c3b1637f58c25df035e67379abe2573421949.png)
This completes the proof.
is a harmonic quadrilateral, so 
.
,![\[\frac{QB}{BB_1}=\frac{2QB}{BX}=\frac{2QC}{CY}=\frac{QC}{CC_1}\]](http://latex.artofproblemsolving.com/a/c/1/ac138f75b85bded88b382e88fdfe7779828ba40e.png)
and 
,
,
is cyclic. Analogously since 
,
,
,
,
is also tangent to 
,
.
inversion at 
.
maps to 
and is tangent to 
,
is tangent to 
.
,
is cyclic. This implies that 
are swapped by this transformation.
.
.
is a line, if 
and 
,
and 
So we can calculate 
Let 
.
Thus 
is angle bisector of 
are collinear where 
it suffices to show 
and since 
is cyclic (follows from inversion at 
and 
.
,
and 
whence 
.
whence 
so 
.
is a symmedian in 
which follows from 
.
,
bisects 
,
and 
intersect on 
,
,
so 
which, combined with 
,
since they are polar. Now note that as 
is a diameter of 
.
is cyclic whence 
,
is mapped to a line after inversion about 
,
Therefore, 
is tangent to 
and we get the following problem
be an Isosceles Triangle with 
.
,
are tangent to 
.
and 
Hence,
Let 
.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/7/e/7/7e76c0d33e5db587d13401a77a2f9e74551ea6a9.png)
be circumcircle of 
and 
be the circumcircle of 
,
.
,
are collinear.
are similar triangles. Using the simillarity results we can say that 
are collinear. Hence, points 
form a cyclic quadrilateral with another circle.With the property of power of point 
,
,
form a cyclic quadrilateral, hence all the 
points lie on a circle.
and 
such that 
are collinear.
wrt lines 
and 
.
wrt lines 
and 
.
is cyclic.
.
,
.
.
.
.
is cyclic.
.
is cyclic.
.
claims, we come to the conclusion that 
is cyclic.
and we are done 
.
,![\[\angle XP'Y=\angle XPY=360^\circ-\angle KPX-\angle KPY=\angle KXY+\angle KYX=180^\circ-\angle XKY,\]](http://latex.artofproblemsolving.com/c/1/9/c19baab1c775ad23850e833e3fd46827bcb91a21.png)
so 
is cyclic. Thus, 
is a symmedian of 
and 
intersect the circumcircle of 
,
and 
,
.
.![\[\angle KXT'=\angle KXY-\angle YXT'=180^\circ-\angle AYK-\angle KAY=\angle YKT',\]](http://latex.artofproblemsolving.com/0/4/e/04e6cc55e4f84f00cb4fbbca03abf556c4623085.png)
and similarly, 
,
.![\[\frac{T'X}{T'Y}=\frac{T'X}{T'K} \cdot \frac{T'K}{T'Y}=\left(\frac{KX}{KY}\right)^2\]](http://latex.artofproblemsolving.com/5/0/4/50499bd259428f83a5088ada1ef4a1eede0e595e.png)
and ![\[\frac{QX}{QY}=\frac{BX}{CY}=\frac{BX}{XY} \cdot \frac{XY}{YC}=\left(\frac{KX}{KY}\right)^2,\]](http://latex.artofproblemsolving.com/a/2/4/a240f3778562fcc92adfaae0b330ef6b326e7771.png)
so 
is harmonic.
is a symmedian of 
is a median. We have 
,
is cyclic, which means 
.
is cyclic and tangent to 
.![\[ \measuredangle XKY = \measuredangle XBP + \measuredangle XCY = \measuredangle YXP + \measuredangle PYX = \measuredangle YPX = \measuredangle XP'Y \]](http://latex.artofproblemsolving.com/5/c/0/5c02d2ff373d2287200e65c3417f82acad46e6c4.png)
gives that it is cyclic.![\[ \measuredangle AXY = \measuredangle BXY = \measuredangle BXP + \measuredangle PXY = \measuredangle BXP + \measuredangle PBX = \measuredangle BPX = \measuredangle YPX \]](http://latex.artofproblemsolving.com/d/d/5/dd550328b47cc7b3ab0abceded9c1a6ed1773fd3.png)
as desired, symmetry suffices. 
,
.
is harmonic.
is cyclic.![\[ \measuredangle EQG = \measuredangle EMG = \measuredangle EP'G = \measuredangle EK'G = 90^\circ \]](http://latex.artofproblemsolving.com/0/3/9/039cb7a2db9d327544d7442a60910930c19fb8ab.png)
,
,
.
holds as well, and thus ![\[ \measuredangle QXA = \measuredangle QGA = \measuredangle QGM = \measuredangle QKM = \measuredangle QKP \]](http://latex.artofproblemsolving.com/1/7/5/175514a68ae9840a5558c4974b6a60eb7e10dad3.png)
as desired.
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where the last follows by the fact that 
where the last line again follows from the fact that 
.
