[/quote]
,
be a circle and 
a number from a set 
is written --- in the first row numbers 
,
and so on. Exactly 
of them have been chosen, no two from the same row or the same column. Let us denote by 
a number chosen from row number 
.![\[ \frac{1^2}{a_1}+\frac{2^2}{a_2}+\ldots +\frac{n^2}{a_n}\geq \frac{n+2}{2}-\frac{1}{n^2+1}\]](http://latex.artofproblemsolving.com/a/d/5/ad5ae6861da79a08d32093f63827016868856a3a.png)
lying outside the circle 
,
and 
.
is chosen on the segment 
.
and 
respectively. The circumcircle of triangle 
intersects 
)
passes through the centroid of triangle 
.
has circumcircle 
,
to 
and meet 
,
to 
meets 
.
.
and 
intersect the perpendicular bisectors of 
and 
,
intersect at 
.
be the set of positive rational numbers. Construct a function 
such that![\[ f(xf(y)) = \frac {f(x)}{y}
\]](http://latex.artofproblemsolving.com/5/9/1/59172e31cf6b7d885295f6e1daf6891a71739247.png)
,
in 
with 
and incenter 
.
be the mixtilinear circle at vertex 
,
and also tangent to lines 
and 
.
passes through points 
and 
and is tangent to 
,
and 
.
and 
with 
,
and 
,
,
and 
.
varies on its side 
and 
are the midpoints of the arcs 
and 
(not containing 
and 
,
passes through a fixed point, independent of the choice of 
are positive integer.
and 
respectively. Let 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
and a constant 
,
and 
are also friendly polynomials. Prove that 
.
Y by Mahmood.sy, megahertz13, itslumi, Eagle42, t131thunder3, amar_04, aansc1729, Lamboreghini, S.Ragnork1729, GeoKing, ehuseyinyigit, deplasmanyollari, magnusarg, Rounak_iitr, ItsBesi, cubres
Let 
be a convex pentagon such that 
. Assume that there is a point 
inside with 
and 
. Let line 
intersect lines 
and 
at points 
and 
, respectively. Assume that the points 
occur on their line in that order. Let line 
intersect and 
at points 
and 
, respectively. Assume that the points 
occur on their line in that order. Prove that the points 
lie on a circle.
be a convex pentagon such that 
. Assume that there is a point 
inside with 
and 
. Let line 
intersect lines 
and 
at points 
and 
, respectively. Assume that the points 
occur on their line in that order. Let line 
intersect and 
at points 
and 
, respectively. Assume that the points 
occur on their line in that order. Prove that the points 
lie on a circle.
.
are concyclic.
,
implying the result. 
and 
are rotations of each other. Since 
,
.
(by the same rotation taking 
is a cyclic quadrilateral. Hence 
,
is cyclic.
and 
again. Evidently the chord cut is parallel to 
.
is cyclic, as desired!
and 
are directly similar with scale factor 
,
so that 
and 
.
be such that 
and
which encapsulates the fact that 
.
For complex numbers 
,
if and only if 
.
,
,
.
.
We compute 
and 
.
so
and 
;
and so
Similarly,
Combining these and writing in terms of 
and 
,
So, we need to show that
is real. The numerator 
and the denominator 
so 
as desired.
.
.
.
.
is cyclic. By angle chasing, 
.
.
meet 
at 
respectively
,
,
,
,
i
are concyclic
are concyclic, 
by Reim, the rest is easy, 
implies 
are concyclic
since their sidelengths are all equal. Now 
and therefore 
.
and therefore
are concylcic. Finally write
and 
![\[ \angle VQU = \angle BTC - \angle TBQ = \angle DTE - \angle TES = \angle VSU \rightarrow SVUQ \text{
cyclic}\]](http://latex.artofproblemsolving.com/c/0/6/c060fbaed347324d7f0ec88ef3fefa41f569f074.png)
.![\[ \frac{TV}{TD} = \frac{TV}{TB} = \frac{TU}{TE} = \frac{TU}{TC} \rightarrow VU || CD || PR \]](http://latex.artofproblemsolving.com/8/7/a/87a8b7bc9d17fddfc6e26c64a45491ac573a0568.png)
and we are done by Reim's theorem.
and 
are congruent (3 equal sides), so there is a spiral similarity (actually rotation) that takes 
and 
are similar isosceles triangles, so 
.
(due to the congruent triangles), so 
and 
is cyclic. So we are left to prove 
,
,
,
and 
are similar. Hence 
,
.
we have 
.
.
and 
since their sides are equal by the problem condition, so 
yields 
.
,
,
- that is, 
.
,
which would follow from 
.
(as 
One can deduce that quadrilateral 
is cyclic by simple angle chasing. Thus, it suffices to show that
Note that 
which implies
as required. 
.
.
we have 
so if 
are 
and 
respectively, then 
so 
so it follows that 
we have 
.
.
and 
.
and 
.
.
(
.
due to 
.
,
.
,
,
cyclic, as desired.
since
Note by LoS
Dividing gives 
,
Which implies the cyclicity 
be the intersection of 
with 
,
be the intersection of 
with 
.![\[ \angle BQC = 180^{\circ} - \angle QBC - \angle TCB = 180^{\circ} - \angle TBC - \angle QBT - \angle TCB = \angle BTC - \angle QBT.\]](http://latex.artofproblemsolving.com/0/6/0/0604afdd1270c0c444c6ab193bc94d036cc7f31b.png)
Since 
,
,
is cyclic. By Reim's, it suffices to prove that 
.
and![\[\angle P_1TB = \angle P_1TC - \angle BTC = \angle R_1TD - \angle ETD = \angle R_1 TE,\]](http://latex.artofproblemsolving.com/e/2/4/e24af8171d9ebe67daf8ae30664ad1145dbc2148.png)
it follows that 
.![\[ \frac{P_1T}{R_1T} = \frac{BT}{TE} = \frac{TD}{TC}, \]](http://latex.artofproblemsolving.com/7/a/b/7abbd2214ed21dbdeedf0e6222e84c535603c68b.png)
as desired.
.
and 
.
,
and 
is cyclic.
,
,
and 
intersect 
and 
is cyclic
since 
and 
are also similar we get that 
impying the result.
and 
.
and 
,
we have 
.
.
is cyclic.![\[\measuredangle NSM = \measuredangle EST = \measuredangle ETD + \measuredangle SET = \measuredangle CTB + \measuredangle TBQ = \measuredangle TQB = \measuredangle NQM \]](http://latex.artofproblemsolving.com/0/1/3/0135046b93e6071c649d6a720ae35e978466bb5b.png)
from which the claim clearly follows.
and 
are parallel.![\[\measuredangle NTE = \measuredangle NTD + \measuredangle DTE = \measuredangle CTM + \measuredangle BTC = \measuredangle BTM\]](http://latex.artofproblemsolving.com/2/2/e/22e0aed6356f1a0a3427ee1aa6d7aa84fdf38fe3.png)
Thus, combined with the condition that 
we conclude that 
.![\[\frac{MT}{NT} = \frac{TB}{TE} = \frac{TD}{TC}\]](http://latex.artofproblemsolving.com/b/d/3/bd306990a0c7acd890069cbf8311f16244172954.png)
so, since 
quite clearly, 
.![\[\measuredangle CNM = \measuredangle TNM = \measuredangle TCD= \measuredangle NCD\]](http://latex.artofproblemsolving.com/c/9/9/c99cd24a5bac51646abfa64024ad86c5a4c81196.png)
which proves the claim.
.
,
is cyclic. Then Note 
is similar to 
,
,
is cyclic, giving 
is parallel to 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(13cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -18.521900412574336, xmax = 21.368136697741033, ymin = -10.73284839332707, ymax = 12.492943448971472; /* image dimensions */
pen xfqqff = rgb(0.4980392156862745,0,1); pen ffccww = rgb(1,0.8,0.4); pen ffwwqq = rgb(1,0.4,0); pen ffxfqq = rgb(1,0.4980392156862745,0); pen ffdxqq = rgb(1,0.8431372549019608,0); pen qqccqq = rgb(0,0.8,0); pen ffqqff = rgb(1,0,1); pen qqzzff = rgb(0,0.6,1);
draw((-2.4620330926517306,6.161603298728669)--(-7.434523977531531,0.6052976945714785)--(-5.502958796936164,-4.822248899827327)--(3.6499876243977587,-4.529153459206897)--(4.547664130336512,0.9841990607473077)--cycle, linewidth(2));
/* draw figures */
draw((-2.4620330926517306,6.161603298728669)--(-7.434523977531531,0.6052976945714785), linewidth(1) + xfqqff);
draw((-7.434523977531531,0.6052976945714785)--(-5.502958796936164,-4.822248899827327), linewidth(1) + ffccww);
draw((-5.502958796936164,-4.822248899827327)--(3.6499876243977587,-4.529153459206897), linewidth(1) + ffwwqq);
draw((3.6499876243977587,-4.529153459206897)--(4.547664130336512,0.9841990607473077), linewidth(1) + ffccww);
draw((4.547664130336512,0.9841990607473077)--(-2.4620330926517306,6.161603298728669), linewidth(1) + ffxfqq);
draw((-2.4620330926517306,6.161603298728669)--(2.4223649371430067,10.702336521201197), linewidth(1) + xfqqff);
draw((2.4223649371430067,10.702336521201197)--(-5.502958796936164,-4.822248899827327), linewidth(1) + ffdxqq);
draw((-7.434523977531531,0.6052976945714785)--(-13.106020700468857,-5.473195980609238), linewidth(1) + xfqqff);
draw((-5.502958796936164,-4.822248899827327)--(-13.106020700468857,-5.473195980609238), linewidth(1) + red);
draw((4.547664130336512,0.9841990607473077)--(12.580788608524934,-4.588102640620278), linewidth(1) + ffxfqq);
draw((12.580788608524934,-4.588102640620278)--(3.6499876243977587,-4.529153459206897), linewidth(1) + red);
draw((-2.4620330926517306,6.161603298728669)--(-7.039193907581469,9.108018628905542), linewidth(1) + ffxfqq);
draw((-7.039193907581469,9.108018628905542)--(3.6499876243977587,-4.529153459206897), linewidth(1) + ffdxqq);
draw((-7.039193907581469,9.108018628905542)--(-5.502958796936164,-4.822248899827327), linewidth(1) + qqccqq);
draw((2.4223649371430067,10.702336521201197)--(3.6499876243977587,-4.529153459206897), linewidth(1) + qqccqq);
draw((-7.039193907581469,9.108018628905542)--(-13.106020700468857,-5.473195980609238), linewidth(1) + red);
draw((2.4223649371430067,10.702336521201197)--(12.580788608524934,-4.588102640620278), linewidth(1) + red);
draw((-7.039193907581469,9.108018628905542)--(2.4223649371430067,10.702336521201197), linewidth(1) + ffwwqq);
draw(circle((-1.1628692923240138,2.706219848718833), 8.68989174880106), linewidth(1) + blue);
draw(circle((-0.3593248081168896,-2.224008000042246), 13.154295066827547), linewidth(1) + linetype("4 4") + red);
draw((-7.434523977531531,0.6052976945714785)--(-1.8022955401431777,2.426825683397818), linewidth(1) + ffqqff);
draw((-1.8022955401431777,2.426825683397818)--(4.547664130336512,0.9841990607473077), linewidth(1) + ffqqff);
draw((-7.434523977531531,0.6052976945714785)--(3.6499876243977587,-4.529153459206897), linewidth(1) + qqzzff);
draw((4.547664130336512,0.9841990607473077)--(-5.502958796936164,-4.822248899827327), linewidth(1) + qqzzff);
/* dots and labels */
dot((-2.4620330926517306,6.161603298728669),dotstyle);
label("$A$", (-2.5867158750332107,6.842722787784495), NE * labelscalefactor);
dot((-7.434523977531531,0.6052976945714785),dotstyle);
label("$B$", (-7.898444054213586,1.0883505936724118), NE * labelscalefactor);
dot((-5.502958796936164,-4.822248899827327),dotstyle);
label("$C$", (-5.737299746017649,-5.68149904645945), NE * labelscalefactor);
dot((3.6499876243977587,-4.529153459206897),dotstyle);
label("$D$", (3.6103003340105606,-5.39508233091541), NE * labelscalefactor);
dot((4.547664130336512,0.9841990607473077),dotstyle);
label("$E$", (4.886156612343102,1.3226915427538992), NE * labelscalefactor);
dot((2.4223649371430067,10.702336521201197),dotstyle);
label("$Q$", (2.5687850047595067,11.138973520945099), NE * labelscalefactor);
dot((-13.106020700468857,-5.473195980609238),dotstyle);
label("$P$", (-13.809043547713317,-5.837726345847108), NE * labelscalefactor);
dot((12.580788608524934,-4.588102640620278),dotstyle);
label("$R$", (13.062051946963877,-4.926400432752435), NE * labelscalefactor);
dot((-7.039193907581469,9.108018628905542),dotstyle);
label("$S$", (-7.4037242728193355,9.498586877374686), NE * labelscalefactor);
dot((-1.8022955401431777,2.426825683397818),linewidth(4pt) + dotstyle);
label("$T$", (-1.7795414948636439,1.2185400098287937), NE * labelscalefactor);
dot((-1.1483638827199552,-2.3065114234796185),linewidth(4pt) + dotstyle);
label("$X$", (-1.2327459470068405,-3.15582437302564), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/8/0/4/8046ed8ccccbc41b6dd0ab7b5507f1b841f98e9e.png)
are concyclic.
Firstly the 
and 
are congruent by SSS congruency. Therefore we get angle condition that 
Now Angle Chasing we get, 
Also we get, 
Now From Angle condition 
Therefore we get, 
are concyclic points by PoP.
are also cyclic points.
We Know that 
Now in 
we get by angle chasing that 
Now in 
we get, 
Equating these we get, 
Now by a little bit Angle Chasing we get that 
are concyclic points.
be the intersections of 
with 
respectively.
are concyclic.
.
.
implying the claim.
are concyclic.
.
by basic exterior angles so 
and then 
by angle chasing. Therefore 
.
so that just finishes.
be the intersections of 
Notice that 
so 
is cyclic.
(
similarity), so 
so 
This finishes.
.![\[ \angle YSX = 180^\circ - \angle TEA - \angle TED - \angle TDE = 180^\circ - \angle TBA - \angle TCB - \angle TBC = \angle YQX. \]](http://latex.artofproblemsolving.com/9/3/6/936d1fe060bf28efa904713c82c9197640ac0152.png)
Then 
is cyclic and by Reim we require ![\[ \dfrac{XT}{YT} = \dfrac{BT}{ET} = \dfrac{CT}{DT} \implies XY \parallel CD \]](http://latex.artofproblemsolving.com/2/4/3/2431a00b32ecf9fa2b008473640854df21aa2243.png)
which finishes.
