be a convex quadrilateral with 
and 
.
and 
be points on segments 
,
intersects lines 
and 
,
.
be 
and the midpoint of 
.
and 
points are marked on it, dividing it into 
and announces this number to Vasya. To win the game, Vasya needs to color all marked points using 
colors, such that each color is assigned to exactly two points, and for each pair of same-colored points, one of the arcs between them contains exactly 
marked points. Find all 
board has some of its cells coloured red, blue, or green. Each cell is coloured with at most one colour, and some cells may remain uncoloured. Additionally, there is at least one cell of each colour. Two coloured cells are said to be friends if they have different colours and lie in the same row or in the same column. The following conditions are satisfied:
,
be positive integers. Prove that ![\[ \sum_{i = 1}^n \min\left(\left\lfloor \frac{x}{i} \right\rfloor, m \right) = \sum_{i = 1}^m \min\left(\left\lfloor \frac{x}{i} \right\rfloor, n \right). \]](http://latex.artofproblemsolving.com/b/7/a/b7aebfe11f646d604d2e8cb9d5bd70ec796949e8.png)
Y by nguyendangkhoa17112003, Adventure10, son7, Mango247, crazyeyemoody907, and 1 other user
Let 
be a triangle. Choose a point 
in its interior. Let 
be a circle passing through 
and and 
be a circle passing through 
and so that the other point of intersection of the two circles lies on 
. Let and intersect side 
at 
and 
, respectively. Denote by 
the intersection of 
, 
and 
the intersection of 
. Show that 
.
be a triangle. Choose a point 
in its interior. Let 
be a circle passing through 
and and 
be a circle passing through 
and so that the other point of intersection of the two circles lies on 
. Let and intersect side 
at 
and 
, respectively. Denote by 
the intersection of 
, 
and 
the intersection of 
. Show that 
.
and 
.
thus 
is cyclic. Analogously, 
is cyclic too.
all lie on a circle. From here we obatin that 
and 
.
be the second point of intersection of the two circles, which lies on 
.
in triangle 
gives us:
.
in triangle 
gives us:
.
is parallel to 
,
- using 
that is equivalent to proving:
,
.
.
.
,
and 
and define 
,
and 
.
and 
,
is the second intersection of 
.
never appears again, because the degree of freedom described by 
.![\[ (DED_1) : -a^2yz-b^2zx-c^2xy + \left( x+y+z \right)\left( \frac{L}{n} z \right) = 0. \]](http://latex.artofproblemsolving.com/c/0/0/c00fc5f09643ba9f24df1dd5879fcdf4acae76de.png)
and dividing through by 
(so that 
)
;![\[ B = (0 : L : na^2 - L). \]](http://latex.artofproblemsolving.com/0/d/f/0df8e8528686ea48736c7f92ffd3d09b2b79a028.png)
Then computing 
,
,![\[
X = \left( -\det \left[ \begin{array}{cc} v & m \\ 0 &n \end{array} \right]
: 0
: \det \left[ \begin{array}{cc} m & n \\ L & na^2 - L \end{array} \right] \right)
\]](http://latex.artofproblemsolving.com/4/7/0/4709359f1a677f645b3c8bb91f5e6cd143e847d9.png)
That is, ![\[ X = \left( -vL: 0 : mna^2-mL-nL \right). \]](http://latex.artofproblemsolving.com/7/c/f/7cf793c0900dd5f6a955a186b3543281cd5c1845.png)
Similarly, 
.
and we're done.
and eq of circles are 
and 
.
.
so intersection point of 
is 
.
,
,
at 
and 
and 
,
is 
and 
,so 
is cyclic. so 
,
,
,
.
be the point on 
such that 
.
and so 
is cyclic. Notice also 
since 
is cyclic. Therefore 
is cyclic. So 
'
so we're done.
.
,
,
,
,
the second intersection of 
Let 
and 
Then using directed angles, we have that
Therefore, we have that 
,
are concyclic. It follows that ![\[\measuredangle D'RS = \measuredangle D'DS = \measuredangle D'DE = \measuredangle D'BE = \measuredangle D'BC.\]](http://latex.artofproblemsolving.com/8/0/a/80abb1174e6531d48c8848768f17237813edf130.png)
Then, since 
,
Since these three lines all concur at one point (that is, 
)
and 
are in perspective. By Desargue's Theorem, it follows that 
are collinear. Therefore, 
lies on 
,
It follows that 
and 
;
is cyclic. This follows quickly from Power of a Point; let the second intersection of the two circles be 
,
.
is cyclic. It suffices to show that 
is cyclic, since the total conclusion follows from symmetry. Remark that 
and 
by the previously established cyclicity.
.
as desire
and 
is cyclic. This is true since 
.
is cyclic. So 
,
.
![[asy]import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-3.715082517158455,xmax=8.57101545322826,ymin=-0.8000376867959583,ymax=6.082084462204611;
pen ubqqys=rgb(0.29411764705882354,0.,0.5098039215686274), uuuuuu=rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666), yqqqqq=rgb(0.5019607843137255,0.,0.), qqffff=rgb(0.,1.,1.);
pair A=(0.4149689204417927,5.0446348930406835), B=(-0.8700481876639894,0.45559198934964207), C=(4.982241647714512,0.4482712566770508), D=(1.0490909090909089,2.889090909090912), X_1=(1.8477622637606381,0.17420124349384086), F=(1.537023192408008,0.4525809412653818), X=(0.5330076578212042,5.4661738956848795), Y=(-0.004570323506018968,5.466846361472242), T=(1.6558161096067847,3.7958842185943857);
D(A--B--C--cycle,blue);
D(A--B,blue); D(B--C,blue); D(C--A,blue); D(CR((0.5886467082123572,1.2787144125300975),1.6749093476359163),gray); D(CR((3.2616521993693435,2.0650642018842325),2.3610267845639625),gray); D(Y--X,blue); D(CR((0.26247397537495465,4.071780606014076),1.420394705571974),yellow); D(T--D,yqqqqq); D((-0.2304653826081835,2.739665029583861)--T); D(A--X_1,linetype("2 2")+qqffff); D(Y--A); D(X--A); D(F--X,yqqqqq); D((2.0452777254599024,0.4519451566971582)--Y,yqqqqq); D((-0.2304653826081835,2.739665029583861)--D,yqqqqq);
D(A,linewidth(3.pt)+blue); MP("A",(0.6,5.017183344460874),NE*lsf,fp+blue); D(B,linewidth(3.pt)+blue); MP("B",(-0.8064720015300348,0.5509562685505738),NE*lsf,fp+blue); D(C,linewidth(3.pt)+blue); MP("C",(5.0425371227937825,0.5509562685505738),NE*lsf,fp+blue); D(D,linewidth(3.pt)+red); MP("D",(1.2120719082230216,2.8238049228394453),NE*lsf,fp+red); D((2.0452777254599024,0.4519451566971582),linewidth(3.pt)+ubqqys); MP("E",(2.149820653698851,0.4873800824166193),NE*lsf,fp+ubqqys); MP("\omega_1",(-1.3468695836686484,1.7589038050957084),NE*lsf,fp+gray); D(X_1,linewidth(3.pt)+uuuuuu); MP("\omega_2",(3.3418741437104984,3.6820834356478307),NE*lsf,fp+gray); D(F,linewidth(3.pt)+ubqqys); MP("F",(1.545846885426283,0.4873800824166193),NE*lsf,fp+ubqqys); D(X,linewidth(3.pt)+ubqqys); MP("X",(0.5922040934169649,5.557580926599487),NE*lsf,fp+ubqqys); D(Y,linewidth(3.pt)+ubqqys); MP("Y",(0.05180651127835134,5.557580926599487),NE*lsf,fp+ubqqys); D((-0.2304653826081835,2.739665029583861),linewidth(3.pt)+uuuuuu); MP("S",(-0.45680297779328494,2.8238049228394453),NE*lsf,fp+uuuuuu); D(T,linewidth(3.pt)+uuuuuu); MP("T",(1.720681397294658,3.8887060405831826),NE*lsf,fp+uuuuuu);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/6/7/5/675a4b58a49dab2db00bc329bfedea877cbb44e3.png)
be the second intersections of 
with 
with 
,
is cyclic. Now we have the following equality of angles:![\begin{align*} \angle SXD=&\angle BXF\\
=&\angle XFC-\angle XBF\\
=&\angle DFC-\angle SBC\\
=&\angle DTA-\angle STA\qquad\text{[since quadrilaterals DFCT and SBCT are cyclic]}\\
=&\angle STD\end{align*}](http://latex.artofproblemsolving.com/4/0/5/405823c997f694fedf6ec8adcd191574194a2231.png)
Thus points 
are concyclic. Similarly it may be shown that 
are concyclic; thus it follows that 
are concyclic (all lie on the circumcircle of 
)
From cyclic quadrilateral 
,
From 
and 
,
.
.
.
for some 
.
,
and 
satisfy these two equations:![\[a^2qr+b^2rp+c^2qp=(p+q+r)(pu+rw)\text{ and }a^2qr+b^2rs+c^2qs=(s+q+r)(su+rw).\]](http://latex.artofproblemsolving.com/b/0/d/b0dad6858a54a72cd0403665a7ac62b28cd830bc.png)
To find 
.
,
Now if 
we get 
.
,![\[\begin{vmatrix} y & 0 & 1-y \\ p & q & r \\ 0 & w & a^2-w \end{vmatrix}=0\iff w(p-y(p+q+r))+a^2qy=0.\]](http://latex.artofproblemsolving.com/9/3/b/93b6242b014bd5fc5a7fa88cee5e83f96cc4c9b2.png)
After multiplying by 
,
and 
and 
.
.
,
This means that 
.
.
and 
.
and thus the equation of radical axis is 
.
.
.
.
Similarly, if 
.
,
intersect 
,
intersect 
.
cyclic by Pop. So, 
.
cyclic. Similarly, 
cyclic, which means that 
,
be a point. Let 
be two points such that 
Let 
Let 
and 
Then 
We need to show 
is the same map as 
so need to check it for three positions. The case 
is obvious. The cases when 
and 
are easy by using the hypothesis 
doesnot lie on the segments (sorry)
and 
is cyclic. Let 
Thus, 
.
,
,
and so we can set 
.
Intersecting the cevians 
and 
we get 
.
and we're clearly done.
![[asy]
size(8cm);
defaultpen(fontsize(9pt));
import geometry;
pair A = dir(110), B = dir(180+30), C = dir(-30), D = (-0.2, 0.05), D1 = 1.3*D - 0.3*A;
path w1 = circumcircle(B, D, D1), w2 = circumcircle(C, D, D1);
pair E = intersectionpoints(line(B, C), w1)[1], F = intersectionpoints(line(B, C), w2)[0], G = intersectionpoints(line(A, B), w1)[0], H = intersectionpoints(line(A, C), w2)[0], X = extension(D, F, A, B), Y = extension(D, E, A, C);
draw(A--B--C--cycle, linewidth(0.9)); draw(w1^^w2, heavygray); draw(circumcircle(B, G, H), dotted); draw(circumcircle(G, H, D), dashed); draw(E--Y^^F--X, gray);
string[] names = {"$A$", "$B$", "$C$", "$D$", "$D_1$", "$E$", "$F$", "$G$", "$H$", "$X$", "$Y$"};
pair[] points = {A, B, C, D, D1, E, F, G, H, X, Y};
pair[] ll = {A, B, C, D, D1, E, F, G, H, X, Y};
for (int i=0; i<names.length; ++i)
dot(names[i], points[i], dir(ll[i]));
[/asy]](http://latex.artofproblemsolving.com/a/d/d/addbd7bc02b8b09be4054e9725d2e8843fef9827.png)
intersect the sides 
as shown. Since![\[
AD \cdot AD_1 = AG \cdot AB = AH \cdot AC,
\]](http://latex.artofproblemsolving.com/d/0/a/d0a02e9cc0511e5524bdeafb34179d5dcc47eba2.png)
is cyclic. The key observation is that 
is cyclic, which would clearly imply the problem (since then 
)![\[
\measuredangle XDH = \measuredangle FDH = \measuredangle BCH = \measuredangle BGH = \measuredangle XGH.
\]](http://latex.artofproblemsolving.com/d/0/4/d0449368d09cf1d461bdde56eb9dfd03e4f87896.png)
and let 
and 
Notice that 
where 
Letting 
we find that 
or 
Hence, 
Let 
and notice that 
so 
Similarly, 
so we are done. 
and 
is cyclic by Pop
Now angle chase finishes the problem
.
is cyclic.
is cyclic.
.
is cyclic.
.
is cyclic 
is cyclic 
and we are done 
and 
respectively. 
respectively. Let 
and 
to 
.
.
and 
.
thus 
are collinear, hence 
.
so 
is cyclic so by simple angle chasing we just need to prove 
is cyclic. we have 
is cyclic. 
is cyclic so now we have 
is cyclic.
,
.
so 
is cyclic. This yields 
which means 
.
are centrally perspective at 
,
,
are collinear, which finishes. 
.
.
,
.
,
must be an inversion at 
,
as desired.
,
Save this equation for now.
,
The equation for line 
is ![$$[q(1-e)-er]x+[p(e-1)]y+epz=0,$$](http://latex.artofproblemsolving.com/8/c/6/8c6d5218e0bf203c3825eab5f771be94310d4f73.png)
which we see 
and 
satsifies. Intersecting this with line 
we see that ![$$[q(1-e)-er]x+epz=0.$$](http://latex.artofproblemsolving.com/e/9/d/e9d15726c00c01dcce3f9f7df1d97195c6c1c6ae.png)
Note that 
By symmetry, we then have 
Therefore, 
is equivalent to 
,
and 
so 
Similarly, 
Thus, we have 
as desired.
,
,
.
and let the other intersection of 
.![\[
-a^2yz-b^2zx-c^2xy+(x+y+z)(u_1x+wz)=0.
\]](http://latex.artofproblemsolving.com/9/c/2/9c2ad77fa329f3764a8d1f11dd2ab0abd5002e19.png)
It follows that
In particular, note that if ![\[
-a^2yz-b^2zx-c^2xy+(x+y+z)(u_2x+vy)=0,
\]](http://latex.artofproblemsolving.com/f/a/8/fa83e78a93a978c767d0e0a12cfaa87229109e65.png)
we have 
by symmetry (the only thing that is different is 
being replaced with 
)
.
so 
.
so![\[
\begin{vmatrix}
x_1 & 0 & x_2\\
p & q & r\\
0 & \frac{w}{a^2} & 1-\frac{w}{a^2}
\end{vmatrix}
=0.
\]](http://latex.artofproblemsolving.com/8/7/8/8783c07452f9fa115aea1cc479348a27be2e7506.png)
Solving for 
,
.
,
.
and 
.
is parallel to 
.
at 
.
lie on a circle. We also have 
.
lie on a circle. Similarly, 
lie on a circle. So 
lie on a circle. Hence 
or 
is cyclic since,![\[AB \cdot AM = AD \cdot AP = AN \cdot AC\]](http://latex.artofproblemsolving.com/f/8/c/f8c4faffd26f6f4335c69ee187bcbecde749738f.png)
Further, we can prove the following key claim.![\[\measuredangle MXD = \measuredangle BXF = \measuredangle BFX + \measuredangle XBF = \measuredangle EFD + \measuredangle MBF = \measuredangle CFD + \measuredangle MBC = \measuredangle CND + \measuredangle MNC = \measuredangle MND \]](http://latex.artofproblemsolving.com/2/4/6/2465c6c1a1948cadbe4eb1a2c144375384d62d47.png)
from which it is clear that 
is cyclic. A similar argument show that 
is also cyclic, proving the claim.
