[/quote]
,
positive integers 
such that for every index 
,
,
is a multiple of 
.
be a triangle. Let 
and 
be the points in which the median and the angle bisector, respectively, at 
meet the side 
.
and 
be the points in which the perpendicular at 
meets 
and 
,
the point in which the perpendicular at 
produced.
is perpendicular to 
and 
such that for all 
if 
then 
Find the smallest of sum of all elements of 
such that if 
and 
are distinct rational numbers satisfying 
or 
,
?
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
,
.
real solutions.
be 
given points in the plane, and let 
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points 
a non-negative real number 
,
,
.
such that 
and 
are connected by an edge for every 
.
is not fixed and the first selected vertex always has to be 
.![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)
otherwise, Alice wins. Depending on 
,
for which Bob has a winning strategy.
,
,
,
,
,
be positive integers and let 
.
divides 
and 
.
,
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
,
:
,
,
Y by Rounak_iitr, S_14159, Philomath_314, GeoKing, MihaiT, GeoGuy3264, hectorleo123
Euclid has a tool called splitter which can only do the following two types of operations :
• Given three non-collinear marked points
it can draw the line which forms the interior angle bisector of 
.
• It can mark the intersection point of two previously drawn non-parallel lines .
Suppose Euclid is only given three non-collinear marked points
in the plane . Prove that Euclid can use the splitter several times to draw the centre of circle passing through 
and 
.
Proposed by Shankhadeep Ghosh
• Given three non-collinear marked points
it can draw the line which forms the interior angle bisector of 
.• It can mark the intersection point of two previously drawn non-parallel lines .
Suppose Euclid is only given three non-collinear marked points
in the plane . Prove that Euclid can use the splitter several times to draw the centre of circle passing through 
and 
.Proposed by Shankhadeep Ghosh
This post has been edited 4 times. Last edited by S.Ragnork1729, Jan 20, 2025, 6:14 PM
of 
.
and 
.
denote the angle bisector of 
.
and mark 
.Clearly 
lies on the opposite side of 
and mark 
.
is the 
which is the 
excentrez of 
is cyclic.So by fact 
we have that 
lie on the circle 
and 
(on which 
is isosceles, we have that 
is perpendicular bisector of 
.Now note that 
is the centre of 
is the perpendicular bisector of 
or 
and mark the intersection of the perpendicular bisectors to get the circumcentre 
to get they meet at 
,
meets 
,
to draw internal and external angle bisectors of 
,
to draw 
which are the 
-
we have that 
is cyclic with diameter 
and now using device on 
to draw those two angle bisectors that meet at 
which means that using device on 
gives the perpendicular bisector of 
.
,
,
and 
.
incenter of 
,
the intersection of angle bisector of 
and 
,
.
by Fact 5 and thus we get the arc midpoint of 
in 
.
using Fact 5, and finally we have 4 points on 
,
be the incenter of 
,
be the incenter of 
and so on. Construct 
such that:![\[ n \geq 3 \quad\text{and}\quad \angle BAI_n < \frac{\angle ABC}{4} \]](http://latex.artofproblemsolving.com/f/8/d/f8d2f030e3b97f6c68fc7c7f9c8ec4ba244b8ef6.png)
Let 
and 
.
meets ![\[ \angle BPX = \frac{\angle ABP + \angle BAQ}{2} = \frac{\frac{\angle ABC}{4} + \angle BAI_n}{2} < \frac{\angle ABC}{4} = \angle ABP \]](http://latex.artofproblemsolving.com/0/b/5/0b58562779681dd1ffa7ee7694d3f834043c8088.png)
Now construct the internal angle bisector of 
and we're done!
are drawn (just replace 
meet at 
,
and 
are drawn, we can draw the external angle bisector of 
.
in 
,
is just what we want.
perpendicular bisectors and intersect them at 
(not saying we can connect the vertices).
and 
(with this we also get lines 
)
e![[asy]
pair A = (-3,5);
pair B = (-5,0);
pair C = (3,0);
pair D = (-1.3905182604822757,2.443738111322006);
pair E = (-3.1835210500174114,2.238743395351199);
pair I = (-2.212542434386075,1.8872006504994152);
pair I_B = (3.4549387988697458,2.5929474543131246);
pair I_C = (-5.454938798869745,1.4834391245672034);
import graph; size(9cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
real xmin = -7, xmax = 7, ymin = -5, ymax = 7;
draw((xmin, -3.952974084481309*xmin-6.858922253443925)--(xmax, -3.952974084481309*xmax-6.858922253443925), linewidth(0.6));
draw((xmin, 0.6770329614269008*xmin + 3.385164807134504)--(xmax, 0.6770329614269008*xmax + 3.385164807134504), linewidth(0.6));
draw((xmin, -0.36204993518133094*xmin + 1.0861498055439927)--(xmax, -0.36204993518133094*xmax + 1.0861498055439927), linewidth(0.6));
draw((xmin, 15.045993930324776*xmin + 50.13798179097433)--(xmax, 15.045993930324776*xmax + 50.13798179097433), linewidth(0.6));
draw((xmin, -1.5882515631671401*xmin + 0.23524531049857966)--(xmax, -1.5882515631671401*xmax + 0.23524531049857966), linewidth(0.6));
draw((xmin, 0.12452565341946036*xmin + 2.1627189428596245)--(xmax, 0.12452565341946036*xmax + 2.1627189428596245), linewidth(0.6)+blue);
draw((xmin, -0.1754523787641738*xmin + 0.5263571362925213)--(xmax, -0.1754523787641738*xmax + 0.5263571362925213), linewidth(0.6)+blue);
draw((xmin, 0.3066784415588851*xmin + 1.5333922077944255)--(xmax, 0.3066784415588851*xmax + 1.5333922077944255), linewidth(0.6)+blue);
draw(circle((-1,2.0381932894401587), 4.489346487532333), linewidth(0.6));
dot((-3,5),linewidth(3pt) + dotstyle);
label("$A$", (-2.74,5.09), NE * labelscalefactor);
dot((-5,0),linewidth(3pt) + dotstyle);
label("$B$", (-5.2,-0.63), NE * labelscalefactor);
dot((3,0),linewidth(3pt) + dotstyle);
label("$C$", (3.06,-0.59), NE * labelscalefactor);
dot((-2.212542434386075,1.8872006504994152),linewidth(3pt) + dotstyle);
label("$I$", (-2.22,2.27), NE * labelscalefactor);
dot((-1.3905182604822757,2.443738111322006),linewidth(3pt) + dotstyle);
label("$D$", (-1.44,2.71), NE * labelscalefactor);
dot((-3.1835210500174114,2.238743395351199),linewidth(3pt) + dotstyle);
label("$E$", (-3.1,2.35), NE * labelscalefactor);
dot((-5.454938798869745,1.4834391245672034),linewidth(3pt) + dotstyle+blue);
label("$I_C$", (-5.86,1.07), NE * labelscalefactor);
dot((3.4549387988697458,2.5929474543131246),linewidth(3pt) + dotstyle+blue);
label("$I_B$", (3.7,2.21), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/f/7/5f7b48703cb268b75539b8477e8993cb31798d4a.png)
and 
and angle bisector of 
to get the 
e
e
and 
to be cyclic with diameter 
.
w.r.t 
.![[asy]
import graph; size(9cm);
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 = -6.6, xmax = 5, ymin = -4.6, ymax = 6; /* image dimensions */
pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
/* draw figures */
draw((xmin, 0.16422203735698718*xmin + 0.8036735382581797)--(xmax, 0.16422203735698718*xmax + 0.8036735382581797), linewidth(0.6)); /* line */
draw((xmin, 0.34415073140891084*xmin + 0.3766029256356433)--(xmax, 0.34415073140891084*xmax + 0.3766029256356433), linewidth(0.6)); /* line */
draw((xmin, -0.17030362141383307*xmin-0.6593927571723338)--(xmax, -0.17030362141383307*xmax-0.6593927571723338), linewidth(0.6)); /* line */
draw(circle((-1,0.6394515009011971), 3.418742637843215), linewidth(0.6));
draw((xmin, 1.1776167667461244*xmin-1.6016743701958989)--(xmax, 1.1776167667461244*xmax-1.6016743701958989), linewidth(0.6)); /* line */
draw((xmin, -0.8491726920321478*xmin-3.6284638289741675)--(xmax, -0.8491726920321478*xmax-3.6284638289741675), linewidth(0.6)); /* line */
draw((xmin, -1936775683387793*xmin-1936775683387788)--(xmax, -1936775683387793*xmax-1936775683387788), linewidth(0.6)); /* line */
draw(circle((-1,0.375), 3.3000946956110218), linewidth(0.6) + blue);
/* dots and labels */
dot((-3,3),linewidth(3pt) + dotstyle);
label("$A$", (-2.979583901938016,3.172663134820939), NE * labelscalefactor);
dot((-4,-1),linewidth(3pt) + dotstyle);
label("$B$", (-4.133298321565583,-1.3822613270852822), NE * labelscalefactor);
dot((2,-1),linewidth(3pt) + dotstyle);
label("$C$", (1.9499231637434085,-1.4421945436893113), NE * labelscalefactor);
dot((-2.140009305907594,0.45223685007912257),linewidth(3pt) + dotstyle);
label("$I$", (-2.1405188694816033,0.6404847333007042), NE * labelscalefactor);
dot((2.3735547844263953,1.1934635407351069),linewidth(3pt) + dotstyle);
label("$I_B$", (2.504305417330681,0.8802175997168211), NE * labelscalefactor);
dot((-4.373554784426399,0.08543946106727772),linewidth(3pt) + dotstyle);
label("$I_C$", (-4.732630487605878,-0.2884801240617488), NE * labelscalefactor);
dot((-1,-2.7792911369420183),linewidth(3pt) + dotstyle);
label("$M_A$", (-2.0356357404245515,-3.195241129357166), NE * labelscalefactor);
dot((-1,0.375),linewidth(3pt) + dotstyle+blue);
label("$O$", (-0.8219881041929548,0.3258353461295508), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/7/1/5/715f5f20fcb12791617c2a2c43a49b2204cfb5c1.png)
and 
to be 
.
.
.
and 
respectively.![[asy]import graph; size(10cm);
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 = -5.6, xmax = 5, ymin = -2.5, ymax = 4; /* image dimensions */
pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
/* draw figures */
draw((xmin, 0.780776406404415*xmin + 2.12310562561766)--(xmax, 0.780776406404415*xmax + 2.12310562561766), linewidth(0.6)); /* line */
draw((xmin, -0.3507810593582122*xmin-0.2984378812835756)--(xmax, -0.3507810593582122*xmax-0.2984378812835756), linewidth(0.6)); /* line */
draw((xmin, -2.9625473478055104*xmin-5.887642043416531)--(xmax, -2.9625473478055104*xmax-5.887642043416531), linewidth(0.6)); /* line */
draw((xmin, -1.4283558314231763*xmin-0.6850674942695293)--(xmax, -1.4283558314231763*xmax-1.2850674942695293), linewidth(0.6)); /* line */
draw((xmin, 0.16422203735698718*xmin + 0.8036735382581797)--(xmax, 0.16422203735698718*xmax + 0.8036735382581797), linewidth(0.6)); /* line */
draw((xmin, 0.34415073140891084*xmin + 0.3766029256356433)--(xmax, 0.34415073140891084*xmax + 0.3766029256356433), linewidth(0.6)); /* line */
draw((xmin, -24.81288160837314*xmin-71.43864482511943)--(xmax, -24.81288160837314*xmax-71.43864482511943), linewidth(0.6)); /* line */
draw((xmin, -0.17030362141383307*xmin-0.6593927571723338)--(xmax, -0.17030362141383307*xmax-0.6593927571723338), linewidth(0.6)); /* line */
/* dots and labels */
dot((-3,3),linewidth(3pt) + dotstyle);
label("$A$", (-2.979583901938019,3.1726631348209313), NE * labelscalefactor);
dot((-4,-1),linewidth(3pt) + dotstyle);
label("$B$", (-4.1332983215655865,-1.38226132708529), NE * labelscalefactor);
dot((2,-1),linewidth(3pt) + dotstyle);
label("$C$", (1.949923163743405,-1.442194543689319), NE * labelscalefactor);
dot((-2.140009305907594,0.45223685007912257),linewidth(3pt) + dotstyle);
label("$I$", (-2.1405188694816064,0.6404847333006966), NE * labelscalefactor);
dot((-1.5427655536088263,0.918550680746443),linewidth(3pt) + dotstyle);
label("$D$", (-1.4812534868372822,1.0150673370758792), NE * labelscalefactor);
dot((2.3735547844263953,1.1934635407351069),linewidth(3pt) + dotstyle);
label("$I_B$", (2.249589246763552,1.494533069908113), NE * labelscalefactor);
dot((-2.9081806282862583,0.7216968003117091),linewidth(3pt) + dotstyle);
label("$E$", (-2.8447341645789526,0.8053010789617769), NE * labelscalefactor);
dot((-4.373554784426399,0.08543946106727772),linewidth(3pt) + dotstyle);
label("$I_C$", (-4.50788092534077,0.38576856273357235), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/4/a/b/4abc00a588b094ca8e940a01af6063aa41ca541f.png)
![[asy]
import graph; size(11cm);
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 = -6.6, xmax = 5, ymin = -4.6, ymax = 6; /* image dimensions */
pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
/* draw figures */
draw((xmin, 0.16422203735698718*xmin + 0.8036735382581797)--(xmax, 0.16422203735698718*xmax + 0.8036735382581797), linewidth(0.6)); /* line */
draw((xmin, 0.34415073140891084*xmin + 0.3766029256356433)--(xmax, 0.34415073140891084*xmax + 0.3766029256356433), linewidth(0.6)); /* line */
draw((xmin, -0.17030362141383307*xmin-0.6593927571723338)--(xmax, -0.17030362141383307*xmax-0.6593927571723338), linewidth(0.6)); /* line */
draw(circle((-1,0.6394515009011971), 3.418742637843215), linewidth(0.6));
draw((xmin, 1.1776167667461244*xmin-1.6016743701958989)--(xmax, 1.1776167667461244*xmax-1.6016743701958989), linewidth(0.6)); /* line */
draw((xmin, -0.8491726920321478*xmin-3.6284638289741675)--(xmax, -0.8491726920321478*xmax-3.6284638289741675), linewidth(0.6)); /* line */
draw((xmin, -1936775683387793*xmin-1936775683387788)--(xmax, -1936775683387793*xmax-1936775683387788), linewidth(0.6)); /* line */
draw(circle((-1,0.375), 3.3000946956110218), linewidth(0.6) + blue);
/* dots and labels */
dot((-3,3),linewidth(3pt) + dotstyle);
label("$A$", (-2.979583901938016,3.172663134820939), NE * labelscalefactor);
dot((-4,-1),linewidth(3pt) + dotstyle);
label("$B$", (-4.133298321565583,-1.3822613270852822), NE * labelscalefactor);
dot((2,-1),linewidth(3pt) + dotstyle);
label("$C$", (1.9499231637434085,-1.4421945436893113), NE * labelscalefactor);
dot((-2.140009305907594,0.45223685007912257),linewidth(3pt) + dotstyle);
label("$I$", (-2.1405188694816033,0.6404847333007042), NE * labelscalefactor);
dot((2.3735547844263953,1.1934635407351069),linewidth(3pt) + dotstyle);
label("$I_B$", (2.504305417330681,0.8802175997168211), NE * labelscalefactor);
dot((-4.373554784426399,0.08543946106727772),linewidth(3pt) + dotstyle);
label("$I_C$", (-4.732630487605878,-0.2884801240617488), NE * labelscalefactor);
dot((-1,-2.7792911369420183),linewidth(3pt) + dotstyle);
label("$M_A$", (-2.0356357404245515,-3.195241129357166), NE * labelscalefactor);
dot((-1,0.375),linewidth(3pt) + dotstyle+blue);
label("$O$", (-0.8219881041929548,0.3258353461295508), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/5/5/3/55351257f410862bb882071c7cfc70c77362f3da.png)
), can mark a point on a given line. So, I thought if the lines AB, BC, CA are given, we can easily construct the excenter, incenter, and by few steps, we can construct the circumcenter. This looked pretty easy, so next target was somehow to eliminate the lines AB, BC and CA, can we construct when only the points A, B and C are given. And then the thought popped up that BI and CI, are constructed, using these lines we can construct excenter, incenter, circumcenter of triangle BIC, and basically giving the midpoint of arc BC, through which we construct the circumcenter of ABC.
. I am sharing the solution that I originally came up with.
which are known to be cyclic, we can construct the circumcenter of the corresponding circle.
and 
,
.
is isosceles, with 
,
,
.
and 
and 
,
,
of 
.
are cyclic. Construct the center of 
using Main Lemma, call it 
in 
,
are cyclic.
(call it 
,
(again by Incenter-Excenter Lemma). So, the points 
are cyclic.
using Main Lemma. 
and 
since after this we can construct the angle bisectors of 
and 
,
at 
.
,say 
,
is the 
to mark the arc midpoint of 
(say 
.
meet 
i.e. the center of 
.
is just the perpendicular bisector of 
. In a similar way we draw the perpendicular bisector of 
is the perpendicular bisector of 
and 
.
is cyclic. We can construct the mipdoint of arc of minor arc 
)
and 
.
and 
to get the midpoint of minor arc 
and 
)
and 
