AoPS Academy
denote the set of all positive integers less than 
that are relatively prime to . Let 
. Is the polynomial 
reducible over the rational numbers, given that it has integer coefficients?
blocks, each weighing at least 
; their total weight is 
. Prove that for every real number 
with 
you can choose a subset of the blocks whose total weight is at least but at most 
.
such that 
.
chessboard. Initially, Azer colors all cells of this chessboard with some colors. Then, Babek takes 
rows and columns and looks at the 
cells in the intersection. Babek wants to have all these cells in a same color, but Azer doesn't. With at least how many colors, Azer can reach his goal?
, , and 
be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of and the right edge of are on the 
-axis, and contains 
as many lattice points as does . The top two vertices of are in 
, and contains 
of the lattice points contained in . See the figure (not drawn to scale). that are in 
is 27 times the fraction of lattice points in that are in 
. What is the minimum possible value of the edge length of plus the edge length of plus the edge length of ?
and 
be positive real numbers with 
. Let 
be the maximum possible value of 
for which the system of equations![\[ a^2+y^2=b^2+x^2=(a-x)^2+(b-y)^2\]](http://latex.artofproblemsolving.com/1/3/4/13432a7d6e5c1465e08d513ee64c6c7ce99d75f9.png)
has a solution in 
satisfying 
and 
. Then 
can be expressed as a fraction 
, where 
and 
are relatively prime positive integers. Find 
.
and 
Let 
be projection of 
on 
and 
implies that 
Therefore, 
and 
are similar (Z is the projection of Q on AN). By Menelaus' theorem for 
cut by the transversal 
we obtain:
But 
due to 
(A,V,Z,N are harmonically separeted).
But 
is the midpoint of 
Consequently, is the midpoint of 
passes through 
and the conclusion follows.
is tangent to the circumcircle 
of 
. Consider the point 
as a degenerated circle, and thus, since the radical axis of , and 
are concurrent at the circles' radical center, we deduce that is the radical center of the three circles. Thus, we are left to show that 
lies on the radical axis of and . But this is immediate, since lies on the polar of wrt. and so, if we denote by 
, 
the intersections of 
with , we have that 
, which is equivalent with 
. lies on 
since 
is the power of wrt. and 
the power of wrt. 
.
. Rays 
and 
meet at 
, and rays 
and 
meet at . Let a line through parallel to meet at . Let 
and 
intersect at , and 
and 
intersect at 
. Show that 
.
. Rays and meet at , and rays and meet at . Let a line through parallel to meet at . Let and intersect at , and and intersect at . Show that .
is in harmonic conjugation as well.
we conclude that 
where 
is the point of intersection of the line segment 
from the line through and parallel to 
and 
The line connecting the circumcenter of a cyclic quadrilateral, with the point of intersection of its diagonals, is perpendicular to the line connecting the points of intersection of its opposite sidelines. 
we conclude that 
and the proof of the proposed problem is completed.
be the intersection of and 
. Then, it is very well- known that is the orthocenter of triangle 
(This result can be proved by using Pole-Polar Theory). Let 
, 
, 
be the intersections of 
, 
, 
, respectively. Consider the inversion through pole 
, power 
, where is the radius of 
, we get, 
, 
, 
, 
. Therefore, 
. Hence, 
and 
. Now, it is easy to notice that 
, which implies 
is concyclic quadrilateral, let 
be the circumcircle of . 
, 
, 
, . Therefore, 
. In the other hand, 
, then 
, hence, 
. Now, it is easy to notice that is the tangent of 
, which implies that 
, which also implies that is also tangent of 
. Therefore, 
internally tangents . Therefore, 
is also a tangent of . Therefore, 
, hence 
, which implies that lies on the bisector of 
, also lies on the bisector of (Because is considered the circumcircle of 
). Therefore, 
is the bisector of , hence, 
.
is the midpoint of 
.
be a triangle. Point 
lies on its sideline such that 
Circle passing through 
intersects 
at 
, respectively. 
meets at .Denote by the midpoint of 
Show that 
lies on its sideline then D lies on between B and C or not ?
meets 
at 
respectively then observe that 
and so 
and the angle condition means 
parallel to 
and we want 
parallel to 
. Now this is trivial by converting these angles into ratios.
![[asy]
import graph;import geometry;import olympiad;size(10cm);
pair A=(6,6),B=(0,0),C=(10,0),Dee,E,F,G,O,M,X,Y,Z;
real beta=degrees(C/A);pair Dprime=rotate(beta,A)*C;
Dee=extension(A,Dprime,B,C);
E=0.65*A;
path p=circumcircle(B,Dee,E);
pair Ghost=E+C-A;
F=extension(E,Ghost,A,Dee);
G=extension(B,F,Dee,E);
M=(A+G)/2;
O=circumcenter(B,Dee,E);
X=extension(E,F,B,C);Z=extension(X,G,A,B);Y=extension(X,G,A,C);
draw(A--B--C--A--Dee,blue);draw(E--X,magenta);draw(E--Dee,magenta);draw(B--F,magenta);draw(Z--Y--C--M,heavycyan);draw(A--X,heavycyan);draw(A--G,magenta);draw(A--O,heavygreen+dotted);draw(p,orange);
dot(A);dot(B);dot(C);dot(Dee);dot(E);dot(F);dot(G);dot(O);dot(M);dot(X);dot(Y);dot(Z);
label("$A$",A,N);label("$B$",B,SW);label("$C$",C,SW);label("$D$",Dee,SE);label("$E$",E,N);label("$F$",F,ENE);label("$G$",G,SSW);label("$O$",O,W);label("$M$",M,NW*0.4);label("$X$",X,NE);label("$Y$",Y,S);label("$Z$",Z,WNW);
[/asy]](http://latex.artofproblemsolving.com/7/4/4/7443903952eeb47b920f51045d6c5973e3000663.png)
. Note that 
, so that 
; let 
be the point at infinity in this direction.
, 
are concurrent cevians and 
, so we have 
. Intersecting this pencil with 
we have 
is the midpoint of 
. But 
is the midpoint of , so 
. But applying 
to the cyclic quadrilateral 
, we 
is the polar of 
w.r.t. 
. Combining this with 
, we arrive at 
, as desired. 
be the circumradius of 
. We want to show, by the perpendicularity lemma, that![\[ CA^2 - OC^2 = MA^2 - MO^2 \]](http://latex.artofproblemsolving.com/2/0/b/20b8e184ed9103d1faa335d386165b9e392b3749.png)
, we have that 
. Then, since 
, we have that 
. Now I claim that 
, which will of course finish the problem. By the parallelogram law, we can compute![\[ MO^2 - AM^2 = \frac{2OG^2 + 2OA^2 - 2AG^2}{4} = \frac{2(OG)(OA)\cos{\angle{GOA}}}{2} = OG(OA)\cos{\angle{GOA}} \]](http://latex.artofproblemsolving.com/9/b/8/9b859530aed7aac8b9678e91926062a667520a96.png)
is the length of the projection from 
onto 
, and this rearranges to lying on the polar of wrt , which is true by Brokard's theorem.
cuts 
at , 
cuts 
at 
, cuts at 
so 
. 
with we have 
is the midpoint of 
and lie on the radical axis of the degenerate circle and circle . lies on the circle since 
is given in the statement., and are conjugates wrt , so the circle 
is orthogonal with the circle having 
as its diameter.
as the triangle where is tangent to 
.Now, it is very much known that any circle passing through 
,
creates a parallel to . Here, similarly we can find that is parallel to .
,then we apply brocard to that circle..Well , let say 
,after applying brocard we see that 
is perpendicular to 
.Now we have to prove that is perpendicular to .So, the problem turn to showing that is parallel to .
be a triangle. Point lies on its sideline such that Circle passing through intersects at , respectively. meets at .Denote by the midpoint of Show that 
. Let 
. Brokard's Theorem implies, 
. Hence we want to show 
.
Hence, 
as desired 
, let 
, 
and reflection of over .
and projecting we get:
Now by brokard theorem we get 
thus we are done.
high-quality questions. This test was created by multiple 130+ scorers, and we have tried our very best to produce the highest quality test possible. We've blended the question types in a good distribution, and maintained a realistic difficulty. (something most AOPS mocks fail at)
and 
are 
,
,
hence 
.
,
.
and 
and 
.
is harmonic, then: 
it follow that 
.
then , 
it follow that 
.
then 
from which follow that 
.
inscribed in circle 
,
and 
and 
meet at 
.
.![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm);
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 = -7.575020763493548, xmax = 15.65983559323034, ymin = -2.020247262712083, ymax = 11.78763381008927; /* image dimensions */
/* draw figures */
draw(circle((1.000000000000000,1.240000000000001), 2.560703028466990));
draw((0.3370288777226243,3.713392263880983)--(-0.3627332875045752,-0.9279847755754601));
draw((-0.3627332875045752,-0.9279847755754601)--(2.448954595478943,-0.8713338391264570));
draw((2.448954595478943,-0.8713338391264570)--(2.040000000000002,3.580000000000004));
draw((2.040000000000002,3.580000000000004)--(0.3370288777226243,3.713392263880983));
draw((2.040000000000002,3.580000000000004)--(-0.3627332875045752,-0.9279847755754601));
draw((2.448954595478943,-0.8713338391264570)--(0.3370288777226243,3.713392263880983));
draw((1.387573930673606,10.68143930897680)--(1.074848395515719,-1.319608899361137));
draw((0.3370288777226243,3.713392263880983)--(1.387573930673606,10.68143930897680));
draw((2.040000000000002,3.580000000000004)--(1.387573930673606,10.68143930897680));
/* dots and labels */
dot((2.040000000000002,3.580000000000004),dotstyle);
label("$D$", (2.127194674704089,3.713922618019377), NE * labelscalefactor);
dot((0.3370288777226243,3.713392263880983),dotstyle);
label("$A$", (0.4298803900075759,3.851542695156932), NE * labelscalefactor);
dot((-0.3627332875045752,-0.9279847755754601),dotstyle);
label("$B$", (-0.2811566752031257,-0.7816665684740874), NE * labelscalefactor);
dot((2.448954595478943,-0.8713338391264570),dotstyle);
label("$C$", (2.540054906116755,-0.7357932094282358), NE * labelscalefactor);
dot((1.159474310641712,1.927966283743139),dotstyle);
label("$E$", (1.255600852832907,2.062481692368716), NE * labelscalefactor);
dot((1.387573930673606,10.68143930897680),dotstyle);
label("$F$", (1.484967648062165,10.82429327012639), NE * labelscalefactor);
dot((1.208053821478528,3.792236981036087),dotstyle);
label("$P$", (1.301474211878759,3.920352733725709), NE * labelscalefactor);
dot((1.074848395515719,-1.319608899361137),dotstyle);
label("$Q$", (1.163854134741204,-1.171590120363827), NE * labelscalefactor);
dot((1.273524120657659,6.304702796359971),dotstyle);
label("$M$", (1.370284250447536,6.443387481247552), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/d/8/e/d8ed7dc38f878490a8896efb96b7733808c3fa59.png)
.
are harmonic since they all lie on the polar of 
w.r.t 
as desired. ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm);
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 = -7.064792895842986, xmax = 7.877105805560314, ymin = -1.380679926520356, ymax = 7.498908442921660; /* image dimensions */
pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000);
draw((-0.9978325801052033,6.136012395039179)--(-4.591330639383286,2.274342838800053)--(1.188517849679934,3.554210678069191)--cycle, zzttqq);
/* draw figures */
draw((-0.9978325801052033,6.136012395039179)--(-4.591330639383286,2.274342838800053), zzttqq);
draw((-4.591330639383286,2.274342838800053)--(1.188517849679934,3.554210678069191), zzttqq);
draw((1.188517849679934,3.554210678069191)--(-0.9978325801052033,6.136012395039179), zzttqq);
draw((-0.9978325801052033,6.136012395039179)--(5.227320696173790,4.448547940050591));
draw((5.227320696173790,4.448547940050591)--(1.188517849679934,3.554210678069191));
draw(circle((-1.522578155329372,2.106693273412406), 3.073328518912635));
draw((0.3517848268045678,4.542285931548758)--(-4.591330639383286,2.274342838800053));
draw((-1.910282692006456,5.155468991204999)--(1.188517849679934,3.554210678069191));
draw((-0.9978325801052033,6.136012395039179)--(-0.2178527383663655,4.280931409256780));
draw((-0.9978325801052033,6.136012395039179)--(-1.522578155329372,2.106693273412406));
draw((-1.110109263303887,5.273882997698336)--(5.227320696173790,4.448547940050591));
draw(circle((-1.777726823439640,3.258937746330608), 2.980904823208507));
draw((-0.9978325801052033,6.136012395039179)--(-1.777726823439640,3.258937746330608));
/* dots and labels */
dot((-0.9978325801052033,6.136012395039179),dotstyle);
label("$A$", (-0.9434819767093537,6.230395818715658), NE * labelscalefactor);
dot((-4.591330639383286,2.274342838800053),dotstyle);
label("$B$", (-4.527767647430974,2.365857358925278), NE * labelscalefactor);
dot((1.188517849679934,3.554210678069191),dotstyle);
label("$D$", (1.254289895461516,3.649120129924374), NE * labelscalefactor);
dot((5.227320696173790,4.448547940050591),dotstyle);
label("$C$", (5.281079969975930,4.534128937509957), NE * labelscalefactor);
dot((0.3517848268045678,4.542285931548758),dotstyle);
label("$F$", (0.4135315282552104,4.637379965061608), NE * labelscalefactor);
dot((-1.910282692006456,5.155468991204999),dotstyle);
label("$E$", (-1.857991077881125,5.242135983578423), NE * labelscalefactor);
dot((-0.2178527383663655,4.280931409256780),dotstyle);
label("$G$", (-0.1617241966754200,4.371877322785933), NE * labelscalefactor);
dot((-1.522578155329372,2.106693273412406),dotstyle);
label("$O$", (-1.459737114467612,2.188855597408162), NE * labelscalefactor);
dot((-0.6078426592357844,5.208471902147980),dotstyle);
label("$M$", (-0.5452280132958403,5.301136570750796), NE * labelscalefactor);
dot((-1.110109263303887,5.273882997698336),dotstyle);
dot((-1.777726823439640,3.258937746330608),dotstyle);
label("$O_2$", (-1.725239756743288,3.354117194062513), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/8/6/3/8637a5f7df44111792ca3dde67fd0d3940aa81a5.png)
Let 
be the circumcenter of 
.
Considering 
(which is 
.
.
)
and after projecting through