[/quote]
,
.
-
intersect its circumcircle 
at 
.
determine a triangle 
,
is relative to 
and so on. Prove that lines 
are concurrent at a point 
and that 
,
be the result.
,
is the Euler's totient theorem, which calculates the number of integers less than 
.
for 
.
be the remainder 
leaves when you divide it by 
.
be your predicted USA(J)MO score.![$\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$](http://latex.artofproblemsolving.com/c/d/3/cd3f7d917b0ec944ec3017e1f737c538e1edd10e.png)
.
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
.
,
.
and 
to be the remainder when 
is divided by 
.
and 
is odd for 
.
or an equivalent formulation. To prove this construction works, we need to show that
is always an integer between 
and 
inclusive.
is divided by 
.
,
since it telescopes.
,![\[\frac{r_{i+1} n - r_i}{2^i} > \frac{r_{i+1} n}{2^i} - 1 \geq \frac{n}{2^i} - 1 \geq \frac{n}{2^k} - 1.\]](http://latex.artofproblemsolving.com/4/2/0/420d26b6099ca264846885e7eef23b2c9c7b2dd0.png)
Thus, each digit is lower bounded by 
,
fails.
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
,
such that 
has no nonreal roots. We can also assume 
case is trivial as the constant term must be nonzero, so fix 
.
.
.
which has the 
for all integers 
.
be the coefficient of the 
term of 
for all 
,
for which 
.
and 
.
and 
.
is the coefficient of the 
term in 
and 
.
,
.
,
)
has 
distinct real roots as well, along with two consecutive nonleading coefficients equal to zero, but one degree lower, by the Power Rule. By doing this repeatedly, we eventually end up with a polynomial where the two consecutive coefficients have degree 
,
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
distinct from 
,
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
,
to 
,
to 
.
to be the set of points strictly outside the circle with diameter 
.
is always acute. Suppose two roads 
and 
cross. Then, 
is a convex quadrilateral. Since 
.
is not acute but there does not exist an 
such that 
are connected by some path. This is trivially true for 
,
are also connected.
is obtuse, we have 
.
,
,
is not mentioned. Do not reward points if the induction argument is incomplete or incorrect.
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
.
intersects line 
.
be the center of 
.
be the intersection of 
and the line through 
,
.
be the foot of the altitude from 
and 
,
is a midline and thus 
.
is a midline of 
,
and thus 
.
(by definition), by either HL congruence or the Pythagorean theorem, we must have 
.
.
and 
to attempt to identify 
.
or 
,
.
.![\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]](http://latex.artofproblemsolving.com/3/c/5/3c5984d0003c1ca2df2e68841dd9e93ed6c2d27c.png)
is an integer for every positive integer 
is divisible by 
,
be a prime power dividing 
.
.
,
if 
and 
otherwise, so for each term, either both the numerator are divisible by 
,
,
,
.
has an inverse modulo 
.![\[\binom ni \equiv \pm 1 \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor} \pmod{p^a}.\]](http://latex.artofproblemsolving.com/0/1/c/01c00beb6efc07ed028fd0273f262b8bfb965d33.png)
modulo 
.
,![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \equiv p \equiv 0 \pmod p.\]](http://latex.artofproblemsolving.com/2/e/c/2ececb6afa8a10e116056cdc57a6c948298d0c29.png)
where 
,
satisfies the induction hypothesis for the prime 
.![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}^k \equiv p\sum_{i=0}^d \binom di^k \pmod {p^a}\]](http://latex.artofproblemsolving.com/6/c/9/6c9a4b6aac62b1befef1511451c9fa30993d2cf7.png)
By the induction hypothesis, 
divides the inner binomial sum, so since we are multiplying it by 
.
.
is incorrect, reward up to 
points total.
.
between the set of people except Evan 
if the total score of the cupcakes in that bubble with respect to 
.
denotes the set of neighbors of 
people.
.
from the graph, noting that no person in 
and 
.
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
intersect 
at 
and 
and 
.
Where 
such that :
.
be reals such that 
Prove that 
Let 
Prove that 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
consecutive beads every color appears at least once. Find the largest value of 
possible.
This easy via the binomial theorem for the quantity is just 
,
Y by jam10307, Davi-8191, yojan_sushi, blacksheep2003, Ultroid999OCPN, HsuAn, Toinfinity, OlympusHero, sugar_rush, Adventure10, megarnie, jhu08, suvamkonar, Lamboreghini, mathmax12, Mango247, ESAOPS, Rounak_iitr, and 6 other users
Let 
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
.
.
on 
.
.
,
,
.
,
.
and 
.
.
is the circumcenter of 
then 
so 
is cyclic. Then 
.
.
is a right triangle.
,
is taken to 
is taken to 
and is the midpoint of 
,
all lie on a circle.
to 
.
to 
.
and 
.
is a symmedian.
is a median, use the lemma again to show that 
,
,
.
and similarly 
,
.
so 
.
and 
.
so 
.
,
to the midpoint of 
and 
is cyclic.
.
and 
is 
and 
.
,
.
.
,
.
intercepts the minor arc 
in the circumcircle of 
.
,
is cyclic.
is directly similar to 
![\[ \angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A
\]](http://latex.artofproblemsolving.com/e/1/3/e13a8654a1709a7a02d1fa646bda4c1385cebc6a.png)
.![\[ \angle ENM = 90 - \angle MNC = 90 - A
\]](http://latex.artofproblemsolving.com/2/4/4/244049c0ce519505cd45904aa098efa03c182832.png)
,
is the medial triangle of 
.
.
since 
.
.![\[ \angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM
\]](http://latex.artofproblemsolving.com/1/6/b/16b4796a6589fb88d210d716ae46f00788d4dc96.png)
.
.
so 
by SAS similarity. Hence![\[ \angle EMO = \angle ENF = \angle ONF
\]](http://latex.artofproblemsolving.com/2/e/c/2ecaec60677d316cdf0e8c0f9bb44e2536dfcdc4.png)
is directly similar to 
.
is directly similar to 
.![\[ \angle EMO = \angle DMO = \angle DPF = \angle OPF
\]](http://latex.artofproblemsolving.com/5/e/a/5ea9ca703fd13b16d514858e04d72f2abd8cc4a7.png)
,
is cyclic. In other words, 
.
,
is cyclic. In other words, 
be an isosceles triangle with 
(
and 
on 
such that 
.
denotes the intersection of 
and 
and the gable(an art term) pass through 
.
,
intersect the line through 
meets 
(
.
and 
we want to show that 
is a harmonic range. This follows from![\[ \frac {AF}{FE} = \frac {AA'}{EE'} = \frac {AC}{EC}
\]](http://latex.artofproblemsolving.com/4/f/7/4f7833c91c0644bccae109bfa016346fb8ac6d43.png)
)
.
intersect at 
,
is a harmonic range and by the problem above, we conclude that 
.
,![\[ \frac {AF'}{F'E} = \frac {PA^2}{PE^2} = \frac {AC}{EC}
\]](http://latex.artofproblemsolving.com/5/b/0/5b024deeae4d226b5fbccb9b8c4ec40e4ade8137.png)
are six points on a circle then 
concur iff 
This result come from a slightly stronger version of the 1988 IMO Longlist Problem 88.
where 
is the median.
which is obviously true.
be the symmedian, and 
be a point on the symmedian such that 
is perpendicular to 
,
and similarly 
,
,
as the reference triangle, with A = (1:0:0), B=(0:1:0) and C = (0:0:1). Calculate the midpoints of BC, AC and AB and we get (0:1/2:1/2), (1/2:0:1/2) and (1/2:1/2:0) the points are M, N and P. So let's calculate the perpendicular bissector of AB and AC. By the Evan's Forgotten Trick (Perpendicularity Criterion), two vectors 
with coordinates 
and 
with coordinate
.
and AC :
are in the form (t :1/2 : 1/2), plugging these we can calculate t and ,furthermore, calculate the coordinates o D and E :
Now, let's calculate the intersection of lines BD and CE :
and points on CE are in the form : 
.
and CE by 
,
equation that is :
and find the constants u,v,w.
.
and 
.
.
which is clearly equal to zero, so, ANFP is cyclic.
.
s![[asy]
import olympiad;
size(10cm);
pair A = dir(120);
pair B = dir(210);
pair C = dir(330);
pair M = (B+C)/2;
pair N = (A+C)/2;
pair P = (A+B)/2;
pair O = circumcenter(A,B,C);
pair D = extension(A,M,P,O);
pair E = extension(N,O,A,M);
pair F = extension(B,D,E,C);
pair Q = extension(A,F,B,C);
draw(A--B--C--A, darkmagenta);
draw(P--N, orange);
draw(N--M, orange);
draw(P--M, orange);
draw(A--M, orange);
draw(P--O, orange);
draw(E--N, orange);
draw(B--D, orange);
draw(C--F, orange);
draw(O--F, orange);
draw(A--Q, fuchsia);
draw(circumcircle(A,P,N), purple + dashed);
markscalefactor = 0.02;
draw(anglemark(D,B,A), deepgreen);
draw(anglemark(B,A,D), deepgreen);
draw(anglemark(A,C,E), cyan);
draw(anglemark(E,A,C), cyan);
markscalefactor = 0.006;
draw(rightanglemark(O,F,A), blue);
markscalefactor = 0.008;
draw(anglemark(B,F,Q));
draw(anglemark(Q,F,C));
markscalefactor = 0.01;
draw(anglemark(B,F,Q));
draw(anglemark(Q,F,C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$P$", P, dir(P));
dot("$O$", O, dir(O));
dot("$D$", D, NE);
dot("$E$", E, dir(E));
dot("$F$", F, NW);
dot("$Q$", Q, dir(Q));
[/asy]](http://latex.artofproblemsolving.com/0/3/4/034c594f58b53d14c17014dd4fda7e8950ace0c1.png)
is the angle bisector of 
.
.
and 
.![\[\frac{\overline{FB}}{\sin(\angle BAF)} = \frac{\overline{AF}}{\sin(\angle ABF)} \qquad \text{and} \qquad \frac{\overline{AF}}{\sin(\angle ACF)} = \frac{\overline{CF}}{\sin(\angle FAC)}.\]](http://latex.artofproblemsolving.com/7/8/a/78a7a08804be7595a64a85ebe7c83f03279a3350.png)
Dividing both equations, eliminating 
and putting 
and 
,![\[\frac{\sin(\angle ACF)}{\sin(\angle ABF)} = \frac{\overline{BF}}{\overline{FC}} \cdot \frac{\sin(\angle FAC)}{\sin(\angle FAB)} = \frac{\sin(\angle MAC)}{\sin(\angle MAB)}.\]](http://latex.artofproblemsolving.com/b/7/f/b7f89c57ff7af1cde76dbeaabbbcc6fc0417f145.png)
Applying ``Ratio Lemma'' on ![\[\frac{\overline{BF}}{\overline{FC}} \cdot \frac{\overline{QC}\cdot \overline{AB}}{\overline{BQ}\cdot \overline{AC}} = \frac{\overline{AB}}{\overline{AC}} \iff \frac{\overline{FB}}{\overline{FC}} = \frac{\overline{QB}}{\overline{QC}}\]](http://latex.artofproblemsolving.com/0/9/3/0931db781296545a84247bddecda8a35fd795e35.png)
By the converse of ``Angle Bisector Theorem'', we get that 
and the claim is proven. 
e
since 
and 
are exterior angle bisectors by symmetry. Due to the above claim, 
is exterior angle bisector of 
.
.![\[\angle OFN = \angle ONA = \angle OPA = 90^\circ\]](http://latex.artofproblemsolving.com/e/c/e/ece57554de860ccc1a91e3e87834f377b8989d78.png)
we get that 
with diameter 
as desired. 
,
are concyclic, where 
be the point on 
such that 
is a symmedian. Then 
in 
,
,
,
.
.
.
.
.
are collinear. Similarly we also get that 
are collinear and we are done.
,
is the intersection of the A-symmedian with the circumcircle. Note that 
,
to 
ratio and it satisfies the circumcircle equation. Thus, letting 
(we make them the same sum so we can just average them), we have that the midpoint of 
Denote this point by 
intersects 
.
and 
,
The equation of 
,
Note that 
and 
This is done so that 
and 
(strictly speaking this is not exactly 
,
which is clearly true. Hence, by EFFT, 
so 
.
.
by definition (and thus 
,
,
and 
,
inversion, then a homothety of scale factor 
centered at 
be the 
be the midpoints of 
and 
respectively. Let 
and 
meet on 
.
of 
and suppose it intersects 
.
,
is the midpoint of arc 
,
.
is right (since 
;
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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 = -6.56, xmax = 5.32, ymin = -4.42, ymax = 7.74; /* image dimensions */
draw((-1.16,6.34)--(-2.88,-0.96)--(4.18,-0.9)--cycle, linewidth(0.4));
draw(arc((4.18,-0.9),0.6,158.05618236276638,180.48692124781314)--(4.18,-0.9)--cycle, linewidth(2) + blue);
draw(arc((-1.16,6.34),0.6,-76.01940149306334,-53.58866260801658)--(-1.16,6.34)--cycle, linewidth(2) + blue);
draw(arc((-1.16,6.34),0.6,-103.25801748053517,-76.01940149306334)--(-1.16,6.34)--cycle, linewidth(2) + red);
draw(arc((-2.88,-0.96),0.6,0.48692124781311374,27.725537235284936)--(-2.88,-0.96)--cycle, linewidth(2) + red);
draw(arc((-2.88,-0.96),0.6,49.50336653199302,76.74198251946484)--(-2.88,-0.96)--cycle, linewidth(2) + red);
draw(arc((4.18,-0.9),0.6,126.41133739198344,148.8420762770302)--(4.18,-0.9)--cycle, linewidth(2) + blue);
/* draw figures */
draw((-1.16,6.34)--(-2.88,-0.96), linewidth(0.4));
draw((-2.88,-0.96)--(4.18,-0.9), linewidth(0.4));
draw((4.18,-0.9)--(-1.16,6.34), linewidth(0.4));
draw((-1.16,6.34)--(0.65,-0.93), linewidth(0.4));
draw((-2.88,-0.96)--(0.2481425288175455,0.6840905057991408), linewidth(0.4));
draw((0.2481425288175455,0.6840905057991408)--(4.18,-0.9), linewidth(0.4));
draw((-2.88,-0.96)--(-0.14104396110317108,2.2472870702873227), linewidth(0.4));
draw((-0.4421832207949546,1.8946563452047962)--(4.18,-0.9), linewidth(0.4));
/* dots and labels */
dot((-1.16,6.34),linewidth(2pt) + dotstyle);
label("$A$", (-1.08,6.42), NE * labelscalefactor);
dot((-2.88,-0.96),linewidth(2pt) + dotstyle);
label("$B$", (-3.28,-0.9), NE * labelscalefactor);
dot((4.18,-0.9),linewidth(2pt) + dotstyle);
label("$C$", (4.26,-0.82), NE * labelscalefactor);
dot((0.65,-0.93),linewidth(1pt) + dotstyle);
label("$M$", (0.74,-0.9), NE * labelscalefactor);
dot((0.2481425288175455,0.6840905057991408),linewidth(2pt) + dotstyle);
label("$Q$", (0.32,0.76), NE * labelscalefactor);
dot((-0.14104396110317108,2.2472870702873227),linewidth(1pt) + dotstyle);
label("$D$", (0.12,2.3), NE * labelscalefactor);
dot((0.015671121441022638,1.6178292525545666),linewidth(1pt) + dotstyle);
label("$E$", (0.16,1.52), NE * labelscalefactor);
dot((-0.4421832207949546,1.8946563452047962),linewidth(2pt) + dotstyle);
label("$F$", (-0.7,1.82), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/b/d/9bd60ac35e48a1b5c5a42e5a74d222ea83eb6fcc.png)
H
and 
then 
sends 
(where 
finishes. 
.
.
,
is collinear. Similarly 
is collinear, hence 
.
and 
.![\[\frac{AD}{DM}\cdot\frac{MB}{BC}\cdot\frac{CF_B}{F_BA}=-1\]](http://latex.artofproblemsolving.com/5/e/5/5e54170ce2ac7024f3333172ae62de301b4e5615.png)
![\[\frac{AE}{EM}\cdot\frac{MC}{CB}\cdot\frac{BF_C}{F_CA}=-1\]](http://latex.artofproblemsolving.com/7/a/1/7a1664eaa620dd943d59f799bb20b5160a416700.png)
![\[\Longrightarrow\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=\frac{DM}{AD}\cdot\frac{AE}{EM}\]](http://latex.artofproblemsolving.com/c/e/1/ce165eb9c69492f598abd43877c54c1ebe4d3daf.png)
![\[\frac{BF_A}{F_AC}\cdot\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=1\]](http://latex.artofproblemsolving.com/a/e/e/aee8d8b39f862aabd18e788fb5e77ad67d0a0b0e.png)
![\[\Longrightarrow\frac{BF_A}{F_AC}=\frac{AD}{DM}\cdot\frac{EM}{AE}=\frac{BD}{DM}\cdot\frac{EM}{CE}=\frac{\sin(\angle FCM)}{\sin(
\angle MBF)}=\frac{BF}{FC}\]](http://latex.artofproblemsolving.com/0/5/5/055bf1dd252f3a766680ede6383b59b84bbee079.png)
Hence, 
.![\[\angle BFC=\angle BAC+\angle FBA+\angle ACF=\angle BAF+\angle FAC+\angle FBA+\angle ACF=2\angle BAC.\]](http://latex.artofproblemsolving.com/f/b/e/fbe29c96f075f66804ea6bc2c775490d29f7ab4d.png)
Hence, 
are concyclic. Moreover, ![\[\angle BAF+\angle FBA=\angle FAC+\angle ACF=\angle BAC=\angle BAF+\angle FAC\Longrightarrow \angle FBA=\angle FAC,\angle BAF=\angle ACF\]](http://latex.artofproblemsolving.com/b/1/5/b1590a555d679dea87082d4811e29284a18bd8f4.png)
This means 
,
are collinear, which would imply by symmetry that 
are collinear, solving the problem.
This sends 
is a parallelogram. Additionally, the perpendicular bisector of 
We wish to show that 
are concyclic. Indeed, 
so we are done.
and 
.
