[/quote]
,
.
-
,
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 
distinct from 
and 
,
such[/center]
is directly similar to either 
.
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 
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 
intersects line 
.
be the center of 
.
be the intersection of 
and the line through 
,
.
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 
,
satisfies 
.
,
.
Y by Generic_Username, brianapa, yojan_sushi, artsolver, JustKeepRunning, v4913, HamstPan38825, jhu08, megarnie, bobjoe123, mathlearner2357, Lamboreghini, OronSH, Adventure10
Let 
be an equilateral triangle, and point 
on its circumcircle. Let 
and 
intersect at 
, 
and 
intersect at 
, and 
and 
intersect at 
. Prove that the area of 
is twice the area of 
.
Proposed by Titu Andreescu, Luis Gonzales, Cosmin Pohoata
be an equilateral triangle, and point 
on its circumcircle. Let 
and 
intersect at 
, 
and 
intersect at 
, and 
and 
intersect at 
. Prove that the area of 
is twice the area of 
.Proposed by Titu Andreescu, Luis Gonzales, Cosmin Pohoata
This post has been edited 3 times. Last edited by v_Enhance, Apr 20, 2017, 12:45 AM
Reason: correct acknow
Reason: correct acknow
,
(circumcircle equation with 
)
,
,
.![\begin{align*} \frac{[DEF]}{[ABC]} &= \frac{1}{(u+v)(v+w)(w+u)} \det \begin{bmatrix} 0 & v & w \\ u & 0 & w \\ u & v & 0 \end{bmatrix} \\ &= \frac{2uvw}{(u+v)(v+w)(w+u)} \\ &= \frac{2uvw}{(u+v+w)(uv+vw+wu)-uvw} \\ &= \frac{2uvw}{-uvw} = -2 \end{align*}](http://latex.artofproblemsolving.com/0/6/6/0664931f0ab4dbad7d4472334d2f15dc4a9701c9.png)
as desired (areas signed).
it gives the correct area ratio! 
.
,
and 
gives that 
,
,
for real numbers 
is not equal to 
,
.
,
,
,
,
are 
,
,
,
,
,
,![$\frac{[\Delta{}DEF]}{[\Delta{}ABC]}$](http://latex.artofproblemsolving.com/6/4/b/64be0d2d134bb065b9b338407e23cfd7cabb7c1c.png)
by taking the absolute value of the following determinant, as shown(I have used Wolfram Alpha here because I was too lazy to calculate it by hand)
![$|\frac{[\Delta{}DEF]}{[\Delta{}ABC]}|=2$](http://latex.artofproblemsolving.com/2/a/0/2a0f51933f91ecf355222b4c925bbbdc24fd6c72.png)
,
is twice the area of 
where 
instead of abstracting it to 
and spam 
as well as the crucial identity 
,
The original matrix has the first column the negative of what it is here, but 
is oriented clockwise so we factor out the 
and remove it because determinants give us the signed area (which would be negative).
submatrix, you get
But since 
,
,
.
Compute 
,
but then both sides are equal to 
,
![\[ \frac{1}{PD} = \frac{1}{PB} + \frac{1}{PC}. \]](http://latex.artofproblemsolving.com/b/e/d/bedb4af7a33c3d558e5e872dc7a2067d9d99cedc.png)
,
.![\[ PD \cdot PA = PB \cdot PC, \]](http://latex.artofproblemsolving.com/e/4/8/e48268d58644ad7c5fcb2a049d7d30459694e97a.png)
we're done. Indeed, note that 
so we obtain that ![\[ \frac{PA}{PB} = \frac{PC}{PD}, \]](http://latex.artofproblemsolving.com/9/b/4/9b4cc1de9fb233df2b71b961c7d830c9c5daef99.png)
which rearranges into the thing we want, as desired.
'
)
,
.![\[ [DEF] = \frac 12 \sin{120^{\circ}}(PD \cdot PE + PE \cdot PF + PF \cdot PD) = \frac 12 \sin{120^{\circ}}(PA^2 + PB^2 + PC^2). \]](http://latex.artofproblemsolving.com/d/e/9/de96576c1d6ba1463af665561cc8f42ae65cb121.png)
Hence it remains to show that 
where 
is the side length. However remember that ![\[ PA^2 + PB^2 + PC^2 = 2(PB^2 + 2 PB \cdot PC + PC^2) = 2BC^2 \]](http://latex.artofproblemsolving.com/7/2/0/720d3e5118832a09ad3a34418fb63b4dbadac22b.png)
where the last equality is just the law of cosines. Thus we're done.
then 
.
.
,
,
(normalized).
(from the barycentric area formula). We can see that this determinant is equal to 
.
(areas are signed).
. 
,
easily follow. Note that by chord intersection formula, each of points/complex numbers 
is of the form 
,![$g, h\in \mathbb{C}[ x ]$](http://latex.artofproblemsolving.com/f/1/9/f191ef031228a21f8a20fcf4186d6ff3f80d2d93.png)
and 
(
are different for different points), and so are 
.
,![$A, B\in \mathbb{C}[ x ]$](http://latex.artofproblemsolving.com/e/2/e/e2ea6a80d965f493a7161642c11a115acc850749.png)
and 
.![$A(p)=2[ABC]B(p)$](http://latex.artofproblemsolving.com/d/f/b/dfb42c74bb6192180b274a2255cd686ebcd1ba3d.png)
,
,
,
(
,
)
.![$[DEF]=\frac{d(D, EF)EF}{2}=\frac{2d(A, BC)AC}{2}=2[ABC]$](http://latex.artofproblemsolving.com/4/a/5/4a5ef50070518111402cff45756680687748bd5c.png)
.
,
(using 
)
where 
are of degree 
(using the fact that there are 
for 
.
are rational functions of 
.
are also 
.
is constant (and the correct constant). This determinant is hence of the form 
is identically 
points 
.
and 
.
.
Which is the required value.
.
,
,
.
,
.
,
,
by the angle bisector theorem. Thus, we have![\begin{align*}
\frac{[DEF]}{[ABC]}&=\frac{[AEF]-[BFD]-[CDE]-[ABC]}{[ABC]} \\
&=\frac{AE}{AC} \cdot \frac{AF}{AB}-\frac{BF}{BA} \cdot \frac{BD}{BC}-\frac{CD}{CB} \cdot \frac{CE}{CA}-1 \\
&=\frac{a}{a-c} \cdot \frac{a}{a-b}-\frac{b}{a-b} \cdot \frac{b}{b+c}-\frac{c}{b+c} \cdot \frac{c}{a-c}-1 \\
&=\frac{a}{b} \cdot \frac{a}{c}-\frac{b}{c} \cdot \frac{b}{a}-\frac{c}{b} \cdot \frac{c}{a}-1 \\
&=\frac{a^3-b^3-c^3}{abc}-1.
\end{align*}](http://latex.artofproblemsolving.com/3/2/c/32c59ff44495f8bb7c120dea41b864d0c8bde055.png)
,
,
,
,![$\tfrac{[DEF]}{[ABC]}=2$](http://latex.artofproblemsolving.com/6/8/5/6851e9953b28e86934f3a1021e64d3cf6fb9fb5c.png)
,
,
through 
and 
are collinear
and 
are both concyclic, and by repeatedly plugging law of sines, we get 
,
,
,
,![$S[ABC]=1*1*\sin{60}$](http://latex.artofproblemsolving.com/4/8/c/48ca8b2b6ee5a7e05e43ea28bad0b849594128b9.png)
,![$S[BCEF]=\frac{\sin{PAF}^2+\sin{CAP}^2+\sin{PAF}\sin{CAP}}{\sin{PAF}\sin{CAP}}*\sin{60}$](http://latex.artofproblemsolving.com/9/f/3/9f3ee314c4e6b90255bbaaa03f24c502f262144b.png)
,![$S[BDF]=\frac{\sin^2{PAF}}{\sin{CAP}\sin{CAP+60}}*\sin{60}$](http://latex.artofproblemsolving.com/8/7/0/870daf37b256106bc1a3eadaca6914ceb6e13ef1.png)
,![$S[CDE]=\frac{\sin^2{CAP}}{\sin{PAF}\sin{PAF+60}}*\sin{60}$](http://latex.artofproblemsolving.com/6/6/6/66655edbfa99060c60af5bff4fb5547a31592a62.png)
,
(I think this would work).
so 
so![\begin{align*}
\frac{[DEF]}{[ABC]}&=\frac{1}{(y+z)(z+x)(x+y)}
\begin{vmatrix}
0 & y & z\\
x & 0 & z\\
x & y & 0
\end{vmatrix}\\
&=\frac{2xyz}{(1-x)(1-y)(1-z)}\\
&=\frac{2xyz}{1-x-y-z+xy+yz+zx-xyz}\\
&=-2,
\end{align*}](http://latex.artofproblemsolving.com/a/8/3/a83d06f4ab25e9f3cf08967fc8534c5f440c6f9a.png)
so we are done. 
for some 
.![\[a^2(yz+zx+xy)=0 \implies yz+zx+xy=0\]](http://latex.artofproblemsolving.com/7/5/7/75718684e4937c291ebfbb6d4bba88e8ab29b0b9.png)
since 
,
.![\[D=(0:y:z)=\left(0,\frac{y}{y+z},\frac{z}{y+z}\right)\]](http://latex.artofproblemsolving.com/e/1/5/e15fdff442973751fa1d2bf930d127e8e3805fc4.png)
Similarly we also have that 
and 
.![\begin{align*}
[DEF] &= [ABC] \begin{vmatrix}
0 & \frac{y}{y+z} & \frac{z}{y+z}\\
\frac{x}{x+z} & 0 & \frac{z}{x+z}\\
\frac{x}{x+y} & \frac{y}{x+z} &0
\end{vmatrix}\\
&= [ABC] \left(\frac{2xyz}{(x+y)(y+z)(x+z)}\right)\\
&= 2[ABC] \left(\frac{xyz}{\sum_{cyc}x^2y + 2xyz}\right)\\
&= 2[ABC] \left(\frac{xyz}{(\sum_{cyc}x^2y +3xyz) - xyz}\right)\\
&= 2[ABC] \left(\frac{xyz}{-xyz}\right)\\
&= -2[ABC]
\end{align*}](http://latex.artofproblemsolving.com/7/5/f/75f8a4ad9a56a2759f4f6ab294261b151b9f53f8.png)
which proves the desired result.
we have![\[ \angle APE = \angle APC + \angle CPE = \angle APB + \angle BPC = \angle FPA. \]](http://latex.artofproblemsolving.com/5/d/6/5d6e445f41035acf4e7a210b38ccd2fe80cd1cff.png)
Moreover, 
gives![\[\angle PEA = 180^\circ - \angle BAC - \angle ABP = 180^\circ - \angle BPA - \angle ABP = \angle PAF. \]](http://latex.artofproblemsolving.com/7/e/6/7e66a335a36c51e0475ab75b081fb02e9f2ddce5.png)
Therefore by AA Similarity we have 
.
.
by angle chasing, now by area decomposition and Law of Sines we have![\[[DEF] = [DPE] + [EPF] + [FPD] = \dfrac{\sqrt{3}}{4} \bigg[ (PD)(PE) + (PE)(PF) + (PF)(PD) \bigg] = \dfrac{\sqrt{3}}{4} \bigg[ PA^2 + PB^2 + PC^2 \bigg] .\]](http://latex.artofproblemsolving.com/4/5/4/4549e8a345c82222956a834593b08ffa79c5af77.png)
WLOG ![$[ABC] = \dfrac{3\sqrt{3}}{4}$](http://latex.artofproblemsolving.com/a/8/f/a8fb1d7b52b3af519a81fb0643040f10ee613069.png)
we wish to show 
.
be represented by 
where 
and let 
as wanted. The end.
,
by circumcircle equation. Now its trivial to find the coordinates of DEF to be 
.![\begin{align*}
\frac{[DEF]}{[ABC]} &= \det \begin{bmatrix} 0 & \frac{y}{y+z} & \frac{z}{y+z} \\ \frac{x}{x+z} & 0 & \frac{z}{x+z} \\ \frac{x}{x+y} & \frac{y}{x+y} & 0 \end{bmatrix} \\
&= \frac{2xyz}{(x+y)(y+z)(x+z)} \\
&= \frac{-2x^2(y+z)}{(y+z)(x+y)(x+z)} \\
&= -2 \frac{x^2}{x^2+xy+xz+yz} \\
&= -2
\end{align*}](http://latex.artofproblemsolving.com/4/6/a/46a3fd01834a1c0033fb656ea21339e83ae0c730.png)
finishing.
be 
.
are 
.
.
,
.
and 
.
.
.
.
.
and as 
we must have 
as desired. It follows that the area of 
