AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be positive integers such that 
and 
. Let 
be the set of values attained by 
as 
runs through the positive integers. Show that is the set of all positive divisors of some positive integer.
be a set of integers (not necessarily positive) such that
with 
;
and 
are elements of (possibly equal), then 
also belongs to . is the set of all integers.
the set of all positive integers. Find all functions 
such that for all positive integers 
and , the integer 
is nonzero and divides 
.
of positive integers such that 
is divisible by 
.
be an odd integer. Determine all functions 
from the set of integers to itself, such that for all integers and the difference 
divides 
for which there exist infinitely many positive integers 
such that ![\[ \frac{a^m+a-1}{a^n+a^2-1} \]](http://latex.artofproblemsolving.com/2/8/a/28a68e409451bbdcaee7c1286341f6aeea9ffdad.png)
is itself an integer.
such that ![\[ a^2b-1 \mid ab^3-1. \]](http://latex.artofproblemsolving.com/7/3/5/735fd61de9a18113846ce2048a00889dcc5ce8a4.png)
, 
be positive integers such that 
is a multiple of 
for all positive integers . Prove that 
. for which ![$\dfrac{n^2+1}{[\sqrt{n}]^2+2}$](http://latex.artofproblemsolving.com/7/e/d/7ed6c67114c0c7d262f67e5b7a733daab7997415.png)
is an integer. Here ![$[r]$](http://latex.artofproblemsolving.com/2/f/c/2fcabe2c2b8ebe2c428830acb9874cb8a8df790f.png)
denotes the greatest integer less than or equal to 
.
of positive integers such that 
is multiple of 
and 
is multiple of 
.
be an integer. Find all positive real solutions to the following system of 
equations:
+1 w, , 
, 
be real numbers such that 
and all zeros 
and 
of the polynomial 
are real. Find the smallest value the product 
can take.
, the polynomial![\[ (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 \]](http://latex.artofproblemsolving.com/f/7/f/f7fb9528a1f63668a3445741ded0a7a712d4c314.png)
can be expressed as 
, where 
is a polynomial with complex coefficients and with degree at most 
. The value of![\[ \left| \sum_{1 \le j <k \le 673} z_jz_k \right| \]](http://latex.artofproblemsolving.com/c/4/2/c42909e045b37ae9673a3f9e3a53f348dfcadddf.png)
can be expressed in the form 
, where and are relatively prime positive integers. Find 
.
, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy 
, 
, and 
. Furthermore 
, 
, and 
. Prove that diagonals 
, 
, and 
are concurrent.
.
.
the unit circle is nicer than, say 
or 
the unit circle (which I'm not sure are even tractable, really).
is the smallest of 
so 
such that 
with 
now we have 
and 
so 
by the angle conditions 
giving 
so 
and they have the same orientation. but 
and 
so 
therefore 
but sincfe 
then 
is a paralelogram. If 
then clearly 
and 
meet at the orthocenter of 
but if not then 
is a translation of 
by the length giving 
so 
so since we have 
but this is a contradiction.
are aligned clockwise. Let 
, 
and 
. The key idea is we will rearrange vectors 
into new hexagon which is easier to work with.![[asy]
unitsize(0.6cm);
defaultpen(fontsize(10pt));
pair A = (1.5,4);
pair C = (0,0);
pair E = (5,0);
pair B = 2*foot(E,C,A)-E;
pair D = 2*foot(A,C,E)-A;
pair F = 2*foot(C,A,E)-C;
pair T =extension(B,C,F,E);
draw(B--A,red);
draw(D--E,red);
draw(B--C,green);
draw(E--F,green);
draw(C--D,blue);
draw(F--A,blue);
draw(C--T--E,dashed+green);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(T);
label("$F$",F,NE);
label("$A$",A,N);
label("$B$",B,NW);
label("$C$",C,SW);
label("$D$",D,S);
label("$E$",E,SE);
label("$T$",T,S);
pair A1 = (12,5);
pair B1 = A1 + A - F;
pair C1 = B1 + D - C;
pair D1 = C1 + C - B;
pair E1 = D1 + F - E;
pair F1 = E1 + E- D;
draw(A1--B1--C1,blue);
draw(C1--D1--E1,green);
draw(E1--F1--A1,red);
draw(A1--C1--E1--cycle,dashed);
dot(A1);dot(B1);dot(C1);dot(D1);dot(E1);dot(F1);
label("$A_1$",A1,N);
label("$B_1$",B1,NW);
label("$C_1$",C1,SW);
label("$D_1$",D1,S);
label("$E_1$",E1,SE);
label("$F_1$",F1,NW);[/asy]](http://latex.artofproblemsolving.com/7/b/a/7baa4dbb5d8d82cf9ad39cfab5c6571d5ef81908.png)
, 
, 
, 
, 
, 
are equal to vectors 
, 
, 
, 
, 
, 
respectively.
, then 
. Thus 
are isosceles with base angle 
. Similarly 
are isosceles with base angle and 
respectively.
thus 
. But by Law of Sine on , we get 
. Thus Law of Sine on 
gives
or 
. Hence 
which is enough to conclude that 
are reflections of 
across 
respectively. Thus 
are concurrent at orthocenter of 
.
, 
, and 
. WLOG assume 
is the origin and 
lies on the positive -axis, then these are fixed.
, 
, and 
, then we see that 
is a linear map from 
, and the possible values of 
is precisely 
, so if the hexagon is not unique up to scaling we would have 
by Rank-Nullity. In particular, 
, so 
are parallel. We will show that this cannot happen by complex numbers., , etc. be 
. Then, if 
, 
, 
, we have: ![\[
a = 1, b = x, c = xz^3, d = xyz^3, e = x^4yz^3, f = x^4yz^4.
\]](http://latex.artofproblemsolving.com/e/9/d/e9d78111168a00831ee1b4620ba9c80510f5b63b.png)
Suppose now by contradiction that 
, so that 
, then ![\[
\frac{a+d}{b+e} = \frac{\frac 1a + \frac 1d}{\frac 1b + \frac 1e} = \frac{be}{ad} \frac{a+d}{b+e},
\]](http://latex.artofproblemsolving.com/d/4/a/d4a1ff654c3d900c6220caf9f1b2832f4ba2a5ca.png)
so that we force 
, or ![\[
xyz^3 = x^5yz^3 = x^5yz^7.
\]](http://latex.artofproblemsolving.com/1/e/0/1e0478c1d897a2ccebdc3c4ba91e98f565ff36a4.png)
Then, 
, but this causes us to conclude 
, which is absurd since 
and 
. 
such that and are reflections over 
and cyclic permutations. In particular, the lines 
concur at the orthocenter of triangle in any good hexagon., and note that as 
, 
, and 
satisfy 
, we may construct a good hexagon 
that has identical angles to . To do so, let 
, 
, 
be the sums of unit vectors in the directions and . Notice that we have 
Now, the non-parallel condition implies that none of 
are nonzero, and an argument computation yields that no two are scalar multiples of each other (say, just set horizontal and compute all the angles in terms of 
.) So , , are linearly independent, implying that 
are in the same ratio as 
. This implies that 
, as needed.
WLOG on the unit circle. It suffices to show that 
,
be the reflection of 
,
,
over 
,![\begin{align*}
(a-b)(c-d)(e-f) &= (b-c)(d-e)(f-a) \Longleftrightarrow f=\frac{(a-b)(c-d)e+(c-b)(e-d)a}{(a-b)(c-d)+(c-b)(e-d)} \\
\frac{(e-a)(c-b)}{(a-b)(c-d)+(c-b)(e-d)} = \frac{f-e}{d-e} &= \left(\frac{c-b}{a-b}\right)^2 \overline{\left(\frac{a-b}{c-b}\right)} = \frac{c-b}{a-b}\cdot\frac{c}{a} \Longleftrightarrow d=\frac{c[(a-b)c+(c-b)e]+a(a-e)(a-b)}{c[(a-b)+(c-b)]} \\
\frac{(a-b)(c-d)+(c-b)(e-d)}{(a-e)(c-d)} = \frac{b-a}{f-a} &= \left(\frac{e-d}{c-d}\right)^2 \overline{\left(\frac{c-d}{e-d}\right)}.
\end{align*}](http://latex.artofproblemsolving.com/f/0/7/f07fb9a179789b16b6609753281501a282c366e0.png)
Note that![\[\frac{e-d}{c-d}=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},\]](http://latex.artofproblemsolving.com/6/7/c/67c7fa7d5f35b20de4dc4f317484bdd536448d70.png)
so plugging into the third equation we have![\begin{align*}
\frac{a(a-b)(2b-a-c)}{c(c-e)(c-b)-a(a-e)(a-b)}
&=\frac{(a-b)+(c-b)\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}}{(a-e)}\\
&=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\overline{\left(\frac{c(c-e)(c-b)-a(a-e)(a-b)}{(a-b)[c(e-c)+a(e-a)]}\right)}\\
&=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{\frac{1}{c}\left(\overline{e}-\frac{1}{c}\right)\frac{b-c}{bc}-\frac{1}{a}\left(\overline{e}-\frac{1}{a}\right)\frac{b-a}{ba}}{\frac{b-a}{ab}\left(\frac{1}{c}\left(\frac{1}{c}-\overline{e}\right)+\frac{1}{a}\left(\frac{1}{a}-\overline{e}\right)\right)}\\
&=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c(a-b)[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}\\
&=\frac{(a-b)[c(e-c)+a(e-a)]^2}{[c(c-e)(c-b)-a(a-e)(a-b)]^2}\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}.
\end{align*}](http://latex.artofproblemsolving.com/5/b/1/5b18ed902ca414b7cf07385be2bc69392ffa23b2.png)
Simplifying, this becomes![\begin{align*}
&ac(2b-a-c)[c(c-e)(c-b)-a(a-e)(a-b)][a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]\\
&=[c(e-c)+a(e-a)]^2[c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)].
\end{align*}](http://latex.artofproblemsolving.com/c/3/d/c3daf03d3b127d535016b7b139bf7a1978d3a42d.png)
Of course, we can also "conjugate" this equation -- a nice way to do this is to note that if![\[x=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},\]](http://latex.artofproblemsolving.com/2/5/d/25d073f6854ccbeedff56f724f2fadf02d835f7c.png)
then![\[\frac{a(2b-a-c)}{c(e-c)+a(e-a)}=\frac{x}{\overline{x}}=\overline{\left(\frac{c(e-c)+a(e-a)}{a(2b-a-c)}\right)}=\frac{\frac{1}{c}\left(\overline{e}-\frac{1}{c}\right)+\frac{1}{a}\left(\overline{e}-\frac{1}{a}\right)}{\frac{1}{a}\left(\frac{2}{b}-\frac{1}{a}-\frac{1}{c}\right)},\]](http://latex.artofproblemsolving.com/d/8/9/d89b95a000227d31fbf2aa9d7a3eb84b6f0b7ffd.png)
whence![\[ac(2b-a-c)[2ac-b(a+c)]=b[c(e-c)+a(e-a)][a^2(c\overline{e}-1)+c^2(a\overline{e}-1)].\]](http://latex.artofproblemsolving.com/a/9/b/a9b744a1cb6e210786bd6c54a6946ddec4b82298.png)
If 
,
,![\[e\in\left\{a+c-\frac{ac}{b},a+\frac{2c(c-b)}{a+c},c+\frac{2a(a-b)}{a+c}\right\}.\]](http://latex.artofproblemsolving.com/b/5/1/b51e707fde84d57574551bce311e4d2ea9bc058e.png)
The first case corresponds to (*) (since 
uniquely determine 
(or equivalently, since 
)
.
,
from the first of the two equations in 
(we actually don't need this, but it tells us that the locus of working 
is a line through the origin). It's easy to compute 
and 
,
,
uniquely determine 
.![\begin{align*}
\frac{a(a-b)(2b-a-c)}{c(c-e)(c-b)-a(a-e)(a-b)}
&=\frac{(a-b)+(c-b)\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}}{(a-e)}\\
&=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\overline{\left(\frac{c(c-e)(c-b)-a(a-e)(a-b)}{(a-b)[c(e-c)+a(e-a)]}\right)}\\
&=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{\frac{1}{c}\left(\frac{1}{c}-\overline{e}\right)\frac{b-c}{bc}-\frac{1}{a}\left(\frac{1}{a}-\overline{e}\right)\frac{b-a}{ba}}{\frac{b-a}{ab}\left(\frac{1}{c}\left(\frac{1}{c}-\overline{e}\right)+\frac{1}{a}\left(\frac{1}{a}-\overline{e}\right)\right)}\\
&=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c(a-b)[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}\\
&=\frac{(a-b)[c(e-c)+a(e-a)]^2}{[c(c-e)(c-b)-a(a-e)(a-b)]^2}\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}.
\end{align*}](http://latex.artofproblemsolving.com/6/f/7/6f7b0396f278f1bcdfa4806e3751436e4bae1024.png)
Simplifying, this becomes![\[\frac{a(2b-a-c)}{c(e-c)+a(e-a)}=\frac{x}{\overline{x}}=\overline{\left(\frac{c(e-c)+a(e-a)}{a(2b-a-c)}\right)}=\frac{\frac{1}{c}\left(\frac{1}{c}-\overline{e}\right)+\frac{1}{a}\left(\frac{1}{a}-\overline{e}\right)}{\frac{1}{a}\left(\frac{2}{b}-\frac{1}{a}-\frac{1}{c}\right)},\]](http://latex.artofproblemsolving.com/3/4/4/3445258b67681afc80bdbfa06b9f677290387b16.png)
whence![\[ac(2b-a-c)[2ac-b(a+c)]=b[c(e-c)+a(e-a)][a^2(1-c\overline{e})+c^2(1-a\overline{e})].\]](http://latex.artofproblemsolving.com/2/1/a/21a5d261685974f61232c61b64e525fd5e6bd494.png)
Thus we have two inequivalent quadratics in 
are as well).
works, so we're done.
up to scaling.
be points such that 
are all parallelograms. Then, we know from the length conditions that 
and 
From the angle conditions we obtain 
,
Therefore, we know that 
,
In other words, 
is fixed up to scaling! Since we've determined the angles of 
,
is fixed up to scaling! From here, it only suffices to construct parallelograms 
,