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.
. Let 
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
is the incenter 
of 
.
sides 
. His vertices are marked with numbers such that sum of numbers marked by any 
consecutive vertices is constant and its value is 
. If we know that 
is marked with 
and 
is marked with 
, determine with which number is marked 
has two isogonal conjugate points 
and 
. The circle 
intersects circle 
at 
, and the circle 
intersects circle 
at 
. Prove that 
and 
are isogonal conjugates in triangle . is the circle with diameter 
, Circle is the circle with diameter 
.
. Points and on the medial perpendicular to 
are such that 
. Prove that is the circumcenter of triangle 
.
denote the set of positive real numbers. Find all functions 
such that![\[x+f(yf(x)+1)=xf(x+y)+yf(yf(x))\]](http://latex.artofproblemsolving.com/1/6/c/16c10ac3b45b6b3f864f9314c3aeae654db514db.png)
for all 
, 
is the midpoint of the hypotenuse 
. Point 
is chosen on the leg 
so that the line segment 
meets 
again at 
(
). Let 
be the reflection of in . The circles and 
meet again at 
(
). Find the measure of 
.
that satisfy the functional equation:![\[
f(f(x)(1 + y)) = f(x) + f(xy), \quad \forall x, y > 0.
\]](http://latex.artofproblemsolving.com/9/f/c/9fcb6b114b87842f23f31a97ddae6c722fe71611.png)
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist.
of positive integers such that 
is an integer.
and 
be positive integers. Prove that there exists a positive integer such that for every odd integer 
, the digits in the base-
representation of 
are all greater than .
, pairwise distinct points 
, 
, and 
are collinear iff 
.
are collinear iff the slope of the lines 
and 
are equal. That is, we need:
which simplifies as:
Now collecting terms and factoring out 
, we have . These steps are reversible, so the proof is complete.
, which would make 
and have 
collinear iff 
.
, let 
, Then for 
, we need that
Noticing that for 
,
For 
, let 
, then
Therefore are collinear if and only if .
; then 
, 
, 
are collinear iff the area of the triangle with these three points as vertices is 0, or by Shoelace![\[ 0 = \begin{vmatrix}1&a&\left(a-\frac{2014}{3}\right)^3\\1&b&\left(b-\frac{2014}{3}\right)^3\\1&c&\left(c-\frac{2014}{3}\right)^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c - 2014),\]](http://latex.artofproblemsolving.com/1/a/b/1ab2f2af01a0f6f1a8804fe4b6983c4e264c6774.png)
iff 
and 
are distinct, as desired.
![\[P_x=\left(x^2-2014x^2, x\right)\]](http://latex.artofproblemsolving.com/b/a/a/baa71c63d857307a0f0da1bbe86c4cc6f002ec39.png)
Then we have that 
are collinear if and only if![\[ 0 = \begin{vmatrix}a^3-2014a^2&a&1\\b^3-2014b^2&b&1\\c^3-2014c^2&c&1\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c - 2014)\]](http://latex.artofproblemsolving.com/c/e/1/ce198e5f3485ca287d559c7a5828cdb79d13cb67.png)
Since 
are distinct it follows that the 3 points are collinear iff QED.
be a projective elliptic curve with nonzero rank and let ![$O = [0: 1 : 0]$](http://latex.artofproblemsolving.com/3/c/f/3cf8e617f09958482e532c68a3bfc771777e395e.png)
be the identity in its group law. and let be a point on 
which is not in the torsion subgroup. Now, we define 
, whose points are all distinct due to the earlier condition.
,then it follows that![\[
P_a + P_b + P_c = (3a + 3b + 3c - 3 \cdot 2014) \cdot P = O.
\]](http://latex.artofproblemsolving.com/b/b/b/bbb4482eea87f566c1cb04a89e95b7c4c6b6010b.png)
and thus are collinear over projective space. Finally, since 
it follows that 
so all our points can just be put in 
.
instead of 
Have each 
Then by shoelace 
so our final answer is 
for each positive integer , we can replace the condition with 
.
works. Indeed, notice that the area determinant![\begin{align*}
[P_aP_bP_c] &= \begin{vmatrix} a & a^3-a^2 & 1 \\ b & b^3-b^2 & 1 \\ c & c^3-c^2 & 1\end{vmatrix} \\
&= \sum_{\mathrm{cyc}} a\left(b^3-b^2\right)-b\left(a^3-a^2\right) \\
&= (a-b)(b-c)(c-a)(a+b+c-1).
\end{align*}](http://latex.artofproblemsolving.com/3/e/9/3e9cdba1168e603ee3491ebe90ca456c2e89d41d.png)
Thus ![$[P_aP_bP_c]=0$](http://latex.artofproblemsolving.com/1/8/3/183f57efecc0ff3487d235f39500816bc8b55838.png)
, i.e. for distinct 
if and only if .
determinant, which we know must have these factors.
works. and , therefore the line they define is ![\[y=(x-a)(2014(a+b)-a^2-ab-b^2)-a^2(a-2014)\]](http://latex.artofproblemsolving.com/c/2/3/c234fdabffb4614bc7df44c221612bcb80555afa.png)
Plugging in 
, we get the equation ![\[0=c^3-2014c^2+c(2014(a+b)-a^2-ab-b^2)+ab(a+b-2014)=(c-a)(c-b)(c-(2014-a-b))\]](http://latex.artofproblemsolving.com/3/2/5/3257b644c0e777074f5535b6fda4d049af916ba5.png)
The conclusion follows.
, rather than only 
, hence the generalisation to 
fixed is also true.
be one of its generators, define the addition of points on it as usual so we need 
, then let 
this works as we get its 0 iff 
, we claim that 
works. One can verify that by Shoelace:![\[ \dfrac{1}{2} \bigg( a f(b) + b f(c) + c f(a) - b f(a) - c f(b) - a f(c) \bigg) = \dfrac{1}{2} \bigg| (a - b)(b - c)(c - a)(a + b + c - 2014) \bigg| \]](http://latex.artofproblemsolving.com/e/d/b/edbede890ec07dac32591d1465c1e23718b08290.png)
which is clearly iff 
are distinct and as needed, finishing.
where 
are functions. Shoelace is a very simple way to encode the collinearity criterion with cyclicity, so in essence, our goal is to force![\[x(a)y(b) + x(b)y(c) + x(c)y(a) - y(a)x(b) - y(b)x(c) - y(c)x(a) \]](http://latex.artofproblemsolving.com/9/3/7/937a5d386d60f9c3fb57512d40d27d0674cf3bb0.png)
to be a cyclic polynomial with 
as a factor. Since 
are guaranteed to be factors of the above, naturally we try to equate this to 
. Upon expanding (in an organized manner) we get the desired functions.
![\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]](http://latex.artofproblemsolving.com/a/c/c/acc097f524dc9fe9bd85e8eb659ea4a75aec475d.png)
in the plane with the following property: For any three distinct integers 
and 
where 
.
-
the North Pole, and 
any vertex of the Bermuda triangle. These points are noncollinear since they lie on the spherical Earth. We use barycentric coordinates with respect to triangle 
,
when 
,
,
are distinct. Back-construct the original matrix after expanding this:![\[ \left( \begin{array}{ccc} 1 & x & x^3 \\ 1 & y & y^3 \\ 1 & z & z^3 \end{array} \right). \]](http://latex.artofproblemsolving.com/0/b/1/0b1d8fe67fe0dfdefe9cc173bfcaaf2b12fd12e3.png)
,
.
.
are all roots of this, so if we want 
to be polynomials then we must have 
.
.
.
works. Then to finish retranslate everything 
units to get 
.
for any real number 
and that you will have the same property.
for polynomials 
and 
Now the thing is that the question "when is 
collinear" is a question that has an ALGEBRAIC answer. What does this mean? It means there will be some polynomial 
So all you do here is pick 
be an elliptic curve in 
,
determine the whole set, it seems like Pappus/projective transformation should work, unless we run into other issues (I'm guessing "other issues" means it won't work)
problems in computational geometry" (available online 
be the line through 
Let the 
,
,
.
to be 
,
and 
.
Putting this into the slope equation for collinearity, we obtain that 
cancels out anyways, so we just take 
to be the coordinates (which can be checked to work).