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.
, where 
, then the number of integers in the range of 
is __
be a triangle and 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
, then which of the following is/are true?
is obtuse
is acute
, where 
is semi perimeter
, where is semi perimeter
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
. Prove that
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
. Prove that
satisfy that: 
and 
. Find the maximum: 
+1 w
, such that 
Prove that :
equality cases : ?
. The point 
is in the interior of . The following ratio equalities hold:![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](http://latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
be the set of positive real numbers. Determine all functions 
such that for all positive real numbers 
and 
![\[f(x+f(xy))+y=f(x)f(y)+1\]](http://latex.artofproblemsolving.com/c/6/a/c6a968a17f6fd5fd637c1ee5d855bdff11e5423a.png)
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
be 
. The tangents to at 
meet at point 
. For a point 
on line 
which is not on the segment , let the midpoint of 
be 
. Lines 
meet again at points 
respectively. Let 
be the midpoint of . Prove that the points 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
, where 
are linearly independent over 
, this is a unique representation.
is invariant for both endpoints. They start at different parity. So they can't interchange positions.
![${\mathbb Z}[\omega]$](http://latex.artofproblemsolving.com/d/1/2/d12fd2e831d0da64de29e065253c03b71f18bad6.png)
where 
. Draw the needle as a directed segment from 
to 
in the plane.
. Rotations around the other endpoint correspond to ![\[ z \mapsto z + \omega^k - \omega^{k-1} \]](http://latex.artofproblemsolving.com/2/e/0/2e04e71d6294a14fa622b3807200a6ed85095f58.png)
for some choice of 
.. To prove this we only need show the following claim, which proves the relevant invariant.
in .![${\mathbb Z}[\omega]/(\omega-1)$](http://latex.artofproblemsolving.com/3/a/0/3a0dc4b76ee55649dcdacc1f56d8a38433a9862d.png)
is not trivial. Write ![${\mathbb Z}[\omega] = {\mathbb Z}[T] / (T^4+1)$](http://latex.artofproblemsolving.com/5/5/2/5528f519ce9305f3ab167cca0e09a578155aae11.png)
, then ![\[ {\mathbb Z}[\omega]/(\omega-1) \cong {\mathbb Z}[T] / \left( T^4+1, T-1 \right) \cong {\mathbb Z} / 2 = {\mathbb F}_2 \]](http://latex.artofproblemsolving.com/8/9/1/891f618efc341f8fd237341a0972a89a0df9eabf.png)
as desired. 
and the other is at 
, and WLOG rotate about the left endpoint first. Suppose a sequence of rotations works. Right before we switch endpoints to rotate around, draw the needle's current position on the plane. Also draw the line joining and . Then the drawn segments clearly form a cycle/polygon. Furthermore, because the endpoints must be interchanged, there must be an odd number of vertices on this graph. On the other hand, every drawn segment has length and is either horizontal, vertical or has slope 
. Because and 
are linearly independent over 
, it then follows that to end at the same position we started, we need an even number of horizontal and an even number of vertical edges. By rotating the argument 
the same is true for edges with slope and edges with slope 
, so there are an even number of edges and thus vertices: contradiction. 
.. A move consists of taking a complex number and adding 
where 
is some integer from to 
. (Effectively what this does is that we rotate the needle by some amount, and then jump to the other endpoint.) We now show that the number of moves must be even, if we return to . More formally, if![\[ \sum_{k = 0}^{7} a_k\omega^k = 0 \]](http://latex.artofproblemsolving.com/a/b/c/abcfbd631cd0597cdacf245fbbdf59dc6e9ead09.png)
for some integers 
through 
, we must have that![\[ \sum_{k = 0}^{7} a_k \equiv 0 \pmod{2}.\]](http://latex.artofproblemsolving.com/4/1/c/41c7d30468c67c6fe0d6a84cc77bf2b8e2ac730f.png)
through are zero, by subtracting pairs of 
while preserving the parity.
. As such 
. Likewise, 
. This then forces 
, so the statement is true.
be a primitive eighth root of unity. Suppose otherwise, and consider the locus of all positions of the segment. Modulo the first and last positions, all such segments 
form an equilateral polygon with an odd number of sides.'s that sum to ; equivalently, there exists a polynomial ![$f \in \mathbb Z[X]$](http://latex.artofproblemsolving.com/3/e/3/3e3d314244e5e514fdf63e59038fc713b6a48ed1.png)
such that 
is odd and 
.
, and hence 
, contradiction!
and the other at . We will let 
and 
and after 
moves, we let 
denote the new endpoint of the needle after a move. Note that if then we can set up the following recurrence relation for 
![\[ z_{n} = \omega^{e_{n}}(z_{n-2}-z_{n-1})-z_{n-2} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2}\]](http://latex.artofproblemsolving.com/4/1/c/41c8e53e652c5e96cd471fdbacb2f92e05d4a376.png)
for an arbitrary sequence of integers 
. This recurrence works because we translate to a new point 
on the unit circle, rotate it by 
and undo the translation, which is exactly the operation we desire. is odd and 
are integers, then ![\[ \omega^{a_1}+\omega^{a_2}+ \cdots + \omega^{a_k} \not = 0 \]](http://latex.artofproblemsolving.com/c/8/3/c8365b7b5921c9a82fe7334cff111fa28799e0f8.png)
then 
is even. Indeed, note that ![\[ \Re(t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = t_1\cdot \Re(\omega^1)+t_2 \cdot \Re(\omega^2) \cdots +t_8 \cdot \Re(\omega^8)=0 \]](http://latex.artofproblemsolving.com/4/9/4/494a80516578c37d743fd357ba5482b01bbee943.png)
![\[ \implies (t_1+t_3-t_5-t_7) \frac{\sqrt{2}}{2} + t_2+t_4 = 0 \implies t_1+t_3-t_5-t_7=0 \]](http://latex.artofproblemsolving.com/f/9/f/f9fbeef15255e342809c0b6d58893034221c02e0.png)
So, 
is even. We also note that ![\[\omega \cdot (t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = 0 \implies t_8\omega^1+t_1\omega^2 \cdots +t_7\omega^8\]](http://latex.artofproblemsolving.com/a/f/0/af0ed3863ec1e2d86c7f2d7597e97424acae65a2.png)
So, 
is even by the same logic. Thus, 
is even as desired. 
denote the number of terms in the expansion of , we can set up the following recursion ![\[ z_{n} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2} = z_{n-1} + \omega^{e_n+4} \cdot z_{n-1} + \omega^{e_n}z_{n-2}\]](http://latex.artofproblemsolving.com/c/6/3/c63c4e287786626a944e450a5e7935f11337c8c1.png)
![\[ \implies A_n = 2A_{n-1} + A_{n-2} \]](http://latex.artofproblemsolving.com/2/8/7/28717d4f0ecbab658c885103bee6adbe595c82fc.png)
Where 
and 
. Note that this recursion implies that for all odd , we have 
. Thus, 
has an odd number of terms in it's expansion. Thus, 
for all odd . So, the two ends of the needle cannot swap.
![$\mathbb{Z}\left[\frac{\sqrt{2}}{2}\right]^2$](http://latex.artofproblemsolving.com/9/c/0/9c00968aacadfe8470d61c5f8f7eecaf71153545.png)
. WLOG the needle has length . We will only focus on one endpoint, and prove that it cannot lie on the starting point of the other endpoint.
around another results in both 
and 
being invariant modulo 
. However, since the two endpoints of the needle have distance , the two endpoints have different parities, so they cannot swap.
are vectors with the same magnitude that sum to and the argument of each vector is a multiple of 
degrees, then is even.. The idea here is that since is irrational, both the integer and parts of both and 
coordinates must be zero. Thus 
appears the same number of times as 
, and 
appears the same number of times as 
. Furthermore, 
in total occur the same number of times as 
in total due to the coordinate. Thus, the total number of vectors is even. degrees. However, in order for it to get back to its original position with the orientation swapped, the number of such vectors must be odd, contradiction.
. Induct gives 
parity unchanged.
satisfying 
and 
Prove that
round any of its endpoints. Is it possible that after several rotations the needle returns to initial position with the endpoints interchanged?
