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is selected inside acute 
so that 
and 
. Point 
is chosen on ray 
so that 
. Let 
be the midpoint of 
.
is tangent to the circumcircle of triangle 
.
+1 w
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
and 
be positive integers. Let 
be the set of integer points 
with 
and 
. A configuration of 
rectangles is called happy if each point in is a vertex of exactly one rectangle, and all rectangles have sides parallel to the coordinate axes. Prove that the number of happy configurations is odd.
be an acute triangle, and let 
and 
denote its 
-excenter, 
-excenter, and circumcenter, respectively. Points and 
are selected on 
such that 
and 
. Similarly, points 
and 
are selected on 
such that 
and 
.
and 
meet at 
. Prove that 
and 
are perpendicular.
plot equally spaced points around a circle. Label one of them 
, and place a marker at . One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of 
distinct moves available; two from each point. Let 
count the number of ways to advance around the circle exactly twice, beginning and ending at , without repeating a move. Prove that 
for all 
.
![[asy]
import geometry;
unitsize(3cm);
draw(circle((0,0),1),linewidth(1.5));
for (int i = 0; i < 7; ++i) {
for (int j = 0; j < i; ++j) {
draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));
}
}
for(int i = 0; i < 7; ++i) {
dot(dir(i * 360/7),5+black);
}
[/asy]](http://latex.artofproblemsolving.com/9/2/e/92ecdecb16bc7556a4fcda2df737b19e1d80013e.png)
, where 
. Then, the desired answer is![\[\sum_{j=1}^7\sum_{k=j+1}^7\left|\omega^j-\omega^k\right|^4=\frac 12\sum_{k=1}^7\sum_{k=1}^7\left|\omega^{j-k}-1\right|^4\left|\omega^k\right|^4=\frac 72\sum_{j=1}^6\left|\omega^j-1\right|^4\]](http://latex.artofproblemsolving.com/7/d/1/7d1b739cb5a665dae8996f9db79cde32440a03e3.png)
In particular, we know![\[|\omega^j-1|^2=(\omega^j-1)\overline{(\omega^j-1)}=(\omega^j-1)\left(\frac 1{\omega^j}-1\right)\]](http://latex.artofproblemsolving.com/f/5/3/f5325e7543c3b6f6c2a5543c993964ce7c106db3.png)
Thus, the sum rewrites itself as![\[\frac72\sum_{j=1}^6(\omega^j-1)^2\left(\frac 1{\omega^j}-1\right)^2\]](http://latex.artofproblemsolving.com/a/c/a/acae28d7b313f9ff2337d3befbffda78dc246088.png)
Define the rational function![\[P(x)=\sum_{j=1}^6(x^j-1)^2\left(\frac 1{x^j}-1\right)^2\]](http://latex.artofproblemsolving.com/8/5/5/855db128186f3642b829d9ea4c8be0f507823189.png)
Note that 
, and 
for 
(as it simply re-shuffles the order of the same summands), we have![\[P(\omega)=\frac 16\sum_{j=1}^7P(\omega^j)\]](http://latex.artofproblemsolving.com/4/b/d/4bd59cb3c87959e4bcaac21134f29ec711607135.png)
Now, as the polynomial with roots is 
, we know![\[\sum_{k=1}^7\omega^{jk}=\begin{cases}7&7\mid j\\0&\mathrm{otherwise}\end{cases}\]](http://latex.artofproblemsolving.com/c/f/d/cfdff33cd6b03da5a04614260572651605517bc8.png)
In particular, this implies we only care about the terms in the expansion of 
with exponents which are multiples of powers of 
, which happens to be the constant terms. However, we note that the constant term is simply 
, so thus![\[\frac 16\sum_{j=1}^7P(\omega^j)=\frac 16\cdot 6\cdot 6\cdot 7=42\]](http://latex.artofproblemsolving.com/2/a/0/2a0410eed6edde419ecda71be90a98187ffce9dc.png)
so the answer is 
, or 
.
. Answer is half of 
. Since terms such as 
sum as 
and all annihilate. Thus the answer is 
.
. Then the answer is 
, since there are edges with length , with length 
, and with length 
.
, we obtain
and 
. Additionally, we know that 
. Adding, we get
so the answer is 
, which is .
so I'd answer C in a hurry.
, where 
. The key idea is that 
Similarly, the fourth powers of the other two lengths are 
and 
. Adding these up and reducing using 
gives 
. Since 
, the resulting sum is equal to 
. Since there are seven segments of each length, the answer is 
.
-gons inscribed in a unit circle. As a warm-up, here are the answers for 
, 
, and 
.
, the summation is just:
, the summation is just:
, the summation is just:
, 
, and 
, so by the principle of Engineer's Induction we find that the answer for general -gons is 
. Applying this for the case of 
gives our final answer of 
over all 
with 
, where 
. This is simply![\[ S: = \frac{1}{2}\sum_{m = 0}^6\sum_{n = 0}^6 |z^m - z^n|^4.\]](http://latex.artofproblemsolving.com/1/b/f/1bfac9fd79ec367f5f477d393c21fd17ce36a3e8.png)
Note that![\[ |z^m - z^n|^2 = (z^m - z^n)(\overline{z}^m - \overline{z}^n) = 2 - z^{m - n} - z^{n- m}.\]](http://latex.artofproblemsolving.com/0/5/f/05fef2ab23d54affdcd516a893905740d6748fa5.png)
Square both sides to get![\[ |z^m - z^n|^4 = 6 - 4(z^{m-n} + z^{n- m}) + z^{2m-2n}+ z^{2n -2m}.\]](http://latex.artofproblemsolving.com/5/2/f/52f369dadf53f6ed5475e74a9f3e2fdbfbd1b961.png)
Sum both sides over the set previously mentioned to get![\[ S = \frac{1}{2}\cdot 6\cdot 7^2 - 4\sum_{m = 0}^6\sum_{n = 0}^6 z^{n-m} + \sum_{m = 0}^6\sum_{n = 0}^6 (z^2)^{n-m}.\]](http://latex.artofproblemsolving.com/e/4/b/e4b3f9bbcc6acc19c0966c159db39348f70d85fe.png)
For obvious reasons 
, so![\[ S = 3\cdot 7^2 = \boxed{147}.\]](http://latex.artofproblemsolving.com/d/8/9/d89e682178cf1f30877bb193f862f07fc63c6fc7.png)
. Then the answer is , since there are edges with length , with length , and with length ., we obtain and . Additionally, we know that . Adding, we getso the answer is , which is .
, where 
.
for all 
. However, since 
for all 
, we have
.
respectively.
Denote to be a complex solution to 
. WLOG let 
.![$2\sin{\frac{2\pi}{7}}=2[\frac{1}{2}(\omega+\omega^{-1})], 2\sin{\frac{4\pi}{7}}=2[\frac{1}{2}(\omega^2+\omega^{-2})], 2\sin{\frac{6\pi}{7}}=2[\frac{1}{2}(\omega^3+\omega^{-3})]$](http://latex.artofproblemsolving.com/6/1/3/613013052904a1f00113a6868d78b8811ef83b9d.png)
.![\begin{align*}
&7[(\omega+\omega^{-1})^4+(\omega^2+\omega^{-2})^4+(\omega^3+\omega^{-3})^4] \\
&=7[(\omega^4-4\omega^2+6-4\omega^{-2}+\omega^{-4})+(\omega^8-4\omega^4+6-4\omega^{-4}+\omega^{-8})+(\omega^{12}-4\omega^6+6-4\omega^{-6}+\omega^{-12})] \\
&=7[-3(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega)+18]=7[-3(-1)+18]=\boxed{147} * \\
\end{align*}](http://latex.artofproblemsolving.com/a/e/9/ae97ab49635b30030848d9dd099cc0b35d9f0c0b.png)
so 
.
