be a convex quadrilateral inscribed in a circle and satisfying 
.
and 
are chosen on sides 
and 
such that 
and 
.
.
be a function such that for each 
,
and 
is equal to distance between 
and 
.
is bijective
.
,
,
,
.
,
,
,
is also a parallelogram. Furthermore, remark that by the definition of 
.
.
,
and 
.
are also collinear. Furthermore, 
implies 
.
.
be the matrix corresponding to 
.
,![\[I=QQ^{-1}=\begin{bmatrix}a^2+b^2 & ac+bd\\ac+bd & c^2+d^2\end{bmatrix}.\]](http://latex.artofproblemsolving.com/1/7/8/178962ebe627bd9d50ca4a90293c2ed6fcf3ec9b.png)
This means we need 
and 
.
,
,
,
for some 
and 
between 
and 
.![\[ac+bd=\sin\alpha\sin\beta+\cos\alpha\cos\beta=\cos(\beta-\alpha)=0\quad\implies\quad \beta-\alpha=\pm\dfrac{\pi}2.\]](http://latex.artofproblemsolving.com/0/3/6/036164f0a9af1839ca47f7a40acffbaee5c0c27c.png)
If 
,
and 
,
and 
.![\[Q=\begin{bmatrix}\sin\alpha&\cos\alpha\\\pm\cos\alpha&\mp\sin\alpha\end{bmatrix},\]](http://latex.artofproblemsolving.com/f/3/8/f38ed558d66032ddfbe2bfaa30e04f19f78ca896.png)
which implies that 
be a finite graph on 
vertices (not necessarily simple), and let 
denote its adjacency matrix. Then the number of closed walks on 
is ![\[f_G(\ell)=\sum_{i=1}^n(A(G)^\ell)_{i,i}=\operatorname{tr}\left(A(G)^\ell\right)=\lambda_1^\ell+\lambda_2^\ell+\cdots+\lambda_n^\ell,\]](http://latex.artofproblemsolving.com/6/0/9/609bcc4e4fdcc62bdca9c7de8ce776ad99658610.png)
where 
is the sequence of eigenvalues of 
are real by the Spectral Theorem.)
vertices and that the number of closed walks of length ![\[8^\ell+2\cdot 3^\ell+3\cdot(-1)^\ell+(-6)^\ell+5\]](http://latex.artofproblemsolving.com/8/7/b/87be345b574f9b30a4b84d9aa63d27843f6c0fe4.png)
for all 
.
be the graph obtained from 
,
,
and 
,
are nonzero complex numbers such that for all positive integers ![\[\alpha_1^\ell+\cdots+\alpha_r^\ell=\beta_1^\ell+\cdots+\beta_s^\ell.\]](http://latex.artofproblemsolving.com/4/c/3/4c33b8a747861a5d3e9a6bc347ce5d1f3c5cbe05.png)
Then 
and the 
and sum on all ![\[\dfrac{\alpha_1x}{1-\alpha_1x}+\cdots+\dfrac{\alpha_rx}{1-\alpha_rx}=\dfrac{\beta_1x}{1-\beta_1x}+\cdots+\dfrac{\beta_sx}{1-\beta_sx}.\]](http://latex.artofproblemsolving.com/f/3/e/f3e4360f0965b0fe5786b24b9d40ed62160fdf3a.png)
This is an identity valid for complex numbers of sufficiently small modulus. By clearing denominators we obtain a polynomial identity. But if two polynomials in 
.
and let 
.
'
,
'
added to each diagonal entry. Thus 
,
,
,
,
,
,
,
,
,![\[\boxed{9+2\cdot 4^\ell+5\cdot 2^\ell+(-5)^\ell+3}.\]](http://latex.artofproblemsolving.com/b/b/8/bb8fdea23059941b3e6066b6126419483f063c2c.png)
is a graph whose vertex set is the disjoint union 
of 
and 
such that every edge of 
.
if 
if 
,
,
,
,
,
and 
for all 
and 
.
,
are the eigenvalues of ![\[\lambda_1^\ell+\cdots+\lambda_n^\ell=0\qquad\text{for all odd }\ell.\]](http://latex.artofproblemsolving.com/c/2/c/c2c5452fd125900aa0a84fc2f443fbb4a69c845f.png)
Now for each positive integer 
,
denote the sum of the 
powers of the eigenvalues and 
denote their 
for all odd 
,
for all 
,![\[P_{i+2}-S_1P_{i+1}+S_2P_i-\cdots+S_{i+1}P_1-(i+2)S_{i+2}=0.\]](http://latex.artofproblemsolving.com/2/9/f/29f42f9817ef9878fd502ca24ef93d409327778b.png)
Consider the expression 
for some arbitrary 
must be odd. If 
from the original problem statement. Thus 
for each ![\[-(i+2)S_{i+2}=0\quad\implies\quad S_{i+2}=0.\]](http://latex.artofproblemsolving.com/c/8/9/c897dafac635345bc800083a4af190a6db8692ad.png)
Hence we're done by induction. ![\[\lambda(A(G))=\sum_{i=0}^na_i\lambda^i\]](http://latex.artofproblemsolving.com/a/a/7/aa7230c08ed6dfc75a56213023056da7587fad3c.png)
be the characteristic polynomial of 
.
,![\[\lambda(A(G))=\begin{cases}a_n\lambda^n+a_{n-2}\lambda^{n-2}+\cdots+a_0,&n\text{ even},\\a_n\lambda^n+a_{n-2}\lambda^{n-2}+\cdots+a_1\lambda,&n\text{ odd}\end{cases}\]](http://latex.artofproblemsolving.com/6/7/f/67f4a48f4001e94104b68a1b896bfe774ac6399a.png)
as desired.
be the complete bipartite graph 
with 
vertices and 
edges, each of degree 
.
(once each).[/quote]
.
through 
(or rather the edge set of 
.
for all 
and 
by our labeling system. Furthermore, note that with respect to the remaining entries of 
and 
entries. Thus, we can conclude that our adjacency matrix has the form ![\[A(H_n)=\begin{bmatrix}0&J-I\\J-I&0\end{bmatrix},\]](http://latex.artofproblemsolving.com/e/2/a/e2a26a1cbe4424be2f66c73e6ad8c1b506f624e7.png)
where 
is the matrix with all ones. For example, when 
,![\[A(H_3)=\begin{bmatrix}0&0&0&0&1&1\\0&0&0&1&0&1\\0&0&0&1&1&0\\0&1&1&0&0&0\\1&0&1&0&0&0\\1&1&0&0&0&0\end{bmatrix}.\]](http://latex.artofproblemsolving.com/8/1/4/814a7f4fdebf592fbb01cc9938a1a22b9c424f10.png)
To compute the eigenvalues of this matrix, recall that if 
,
are square blocks, then ![\[\det\begin{bmatrix}A&B\\C&D\end{bmatrix}=\det(AD-BC).\]](http://latex.artofproblemsolving.com/b/8/d/b8d4d9c6c2a3330125112ce4015dc1941498967d.png)
(I seriously need to find the proof for this....) Now for any 
,![\[\det[A(H_n)-\lambda I]=\det\begin{bmatrix}-\lambda I&J-I\\J-I&-\lambda I\end{bmatrix}=\det(\lambda^2 I-(J-I)^2).\]](http://latex.artofproblemsolving.com/b/3/7/b370ae17dcf086beed2fcc66a08be1775733f5df.png)
Using the fact that 
and 
(why?), remark that ![\[(J-I)^2=J^2-2IJ+I^2=nJ-2J+I=(n-2)J+I.\]](http://latex.artofproblemsolving.com/c/1/6/c1617138d9534169849f87c324b1c04db275f5b7.png)
Thus ![\[\det[\lambda^2I^2-(I-J)^2]=0\iff \det[J(n-2)-(\lambda^2-1)I]=(n-2)\det\left[J-\left(\dfrac{\lambda^2-1}{n-2}\right)I\right]=0.\]](http://latex.artofproblemsolving.com/5/f/c/5fc0303606b034131782d5bef51af3b11ccc1cb4.png)
(This obviously cannot be done for 
,
,
,
.
,
is an eigenvalue of 
with multiplicity 
with multiplicity ![\[0=\dfrac{\lambda^2-1}{n-2}\quad\implies\quad \lambda=\pm 1\]](http://latex.artofproblemsolving.com/f/d/4/fd4cb25bd43145c02075662040c0c6534a42c021.png)
or ![\[n=\dfrac{\lambda^2-1}{n-2}\quad\implies\quad \lambda=\pm (n-1)\]](http://latex.artofproblemsolving.com/8/b/6/8b6fc12824058a8daec4b1f152759838178c2454.png)
and it's not hard to show that these roots have the desired multiplicities, so we're done.
denote the complete graph with 
denote 
removed, i.e. choose 
vertices of 
of closed walks in 
of length 
in an advantageous manner and then consult its adjacency matrix. WLOG number the vertices 
such that the three vertices not chosen in 
are 
.![\[A(\Gamma)=\begin{bmatrix}1&1&1&1&\cdots&1\\1&1&1&1&\cdots&1\\1&1&1&1&\cdots&1\\1&1&1&0&\cdots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\1&1&1&0&\cdots&0\end{bmatrix}.\]](http://latex.artofproblemsolving.com/5/d/6/5d612a821e14544e1b46b9e00e6c6b80ad1e61a5.png)
In other words, 
consists of two bands of 
,
are the squares of the eigenvalues of the original matrix 
.![\[(A(\Gamma)^2)_{ii}=\sum_jA(\Gamma)_{ij}A(\Gamma)_{ji}.\]](http://latex.artofproblemsolving.com/4/a/b/4aba38c4fceac3aa8ecdc88459315541be9f3993.png)
If 
,
or 
are equal to 
.![\[A(\Gamma)^2=\begin{bmatrix}21&*&*&*&\cdots&*\\ *&21&*&*&\cdots&*\\ *&*&21&*&\cdots&*\\ *&*&*&3&\cdots&*\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ *&*&*&*&\cdots&3\end{bmatrix},\]](http://latex.artofproblemsolving.com/c/6/d/c6dc774563d73ce0efdfcb3dcf8584d8c072c859.png)
meaning that 
be the two nonzero eigenvalues of 
trivially, we get the system of equations ![\[\begin{cases}\lambda_1+\lambda_2&=3,\\\lambda_1^2+\lambda_2^2&=117.\end{cases}\]](http://latex.artofproblemsolving.com/f/0/8/f08a1408ad8ac1d4a9360467081500d5c0f343cb.png)
Solving yields 
.
be the maximum degree of any vertex of 
.
edges. Show that 
.
be an eigenvector of 
be the largest entry in ![\[(A\vec{x})_i=\sum_jA_{ij}x_j\leq\sum_jA_{ij}x_i=x_i\sum_jA_{ij}\leq\Delta x_i.\]](http://latex.artofproblemsolving.com/e/0/5/e056c8c176b3eefc0cec784f90843cbfe041dd84.png)
But since 
.
for any eigenvalue 
.
,
.
,
,
the entries of ![\[\sum_{i=1}^n\left(\sum_jA_{ij}x_j\right)^2\leq 2q\sum_{j=1}^nx_j^2.\]](http://latex.artofproblemsolving.com/6/5/c/65c4cb4d0e6998e5c58523f188593f1633faf98b.png)
Once this is proven, the LHS becomes 
while the right becomes 
,![\[\sum_{j=1}^nA_{ij}^2\sum_{j=1}^nx_{ij}^2\geq\left(\sum_{j=1}^nA_{ij}x_j\right)^2.\]](http://latex.artofproblemsolving.com/0/3/b/03b7115e1f3dccb7e59531468db915aed07ee26a.png)
Summing over all ![\[\sum_{i=1}^n\left(\sum_{j=1}^nA_{ij}x_j\right)^2\leq\sum_{i=1}^n\left(\sum_{j=1}^nA_{ij}^2\sum_{j=1}^nx_j^2\right)\leq\left(\sum_{i,j}A_{ij}^2\right)\left(\sum_{j=1}^nx_j^2\right).\]](http://latex.artofproblemsolving.com/b/c/5/bc5ee2b5ffbf393bb8cbfb01ded4045cee3b4655.png)
Finally, remark that since all entries in 
simply equals the sum of the entries in 
Y by Exponent11, Alex-131, Pengu14, squareman, Danielzh, NaturalSelection, EpicBird08, ESAOPS, Everestbaker, scannose, KenWuMath, DhruvJha, crazyeyemoody907
Hello to all creative problem solvers,
Do you want a life changing math experience?
Do you want to see me in real life?
Check out the
Rutgers Expo in Problem Solving (REPS) by OMMC!
OMMC is presenting to you its next major event: in-person this time! This spring, OMMC is hosting its THIRD IN-PERSON event, where we will be presenting various speakers of math, holding breakout sessions, games and friendly competitions, and providing a math hub for people all over to learn and enjoy. No math experience is needed, and elementary, middle and high schoolers can all register!
The Rutgers Expo in Problem Solving will take place on June 7th, 1 PM.
The venue is the Science Engineering Complex at Rutgers New-Brunswick. This is 96 Frelinghuysen Rd, Piscataway, NJ 08854.
Event includes:
This event is completely FREE to all students!
Fill out the registration form linked below to sign-up for this event and answer some important questions.
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Students from anywhere can attend, as long as you can commute to the venue. Email us at ommcofficial@gmail.com for any questions or concerns.
We can’t wait to see you there!
- REPS Team
OMMC’S 2025 EVENTS ARE SPONSORED BY:
Do you want a life changing math experience?
Do you want to see me in real life?
Check out the
Rutgers Expo in Problem Solving (REPS) by OMMC!
OMMC is presenting to you its next major event: in-person this time! This spring, OMMC is hosting its THIRD IN-PERSON event, where we will be presenting various speakers of math, holding breakout sessions, games and friendly competitions, and providing a math hub for people all over to learn and enjoy. No math experience is needed, and elementary, middle and high schoolers can all register!
The Rutgers Expo in Problem Solving will take place on June 7th, 1 PM.
The venue is the Science Engineering Complex at Rutgers New-Brunswick. This is 96 Frelinghuysen Rd, Piscataway, NJ 08854.
Event includes:
- Math speakers, including Richard Rusczyk, Dr. Christian Yongwhan Lim, and Dr. Yang Liu.
- Activities including estimathon and mini math competition WITH PRIZES
- Itinerary coming soon!
This event is completely FREE to all students!
Fill out the registration form linked below to sign-up for this event and answer some important questions.
https://docs.google.com/forms/d/e/1FAIpQLSeAr2Nul9MBO_adVzHN9Rsrc8yQEjzxPXZHZ-LUFf-zWcwR7A/viewform?usp=preview
Students from anywhere can attend, as long as you can commute to the venue. Email us at ommcofficial@gmail.com for any questions or concerns.
We can’t wait to see you there!
- REPS Team
OMMC’S 2025 EVENTS ARE SPONSORED BY:
- Nontrivial Fellowship
- Citadel
- Jane Street
This post has been edited 3 times. Last edited by DottedCaculator, May 16, 2025, 2:43 AM
