and 
.
Doing this, we get that 
Now, we can solve this by factoring. By using RRT, we get that 
is a factor. Upon division, we get 
is the factored form. Now, we know that our only positive solutions are 
so our values of 
Summing these gives our answer of 
in or something like that idk my friend explained it pretty wonkily idk if he even solved it right but anyhow we press on. Transformations suck.
Furthermore, note that 
Thus, the answer is just 
so 
Cross multiplying and solving this quadratic, we get that our solution is 
Thus, 
We know from 
that 
so substituting this gives us 
Now, plugging this in to 
and 
gives us 
Solving these equations gives us 
Thus, by Vieta's, the sum of the roots is 
thus by median properties (the 2:1 ratio split thingy), we know that 
(For reference, I missed this part in-contest) Now, we can calculate 
by the Pythagorean Theorem to be 
and thus 
Now, by Pythagorean theorem, we know that the altitude from 
so our area is just 
Thus, simplifying and substituting 
we get that 
Simple guess and check gives us 
as our solutions. My dumb self forgot the former solution. Thus, 
giving us 
as our sum.
we get that 
Now if we try finding 
we will realize that 
this 
pattern continues until we once again reach 
thus 
Now return to the problem statement; 
but each inner function will just become 
hence the answer is 
Thus, simplifying, we get that 
or that 
Plugging this in to the first equation, we get that 
so this means that 
Now, 
thus 
and so by Special Right Triangles we do NOT in fact get 
but rather 
.
to be an integer. Plugging in factors of 
we see that there are 
integers that work.
Plugging in numbers gives us 
as our smallest value.
Furthermore, 
so 
meaning that 
.
and thus 
giving us 
Furthermore, we know that 
so our answer is just 
in the summation. Essentially, we have 
.
In other words, we want to find 
.
.
in such a polynomial is 
.
,
.
.
,
.
and 
and divide by 
.
and 
in our sum, subtract by 
.
term look kind of weird? Well, we actually notice 
,
.
,
,
is our answer!
,
terms. Drawing it out on the unit circle (which I heavily advise), it cycles between 
and 
.
,
,
.
,
for when 
terms! Anyways, the answer is 
.
so our coefficient is just 
Thus, our answer is just 
.
etc.
we get that it is 
Thus, we note that for an integer 
Since 
we know that all values of 
satisfy our conditions. Thus, our answer is just 
is our solution. For these Pell Equations, all solutions are of the form 
for 
Thus, we plug in 
giving us 
as our answer.
and let 
be our three-digit number that we remove. Thus, we can write 
By our condition, we know that 
or that 
Furthermore, we know that 
does not work, so we try 
We know that 
by the fact that it must be divisible by a 
Noting that 
doesn't work, we try 
or that 
This, miraculously, works, and thus our value of 
.![\[\pm 1\pm 2\cdots\pm n \equiv 1+2+\cdots + n\equiv \dfrac{n(n+1)}2\equiv 1\pmod 2.\]](http://latex.artofproblemsolving.com/e/2/d/e2dc53ce3face4faee04869223dafbaa6336aaab.png)
Hence 
,
.
,
where ![\[f(t+\pi) = \prod_{j=1}^n\cos j(t+\pi) = \prod_{j=1}^n\cos(jt+j\pi) = -\prod_{j=1}^n\cos jt = -f(t).\]](http://latex.artofproblemsolving.com/2/6/1/2615c8a9bcdd4e703b7e47de0998ab1fb2370aeb.png)
As a result, ![\[\int_0^{2\pi} f(t)\,dt = \int_0^\pi f(t)\,dt + \int_\pi^{2\pi}f(t)\,dt = \int_0^\pi f(t)\,dt - \int_0^\pi f(t)\,dt = 0\]](http://latex.artofproblemsolving.com/4/4/e/44e0da0fa4d5225e1c143d8ac20829665ec79f2a.png)
as desired.
.![\[F_n(x)=\left(x+\dfrac 1x\right)\left(x^2+\dfrac{1}{x^2}\right)\cdots\left(x^n+\dfrac{1}{x^n}\right).\]](http://latex.artofproblemsolving.com/1/e/d/1edcbd786dc03d7080c7c0fbeccd2d96e42213ba.png)
Note that through Laurent polynomial stuff 
.
term.) Now let 
denote the unit circle of radius one in 
,![\[\oint_\gamma F_n\]](http://latex.artofproblemsolving.com/7/6/5/7656973cbb0f745c83e7641be01b7a94d486811e.png)
in two different ways.
.![\[F_n(x) = S(n) + \sum_{k=1}^m\alpha_k\left(x^k+\dfrac 1{x^k}\right)\]](http://latex.artofproblemsolving.com/a/2/9/a2964d0709d74f87f06c0e2f43a66efcf5122964.png)
for some 
.
Now note that because 
,
term does not appear on the RHS. Hence, all the integrals on the right go to zero due to cancellation. (The reason 
fails is because the function 
has a simple pole at the origin and so by the Cauchy Residue Theorem the integral doesn't actually cancel out.) Hence we can simplify everything to ![\[\oint_\gamma F_n = \oint_\gamma S(n) = 2\pi S(n).\]](http://latex.artofproblemsolving.com/8/6/c/86caed330fca8c507572b83960713215da397b51.png)
This is one expression for the desired integral.
via the parametrization ![\[\Sigma = \left[[0,2\pi],\varphi\right],\quad\text{where}\quad \varphi(t) = \cos t + i\sin t\quad\text{for all } 0\leq t\leq 2\pi.\]](http://latex.artofproblemsolving.com/9/9/c/99c3d978281fd98b0084b3742b53739fdd5218d4.png)
It is clear that 
,
here, but meh.) Now we compute ![\[|\varphi'(t)| = |-\sin t + i\cos t| = \sin^2 t + \cos^2 t = 1.\]](http://latex.artofproblemsolving.com/8/7/8/878a5572c0e95273e474b6821a474011822c147c.png)
Hence by making a change of variables we may write 
Setting the two derived expressions equal to each other gives the desired equality.[/list]
,
,
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 
.
'
'
,
,
,
,
,![\[\boxed{9+2\cdot 4^\ell+5\cdot 2^\ell+(-5)^\ell+3}.\]](http://latex.artofproblemsolving.com/b/b/8/bb8fdea23059941b3e6066b6126419483f063c2c.png)
,
,
,
,
and 
for all 
.
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.
.
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 ![\[\det\begin{bmatrix}A&B\\C&D\end{bmatrix}=\det(AD-BC).\]](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.
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 an eigenvector of ![\[(A\vec{x})_i=\sum_jA_{ij}x_j\leq\sum_jA_{ij}x_i=x_i\sum_jA_{ij}\leq\Delta x_i.\]](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 ![\[P[T]=P[T|A]P[A]+P[T|B]P[B].\]](http://latex.artofproblemsolving.com/2/6/b/26b333457050ae6000ca82b094e67acf00964f2d.png)
To simplify this, we remark that by Bayes' Theorem, ![\[P[T|A]=\dfrac{P[A|T]P[T]}{P[A]},\]](http://latex.artofproblemsolving.com/0/c/8/0c8434d951270ba73354200dce6816e934db4cd6.png)
so the expression above becomes ![\[P[T]=P[A|T]P[T]+P[T|B]P[B].\]](http://latex.artofproblemsolving.com/0/2/7/0276ac7fefbef36e7fd6e7dbf6a899a76804ccff.png)
I now claim that ![$P[A|T]=\tfrac12$](http://latex.artofproblemsolving.com/c/b/b/cbbc949aff1ed0f7e908521b29ecd2f762bf6c9b.png)
.
be the set of strings of votes for which 
similarly. Consider a string 
.
begins with an 
which consists of the sequence of votes applied in reverse. Now the string starts with a 
.
and reverse it to get a string 
,
.![$P[T|B]=1$](http://latex.artofproblemsolving.com/a/e/b/aeb96f2aa80c38870ba41b79d4bc5c022c726dc6.png)
since ![$P[B]=\tfrac{b}{b+a}$](http://latex.artofproblemsolving.com/0/4/f/04f8db710362aedc2666a157193d616a2e8967cc.png)
.![\[P[T]=\frac12P[T]+\dfrac{b}{a+b}\quad\implies P[T]=\boxed{\dfrac{2b}{a+b}}.\]](http://latex.artofproblemsolving.com/0/9/a/09a01c90c88f8de7bbc865343eda51987cb429d9.png)
denotes the set of all words of finite length with elements in 
,![\[\Sigma=\{0,1,2,3,4,5,6,7,8,9,x,+,-,\times, \div, \,^\wedge\}.\]](http://latex.artofproblemsolving.com/2/2/6/2264821eb3e80f01f260ae7ee0c0cd22faac95f4.png)
Remark that every polynomial in 
can be written as x^3+22÷7*x-1. Note that at the same time, multiple strings can be used to encode the same polynomial (due to semantics and term ordering and what-not). Thus there exists a surjection from ![$|Q[x]|\leq |\Sigma^*|$](http://latex.artofproblemsolving.com/1/4/3/1431413240a12e066d62bc2309bb1410548fb345.png)
.
be the 
to denote the radical center of circles 
,
,
.
and 
,
,
is any integer between 
and of 
to be the sequence of real numbers such that 
for all 
and 
.
.![\[2\sum_{i=1}^{n-1}i^2+n^2=\dfrac{(n-1)n(2n-1)}3+n^2=\dfrac{n(2n^2+1)}3.\]](http://latex.artofproblemsolving.com/f/6/7/f672f703e71f17e49ccf13a1db9f0bf42ca42793.png)
![\begin{align*}\left[\sum_{j=1}^{n-1}jb_j+nb_n+\sum_{j=n+1}^{2n-1}(2n-j)b_j\right]^2&\leq\left[\sum_{i=1}^{2n-1}b_i^2\right]\left[\sum_{i=1}^{n-1}i^2+n^2+\sum_{i=n+1}^{2n-1}(2n-i)^2\right]\\&=\dfrac{n(2n^2+1)}3\\\implies S&\leq\boxed{\sqrt{\dfrac{n(2n^2+1)}3}}.\end{align*}](http://latex.artofproblemsolving.com/5/6/5/565021f592b8046cf07e43a80866e602757827f9.png)
This is attainable by considering the equality condition of C-S.![\[\sum_{i=1}^{3}(a_{i+1}-a_i)^2\geq\left[\sum_{i=1}^{3}(a_{i+1}-a_i)\right]^2=(a_{4}-a_1)^2.\]](http://latex.artofproblemsolving.com/f/2/5/f2578743ebec65e8f783faff4463fa3cfec9022b.png)
.
with 
and let 
and 
respectively. The homothety centered at 
respectively so it suffices to show 
,
,
are all cyclic quads from the right angles, so from the givens 
and 
.
.
.
,
.
,
.
be 
,
respectively. From Ratio Lemma we get![\[\dfrac{B_1M}{MC_1}=\dfrac{A_1B_1\sin\angle B_1A_1D}{AC_1\sin\angle DA_1C_1}=\dfrac{A_1B_1\sin\angle DCB_1}{A_1C_1\sin\angle C_1AD}.\]](http://latex.artofproblemsolving.com/a/1/7/a175f489c872b18849c58e3332f3de32af3fbd82.png)
gives 
and similarly 
.![\[\dfrac{B_1M}{MC_1}=\dfrac{A_1B_1}{A_1C_1}\left(\dfrac{DY}{DZ}\right)\left(\dfrac{\sin\angle B}{\sin\angle C}\right)=\dfrac{DY}{DZ}\left(\dfrac{A_1B_1/\sin\angle C}{A_1C_1/\sin\angle B}\right).\]](http://latex.artofproblemsolving.com/f/f/8/ff8dca0fa1c69320a614e868fbb6b659b7bb659e.png)
and 
have the same circumradius, so do 
and 
.![\[\dfrac{MB_1}{MC_1}=\dfrac{DY}{DZ}=\dfrac{B_1N}{C_1P}.\]](http://latex.artofproblemsolving.com/9/3/7/93725589f6b980441a6c941df6bb510f5418f5ca.png)
by SAS similarity as desired.
ways to fill the remaining 
ways to fill the remainder of the board with this piece at the front. Therefore 
.
since there is one way to tile a 
board (which is to not tile it at all!) and 
.
for all 
board with 
.
ways to tile the rest of the board. Otherwise, a 
squares left to tile, and they can be tiled in 
ways. If not, however, then a 
ways to tile the remaining "leftover" board. We have exhausted all possible cases, and adding all them up gives us 
,
board into two 
ways. Otherwise, the boards are "connected", and essentially what I mean by this is that a 
boards, one on each side, which can also be tiled independently. This gives 
ways to tile the board in this case. Therefore 
,
ways to "place" them, and the remaining squares can be filled with 
board. The reasoning for this is simple: I can take each of the 
squares. Now that the bijection is established, it is easy to see that the number of ways to place 
.
,
,![\[\dbinom n0+\dbinom{n-1}1+\dbinom{n-2}2+\cdots=f_n.\qquad\blacksquare\]](http://latex.artofproblemsolving.com/e/1/d/e1d744ba73fff79cc9c588f43bc1c81e489dedc3.png)
such that 
for all ![\[\sum_S1^{b_1+1} 2^{b_2+1} \cdots n^{b_n+1}=n!\sum_S1^{b_1}2^{b_2}\cdots n^{b_n}.\]](http://latex.artofproblemsolving.com/c/4/3/c430571442aae1cc6dd78ec295b9b31331f18072.png)
board with 
is the number of ways to color the board with at most 
is the number of ways to color the board with at most 
ways to choose the 
is the number of ways to color the board with at most 
colors, and so on. Alternating adding and subtracting these expressions gives the required PIE calculation, and also unsurprisingly gives us the expression on the right.
come in. Consider analyzing the number of possible colors in the following way: 
denotes the number of squares for which only three colors can be used, and so on. It is clear that this agrees with the concept described earlier, but there is a missing caveat: the 
squares. This way, we will guarantee the fulfillment of the condition that every color is to be used at least once. This eliminates the first square from any dispute in terms of its color, and leaves us with 
squares to worry about. Now, note that there is one possible color for the squares collected under 
possible colorings. However, there are 
possible colorings related to that sequence. Summing over all possible sequences gives us the total number of ways to color the board with our restrictions. But this is simply the left hand side of our equality as well!
![\[\sum_{k\leq n, 2\nmid k}\dbinom nk 3^{n-k}=2^{2n-1}-2^{n-1}.\]](http://latex.artofproblemsolving.com/6/e/6/6e6e9f9e02a65ceb1125c534f31d4cbf6ec51fe1.png)
and 
.![\[2X=2^{2n}-2^n\implies X=2^{2n-1}-2^{n-1}.\qquad\blacksquare\]](http://latex.artofproblemsolving.com/9/f/3/9f3e99af64d22e5c16dbd6def02c3b8a6ca835bf.png)
,
must be equal to one and the rest must be equal to zero. If any 
that satisfies the conditions 
.
.
.
,
.
and our desired trivially follows. 
,![\[\sum_{k=1}^{n+1}\dfrac{(-1)^{k-1}}k\dbinom {n+1}k=\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk+\dfrac1{n+1}.\]](http://latex.artofproblemsolving.com/3/4/2/342a2e2b5b0fec84047cb76c65ff359eff7d6cf9.png)
Splitting up the sum on the left and simplifying, we have![\begin{align*}\sum_{k=1}^{n}\dfrac{(-1)^{k-1}}k\dbinom {n+1}k+\dfrac{(-1)^n}{n+1}&=\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk+\dfrac1{n+1}
\\\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\left[\dbinom{n+1}k-\dbinom nk\right]&=\dfrac{1-(-1)^n}{n+1}
\\\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom n{k-1}&=\dfrac{1-(-1)^n}{n+1}.\end{align*}](http://latex.artofproblemsolving.com/f/b/8/fb830a057f884af1b5f9ee443251ef58c8a6a940.png)
Now multiplying both sides by 
Now it is not hard to show that this is true. In the interest of time, I will say that it is easy to move everything to one side and condense using Justin Stevens to show that it is true. 
.![\[A_i<A_j\text{ if } \min(A_i)<\max(A_j)\text{ and }A_i\cap A_j\neq\emptyset.\]](http://latex.artofproblemsolving.com/9/9/2/9926c0113dcbddd6ec365a165bf0a5a744fc4c5d.png)
In this case, antichains are formed by pairwise disjoint intervals and chains are formed by intervals that are not pairwise disjoint. I shall first prove that there are either eight pairwise disjoint intervals or eight intervals which pairwise have nonempty intersection. Suppose not. Then the maximum length of both the chains and the antichains are both 
.
,
,
,
,
,
.
.
passes through each of the other seven intervals, and we are done. 
be the midpoints of 
,
respectively. It is clear that we no longer have to focus on the centroids but rather these midpoints. Note that by the Ratio Lemma,![\[\dfrac{M_aC_1}{M_aB_1}=\dfrac{AB_1\sin\angle M_aAB}{AC_1\sin\angle M_aAC}=1.\]](http://latex.artofproblemsolving.com/2/d/1/2d13074917cc5c45beaea48dbf940f4dd68cc7b9.png)
We can derive similar equivalences concerning 
and 
.![\[\left(\dfrac{AB_1}{AC_1}\cdot\dfrac{BC_1}{BA_1}\cdot\dfrac{CA_1}{CB_1}\right)\left(\dfrac{\sin\angle M_aAB_1}{\sin\angle M_aAC_1}\cdot\dfrac{\sin\angle M_bBC_1}{\sin\angle M_bBA_1}\cdot\dfrac{\sin\angle M_cCA_1}{\sin\angle M_cCB_1}\right)=1.\]](http://latex.artofproblemsolving.com/e/7/e/e7ed51791e1d7a404bcb3063d2c2210e224d4185.png)
Now it is clear that normal Ceva for 
,
.
w.r.t line 
gives![\[\dfrac{BE}{EX}\cdot\dfrac{XG}{GA}\cdot\dfrac{AL}{LB}=1.\]](http://latex.artofproblemsolving.com/0/0/3/003e6f8c9c8c0a11eba7ffcfc0f4e87c6d293339.png)
Similarly, applying Menelaus on 
givese![\[\dfrac{BE}{EX}\cdot\dfrac{XF}{FA}\cdot\dfrac{AK}{KB}=1.\]](http://latex.artofproblemsolving.com/e/5/3/e53726a41921d4626630812a93f76f16eee9b0f0.png)
Multiplying these together and rearranging gives![\[\left(\dfrac{BE^2}{GA\cdot FA}\right)\left(\dfrac{XG\cdot XF}{EX^2}\right)\left(\dfrac{AL\cdot AK}{BL\cdot BK}\right)=1.\]](http://latex.artofproblemsolving.com/4/5/e/45e05fe6ca204313251fe4f514fcdb48607f3440.png)
Note that by PoP we have 
and 
.
Rearranging gives
Adding 
to this equality gives 
,
.
and 
be 
and 
,
,
.
.![\[\dfrac{YP}{YQ}=\dfrac{YP/YA}{YQ/YA}=\dfrac{AD/AX}{AE/AX}=\dfrac{AD}{AE}=\dfrac{AT}{AV}=\dfrac{AB}{BC}.\]](http://latex.artofproblemsolving.com/e/e/d/eedc8d28b43f64082683007628ca1e025632e6b3.png)
Since the distances from 
,
for all 
.
Multiplying this by 
and simplifying gives![\begin{align*}2(n-1)\sum_{i=1}^nx_i&=n(n-1)^2=\sum_{i=1}^n(n-1)^2\\\sum_{i=1}^n\left[(n-1)^2-2(n-1)x_i\right]&=0\\\sum_{i=1}^n\left[(n-1)^2-2(n-1)x_i+x_i^2\right]&=\sum_{i=1}^nx_i^2\\\sum_{i=1}^nx_i^2&=\sum_{i=1}^n[(n-1)-x_i]^2=\sum_{i=1}^ny_i^2\end{align*}](http://latex.artofproblemsolving.com/a/4/3/a4345eeb568a3ab727f3e06cbb54969704fd5afa.png)
as desired. 
Y by
In a board, James randomly writes A , B or C in each cell. What is the probability that, for every row and every column, the number of A 's modulo 3 is equal to the number of B's modulo 3?
G
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k a April Highlights and 2025 AoPS Online Class Information
jlacosta 0
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0 replies
Find all tuples of real numbers
nhathhuyyp5c 0
Let 
be an integer. Find all tuples of real numbers 
such that
![\[
2a_1 - 3a_2,\ 2a_2 - 3a_3,\ \ldots,\ 2a_n - 3a_1
\]](//latex.artofproblemsolving.com/c/0/9/c098bda64393b12049d319db49815947b2a0b7a4.png)
is a permutation of 
.
be an integer. Find all tuples of real numbers 
such that![\[
2a_1 - 3a_2,\ 2a_2 - 3a_3,\ \ldots,\ 2a_n - 3a_1
\]](http://latex.artofproblemsolving.com/c/0/9/c098bda64393b12049d319db49815947b2a0b7a4.png)
is a permutation of 
.
0 replies
NC State Math Contest Wake Tech Regional Problems and Solutions
mathnerd_101 9
N
by BackToSchool
Problem 1: Determine the area enclosed by the graphs of 
Hint
Problem 2: Calculate the sum of the real solutions to the equation
Hint
Problem 3: List the two transformations needed to convert the graph
to 
Hint
Problem 4: Let
be positive integers such that 
and 
Determine the value of 
Hint
Problem 5: Eve and Oscar are playing a game where they roll a fair, six-sided die. If an even number occurs on two consecutive rolls, then Eve wins. If an odd number is immediately followed by an even number, Oscar wins. The die is rolled until one person wins. What is the probability that Oscar wins?
Hint
Problem 6: In triangle
is on point 
such that 
and 
and 
is a point on 
such that 
and 
Given that 
calculate 
Hint
Problem 7: Determine the sum of the zeroes of the quadratic of polynomial
given that 
Hint
Problem 8:
Hint
Problem 9:
Find the sum of all real solutions to
Hint
Problem 10:
Define the function
![\[f(x) =
\begin{cases}
x - 9, & \text{if } x > 100 \\
f(f(x + 10)), & \text{if } x \leq 100
\end{cases}\]](//latex.artofproblemsolving.com/4/e/5/4e5b4486030a59e74df648819a1a59f82622b3b7.png)
Calculate
.
Hint
Problem 11:
Let
be real numbers such that 
Find 
Hint
Problem 12: Points
are on circle 
such that 
and 
Determine the path length from 
to 
formed by segment 
and arc 
Hint
Problem 13: Determine the number of integers
such that the expression 
is also an integer.
Hint
Problem 14: Determine the smallest positive integer
such that 
is a multiple of 
Hint
Problem 15: Suppose and 
are real numbers such that 
and 
Calculate 
Funnily enough, I guessed this question right in contest.
Hint
Problem 16: A sequence of points
will follow the rules such that
![\[
p_1 = (0,0), \quad p_{i+1} = (x_i + 1, y_i) \text{ or } (x_i, y_i + 1), \quad p_{10} = (4,5).
\]](//latex.artofproblemsolving.com/6/0/3/6036ef2619ffb1e91db0ab310388217e472490f2.png)
How many sequences 
are possible such that 
is the only point with equal coordinates?
Hint
Problem 18: (Also stolen from akliu's blog post)
Calculate
Hint
Problem 19: Determine the constant term in the expansion of
Hint
Problem 20:
In a magical pond there are two species of talking fish: trout, whose statements are always true, and \emph{flounder}, whose statements are always false. Six fish -- Alpha, Beta, Gamma, Delta, Epsilon, and Zeta -- live together in the pond. They make the following statements:
Alpha says, "Delta is the same kind of fish as I am.''
Beta says, "Epsilon and Zeta are different from each other.''
Gamma says, "Alpha is a flounder or Beta is a trout.''
Delta says, "The negation of Gamma's statement is true.''
Epsilon says, "I am a trout.''
Zeta says, "Beta is a flounder.''
How many of these fish are trout?
Hint
SHORT ANSWER QUESTIONS:
1. Five people randomly choose a positive integer less than or equal to
The probability that at least two people choose the same number can be written as 
Find 
Hint
2. Define a function
on the positive integers using the rule that for 
For all prime , 
and for all other 
Find the smallest possible value of such that 
Hint
3. How many integers
can be written as the sum of two distinct, non-negative integer powers of 
Huge shoutout to OTIS for teaching me how to solve problems like this.
Hint
4. Let
be the set of positive integers of such that 
for some other positive integer 
Find the only three-digit value of in 
Hint
5. Let be a positive integer and let be the integer that is formed by removing the first three digits from 
Find the value of with least value such that 
Hint
Hint
Problem 2: Calculate the sum of the real solutions to the equation
Hint
Problem 3: List the two transformations needed to convert the graph
to 
Hint
Problem 4: Let
be positive integers such that 
and 
Determine the value of 
Hint
Problem 5: Eve and Oscar are playing a game where they roll a fair, six-sided die. If an even number occurs on two consecutive rolls, then Eve wins. If an odd number is immediately followed by an even number, Oscar wins. The die is rolled until one person wins. What is the probability that Oscar wins?
Hint
Problem 6: In triangle
is on point 
such that 
and 
and 
is a point on 
such that 
and 
Given that 
calculate 
Hint
Problem 7: Determine the sum of the zeroes of the quadratic of polynomial
given that 
Hint
Problem 8:
Hint
Problem 9:
Find the sum of all real solutions to
Hint
Problem 10:
Define the function
![\[f(x) =
\begin{cases}
x - 9, & \text{if } x > 100 \\
f(f(x + 10)), & \text{if } x \leq 100
\end{cases}\]](http://latex.artofproblemsolving.com/4/e/5/4e5b4486030a59e74df648819a1a59f82622b3b7.png)
Calculate
.Hint
Problem 11:
Let
be real numbers such that 
Find 
Hint
Problem 12: Points
are on circle 
such that 
and 
Determine the path length from 
to 
formed by segment 
and arc 
Hint
Problem 13: Determine the number of integers
such that the expression 
is also an integer.Hint
Problem 14: Determine the smallest positive integer
such that 
is a multiple of 
Hint
Problem 15: Suppose
and 
are real numbers such that 
and 
Calculate 
Funnily enough, I guessed this question right in contest.
Hint
Problem 16: A sequence of points
will follow the rules such that![\[
p_1 = (0,0), \quad p_{i+1} = (x_i + 1, y_i) \text{ or } (x_i, y_i + 1), \quad p_{10} = (4,5).
\]](http://latex.artofproblemsolving.com/6/0/3/6036ef2619ffb1e91db0ab310388217e472490f2.png)
How many sequences 
are possible such that 
is the only point with equal coordinates?Hint
Problem 18: (Also stolen from akliu's blog post)
Calculate
Hint
Problem 19: Determine the constant term in the expansion of
Hint
Problem 20:
In a magical pond there are two species of talking fish: trout, whose statements are always true, and \emph{flounder}, whose statements are always false. Six fish -- Alpha, Beta, Gamma, Delta, Epsilon, and Zeta -- live together in the pond. They make the following statements:
Alpha says, "Delta is the same kind of fish as I am.''
Beta says, "Epsilon and Zeta are different from each other.''
Gamma says, "Alpha is a flounder or Beta is a trout.''
Delta says, "The negation of Gamma's statement is true.''
Epsilon says, "I am a trout.''
Zeta says, "Beta is a flounder.''
How many of these fish are trout?
Hint
SHORT ANSWER QUESTIONS:
1. Five people randomly choose a positive integer less than or equal to
The probability that at least two people choose the same number can be written as 
Find 
Hint
2. Define a function
on the positive integers using the rule that for 
For all prime , 
and for all other 
Find the smallest possible value of such that 
Hint
3. How many integers
can be written as the sum of two distinct, non-negative integer powers of 
Huge shoutout to OTIS for teaching me how to solve problems like this.
Hint
4. Let
be the set of positive integers of such that 
for some other positive integer 
Find the only three-digit value of in 
Hint
5. Let
be a positive integer and let be the integer that is formed by removing the first three digits from 
Find the value of with least value such that 
Hint
9 replies
geometry parabola problem
smalkaram_3549 9
N
by smbellanki
How would you solve this without using calculus?
9 replies
Calculus and Combinatorics
djmathman 0
It looks like the two can mix, interesting....
[quote="Putnam and Beyond #876"]
For a positive integer, denote by 
the number of choices of the signs "+" or "-" such that ![\[\pm 1\pm2\pm\cdots\pm n = 0.\]](//latex.artofproblemsolving.com/8/4/3/8431a74bb81e51162e98d87313b3dfa6115c38c7.png)
Prove that ![\[S(n)=\dfrac{2^{n-1}}\pi\int_0^{2\pi}\cos t\cos 2t\cdots\cos nt\,dt.\]](//latex.artofproblemsolving.com/2/8/d/28de904c2c8f403e3a5c1dd5442a46e843636b3a.png)
[/quote]
EDIT: Huzzah this is resolved now. I still don't know why this works and the first solution doesn't though rip
[quote="Putnam and Beyond #876"]
For a positive integer
, denote by 
the number of choices of the signs "+" or "-" such that ![\[\pm 1\pm2\pm\cdots\pm n = 0.\]](http://latex.artofproblemsolving.com/8/4/3/8431a74bb81e51162e98d87313b3dfa6115c38c7.png)
Prove that ![\[S(n)=\dfrac{2^{n-1}}\pi\int_0^{2\pi}\cos t\cos 2t\cdots\cos nt\,dt.\]](http://latex.artofproblemsolving.com/2/8/d/28de904c2c8f403e3a5c1dd5442a46e843636b3a.png)
[/quote]EDIT: Huzzah this is resolved now. I still don't know why this works and the first solution doesn't though rip
0 replies
Algebraic Combinatorics
djmathman 4
N
by djmathman
aka a way to use algebra to mask the fact that actual combo is hard
I haven't done a true math post in a while, so here goes....
Chapter 1 Key Idea: Let
be a finite graph on vertices (not necessarily simple), and let 
denote its adjacency matrix. Then the number of closed walks on of length 
is ![\[f_G(\ell)=\sum_{i=1}^n(A(G)^\ell)_{i,i}=\operatorname{tr}\left(A(G)^\ell\right)=\lambda_1^\ell+\lambda_2^\ell+\cdots+\lambda_n^\ell,\]](//latex.artofproblemsolving.com/6/0/9/609bcc4e4fdcc62bdca9c7de8ce776ad99658610.png)
where 
is the sequence of eigenvalues of . (Note that all the 
are real by the Spectral Theorem.)
This isn't too hard to prove (and is probably made easier based on the wording of the statement). Now on to the problems I guess?
[quote="Stanley Chapter 1 Exercise 2"]
Suppose that the graph has 
vertices and that the number of closed walks of length in is ![\[8^\ell+2\cdot 3^\ell+3\cdot(-1)^\ell+(-6)^\ell+5\]](//latex.artofproblemsolving.com/8/7/b/87be345b574f9b30a4b84d9aa63d27843f6c0fe4.png)
for all 
. Let 
be the graph obtained from by adding a loop at each vertex (in addition to whatever loops are already there). How many closed walks of length are there in ?[/quote]
[quote="Stanley Chapter 1 Exercise 3"]A bipartite graph with vertex bipartition 
is a graph whose vertex set is the disjoint union 
of and 
such that every edge of is incident to one vertex in and one vertex in . Show that the nonzero eigenvalues of come in pairs 
. Equivalently, prove that the characteristic polynomial of has the form 
if has an even number of vertices or 
if has an odd number of vertices for some polynomial .[/quote]
[quote="Stanley Chapter 1 Exercise 5"]
Let
be the complete bipartite graph 
with vertex-disjoint edges removed. Thus has 
vertices and 
edges, each of degree 
. Show that the eigenvalues of are 
( times each) and 
(once each).[/quote]
[quote="Stanley Chapter 1 Problem 11"]
Let
denote the complete graph with vertices, with one loop at each vertex. Let 
denote with the edges of 
removed, i.e. choose 
vertices of and remove all edges between these vertices (including loops). Find the number 
of closed walks in 
of length .[/quote]
[quote="Stanley Chapter 1 Exercise 12"]
[list=a]
[*]Let be a finite graph and let 
be the maximum degree of any vertex of . Let 
be the largest eigenvalue of the adjacency matrix . Show that 
.
[*]Suppose that is simple (no loops or multiple edges) and has a total of 
edges. Show that 
.
[/list][/quote]
I haven't done a true math post in a while, so here goes....
Chapter 1 Key Idea: Let
be a finite graph on vertices (not necessarily simple), and let 
denote its adjacency matrix. Then the number of closed walks on of length 
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 . (Note that all the 
are real by the Spectral Theorem.)This isn't too hard to prove (and is probably made easier based on the wording of the statement). Now on to the problems I guess?
[quote="Stanley Chapter 1 Exercise 2"]
Suppose that the graph
has 
vertices and that the number of closed walks of length in is ![\[8^\ell+2\cdot 3^\ell+3\cdot(-1)^\ell+(-6)^\ell+5\]](http://latex.artofproblemsolving.com/8/7/b/87be345b574f9b30a4b84d9aa63d27843f6c0fe4.png)
for all 
. Let 
be the graph obtained from by adding a loop at each vertex (in addition to whatever loops are already there). How many closed walks of length are there in ?[/quote][quote="Stanley Chapter 1 Exercise 3"]A bipartite graph
with vertex bipartition 
is a graph whose vertex set is the disjoint union 
of and 
such that every edge of is incident to one vertex in and one vertex in . Show that the nonzero eigenvalues of come in pairs 
. Equivalently, prove that the characteristic polynomial of has the form 
if has an even number of vertices or 
if has an odd number of vertices for some polynomial .[/quote][quote="Stanley Chapter 1 Exercise 5"]
Let
be the complete bipartite graph 
with vertex-disjoint edges removed. Thus has 
vertices and 
edges, each of degree 
. Show that the eigenvalues of are 
( times each) and 
(once each).[/quote][quote="Stanley Chapter 1 Problem 11"]
Let
denote the complete graph with vertices, with one loop at each vertex. Let 
denote with the edges of 
removed, i.e. choose 
vertices of and remove all edges between these vertices (including loops). Find the number 
of closed walks in 
of length .[/quote][quote="Stanley Chapter 1 Exercise 12"]
[list=a]
[*]Let
be a finite graph and let 
be the maximum degree of any vertex of . Let 
be the largest eigenvalue of the adjacency matrix . Show that 
.[*]Suppose that
is simple (no loops or multiple edges) and has a total of 
edges. Show that 
.[/list][/quote]
4 replies
My Life has been a Lie
djmathman 0
[quote="15-251 Homework 8 Problem 6"]
Donald Trump is running in a race against someone whose name no one can remember, so we will call him candidate "B". Trump has
votes and the other candidate has 
votes, naturally 
. Trump wants to make sure there is no confusion about who won. He wants to know, once the counting starts, what is the probability that at some point in the counting the two candidates have the same number of votes. Here we assume that the votes are counted one by one in a uniformly random order.[/quote]
The natural instinct here is to use the Catalan-like idea of taking a polyline and reflecting it over the-axis to get something cool. However, there is a second (intended) solution which uses nothing more than conditional probability. I originally stuck with the Catalan idea, but then someone else showed me this solution during OH - and my mind was blown.
Donald Trump is running in a race against someone whose name no one can remember, so we will call him candidate "B". Trump has
votes and the other candidate has 
votes, naturally 
. Trump wants to make sure there is no confusion about who won. He wants to know, once the counting starts, what is the probability that at some point in the counting the two candidates have the same number of votes. Here we assume that the votes are counted one by one in a uniformly random order.[/quote]The natural instinct here is to use the Catalan-like idea of taking a polyline and reflecting it over the
-axis to get something cool. However, there is a second (intended) solution which uses nothing more than conditional probability. I originally stuck with the Catalan idea, but then someone else showed me this solution during OH - and my mind was blown.0 replies
Using Computer Science to do Math
djmathman 1
N
by donot
Here's a pretty cool proof mentioned in 15-251 lecture ("Great Theoretical Ideas in Computer Science") a few days ago that I thought was worth sharing.
[quote]
Let![$Q[x]$](//latex.artofproblemsolving.com/7/1/d/71d78263a2218c538c7d41e125015c7c09855d9b.png)
denote the set of polynomials of finite degree with rational coefficients. Show that is countable.
[/quote]
[quote]
Let
![$Q[x]$](http://latex.artofproblemsolving.com/7/1/d/71d78263a2218c538c7d41e125015c7c09855d9b.png)
denote the set of polynomials of finite degree with rational coefficients. Show that is countable.[/quote]
1 reply
A Few Problems
djmathman 0
[quote="MOSP Homework 2013"]
Let be a positive integer. Given non-overlapping circular discs on a rectangular piece of paper, prove that one can cut the piece of paper into convex polygonal pieces each of which contains exactly one disc.[/quote]
[quote="MOSP Homework 2013"]
Let
, 
be real numbers such that 
. Determine the maximum value of
![\[(a_{n+1}+a_{n+2}+\cdots+a_{2n})-(a_1+a_2+\cdots+a_n).\]](//latex.artofproblemsolving.com/6/d/8/6d86f12525865cbc74c9729105905015ea503d46.png)
[/quote]
[quote="Ultimate Elimination Tourney G2"]
Let
be a point in the interior of triangle 
such that 
is isosceles and its base angles 
and 
measures the same as angle . Reflect about 
and to get 
and 
respectively. Suppose that the perpendicular bisector of 
intersects 
at 
. Also, let 
intersect at 
and 
intersect at 
. Prove that 
are collinear.[/quote]
Let
be a positive integer. Given non-overlapping circular discs on a rectangular piece of paper, prove that one can cut the piece of paper into convex polygonal pieces each of which contains exactly one disc.[/quote][quote="MOSP Homework 2013"]
Let
, 
be real numbers such that 
. Determine the maximum value of![\[(a_{n+1}+a_{n+2}+\cdots+a_{2n})-(a_1+a_2+\cdots+a_n).\]](http://latex.artofproblemsolving.com/6/d/8/6d86f12525865cbc74c9729105905015ea503d46.png)
[/quote][quote="Ultimate Elimination Tourney G2"]
Let
be a point in the interior of triangle 
such that 
is isosceles and its base angles 
and 
measures the same as angle . Reflect about 
and to get 
and 
respectively. Suppose that the perpendicular bisector of 
intersects 
at 
. Also, let 
intersect at 
and 
intersect at 
. Prove that 
are collinear.[/quote]0 replies
A Slick Combinatorics Problem
ssilwa 2
N
by viperstrike
The goal is to find a solution by combinatorial arguments only.
[quote]Define the sequence
by by 
. Prove 
.[/quote]
[quote]Define the sequence
by by 
. Prove 
.[/quote]
2 replies
Combinatorially Thinking?
djmathman 0
Last April, I went to a mathematics "convention" (or meeting, whatever) held by the NJ Section of the MAA. I was the only kid there (er, the only one actually paying attention/not accompanied by parents that were paying attention/having parents that probably wouldn't pay attention). (By the way, do any AoPSers actually attend these things? Or MAA MathFest for that matter?) There were a couple of interesting lectures, including some about doodling (not by Vi Hart, surprisingly) and about the Netflix ![$\$[/dollar]1,000,000$](//latex.artofproblemsolving.com/d/b/9/db96a95bccdd62c68b3a9491f78029c80f9e63f3.png)
prize. At one point, though, I got to go to a workshop entitled "Combinatorially Thinking." And as all of you know, combinatorics is obviously my strength and not geometry, right?
So our essential assignment was to prove algebraic statements involving Fibonacci numbers, factorials, and binomial coefficients (as well as Stirling numbers of the first and second kinds, Lucas numbers, and other things, but we didn't have time to get to those) WITHOUT ALGEBRA.
I haven't touched this since April, so I decided to go back to it and see if I was any better at this stuff.
Here's a warm-up problem to get a sense as to the methods we were required to use.
[quote]Let
be the th Fibonacci number, with the alternate definition 
and 
for all 
. Consider the number of ways to tile a 
board with 
squares and 
dominoes. Prove that this number is .[/quote]
This is a standard recursion problem, right? Well, things got a bit hairier after that.
[quote]
Prove that for all
, ![\[f_0+f_2+f_4+\cdots f_{2n}=f_{2n+1}.\]](//latex.artofproblemsolving.com/7/f/7/7f7f92ab0d0ee5c31605af96a077e15569db84cb.png)
[/quote]
[quote]Prove that for, ![\[f_{n-1}^2+f_n^2=f_{2n}.\]](//latex.artofproblemsolving.com/7/4/1/741b5b3a6646974a42f35d30b3c21598b08524e9.png)
[/quote]
[quote]Prove that for any, ![\[\dbinom n0+\dbinom{n-1}1+\dbinom{n-2}2+\cdots=f_n.\]](//latex.artofproblemsolving.com/e/8/2/e822786946a80dde5c74ce6c9a01fd5e40dce9db.png)
[/quote]
And these were all the easy ones.... Unfortunately, I have not been able to get any more of the problems. Here are some of the other ones in the packet (that I have not solved yet), if anybody is interested.
[quote]For, ![\[\sum_{k=0}^nf_k^2=f_nf_{n+1}.\]](//latex.artofproblemsolving.com/5/8/8/58804c716fe7f5b5ae5b67cd4e2e83af9f2a8fb9.png)
[/quote]
[quote]For, ![\[f_{2n}=\sum_{k\geq 0}\dbinom nk f_k\qquad\text{and}\qquad2^nf_n=\sum_{k\geq 0}\dbinom nk f_{3k}.\]](//latex.artofproblemsolving.com/a/c/e/aced00d22514bc1605b0df684eae91f0c171e32a.png)
[/quote]
[quote]For, ![\[\sum_{i\geq 0}\sum_{j\geq 0}\dbinom{n-i}j\dbinom{n-j}i=f_{2n+1}.\]](//latex.artofproblemsolving.com/d/a/8/da8632bbb30d1f7ebe9c7f26c60c72fdbd925c82.png)
[/quote]
I guess I should end with a problem that was NOT from this worksheet, but rather from Combinatorical Arguments at AMSP. Our teacher presented this problem during his lecture, and essentially nobody saw it coming. I was able to grab the main idea of his solution, but there were several key parts that were missing (which was essentially the theme of the course, I guess). I decided to try to piece together this problem using the teacher's (incredibly sketchy) solution sketch as a guide. As per the theme of this blog post, it involves proving algebraic identities using combinatorics.
The solution to this somewhat-contrived problem might seem out of the blue to you, but don't worry, it does to me too.
[quote="USA TST 2010.8"]
Let
be positive integers with 
, and let be the set of all -term sequences of positive integers 
such that 
. Show that
Hmm, I've spent at least ninety minutes working on this blog entry. I'm tired.
EDIT: Also, I guess this is a good
post.
![$\$[/dollar]1,000,000$](http://latex.artofproblemsolving.com/d/b/9/db96a95bccdd62c68b3a9491f78029c80f9e63f3.png)
prize. At one point, though, I got to go to a workshop entitled "Combinatorially Thinking." And as all of you know, combinatorics is obviously my strength and not geometry, right?So our essential assignment was to prove algebraic statements involving Fibonacci numbers, factorials, and binomial coefficients (as well as Stirling numbers of the first and second kinds, Lucas numbers, and other things, but we didn't have time to get to those) WITHOUT ALGEBRA.
I haven't touched this since April, so I decided to go back to it and see if I was any better at this stuff.
Here's a warm-up problem to get a sense as to the methods we were required to use.
[quote]Let
be the th Fibonacci number, with the alternate definition 
and 
for all 
. Consider the number of ways to tile a 
board with 
squares and 
dominoes. Prove that this number is .[/quote]This is a standard recursion problem, right? Well, things got a bit hairier after that.
[quote]
Prove that for all
, ![\[f_0+f_2+f_4+\cdots f_{2n}=f_{2n+1}.\]](http://latex.artofproblemsolving.com/7/f/7/7f7f92ab0d0ee5c31605af96a077e15569db84cb.png)
[/quote][quote]Prove that for
, ![\[f_{n-1}^2+f_n^2=f_{2n}.\]](http://latex.artofproblemsolving.com/7/4/1/741b5b3a6646974a42f35d30b3c21598b08524e9.png)
[/quote][quote]Prove that for any
, ![\[\dbinom n0+\dbinom{n-1}1+\dbinom{n-2}2+\cdots=f_n.\]](http://latex.artofproblemsolving.com/e/8/2/e822786946a80dde5c74ce6c9a01fd5e40dce9db.png)
[/quote]And these were all the easy ones.... Unfortunately, I have not been able to get any more of the problems. Here are some of the other ones in the packet (that I have not solved yet), if anybody is interested.
[quote]For
, ![\[\sum_{k=0}^nf_k^2=f_nf_{n+1}.\]](http://latex.artofproblemsolving.com/5/8/8/58804c716fe7f5b5ae5b67cd4e2e83af9f2a8fb9.png)
[/quote][quote]For
, ![\[f_{2n}=\sum_{k\geq 0}\dbinom nk f_k\qquad\text{and}\qquad2^nf_n=\sum_{k\geq 0}\dbinom nk f_{3k}.\]](http://latex.artofproblemsolving.com/a/c/e/aced00d22514bc1605b0df684eae91f0c171e32a.png)
[/quote][quote]For
, ![\[\sum_{i\geq 0}\sum_{j\geq 0}\dbinom{n-i}j\dbinom{n-j}i=f_{2n+1}.\]](http://latex.artofproblemsolving.com/d/a/8/da8632bbb30d1f7ebe9c7f26c60c72fdbd925c82.png)
[/quote]I guess I should end with a problem that was NOT from this worksheet, but rather from Combinatorical Arguments at AMSP. Our teacher presented this problem during his lecture, and essentially nobody saw it coming. I was able to grab the main idea of his solution, but there were several key parts that were missing (which was essentially the theme of the course, I guess). I decided to try to piece together this problem using the teacher's (incredibly sketchy) solution sketch as a guide. As per the theme of this blog post, it involves proving algebraic identities using combinatorics.
The solution to this somewhat-contrived problem might seem out of the blue to you, but don't worry, it does to me too.
[quote="USA TST 2010.8"]
Let
be positive integers with 
, and let be the set of all -term sequences of positive integers 
such that 
. Show that\[\begin{align*}&\sum_S 1^{a_1} 2^{a_2} \cdots n^{a_n}\\& =
{n \choose n} n^m - {n \choose n-1} (n-1)^m + \cdots +
(-1)^{n-2} {n \choose 2} 2^m + (-1)^{n-1} {n \choose 1}.\end{align*}\][/quote]Hmm, I've spent at least ninety minutes working on this blog entry. I'm tired.
EDIT: Also, I guess this is a good
post.
0 replies
AMSP Mock AIME/Obligatory Week 2 Post
djmathman 1
N
by SodaKing1
So yea. We just found out that AMSP is holding a mock AIME on Monday. The questions are going to be written/stolen from obscure contests by 1=2 and dannyhamtx. I'm actually looking forward to it.
In other news, I probably got two out of five questions on each of the tests today. I'm too lazy to type up the solutions, so I'll just post the four problems I got right.
[quote="Combo Problem 1"]
You want to color the numbers from
with some colors such that no number is divisible by a different number of the same color. What is the smallest possible number of colors you must have?[/quote]
[quote="Combo Problem 2"]
We are given a polygon
, a line 
and a point 
on in general position: all lines containing a side of the polygon intersect in distinct points different from . We mark each vertex of the polygon if the sides from it extended will cut the line in two points such that is between them. Show that lies inside the polygon if and only if on each side of there are an odd number of marked vertices.
[/quote]
[quote="Geometry Problem 1"]
Let
be the points of tangency of the incircle of a triangle with its sides 
respectively. Then, the triangle is equilateral if and only if the centroids of 
and are isogonal conjugates with respect to triangle .[/quote]
[quote="Geometry Problem 2"]
Let
be the midpoints of the arcs of triangle which contain the vertices of the triangle. Prove that the Simson lines of with respect to are concurrent. (Hint: these Simson lines are very special cevians in the medial triangle of .)[/quote]
In other news, I probably got two out of five questions on each of the tests today. I'm too lazy to type up the solutions, so I'll just post the four problems I got right.
[quote="Combo Problem 1"]
You want to color the numbers from
with some colors such that no number is divisible by a different number of the same color. What is the smallest possible number of colors you must have?[/quote][quote="Combo Problem 2"]
We are given a polygon
, a line 
and a point 
on in general position: all lines containing a side of the polygon intersect in distinct points different from . We mark each vertex of the polygon if the sides from it extended will cut the line in two points such that is between them. Show that lies inside the polygon if and only if on each side of there are an odd number of marked vertices.[/quote]
[quote="Geometry Problem 1"]
Let
be the points of tangency of the incircle of a triangle with its sides 
respectively. Then, the triangle is equilateral if and only if the centroids of 
and are isogonal conjugates with respect to triangle .[/quote][quote="Geometry Problem 2"]
Let
be the midpoints of the arcs of triangle which contain the vertices of the triangle. Prove that the Simson lines of with respect to are concurrent. (Hint: these Simson lines are very special cevians in the medial triangle of .)[/quote]
1 reply
Obligatory AMSP Week 1 Post
djmathman 3
N
by AwesomeToad
Darn, AMSP is hard.
Even though it's only been a week I've learned a good deal, including:
[list]
[*]Don't let Nicky Sun steal your name tag on the first day; if he does, threaten him with push-ups later for an easy win
[*]Cosmin is too OP[/list]
Here are some of the problems I've conquered on the worksheets and tests this week. If anybody knows the sources to these problems, let me know!
[quote]
Let
be the number of -tuples 
with values in 
and 
for which the sum 
is odd, and 
the number of -tuples for which the sum is even. Prove that
![\[\dfrac{O_n}{E_n}=\dfrac{2^n-1}{2^n+1}.\]](//latex.artofproblemsolving.com/0/6/1/06122299edef254ebcd50b02ddcddf246e04996a.png)
[/quote]
[quote]
Let be any positive integer. Show that ![\[\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk=1+\dfrac12+\ldots+\dfrac1n.\]](//latex.artofproblemsolving.com/e/3/2/e32b682c54178b3583ba6375505321852e5a39bb.png)
[/quote]
[quote]
We are given
intervals on the real line. Prove that there either exist 
intervals which are pairwise disjoint or intervals with nonempty intersection.[/quote]
[quote]
Points
are chosen on the sides respectively of a triangle . Denote by 
the centroids of triangles 
, 
, 
respectively. Prove that the lines 
, 
, 
are concurrent if and only if lines 
are concurrent.[/quote]
[quote]
Let be an isosceles triangle with 
. Its incircle touches in and in . A line distinct of 
goes through and intersects the incircle in and . Line intersects lines ad in 
and 
. Prove that 
.[/quote]
[quote]
Let be a triangle and let 
and 
be the squares erected on the sides which are directed towards the exterior of the triangle. Let be the circumcenter of triangle 
. Prov that 
is the -symmedian of triangle .[/quote]
Even though it's only been a week I've learned a good deal, including:
[list]
[*]Don't let Nicky Sun steal your name tag on the first day; if he does, threaten him with push-ups later for an easy win
[*]Cosmin is too OP[/list]
Here are some of the problems I've conquered on the worksheets and tests this week. If anybody knows the sources to these problems, let me know!
[quote]
Let
be the number of -tuples 
with values in 
and 
for which the sum 
is odd, and 
the number of -tuples for which the sum is even. Prove that![\[\dfrac{O_n}{E_n}=\dfrac{2^n-1}{2^n+1}.\]](http://latex.artofproblemsolving.com/0/6/1/06122299edef254ebcd50b02ddcddf246e04996a.png)
[/quote][quote]
Let
be any positive integer. Show that ![\[\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk=1+\dfrac12+\ldots+\dfrac1n.\]](http://latex.artofproblemsolving.com/e/3/2/e32b682c54178b3583ba6375505321852e5a39bb.png)
[/quote][quote]
We are given
intervals on the real line. Prove that there either exist 
intervals which are pairwise disjoint or intervals with nonempty intersection.[/quote][quote]
Points
are chosen on the sides respectively of a triangle . Denote by 
the centroids of triangles 
, 
, 
respectively. Prove that the lines 
, 
, 
are concurrent if and only if lines 
are concurrent.[/quote][quote]
Let
be an isosceles triangle with 
. Its incircle touches in and in . A line distinct of 
goes through and intersects the incircle in and . Line intersects lines ad in 
and 
. Prove that 
.[/quote][quote]
Let
be a triangle and let 
and 
be the squares erected on the sides which are directed towards the exterior of the triangle. Let be the circumcenter of triangle 
. Prov that 
is the -symmedian of triangle .[/quote]3 replies
A Random Combo Problem
djmathman 1
N
by NewAlbionAcademy
[quote="AwesomeMath Combo Problem"]
A number of tennis players take part in a tournament in which each of them plays exactly one game with each of the others. If 
and 
denote the number of wins and losses, respectively, of the 
th player, prove that
![\[x_1^2+x_2^2+\cdots + x_n^2=y_1^2+y_2^2+\cdots+y_n^2.\]](//latex.artofproblemsolving.com/e/6/b/e6b624d8ad2d5b7ab9674c676fc8d14df7eaff1d.png)
[/quote]
A number of
tennis players take part in a tournament in which each of them plays exactly one game with each of the others. If 
and 
denote the number of wins and losses, respectively, of the 
th player, prove that![\[x_1^2+x_2^2+\cdots + x_n^2=y_1^2+y_2^2+\cdots+y_n^2.\]](http://latex.artofproblemsolving.com/e/6/b/e6b624d8ad2d5b7ab9674c676fc8d14df7eaff1d.png)
[/quote]1 reply
Probability
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N Quick Reply
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