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.
be the positive real number. Prove that
, such that for all positive reals 
, the following is true:
grid. Alice goes first, and they take turns drawing a black point from the coordinate set![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](http://latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers 
such that Alice has a winning strategy. grid, where 
. Prove that the remainder can be covered by copies of the figures of 
or 
unit squares depicted in the drawing below.
does the sequence 
defined by the initial condition 
and the recursion 
have 
for all 
for which there exists a positive integer 
with the property that is divisible by and 
is divisible by 
.
denote the set of real numbers. Find all functions 
such that![\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]](http://latex.artofproblemsolving.com/5/b/e/5be34cb26d276bae3c60fb299ad66d83bc158a01.png)
.
be a function that satisfies
for all positive integer pairs 
Find all possible functions 
as the set of all positive integers.)
be a triangle with orthocenter 
and circumcenter 
. Let 
be a point in the plane such that 
. Let 
and 
be the reflections of in the lines 
and 
, respectively. Let 
be the orthogonal projection of onto . Let 
be the orthogonal projection of onto . Assume that 
and 
. Prove that 
.
, an imaginary quadratic field.
which contains 
such that 
? Here is the quaternion group with 8 elements 
, a finite subgroup of the group of units 
of the ring 
of all Hamiltonian quaternions.
satisfying
for all integers .
be a cyclic quadrilateral. A circle centered at passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar.
, 
is biased so that, when tossed, it has probability 
of falling heads. If the coins are tossed, what is the probability that the number of heads is odd? Express the answer as a rational function .
![$ = \frac{1}{2} \left [ \left ( \frac{1}{2k+1} + \frac{2k}{2k+1} \right )^{n} + (-1)^{n+1} \left ( \frac{1}{2k+1} - \frac{2k}{2k+1} \right )^{n} \right ]$](http://latex.artofproblemsolving.com/5/5/7/557de3c48c6411218705b4a13e01e5ed3750b95e.png)
denote the probability that given 
coins we have an odd number coming up heads, and 
similarly except for even.
and 
.
and 
and 
and 
, hence the answer is 
.
coins 
, and the probability that 
when flipped will land heads is ![$\Pr[C_k = H] = 1/(2k+1)$](http://latex.artofproblemsolving.com/6/6/f/66f04110a227c213a6ca88771c52a3a7f4509ee1.png)
for each 
. Therefore, you cannot regard distinct 
-subsets of the as having equivalent probabilities of showing heads.
.
, 
.
, the probability that the number of heads is odd among the 
tosses is 
. Then, correspondingly, the probability that the number of heads is even is 
.
. Consider the -th toss. If the number of heads is odd for the first tosses, we need the -th toss be a tail; otherwise, we need it to be a head. So the probability that after -th toss the total number of heads is odd is ![\[\dfrac{k-1}{2k-1}\cdot(1 - \dfrac{1}{2k + 1}) + \dfrac{k}{2k-1}\cdot\dfrac{1}{2k+ 1} = \dfrac{2k^2 - k}{(2k-1)(2k+1)} = \dfrac{k}{2k+1}\]](http://latex.artofproblemsolving.com/b/3/8/b3860ed78d392db59f1b0b2ae78139584af9fa06.png)
This completed the inductive proof.
) is indeed equal to the product of probabilities that those respective coins are even/odd. This statement wouldn't be too difficult to prove by induction though.
heads is the coefficient of 
in 
denote the probability of tossing an odd number of heads the first tosses. Then, 
with 
. Computing the first few values of , they are 
, 
, 
, and 
. We now try to prove that 
by induction.: we have 
which is true., we show that the statement holds for 
:
we have .
be the probability that the number of heads is odd for coins and similarly define 
for an odd number of heads. Then, by considering 
we have the recurrence 
and 
. Note that for every coefficient of a term 
in the expansion of this function, this represents the number of ways to flip the coins so that the number of heads displayed is equal to . Additionally, note that by setting 
, we have that the exponents of all terms are positive, while setting 
makes only the coefficients of the terms with an even exponent of 
positive, while the coefficients of the terms with an odd exponent of are negative. Thus, our desired probability becomes
where we divide by 
to account for each of the coefficients with an odd exponent of being added twice. Note that
Thus, our expression becomes
![$\mathcal{E}_n := \mathbb{P}[\text{even number of heads for a toss of n coins}]$](http://latex.artofproblemsolving.com/1/5/b/15b9f3ebb2e6c8aebfa71d8257badffa771c2bc8.png)
![$\mathcal{O}_n := \mathbb{P}[\text{odd number of heads for a toss of n coins}]$](http://latex.artofproblemsolving.com/b/8/4/b844b81771c136e4b155b8cc23f3a62497482ce3.png)
.
Subtracting above two equations,
Interating we have, 
Here 
and 
.![$\mathcal{O}_n=\frac{1}{2}[(\mathcal{E}_n+\mathcal{O}_n)-(\mathcal{E}_n-\mathcal{O}_n)]=\frac{1}{2}\left(1-\frac{1}{2n+1}\right) = \boxed{\frac{n}{2n+1}}$](http://latex.artofproblemsolving.com/6/f/4/6f4d7e8e18f7930eadd776a6b2426d46a3d73f2e.png)
The coefficient of gives the probability of achieving heads. Thus the desired quantity is the sum of the odd degree coefficients, which is given by 
. We can easily see that 
. On the other hand, 
, which telescopes to . Thus the answer is 
denote the required probability.
, we get 
.
![\[
\mathbb{P}(\text{H}) = \frac{1}{2k+1}, \quad \mathbb{P}(\text{T}) = \frac{2k}{2k+1}
\]](http://latex.artofproblemsolving.com/c/9/b/c9bd64a81fcb5a54313af26f7027e8cb65304d3f.png)
![\[
\left( \frac{1}{2k+1} + \frac{2k}{2k+1} \right)^n \quad \text{and} \quad \left( \frac{2k}{2k+1} - \frac{1}{2k+1} \right)^n
\]](http://latex.artofproblemsolving.com/b/b/4/bb46d190139d4c60f2ddfd8322bac1d974879734.png)
![\[
F_1^k = \left( \frac{1}{2k+1} + \frac{2k}{2k+1} \right), \quad
F_2^k = \left( \frac{2k}{2k+1} - \frac{1}{2k+1} \right)
\]](http://latex.artofproblemsolving.com/7/6/0/76003c79d9d983c9a5c2629e0b6d0a59909e8703.png)
![\[
\frac{\prod_{k=1}^{n}F_1^k - \prod_{k=1}^{n}F_2^k}{2}
\]](http://latex.artofproblemsolving.com/0/a/e/0ae7abe73aa7c8d0a54ccbee1adffb34441990f0.png)
![\[
\prod_{k=1}^{n} F_1^k = 1
\]](http://latex.artofproblemsolving.com/e/8/a/e8ab200d45e3126c1614250482c1d5282e664dbe.png)
![\[
\prod_{k=1}^{n} F_2^k = \prod_{k=1}^{n} \frac{2k - 1}{2k + 1} = \frac{1}{3} \cdot \frac{3}{5} \cdot \frac{5}{7} \cdots \frac{2n - 1}{2n + 1}
= \frac{1}{2n + 1}
\]](http://latex.artofproblemsolving.com/c/4/5/c450fdab6ad6728fbe012d349202752d6761ca5b.png)
![\[
\frac{1 - \frac{1}{2n + 1}}{2} = \frac{n}{2n + 1}
\]](http://latex.artofproblemsolving.com/d/f/8/df8fb1c18e9f612960dbceb2a2a18f4151d61e52.png)
Latexed by Leviee....solved by me....thanks Leviee
and![\[E-O=\left(\frac{2}{3}-\frac{1}{3}\right)\cdot\left(\frac{4}{5}-\frac{1}{5}\right)\cdots\left(\frac{2n}{2n+1}-\frac{1}{2n+1}\right)=\frac{1}{3}\frac{3}{5}\cdots\frac{2n-1}{2n+1}=\frac{1}{2n+1},\]](http://latex.artofproblemsolving.com/0/1/9/01971cdafd04942b458e75522282031c4d3dd08b.png)
so 
.![\[E-O=\left(\frac{2}{3}-\frac{1}{3}\right)\cdot\left(\frac{4}{5}-\frac{1}{5}\right)\cdots\left(\frac{1}{2n+1}-\frac{1}{2n+1}\right)=\frac{1}{3}\frac{3}{5}\cdots\frac{2n-1}{2n+1}=\frac{1}{2n+1},\]](http://latex.artofproblemsolving.com/0/4/3/0438e4ad7df6378df442af9ad9585c6cb6035bb5.png)
so 
instead of 
![\[
O(n) = \mathbb{P}(\text{odd number of heads in } n \text{ tosses})
\]](http://latex.artofproblemsolving.com/4/c/4/4c415436aa6f27721ebf548d83dc4f1d9746955b.png)
![\[
E(n) = \mathbb{P}(\text{even number of heads in } n \text{ tosses})
\]](http://latex.artofproblemsolving.com/2/b/d/2bda81921e1a0a68f2c3f2a59aba9434cbceddc9.png)
![\[
O(n) + E(n) = 1
\]](http://latex.artofproblemsolving.com/b/8/8/b8825d581dd0efded9c6a5f7299cf182afafb1ca.png)
![\[
\text{Observe that:}
\]](http://latex.artofproblemsolving.com/e/0/3/e036292ed4e4ce7f008a1233fa4c921cb19e1e96.png)
![\[
O(n) = O(n-1) \cdot \mathbb{P}(\text{last toss is tails}) + E(n-1) \cdot \mathbb{P}(\text{last toss is heads})
\]](http://latex.artofproblemsolving.com/4/9/6/496f574ca067f23b5fc2c54209c3654f1cee9685.png)
![\[
\Longrightarrow O(n) = O(n-1) \cdot \frac{2n}{2n+1} + E(n-1) \cdot \frac{1}{2n+1}
\]](http://latex.artofproblemsolving.com/8/6/6/8665ee81a9f65806743fbd7b93aa580cf1ddfba6.png)
![\[
\Longrightarrow \boxed{O(n) = O(n-1) \cdot \frac{2n}{2n+1} + (1 - O(n-1)) \cdot \frac{1}{2n+1}}
\]](http://latex.artofproblemsolving.com/a/5/7/a57e1f7256a027eb73fe308385f8823fd3595121.png)