For any positive integer $k$, let $f(k)$ be the number of elements in the set
$\{k + 1, k + 2,\dots, 2k\}$ whose base 2 representation has precisely three $1$s.

a) Surjectivity of f
For space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation.
It's easy to see that (one has to place two 1s in the less n significative bit of a (n+1)-bit number)
Now consider $k=2^n + 2$ with $n \ge 2$ and the corrisponding set $S(2^n+2)=\{2^n+3,...,2^{n+1},2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\}$
whose subset $\{2^n+3,...,2^{n+1}\}$ contains $f(2^n)$ T-numbers since $S(2^n)=\{2^n+1,...,2^{n+1}\}=\{2^n+1,2^n+2\} \cup \{2^n+3,...,2^{n+1}\}$ contains $f(2^n)$ T-numbers by definition but $\{2^n+1,2^n+2\}$ has none. So $S(2^n+2)=\{2^n+3,...,2^{n+1}\} \cup \{2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\}$ but the last set has the only T-number $2^{n+1}+3$. We conclude that:
Now consider $k=2^n +2^r + 2^s$ with $n>r>s\ge 0$ and $n\ge 2$.
We explicitly calculate f(k) for such numbers.
So we have to calculate how many T-numbers are in the set $S(k)=\{2^n +2^r + 2^s+1;...;2^{n+1} +2^{r+1} + 2^{s+1}\}$.
The T-numbers less than $2^{n+1}$ in $S(k)$ are $T_1 = \{2^n +2^j + 2^i : (j=r \wedge r>i>s) \vee  (n>j>r \wedge j>i>0) \}$
whence $\# (T_1)= r-s-1 + \sum_{h=r+1}^{n-1} h$.
The T-numbers greater than $2^{n+1}$ in $S(k)$ are $T_2 = \{2^{n+1} +2^j + 2^i : (j=r+1 \wedge s+1 \ge i\ge 0) \vee  (r\ge j \ge 1 \wedge j>i \ge 0) \}$
whence $\# (T_2)= s+2 + \sum_{h=1}^{r} h$.
Therefore  $S(k)$ contains $\# (T_1) + \# (T_2)=r-s-1 + s+2 + \sum_{h=1}^{r}h  +\sum_{h=r+1}^{n-1} h = \sum_{h=1}^{n-1} h + r + 1$ T-numbers and we have:
Summarizing
That's to say that f takes all the values from $f(2^n)$ to $f(2^{n+1})$ for every $n \ge 2$ and
then takes any positive integer value since $f(4)=1$ and $f(2^n)$ becomes arbitrarily large.
b) one-from-one values
By the fact that $f(2^n +2^r + 2^s)=f(2^n) + r +1$ where $r=1,2,...,n-1$ and $r > s \ge 0$
it follows that each of the values $f(2^n) + r +1$ where $r=2,...,n-1$  come from at least two different k because there are at least two different choices for $s$. These values are $f(2^n)+3,f(2^n)+4,...,f(2^n)+n=f(2^{n+1})$.
Thus the only possible one-from-one values are $f(2^n)+1$ that come from $k=2^n + 2$ and $f(2^n)+2$ that come from $k=2^n + 3$.
If $k=2^n + 3$ we have $f(k)=f(k+1)$ because k+1 is not T and $2k+1=2^{n+1} + 7$ and $2k+2=2^{n+1} + 8$ are not T so f maps $2^n + 3$ and $2^n + 4$ to $f(2^n)+2$.
If $k=2^n + 2$ then  $f(k+1)=f(2^n)+2>f(k)=f(2^n)+1>f(k-1)=f(2^n+1)=f(2^n)$.
The function f is non-decreasing.
It is sufficient to prove that $f(k+1) \ge f(k)$ but this follows by the fact that if k+1 is T then 2k+2 is T too.
By the monotonicity of f $f( 2^n + 2)=f(2^n)+1=\tbinom n2 +1= n(n-1)/2 + 1$ with $n \ge 2$ are the only one-from-one values of f.

a) Surjectivity of f
For space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation and define the sets $S(k)=\{k+1;k+2;...;2k\}$.
So $f(k)$ is the number of T-numbers in $S(k)$.
A positive even integer 2n is T iff n is T.(Fundamental theorem of T numbers)(FTT)
The function f is non-decreasing.
It is sufficient to prove that $f(k+1) \ge f(k)$. This follows by (FTT) if k+1 is T then 2k+2 is T too; so passing from $S(k)$ to $S(k+1)$ we can loose a T-number $k+1$ but we regain it with $2k+2$ in $S(k+1)$ so $f(k+1)$ cannot be less than $f(k)$. We also have $f(k+1) \le f(k) + 1$. This would be false if $k+1$ were not T and both $2k+1$ and $2k+2$ were T. But if $2k+2$ were T then $k+1$ would be T too by the (FTT).
Then we have $f(k) \le f(k+1) \le f(k) + 1$ that is $f(k+1)=f(k)$ or $f(k+1)=f(k)+1$.
But $f(4)=1$ and $f(2^n)= \tbinom n2$ so f takes all of the positive integer values because starting from 1 with step of 1 reaches arbitrarily large integer values.
b) one-from-one values of f
The one-from-one values $f(k)$ are such that $f(k)=f(k-1)+1$ and $f(k+1)=f(k)+1$ since f is monotone non-decreasing and has step 1.
By the condition $f(k+1)=f(k)+1$ since $k+1$ is T iff $2k+2$ is T we have that $2k+1$ must be T.
By the condition $f(k)=f(k-1)+1$ since $k$ is T iff $2k$ is T we have that $2k-1$ must be T.
Then $2k-1$ and $2k+1$ must have the form $2^{n+1}+2^{r+1}+1$ with $n>r\ge 0$ since they are odd and $n \ge 2$ since there is only 1 T-number less than 8 and they must have the same number of bits since their difference is 2 and both are T (the only two binary numbers which differs by 2 and have different number of bits are $2^n+1$ and $2^n-1$ or $2^n$ and $2^n-2$ which are evidently not T). Let $2k+1=2^{n+1}+2^{j+1}+1$ and $2k-1=2^{n+1}+2^{i+1}+1$ with $j>i \ge 0$. Then $2k+ 1-(2k-1)=2^{j+1}-2^{i+1}=2$ that is $j=1$ and $i=0$.
We conclude that $f(k)$ is one-from-one for $k=2^n+2^j=2^n+2$ with $n \ge 2$.
Since $f(2^n+1)=f(2^n)=\tbinom n2$ we have that $f(2^n+2)=\tbinom n2 +1= n(n-1)/2 + 1$ with $n \ge 2$ are the only one-from-one values of f.


Let $S$ denote the set of nonnegative integers.  Find all functions $f$ from $S$ to itself such that
$f(m+f(n))=f(f(m))+f(n)$ $\forall m,n \in S$

Solution: This problem needs a solution. If you have a solution for it, please help us out by adding it.


For any positive integer $n$, let $d(n)$ denote the number of positive divisors
of $n$ (including 1 and $n$ itself). Determine all positive integers $k$ such that
$d(n^2)/d(n) = k$ for some $n$.

Solution:
First we must $d$etermine gener$a$l values for $d(n)$:
Let $n=p1 ^ a1 * p2 ^ a2 * .. * pc ^ ac$, if $d$ is an ar$b$itr$a$ry divisor of $n$ then $d$ must have the same prime factors of $n$, each with an exponent $b_i$ being: $0\leq b_i\leq a_i$.
Hence there are $Ai + 1$ choices for each exponent of Pi in the number d => there are $(a_1 + 1)(a_2 + 1)..(a_c + 1)$ such d
$\implies d(n) = (a_1 + 1)(a_2+1)..(a_c+1)$ where $a_i$ are exponents of the prime numbers in the prime factorisation of $n$.
$\implies d(n^2)/d(n) = {(2a_1 + 1)(2a_2 + 1)..(2a_c + 1)}/{(a_1+1)..(a_c+1)}$
So we want to find all integers $k$ that can $b$e represented by the product of fractions of the form $(2n+1)/(n+1)$
Obviously $k$ is odd as the numerator is always odd.
It's possible for 1 (1/1) and 3 $(5/3 * 9/5)$, which suggests that it may be possible for all odd integers, which we can show by induction.
$P(k)$: It's possible to represent $k$ as the product of fractions $(2n+1)/(n+1)$
Base case: $k = 1: (2(0) + 1) / (0 + 1)$
Now assume that for $k\geq 3$ it's possible for all odds < $k$.
Since $k$ is odd, $k+1 = 2^zy$ where $y$ is odd and $y$ < $k$
Let there be a number $x$ s.t $k-y=x\implies k+1 = 2^z(k-x)\implies (2^zx+1)/(2^z-1)=k$
Also consider $k/y$. ISTS $k/y$ can be represented by a product of fractions of the form $2n+1/n+1$ in order to show $k$ can be represented by product of fractions $2n+1/n+1$, since $y$ can be represented in such a manner too.
$k/y = k/(k-x) = 1/(1 - x/k)$
Using our definition of $k$ in terms of $x$:
$k/y = 1/({1 - {2^z-x}/{2^zx+1}}) = {2^zx+1}/{x+1}$
And that is simply the product of fractions: ${2x+1}/{x+1} * {4x+1}/{2x+1} * .. * {2^zx+1}/{2^{z-1}x}$.
We have shown that $k/y$ can be written s.t it's a product of fractions of the form ${2n+!}/{n+1}\implies k$ can be written in such a way too.
Hence we have shown that all odds less than $k$ satisfies $P(n)\implies P(k)$ is true. Since we have shown P(1) is true, it must hence be true for all odd integers.
Therefore, $d(n^2)/d(n) = k\iff k$ is odd, for some n. I.E all odd $k$ satisfy the condition posed in the question.∎

-dabab_kebab (wrote this solution)


Consider an $n \times n$ square board, where $n$ is a fixed even positive integer.  The board is divided into $n^{2}$ units squares.  We say that two different squares on the board are adjacent if they have a common side.
$N$ unit squares on the board are marked in such a way that every square (marked or unmarked) on the board is adjacent to at least one marked square.
Determine the smallest possible value of $N$.


Let $n \ge 2$ be a positive integer and $\lambda$ a positive real number. Initially there are $n$ fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points $A$ and $B$ to the left of $B$, and letting the flea from $A$ jump over the flea from $B$ to the point $C$ so that $\frac{BC}{AB}=\lambda$.
Determine all values of $\lambda$ such that, for any point $M$ on the line and for any initial position of the $n$ fleas, there exists a sequence of moves that will take them all to the position right of $M$.


Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.

Solution:
For each girl, we know that there is a boy who solved a problem in common with her. Since a girl solves at most six problems, for a girl and her set of six problems there are at least 11 boys who solved a problem in common with the girl such that at least three boys solved that common problem. (This is a simple application of the pigeon hole principle where the problems solved by the girl are the holes and the boys are the pigeons, and the guy/pigeon enters a hole/problem which he solved in common with the girl).
The boys have solved a total of at most $21\times 6$ problems.We view this as a set with repeated elements i.e. if a problem is solved by more than one boy it appears as many times as the number of boys who have solved it. Let this set of problems be A. Clearly, A has size $21\times 6$.
We mark each problem in A which has been solved by at least three boys and a girl. And we mark it the number of times that is the same as the number of girls that has solved it. Since there are a total of $21\times 11$ marks (since there are at least 11 marks for each girl, by the discussion above), either a problem is marked at least thrice, in which case we're done since then it has been solved by at least three boys and three girls. Or each problem has been marked at most twice. In this case it is clear that more than $21\times 5$ problems in A have been marked since $21\times 5 \times 2<21\times 11$ (there are a total of $21\times 11$ marks. This means that there is a boy such that all six of his problems have been marked. But then by our discussion in the first paragraph we know there must a problem that this boy has solved which has been solved by at least three girls. Therefore it must be true that there is a problem such that it has been solved by at least three people of each gender.


Find all pairs of positive integers $m,n \ge 3$ for which here exist infinitely many positive integers $a$ such that

is itself an integer


Each pair of opposite sides of convex hexagon has the property that the distance between their midpoints is $\frac{\sqrt{3}}{2}$ times the sum of their lengths.  Prove that the hexagon is equiangular.


Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.

Determine all $m \times n$ rectangles that can be covered without gaps and without overlaps with hooks such that;
(a) the rectangle is covered without gaps and without overlaps,
(b) no part of a hook covers area outside the rectangle.


Let $x, y, z > 0$ satisfy $xyz\ge 1$. Prove that


Determine the least real number $M$ such that the inequality  holds for all real numbers $a,b$ and $c$


In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size. Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.


Prove that there are infinitely many positive integers $n$ such that $n^{2} + 1$ has a prime divisor greater than $2n + \sqrt {2n}$.

Solution 1
The main idea is to take a gaussian prime $a+bi$ and multiply it by a "twice as small" $c+di$ to get $n+i$. The rest is just making up the little details.
For each sufficiently large prime $p$ of the form $4k+1$, we shall find a corresponding $n$ such that $p$ divides $n^2+1$ and $p>2n+\sqrt{2n}$. Since there exist infinitely many such primes and, for each of them, $n \ge \sqrt{p-1}$, we will have found infinitely many distinct $n$ satisfying the hypothesis.
Take a prime $p$ of the form $4k+1$ and consider its "sum-of-two squares" representation $p=a^2+b^2$, which we know to exist for all such primes. As $a\ne b$, assume without loss of generality that $b>a$. If $a=1$, then $n=b$ is what we are looking for, and $p=n^2+1 > 2n+\sqrt{2n}$ as long as $p$ (and hence $n$) is large enough. Assume from now on that $b>a>1$.
Since $a$ and $b$ are (obviously) co-prime, there must exist integers $c$ and $d$ such that

In fact, if $c$ and $d$ are such numbers, then $c\pm ma$ and $d\mp mb$ work as well for any integer $m$, so we can assume that $c \in \left[-\frac{a}{2}, \frac{a}{2}\right]$.
Define $n=|ac-bd|$ and let's see why this is a good choice. For starters, notice that $(a^2+b^2)(c^2+d^2)=n^2+1$.
If $c=\pm\frac{a}{2}$, from (1) we see that $a$ must divide $2$ and hence $a=2$. This implies, $d=-\frac{b-1}{2}$ and $n=\frac{b(b-1)}{2}-2$. Therefore,
$\left(b-\frac{1}{2}\right)^2 = 1/4 + 2(n+2) > 2n$, so $b > \sqrt{2n}+\frac{1}{2}$. Finally, $p=b^2+2^2 > 2n+\sqrt{2n}$ and the case $c=\pm\frac{a}{2}$ is cleared.
We can safely assume now that

As $b>a>1$ implies $b>2$, we have

so
Therefore,
Before we proceed, we would like to show first that $a+b-1 > \sqrt{p}$. Observe that the function $x+\sqrt{p-x^2}$ over $x\in(2,\sqrt{p-4})$ reaches its minima on the ends, so $a+b$ given $a^2+b^2=p$ is minimized for $a = 2$, where it equals $2+\sqrt{p-2^2}$. So we want to show that
which is not hard to show for large enough $p$.
Now armed with $a+b-1>\sqrt{p}$ and (2), we get

where $u=\sqrt{p}.$
Finally,

The proof is complete.
Solution 2
We begin with a lemma.
Lemma: For every prime $p \equiv 1 \mod{4}$, there exists a positive integer $n \le \dfrac{p - \sqrt{p-4}}{2}$ such that $p\mid n^2 + 1$.
Proof:
We know that there must be a solution $x$ to the equation

with $1 \le x \le p-1$.  However, if $x$ is a solution then so is $p-x$, so there must be a solution $x \le p-1$.  Let $n$ denote this solution and suppose that $n = \frac{p-k}{2}$.  Then, we have

Since $k$ is a positive integer, it follows that $k^2 \ge p-4$, so we have

as desired.  The lemma is proven.

Suppose for sake of contradiction that there are only finitely many integers $n_1,n_2,\dots,n_k$ that work.  Let $p \equiv 1 \mod{4}$ be a prime with $p>20$ and such that $p$ is relatively prime to $n_i^2+1$ for all $i$ (the existence of $p$ is guaranteed by the existence of infinitely many primes of the form $4k+1$).  Then, we know that there exists an $N$ such that $N\le \dfrac{p - \sqrt{p-4}}{2}$ and $p | N^2 + 1$ (this last condition shows that $N$ is not among $n_1,n_2,\dots,n_k$.  We want to show that

But, we know that $p-2N \ge \sqrt{p-4}$, so it suffices to show that

Once again, it suffices to show that

which follows from $p>20$.
Thus, $N$ satisfies the required condition and it follows that there exist infinitely many values of $n$ that satisfy the given condition, as desired.


Suppose that $s_1,s_2,s_3,\ldots$ is a strictly increasing sequence of positive integers such that the subsequences
are both arithmetic progressions. Prove that the sequence $s_1,s_2,s_3,\ldots$ is itself an arithmetic progression.
Author: Gabriel Carroll, USA

Solution:
then
i.s.w.
put S(3) = b, S(2) = a, S(1) = k
therefore,
every S(n) is an arithmetic sequence.
Q.E.D.


Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that $\left(g(m)+n\right)\left(g(n)+m\right)$ is a perfect square for all $m,n\in\mathbb{N}.$
Author: Gabriel Carroll, USA

Solution:
Suppose such function $g$ exist then:
Lemma 1) $g(m) \ne g(m+1)$
but  $\left(g(m)+m\right)^2<\left(g(m+1)+m\right)\left(g(m)+m+1\right)<\left(g(m)+m+1\right)^2$.
Lemma 2) $|g(m)-g(m+1)| = 1$ (we have show that it can't be 0)
Then there must exist a prime number $p$ such that $g(m)$ and $g(m+1)$ are in the same residue class modulo $p$.
If $|g(m)-g(m+1)| = p^aq$ where $q$ is not divisible by $p$.

Hence, that number is not a perfect square.

If $g(m)-g(m+1) = 1$, then $g(x) = -x + k$,  $k\in\mathbb{N}$.
If $g(m)-g(m+1) = -1$, then $g(x) = x + k$,  $k\in\mathbb{N}$.


Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.
Proposed by Alexander A. Polyansky, Russia

Solution:
Let the excenters opposite $A,B,C$ be $I_a,I_b,I_c$. Let the midpoint of $\overline{I_bI_c}$ be $M_a$, which lies on $(ABC)$, the nine-point circle of $\triangle I_aI_bI_c$; analogously define $M_b,M_c$.
$M_aB=M_aC$ and $BC_1=s-a=B_1C$, so $\triangle M_aBC_1\cong\triangle M_aCB_1$ (SAS), thus $M_a$ is equidistant from $B_1,C_1$, with analogous results for $M_b,M_c$. It follows that the circumcentre of $\triangle A_1B_1C_1$ is one of $M_a,M_b,M_c$; WLOG, suppose it is $M_a$.
By isogonal conjugacy, $I_aA_1,I_bB_1I_cC_1$ concur at the Bevan point $V$ of $\triangle ABC$. $M_aM_b$ is the common perpendicular bisector of $\overline{C_1A_1}$ and $\overline{I_cC}$, so $C_1A_1\parallel I_cC$. $(A_1C_1M_b)$ is the circle on diameter $\overline{VB}$, so by Reim's theorem, $V \in (I_bI_cBC)$.
Hence $\angle I_cI_aI_b=\tfrac{1}{2}\angle I_cVI_b=45^{\circ}\implies\angle CAB=180^{\circ}-2\angle I_cI_aI_b=90^{\circ}$, as required.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Convex quadrilateral $ABCD$ has $\angle{ABC}=\angle{CDA}=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside $\triangle{SCT}$ and
Prove that line $BD$ is tangent to the circumcircle of $\triangle{TSH}.$

Solution:
Denote $\angle{HSB}=v$, $\angle{HTD}=u$, $\angle{HSC}=s$, $\angle{HTC}=t$, $\angle{HCS}=x$, $\angle{HCT}=y$, $\angle{AHS}=z$, $\angle{AHT}=w$. Since $\angle{CHS}-\angle{CSB} =90$ and $\angle{CHT}-\angle{CTD}=90$, we have $\angle{CSA}=x+90$, $\angle{CTA}=y+90$.
Since $\angle{CHS} = \angle{CSB}+90$, the tangent of the circumcircle of $\triangle{CSH}$ at point $S$ is perpendicular to $SB$; therefore, the circumcenter of $\triangle{CSH}$ (point $K$) is on $AB$. Similarly, the circumcenter of $\triangle{CTH}$ (point $L$) is on $AD$. In addition, $KL$ is the perpendicular bisector of $CH$.
Extend $CB$ to meet circumcircle of  $\triangle{CSH}$ at $P$, and extend $CD$ to meet circumcircle of  $\triangle{CTH}$ at $Q$. Then, since $\angle{ADC}=\angle{ABC}=90$, $AD$ and $AB$ are the perpendicular bisector of $CQ$ and $CP$, respectively; hence $A$ is the circumcenter of $\triangle{PCQ}$. Since $B$ and $D$ are midpoints on $CP$ and $CQ$, $PQ \parallel BD$; also, $AH \perp BD$, so $AH \perp PQ$. Since $A$ is the circumcenter, $AH$ is also the perpendicular bisector of $PQ$. Hence,
We have

Hence, $u+v = -w-z+\angle{DHC}+\angle{BHC} = 180 -(w+z)$, or
Since quadrilaterals $QTHC$ and $PSHC$ are cyclic, we have $\angle{THQ}=\angle{TCQ}=90-u$, $\angle{SHP}=\angle{SCP}=90-v$; so,

Hence,

Similarly,
Now we apply law of Sines repeatedly on pairs of triangles.
For $\triangle{QCH}$ and $\triangle{PCH}$, $\angle{HQC}=\angle{HTC}=t$, $\angle{HCQ}=y+z$, $\angle{HPC}=s$, $\angle{HCP}=x+w$; hence,

For $\triangle{LHK}$, $\angle{HLK}=\frac{1}{2}\angle{HLC}=t$, $\angle{HKL}=\frac{1}{2}\angle{HKC}=s$; hence,

For $\triangle{LAK}$, $\angle{ALK}=90-\angle{DML}=90-\angle{CMK}=\angle{MCH}=y+z$, and similarly, $\angle{AKL}=w+x$; hence,

Combining $(1), (2), (3)$, we have

Therefore, $TS \parallel KL$, and $\angle{ATS} = \angle{ALK}=y+z$.
Let the circumcircle of $\triangle{THS}$ meets $AH$ at $G$. We have,

And,

This proves $HG$ is the diameter of the circle and the center of the circle is on AH. $\square$
Solution by $Mathdummy$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Let $P = A_1A_2 \cdots A_k$ be a convex polygon in the plane. The vertices $A_1,A_2,\dots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.

Solution:
Note that $2S$ is always an integer for any lattice polygon, so it remains to show that it is divisible by $n$. It clearly suffices to prove the problem for when $n=p^m$ is a prime power. We proceed using induction on $k$, with the base case of $k=3$ settled by Heron's formula: If $a,b,c$ are the side lengths of the triangle, then the square of the area is $S^2=\frac{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}{16}$. As $n\mid a^2,b^2,c^2$, we have that $n^2\mid S^2\implies n\mid S$, as desired.
For the inductive step, we claim that there exists a diagonal whose length squared is also divisible by $n$. Then, we may split $P$ into two polygons with less vertices and areas divisible by $n$ by assumption. Let $3\le i\le k-1$ be such that $v_p(A_1A_i^2)=q$ is minimized. By Ptolemy's Theorem on cyclic quadrilateral $A_1A_{i-1}A_iA_{i+1}$, we have that $\sqrt{A_1A_{i-1}^2\cdot A_iA_{i+1}^2}+\sqrt{A_1A_{i+1}^2\cdot A_iA_{i-1}^2}=\sqrt{A_1A_i^2\cdot A_{i-1}A_{i+1}^2}$, or $\sqrt{\frac{A_1A_{i-1}^2\cdot A_iA_{i+1}^2}{np^q}}+\sqrt{\frac{A_1A_{i+1}^2\cdot A_iA_{i-1}^2}{np^q}}=\sqrt{\frac{A_1A_i^2\cdot A_{i-1}A_{i+1}^2}{np^q}}$. As we have $p^q\mid AA_{i-1}^2,AA_{i+1}^2$ and $n\mid A_iA_{i+1}^2,A_iA_{i-1}^2$, the terms under the square roots on the LHS are integers, so the LHS is an algebraic integer. This implies that the term under the square root on the RHS is also an integer, so $np^q\mid A_1A_i^2\cdot A_{i-1}A_{i+1}^2\implies n\mid A_{i-1}A_{i+1}^2$, as desired.


A hunter and an invisible rabbit play a game in the Euclidean plane.  The rabbit's starting point, $A_0$, and the hunter's starting point, $B_0$, are the same.  After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$ and the hunter is at point $B_{n-1}$.  In the nth round of the game, three things occur in order.
(i) The rabbit moves invisibly to a point $A_n$ such that the distance between $A_{n-1}$ and $A_n$ is exactly 1.
(ii) A tracking device reports a point $P_n$ to the hunter.  The only guarantee provided by the tracking device is that the distance between $P_n$ and $A_n$ is at most 1.
(iii) The hunter moves visibly to a point $B_n$ such that the distance between $B_{n-1}$ and $B_n$ is exactly 1.
Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after $10^9$ rounds she can ensure that the distance between her and the rabbit is at most 100?


An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from $1$ to $10$

Does there exist an anti-Pascal triangle with $2018$ rows which contains every integer from $1$ to $1 + 2 + 3 + \dots + 2018$?

Solution:
Trivially it is required that every positive integer from $1$ to $1+2+3+\cdots+2018$ appears exactly once.
Let $M_n$ denote the maximum number in the $n$th row and let $m_n$ denote the minimum number in the $n$th row.
Now assume $n\leq 2017$ and consider the numbers directly below $M_n$. Call these $a$ and $b$ where w.l.o.g. $a>b$. Then $a-b=M_n$. Since $a\leq M_{n+1}$ and $b\geq m_{n+1}$, we obtain that $M_{n+1}\geq M_n+m_{n+1}$.
Thus, for $1\leq i<j\leq 2018$,
In particular, since $M_1=m_1$,
Thus $M_{2018}$ is a sum of 2018 distinct positive integers. However, $M_{2018}\leq 1+2+3+\cdots+2018$. Thus $M_{2018}=1+2+3+\cdots+2018$ and $\{m_1,m_2,m_3,\dots,m_{2018}\}$ is a permutation of $\{1,2,3,\dots,2018\}$. Also, this implies that the other inequalities are also equalities and, for any $1\leq j\leq 2018$,
Now let any positive integer $n\leq 2018$ be "small" and let any positive integer $1+2+3+\cdots+2017\leq n\leq 1+2+3+\cdots+2018$ be "large". Since $\{m_1,m_2,m_3,\dots,m_{2018}\}$ is a permutation of $\{1,2,3,\dots,2018\}$, there is exactly one small number in each row.
If $n\leq 1954$, we have
so the $n$th row cannot contain any large numbers.
If $1955\leq n\leq 2017$, let $l$ be a large number in the $n$th row. Let the numbers directly below $l$ be $a$ and $b$ where w.l.o.g. $a>b$. We have $b=a-l$ and $a\leq 1+2+3+\cdots+2018$ so, because $l$ is large, $b\leq 2018$ so $b$ is small. Thus $b=m_{n+1}$ so $l$ is directly above $m_{n+1}$. Thus there are at most 2 large numbers in the $n$th row.
Thus there are at most 126 large numbers outside the bottom row. Since there are 2019 large numbers, there are at least 1893 large numbers in the bottom row so at most 125 non-large numbers in the bottom row. Now there are 2017 pairs of adjacent large numbers in the bottom row. We remove the pair directly beneath $m_{2017}$ and at most 250 other pairs containing a non-large number. Thus we can find a pair of adjacent large numbers in the bottom row, not directly beneath $m_{2017}$. However, their difference is small and in the 2017th row but not $m_{2017}$, which is a contradiction. Thus there is no such anti-Pascal triangle.


A social network has $2019$ users, some pairs of whom are friends. Whenever user $A$ is friends with user $B$, user $B$ is also friends with user $A$. Events of the following kind may happen repeatedly, one at a time:
Three users $A$, $B$, and $C$ such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends, change their friendship statuses such that $B$ and $C$ are now friends, but $A$ is no longer friends with $B$, and no longer friends with $C$. All other friendship statuses are unchanged.
Initially, $1010$ users have $1009$ friends each, and $1009$ users have $1010$ friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.


Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB= \angle CAD$. The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$, the point $F$ on the segment $AB$ satisfies $\angle FDA= \angle DBC$, and the point $X$ on the line $AC$ satisfies $CX=BX$. Let $O_1$ and $O_2$ be the circumcentres of the triangles $ADC$ and $EXD$ respectively. Prove that the lines $BC$, $EF$, and $O_1 O_2$ are concurrent.

Solution:
We prove that circles $ACD, EXD$ and $\Omega_0$ centered at $P$ (the intersection point $BC$ and $EF)$ have a common chord.
Let $P$ be the intersection point of the tangent to the circle $\omega_2 = BDC$ at the point $D$ and the line $BC, A'$ is inverse to $A$ with respect to the circle $\Omega_0$ centered at $P$ with radius $PD.$
Then the pairs of points $F$ and $E, B$ and $C$  are inverse with respect to $\Omega_0$, so the points $F, E,$ and $P$ are collinear. Quadrilaterals containing the pairs of inverse points $B$ and $C, E$ and $F, A$ and $A'$ are inscribed, $FE$ is antiparallel to $BC$ with respect to angle $A$ (see $\boldsymbol{Claim}$).
Consider the circles $\omega = ACD$ centered at $O_1, \omega' = A'BD,$
$\omega_1 = ABC, \Omega = EXD$ centered at $O_2 , \Omega_1 = A'BX,$ and $\Omega_0.$
Denote $\angle ACB = \gamma$. Then $\angle BXC =  \angle BXE = \pi – 2\gamma,$
$\angle AA'B = \gamma (AA'CB$ is cyclic),
$\angle AA'E =  \pi –  \angle AFE = \pi – \gamma (AA'EF$ is cyclic, $FE$ is antiparallel),
$\angle BA'E =   \angle AA'E –  \angle AA'B = \pi – 2\gamma =  \angle BXE \implies$
$\hspace{13mm}E$ is the point of the circle $\Omega_1.$
Let the point $Y$ be the radical center of the circles $\omega, \omega', \omega_1.$ It has the same power $\nu$ with respect to these circles. The common chords of the pairs of circles $A'B, AC, DT,$ where $T = \omega \cap \omega',$  intersect at this point.
$Y$ has power $\nu$ with respect to $\Omega_1$ since $A'B$ is the radical axis of $\omega', \omega_1, \Omega_1.$
$Y$ has power $\nu$ with respect to $\Omega$ since $XE$ containing $Y$ is the radical axis of $\Omega$ and $\Omega_1.$
Hence $Y$ has power $\nu$ with respect to $\omega, \omega', \Omega.$
Let $T'$ be the point of intersection $\omega \cap \Omega.$ Since the circles $\omega$ and $\omega'$ are inverse with respect to $\Omega_0,$ then $T$ lies on $\Omega_0,$ and $P$ lies on the perpendicular bisector of $DT.$ The power of a point $Y$ with respect to the circles  $\omega, \omega',$ and $\Omega$ are the same, $DY \cdot YT = DY \cdot YT' \implies$ the points $T$ and $T'$ coincide.
The centers of the circles  $\omega$ and  $\Omega$ ($O_1$ and $O_2$) are located on the perpendicular bisector $DT'$, the point $P$ is located on the perpendicular bisector $DT$ and, therefore, the points $P, O_1,$ and $O_2$ lie on a line, that is, the lines $BC, EF,$ and $O_1 O_2$ are concurrent.
$\boldsymbol{Claim}$
Let $AK$ be bisector of the triangle $ABC$, point $D$ lies on $AK.$ The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$. The point $F$ on the segment $AB$ satisfies $\angle ADF= \angle CBD.$ Let $P$ be the intersection point of the tangent to the circle $BDC$ at the point $D$ and the line $BC.$ Let the circle $\Omega_0$ be centered at $P$ and has the radius $PD.$
Then the pairs of points $F$ and $E, B$ and $C$  are inverse with respect to $\Omega_0$ and $EF$ and $BC$ are antiparallel with respect to the sides of an angle $A.$
$\boldsymbol{Proof}$
Let the point $E'$ is symmetric to $E$ with respect to  bisector $AK, E'L || BC.$
Symmetry of points $E$ and $E'$ implies $\angle AEL = \angle AE'L.$

Similarly, we prove that $FL$ and $BC$ are antiparallel with respect to angle $A,$ and the points $L$ in triangles $\triangle EDL$ and $\triangle FDL$ coincide. Hence, $FE$ and $BC$ are antiparallel and $BCEF$ is cyclic.
Note that $\angle DFE =  \angle DLE –  \angle FDL =  \angle AKC –  \angle CBD$ and
$\angle PDE = 180^o –  \angle CDK –  \angle CDP –  \angle LDE = 180^o – (180^o –  \angle AKC –  \angle BCD) –  \angle CBD –  \angle BCD$
$\angle PDE  =  \angle AKC –  \angle CBD = \angle DFE,$ so $PD$ is tangent to the circle $DEF.$
$PD^2 = PC \cdot PB = PE \cdot PF,$ that is, the points $B$ and $C, E$ and $F$ are inverse with respect to the circle $\Omega_0.$

vladimir.shelomovskii@gmail.com, vvsss


Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around a circle such that the product of any two neighbours is of the form $x^2 + x + k$ for some positive integer $x$.

Solution:
https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]
https://youtu.be/_kF9uXCZ6l4 [Video Solution by little fermat]


For each integer $k \geqslant 2$, determine all infinite sequences of positive integers $a_1, a_2, \ldots$ for which there exists a polynomial $P$ of the form $P(x)=x^k+c_{k-1} x^{k-1}+\cdots+c_1 x+c_0$, where $c_0, c_1, \ldots, c_{k-1}$ are non-negative integers, such that

for every integer $n \geqslant 1$.

Solution:
https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]
https://www.youtube.com/watch?v=CmJn5FKxpPY [Video contains another solution to problem 3]
Let $f(n)$ and $g(j)$ be functions of positive integers $n$ and $j$ respectively.
Let $a_{n}=a_{1}+f(n)$, then $a_{n+1}=a_{1}+f(n+1)$, and $a_{n+k}=a_{1}+f(n+k)$
Let $P=\prod_{j=1}^{k}\left ( a_{n+j} \right ) = \prod_{j=1}^{k}\left ( a_{n}+g(j)) \right )$
If we want the coefficients of $P(a_{n})$ to be positive, then $g(j)\geq 0$ for all $j$ which will give the following value for $P$:
$P=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+\prod_{j=1}^{k} g(j) = P(a_{n})$
Thus for every $j$ and $n$ we need the following:
$a_{n}+g(j)=a_{n+j}=a_{1}+f(n+j)$
Solving for $g(j)$ we get:
$g(j)=a_{1}+f(n+j)-a_{n}=a_{1}+f(n+j)-a_{1}-f(n)$
$g(j)=f(n+j)-f(n)\geq 0$ for all $n$ and $j$ because $g(j)$ needs to be greater than or equal to zero for all coefficients to be non-negative.
This means that $f(n)$ needs to be increasing with $n$, or staying constant, and also with $f(1)=0$ because $a_{1}=a_{1}+f(1)$.
In addition, since we need all coefficients to be integer, then all $f(n)$ and $g(j)$ must also be integers.  We also need $g(j)$ to not be dependent of $n$, so in the expression $f(n+j)-f(n)$, the $n$ needs to cancel.  This mean that the rate of change for $f(n)$ with respect to $n$ needs to be constant.  This can only be achieved with $f(n)$ be the equation of a line with slope being either zero or positive integer.
So, we set $f(n)$ to be the equation of a line as $f(n)=mn+b$ with $m$ being the slope with a non-negative value and with $b$ the intercept at $n=0$.  We know that $f(1)=0$ so $f(1)=m+b=0$ which means that $b=-m$ and our function becomes $f(n)=mn-m=(n-1)m$. Since $f(n)$ needs to be non-negative integer then $m\geq 0 \mid m \in \mathbb{Z}$ then $f(n)$ is increasing or constant, with $f(1)=0$
Then, $g(j)=f(n+j)-f(n)=(n+j-1)m-(n-1)m=jm$
This gives:
$\prod_{j=1}^{k}\left ( a_{n}+jm \right )=P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!m^{k}$
with $C_{0}=k!m^{k}$ and coefficients of polynomial $\geq 0$
Then, $a_{n}=a_{1}+f(n)$
Which provides the solution of all infinite sequences of positive integers as:
$a_{n}=a_{1}+(n-1)m$, $\forall m\geq 0 \mid m \in \mathbb{Z}$ and $a_{1} \geq 1 \mid a_{1} \in \mathbb{Z}$
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Construct a right triangle with a given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.

We denote the  catheti of the triangle as $a$ and $b$.  We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse.  (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)

The conditions of the problem require that
However, we notice that twice the area of the triangle $abc$ is $ab$, since $a$ and $b$ form a right angle.  However, twice the area of the triangle is also the product of $c$ and the altitude to $c$.  Hence the altitude to $c$ must have length $\frac{c}{4}$.  Therefore if we construct a circle with diameter $c$ and a line parallel to $c$ and of distance $\frac{c}{4}$ from $c$, either point of intersection between the line and the circle will provide a suitable third vertex for the triangle.  Q.E.D.

We denote the angle between $b$ and $c$ as $\alpha$.  The problem requires that
or, equivalently, that
However, since $\frac{a}{c} = \sin{\alpha};\; \frac{b}{c} = \cos{\alpha}$, we can rewrite the condition as
or, equivalently, as
From this it becomes apparent that $2\alpha = \frac{\pi}{6}$ or $\frac{5\pi}{6}$; hence the other two angles in the triangle must be $\frac{ \pi }{12}$ and $\frac{ 5 \pi }{12}$, which are not difficult to construct.  Q.E.D.
Note.  It is not difficult to reconcile these two constructions.  Indeed, we notice that the altitude of the triangle is of length $c \sin{\alpha}\cos{\alpha}$, which both of the solutions set equal to $\frac{c}{4}$ .

If we let the legs be $a$ and $b$ with $a < b$, then $c^2 = a^2 + b^2$. Because $c^2 = 4 a b$ as well, we immediately deduce via some short computations through quadratic formula that $a = (2 - \sqrt{3})b$ and $a = (2 + \sqrt{3})b$. Thus, $\frac{a}{b} = \tan 15^\circ$ and $\frac{a}{b} = \tan 75^\circ$, and so one of the angles of the triangle must be $15^\circ$ and the other must be $75^\circ$. A $15^\circ$ angle is easily constructed by bisecting a $30^\circ$ angle (which is formed by constructing the altitude of an equilateral triangle), and $75^\circ$ angle is constructed by constructing a 15 degree angle on top of the 60 degree point.

We draw perpendicular $h$ on $c$ and denote the angle between the median and $c$ as $\alpha$. The area of the triangle is $\frac{ab}{2}$ or $\frac{ch}{2}$
We can write $c$ as $2\sqrt{ab}$ So we have,

Which simplifies to be
Or

So, we draw $c$ and from it's midpoint, we draw a segment $\frac{c}{2}$ forming an angle $\frac{\pi}{6}$
Join the vertices to get the required triangle.

This is a geometric solution. We start with some observations in the first paragraph and describe the construction in the next paragraph. Consider any right triangle with hypotenuse c. Construct square with side length c on the opposite side of c from the triangle. Construct 3 more triangles congruent to the original triangle on the outside of each of the three other sides of the square in such a way that a larger square with side length a+b is formed. The area of this larger square is c^2+2ab = 1.5c^2 so a+b = (3/2)^(1/2) *c
Now go back to the given c. Construct a square with side length c first. Let two adjacent corners of it be M and N. Construct the square's diagonals and let center of the square be O. Construct a circle around O with radius (3/2)^(1/2) * OM. To do so, take OM, first draw a 2:1 right triangle with OM being the short side to obtain a hypothenuse with length 3^(1/2) OM, then draw a 1:1 right triangle from that side to obtain a hypothenuse with length 6^(1/2)*OM. then bisect it for desired length. Next draw the circular locus of points X such that MXO is 45 degrees. To accomplish this, simply find the point on the perpendicular bisector of OM that is 1/(2OM) from OM. There are two such points, pick the one closer to N. Draw circle around this new point going through O and M. The intersection of the two circles is the desired third vertex of the triangle with the given hypotenuse c.


Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

Solution:
Let $M_a$, $M_b$, and $M_c$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let $H_a$, $H_b$, and $H_c$ be the feet of the altitudes from $A$, $B$, and $C$ to their opposite sides, respectively. Since $\triangle ABC\sim\triangle M_bM_aC$, with $M_bM_a=\frac12 AB$, the distance from $M_a$ to side $\overline{AC}$ is $\frac{h_b}{2}$.
Construct $AM_a$ with length $m_a$. Draw a circle centered at $A$ with radius $h_a$. Construct the tangent $l_1$ to this circle through $M_a$. $\overline{BC}$ lies on $l_1$.
Draw a circle centered at $M_a$ with radius $\frac{h_b}{2}$. Construct the tangent $l_2$ to this circle through $A$. $\overline{AC}$ lies on $l_2$. Then $C=l_1\cap l_2$.
Construct the line $l_3$ parallel to $l_2$ so that the distance between $l_2$ and $l_3$ is $h_b$ and $M_a$ lies between these lines. $B$ lies on $l_3$. Then $B=l_1\cap l_3$.


In the interior of triangle $P_1P_2P_3$ a point $P$ is given.  Let $Q_1,Q_2,Q_3$ be the intersections of $PP_1, PP_2,PP_3$ with the opposing edges of triangle $P_1P_2P_3$.  Prove that among the ratios $\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}$ there exists one not larger than $2$ and one not smaller than $2$.

Let $[ABC]$ denote the area of triangle $ABC$.
Since triangles $P_1P_2P_3$ and $PP_2P_3$ share the base $P_2P_3$, we have $\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_1}{P_1Q_1}$.
Similarly, $\frac{[PP_1P_3]}{[P_1P_2P_3]}=\frac{PQ_2}{P_2Q_2}, \frac{[PP_1P_2]}{[P_1P_2P_3]}=\frac{PQ_3}{P_3Q_3}$.
Adding all of these gives $\frac{[PP_1P_3]}{[P_1P_2P_3]}+\frac{[PP_1P_2]}{[P_1P_2P_3]}+\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_2}{P_2Q_2}+\frac{PQ_3}{P_3Q_3}+\frac{PQ_1}{P_1Q_1}=1$.
We see that we must have at least one of the three fractions not greater than $\frac{1}{3}$, and at least one not less than $\frac{1}{3}$. These correspond to ratios $\frac{PP_i}{PQ_i}$ being less than or equal to $2$, and greater than or equal to $2$, respectively, so we are done.

Let $K_1=[P_2PP_3], K_2=[P_3PP_1],$ and $K_3=[P_1PP_2].$  Note that by same base in triangles $P_2PP_3$ and $P_2P_1P_3,$   Thus,
Without loss of generality, assume $K_1\leq K_2\leq K_3.$ Hence,  and  as desired. $\blacksquare$


Solve the equation $\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1$.

Solution:
First, note that we can write the left hand side as a cubic function of $\cos^2 x$. So there are at most $3$ distinct values of $\cos^2 x$ that satisfy this equation. Therefore, if we find three values of $x$ that satisfy the equation and produce three different $\cos^2 x$, then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that $\frac{\pi}2$, $\frac{\pi}4$, and $\frac{\pi}6$ all satisfy the equation, and produce three different values of $\cos^2 x$, namely $0$, $\frac12$, and $\frac34$. So we solve $\cos^2 x = \text{each of these}$. Therefore, our solutions are:
$x = \frac{(2k+1)\pi}2,\, \frac{(2k+1)\pi}4,\, \frac{(6k+1)\pi}6,\, \frac{(6k+5)\pi}6 \quad \forall k\in Z$


Find all solutions $x_1,x_2,x_3,x_4,x_5$ of the system

where $y$ is a parameter.

Solution:
Notice: The following words are Chinese.
首先,我们可以将以上5个方程相加,得到:
$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$
当$x_1+x_2+x_3+x_4+x_5=0$时,因为$x_1,x_2,x_3,x_4,x_5$关于原方程组轮换对称,所以
$x_1=x_2=x_3=x_4=x_5=0$
若反之,则方程两边同除以$(x_1+x_2+x_3+x_4+x_5)$,得到$y=2$,显然解为
$x_1=x_2=x_3=x_4=x_5$
综上所述,若$y=2$,最终答案为$x_1=x_2=x_3=x_4=x_5$,否则答案为$x_1=x_2=x_3=x_4=x_5=0$
The solution in English (translated by Google Translate):
First of all, we can add the five equations to get:
$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$
When $x_1+x_2+x_3+x_4+x_5=0$, Because $x_1,x_2,x_3,x_4,x_5$ is symmetric in the original equations,
$x_1=x_2=x_3=x_4=x_5=0$
Otherwise, dividing both sides by $(x_1+x_2+x_3+x_4+x_5$, we get $y=2$, and clearly
$x_1=x_2=x_3=x_4=x_5$
Summarizing, if $y=2$, then the answer is of the form $x_1=x_2=x_3=x_4=x_5$. Otherwise, $x_1=x_2=x_3=x_4=x_5=0$.


Seventeen people correspond by mail with one another - each one with all
the rest. In their letters only three different topics are discussed. Each pair
of correspondents deals with only one of these topics. Prove that there are
at least three people who write to each other about the same topic.

Solution:
Lemma: Consider a complete graph with 6 vertices colored with 2 colors. There exists a monochromatic triangle.
Proof: Consider one vertex and all connections leading out from it. Call it $V_1$. It has 5 edges coming out from it. By the Pigeonhole Principle, there are at least 3 of the same color. Call this color red. Call those vertices $V_2$, $V_3$ and $V_4$. If any of the segments $V_2V_3$, $V_2V_4$, or $V_3V_4$ are red, then we have a monochromatic triangle with vertices $V_1$ and the other two that are also red. If they are all the other color, then we have a monochromatic triangle with vertices $V_2$,$V_3$, and $V_4$. $\blacksquare$

Main Problem: Represent these people as vertices on a connected graph with 17 vertices and colored with 3 colors, one corresponding with each topic. So this problem is reduced to showing that on a connected graph with 17 vertices and colored with three colors, there exists some monochromatic triangle. Look at an arbitrary vertex. Call it $V_1$. Look at the 16 other vertices that it is connected to. By the Pigeonhole Principle, there are at least 6 vertices connected to $V_1$ that are all one color. Call this color 1. If any of the connections inbetween these six vertices are in color 1, then we are done. If none of them are color 1, we know that that there are only 2 colors in those 6 vertices. By Lemma 1, we know that there is a monochromatic triangle in those 6 vertices. So we are done.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Find all sets of four real numbers $x_1$, $x_2$, $x_3$, $x_4$ such that the sum of any one and the product of the other three is equal to $2$.

Solution:
Let $P = x_1x_2x_3x_4$ be the product of the four real numbers.
Then, for $i = 1,2,3,4$ we have: $x_i + \prod_{j \neq i}x_j = 2$.
Multiplying by $x_i$ yields:
$x^2_i + P = 2x_i \Longleftrightarrow x^2_i-2x_i+1 = (x_i-1)^2 = 1-P \Longleftrightarrow x_i = 1 \pm t$ where $t = \pm \sqrt{1-P} \in \mathbb{R}$.
If $t=0$, then we have $(x_1,x_2,x_3,x_4)=(1,1,1,1)$ which is a solution.
So assume that $t \neq 0$. WLOG, let at least two of $x_i$ equal $1+t$, and $x_1 \ge x_2 \ge x_3 \ge x_4$ OR $x_1 \le x_2 \le x_3 \le x_4$.
Case I: $x_1 = x_2 = x_3 = x_4 = 1+t$
Then we have:
$(1+t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+4t = 0 \Longleftrightarrow t(t^2+3t+4) = 0$
Which has no non-zero solutions for $t$.
Case II: $x_1 = x_2 = x_3 = 1+t$ AND $x_4 = 1-t$
Then we have:
$(1-t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+2t = 0$ $\Longleftrightarrow t(t+1)(t+2) = 0 \Longleftrightarrow t \in \{0,-1,-2\}$
AND
$(1+t)+(1-t)(1+t)^2 = 2 (1+t)+(1-t)(1+t)^2 = 2 -t^3-t^2+2t = 0$ $\Longleftrightarrow -t(t-1)(t+2) = 0 \Longleftrightarrow t \in \{0,1,-2\}$
So, we have $t = -2$ as the only non-zero solution, and thus, $(x_1,x_2,x_3,x_4) = (-1,-1,-1,3)$ and all permutations are solutions.
Case III: $x_1 = x_2 = 1+t$ AND $x_3 = x_4 = 1-t$
Then we have:
$(1-t)+(1-t)(1+t)^2 = 2 \Longleftrightarrow -t^3-t^2 = 0$ $\Longleftrightarrow -t^2(t+1) = 0 \Longleftrightarrow t \in \{0,-1\}$
AND
$(1+t)+(1+t)(1-t)^2 = 2 \Longleftrightarrow t^3-t^2 = 0$ $\Longleftrightarrow t^2(t-1) = 0 \Longleftrightarrow t \in \{0,1\}$
Thus, there are no non-zero solutions for $t$ in this case.
Therefore, the solutions are: $(1,1,1,1)$; $(3,-1,-1,-1)$; $(-1,3,-1,-1)$; $(-1,-1,3,-1)$; $(-1,-1,-1,3)$.


Prove that for every natural number $n$, and for every real number $x \neq \frac{k\pi}{2^t}$ ($t=0,1, \dots, n$; $k$ any integer)

Solution:
Assume that $\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^{n}x}}=\cot{x}-\cot{2^{n}x}$ is true, then we use $n=1$ and get $\cot x - \cot 2x = \frac {1}{\sin 2x}$.
First, we prove $\cot x - \cot 2x = \frac {1}{\sin 2x}$
LHS=$\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}$
$= \frac{2\cos^2 x}{2\cos x \sin x}-\frac{2\cos^2 x -1}{\sin 2x}$
$=\frac{2\cos^2 x}{\sin 2x}-\frac{2\cos^2 x -1}{\sin 2x}$
$=\frac {1}{\sin 2x}$
Using the above formula, we can rewrite the original series as
$\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x \cdot \cdot \cdot + \cot 2^{n-1} x - \cot 2^n x$
Which gives us the desired answer of $\cot x - \cot 2^n x$


Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.

Solution:
Let the edges of one of the faces of the tetrahedron have lengths $a$, $b$, and $c$. Let $d$, $e$, and $f$ be the lengths of the sides that are not adjacent to the sides with lengths $a$, $b$, and $c$, respectively.
Without loss of generality, assume that $\max(a,b,c,d,e,f)=a$. I shall now prove that either $b+f>a$ or $c+e>a$, by proving that if $b+f\leq a$, then $c+e>a$.
Assume that $b+f\leq a$. The triangle inequality gives us that $e+f>a$, so $e$ must be greater than $b$. We also have from the triangle inequality that $b+c>a$. Therefore $e+c>b+c>a$. Therefore either $b+f>a$ or $c+e>a$.
If $b+f>a$, then the vertex where the sides of length $a$, $b$, and $f$ meet satisfies the given condition. If $c+e>a$, then the vertex where the sides of length $a$, $c$, and $e$ meet satisfies the given condition. This proves the statement. $\blacksquare$


A semicircular arc $\gamma$ is drawn with $AB$ as diameter. $C$ is a point on $\gamma$ other than $A$ and $B$, and $D$ is the foot of the perpendicular from $C$ to $AB$. We consider three circles, $\gamma_1, \gamma_2, \gamma_3$, all tangent to the line $AB$. Of these, $\gamma_1$ is inscribed in $\triangle ABC$, while $\gamma_2$ and $\gamma_3$ are both tangent to $CD$ and $\gamma$, one on each side of $CD$. Prove that $\gamma_1, \gamma_2$, and $\gamma_3$ have a second tangent in common.

Solution:
Denote the triangle sides $a = BC, b = CA, c = AB$. Let $\omega$ be the circumcircle of the right angle triangle $\triangle ABC$ centered at the midpoint $O$ of its hypotenuse $c = AB$. Let $R, S, T$ be the tangency points of the circles $K_1, K_2, K_3$ with the line AB. In an inversion with the center $A$ and positive power $r_A^2 = AC^2 = b^2$ ($r_A$ being the inversion circle radius), the line AB is carried into itself, the circle $\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\omega$. This implies that the circle $K_3$ intersecting the inversion circle $A$ is carried into itself, but this is possible only if the circle $K_3$ is perpendicular to the inversion circle $A$. It follows that the tangency point $T$ of the circle $K_3$ is the intersection of the inversion circle $(A, r_A = b)$ with the line $AB$. Similarly, in an inversion with the center B and positive power $r_B^2 = BC^2 = a^2$ ($r_B$ being the inversion circle radius), the line AB is carried into itself, the circle $\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\omega$. This implies that the circle $K_2$ intersecting the inversion circle $B$ is carried into itself, but this is possible only if the circle $K_2$ is perpendicular to the inversion circle $B$. It follows that the tangency point S of the circle $K_2$ is the intersection of the inversion circle $(B, r_B = a)$ with the line $AB$.
The distance between the tangency points S, T is the equal to ST = AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle $K_1$ of the right angle triangle $\triangle ABC$ is equal to
$r = \frac{|\triangle ABC|}{s} = \frac{ab}{a + b + c} = \frac{a + b - c}{2} = s - c$
where $|\triangle ABC|$ and s are the area and semiperimeter of the triangle $\triangle ABC$, for example, because of an obvious identity
$(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab$
or just because the angle $\angle C = 90^\circ$ is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then
$AR' = AT - \frac{ST}{2} = b - r = b - \frac{a + b - c}{2} = \frac{c + b - a}{2} = s - a = AR$
Therefore, the points $R' \equiv R$ are identical and the midpoint of the segment ST is the tangency point R of the incircle $K_1$ with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles $K_2, K_3$ are tangent to the incircle $K_1$. Radii $r_2, r_3$ of the circles $K_2, K_3$ are now easily calculated:
$r_2 = SD = BS - BD = a - \frac{a^2}{c}$
$r_3 = TD = AT - AD = b - \frac{b^2}{c}$
Denote $I, I_2, I_3$ the centers of the circles $K_1, K_2, K_3$. The line $I_2I_3$ cuts the midline RI of the trapezoid $STI_3I_2$ at the distance from the point R equal to
$\frac{SI_2 + TI_3}{2} = \frac{r_2 + r_3}{2} = \frac{a + b}{2} - \frac{a^2 + b^2}{2c} = \frac{a + b - c}{2} = r = RI$
As a result, the centers $I_2, I, I_3$ are collinear (in fact, I is the midpoint of the segment $I_2I_3$). The common center line $I_2I_3$ and the common external tangent AB of the circles $K_1, K_2, K_3$ meet at their common external homothety center $H \equiv I_2I_3 \cap AB$ and the other common external tangent of the circles $K_2, K_3$ from the common homothety center H is a tangent to the circle $K_1$ as well.
The above solution was posted and copyrighted by yetti. The original thread can be found here: [1]


Find the set of all positive integers $n$ with the property that the set $\{ n, n+1, n+2, n+3, n+4, n+5 \}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.

Solution: The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be $\{ 1, 2, 3, 4, 5, 6 \}$, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets.

As in the previous solution, none of the six consecutive numbers can be multiples of $7$.  This means that together, they take on the values $\{ 1, 2, 3, 4, 5, 6, \} \mod 7$.  The product of all the numbers in this set, then, is $-1 \mod 7$, by Wilson's Theorem.  However, $-1$ is not a quadratic residue $\mod 7$, which means that we cannot partition the original set into two sets of equal product.  Thus, no such $n$ exist.


All the faces of tetrahedron $ABCD$ are acute-angled triangles. We consider all closed polygonal paths of the form $XYZTX$ defined as follows: $X$ is a point on edge $AB$ distinct from $A$ and $B$; similarly, $Y, Z, T$ are interior points of edges $BC, CD, DA$, respectively. Prove:
(a) If $\angle DAB + \angle BCD \neq \angle CDA + \angle ABC$, then among the polygonal paths, there is none of minimal length.
(b) If $\angle DAB + \angle BCD = \angle CDA + \angle ABC$, then there are infinitely many shortest polygonal paths, their common length being $2AC \sin(\alpha / 2)$, where $\alpha = \angle BAC + \angle CAD + \angle DAB$.

Solution:
Rotate the triangle $BCD$ around the edge $BC$ until $ABCD$ are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting $X$ and $Z$. Therefore, $XYB=ZYC$.
Summing the four equations like this, we get exactly $\angle ABC+\angle ADC=\angle BCD+\angle BAD$.
Now, draw all four faces in the plane, so that $BCD$ is constructed on the exterior of the edge $BC$ of $ABC$ and so on with edges $CD$ and $AD$.
The final new edge $AB$ (or rather $A'B'$) is parallel to the original one (because of the angle equation). Call the direction on $AB$ towards $B$ "right" and towards $A$ "left". If we choose a vertex $X$ on $AB$ and connect it to the corresponding vertex $X'$ on A'B'. This works for a whole interval of vertices $X$ if $C$ lies to the left of $B$ and $D$ and $D$ lies to the right of $A$. It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.
Finally, regard the sine in half the isosceles triangle $ACA'$ which gives the result with the angles around $C$ instead of $A$, but the role of the vertices is symmetric.


A soldier needs to check on the presence of mines in a region having the
shape of an equilateral triangle. The radius of action of his detector is equal
to half the altitude of the triangle. The soldier leaves from one vertex of the
triangle. What path shouid he follow in order to travel the least possible
distance and still accomplish his mission?

Solution:
Let our triangle be $\triangle ABC$, let the midpoint of $AB$ be $D$, and let the midpoint of $CD$ be $E$. Let the height of the triangle be $h$. Draw circles around points $B$ and $C$ with radius $\frac{h}{2}$, and label them $c_1$ and $c_2$. Let the intersection of $BE$ and $c_1$ be $F$.
The path that is the solution to this problem must go from $A$ to a point on $c_2$ to a point on $c_1$. Let us first find the shortest possible path. We will then prove that this path fits the requirements.
The shortest path is in fact the path from $A$ to $E$ to $F$. We will prove this as follows:
Suppose that a different point on $c_2$ is the optimal point to go to. Let this point be $P$. Then, the optimal point on $c_1$ would be the intersection of $BP$ and $c_1$ (let this point be $R$). Let the line through $E$ parallel to $AB$ be $l$, and the intersection of $AP$ and $l$ be $Q$. Then, we have

The second inequality is true due to the triangle inequality, and we can prove the first to be true as follows:
Suppose we have a point on $l$, $X$. Reflect $B$ across $l$ to get $B'$. Then, $AX + BX = AX + B'X$, which is obviously minimized at the intersection of the two lines, $E$.
By $\textbf{(1)}$, we have

and we have proved the claim.
This path easily covers the whole triangle. This is because if you draw a line perpendicular to $AB$ from any point in the triangle, this line will hit the path in a distance less than or equal to $\frac{h}{2}$. We can compute the length of the path to be

~mathboy100


Consider decompositions of an $8\times8$ chessboard into $p$ non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If $a_i$ is the number of white squares in the $i$-th rectangle, then $a_1<a_2<\cdots<a_p.$
Find the maximum value of $p$ for which such a decomposition is possible. For this value of $p,$ determine all possible sequences $a_1, a_2, \cdots, a_p.$

Solution:
Since each rectangle has the same number of black squares as white squares, $a_1+a_2+\ldots +a_p=\frac{64}{2}=32$. Clearly $a_i\ge i$ for $i=1$ to $i=p$ so $32=a_1+a_2+\ldots + a_p\ge 1+2+\ldots +p=\frac{p(p+1)}{2}$ so this forces $p\le 7$. It is possible to decompose the board into $7$ rectangles, as we will show later. But first let us find all such sequences $a_i$.
Now $32-a_7=a_1+a_2+\ldots +a_6\ge 1+2+\ldots +6=21\implies 11\le a_7$. For a rectangle to have $11$ white squares, it will have an area of $22$ so it's dimensions are either $1\times 22$ or $2\times 11$ - neither of which would fit on a $8\times 8$ board. So $a_7\not= 11\implies a_7\le 10$.
If $a_7=10$ (which could fit as a $4\times 5$ rectangle) then $a_1+a_2+\ldots a_6=22$. Then $22-a_6\ge 1+2+\ldots +5=15$ so $7\ge a_6$. So $a_1,a_2,\ldots ,a_6$ are 6 numbers among 1-7. If $1\le k\le 7$ is the number that is not equal to any $a_i$, then $22=a_1+a_2+\ldots +a_7=1+2+\ldots +7-k=28-k$ so $k=6$. Then $a_1=1,a_2=2,a_3=3,a_4=4,a_5=5,a_6=7,a_7=10$. Such a decomposition is possible. Take a $4\times 5$ rectangle on the top left corner, where there are $4$ squares horizontally and $5$ vertically. Then directly below use a $7\times 2,1\times 2$ and a $8\times 1$ rectangle to cover the 3 rows below it. It's simple from there.
Similarly, you can find the other possibilities as $\{a_1,a_2,\ldots ,a_7\}=\{1,2,3,4,5,8,9\}$ or $\{1,2,3,4,6,7,9\}$ or $\{1,2,3,5,6,7,8\}$. Tilings are not hard to find.
The above solution was posted and copyrighted by WakeUp. The original thread for this problem can be found here: [1]
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


When $4444^{4444}$ is written in decimal notation, the sum of its digits is $A$. Let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$. ($A$ and $B$ are written in decimal notation.)

Solution:
Note that
Therefore $4444^{4444}$ has fewer than 17776 digits. This shows that $A<9\cdot 17776=159984$. The sum of the digits of $A$ is then maximized when $A=99999$, so $B\leq 45$. Note that out of all of the positive integers less than or equal to 45, the maximal sum of the digits is 12.
It's not hard to prove that any base-10 number is equivalent to the sum of its digits modulo 9. Therefore $4444^{4444}\equiv A\equiv B(\bmod{9})$. This motivates us to compute $X$, where $1\leq X \leq 12$, such that $4444^{4444}\equiv X(\bmod{9})$. The easiest way to do this is by searching for a pattern. Note that

and since $4444=3\times 1481+1$,

Thus, $X=7$, which means that the sum of the digits of $B$ is $\boxed{7}$.
~minor edits by KevinChen_Yay
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is $1976.$

Since $3*3=2*2*2+1$, 3's are more efficient than 2's. We try to prove that 3's are more efficient than anything:
Let there be a positive integer $x$. If $3$ is more efficient than $x$, then $x^3<3^x$. We try to prove that all integers greater than 3 are less efficient than 3:
When $x$ increases by 1, then the RHS is multiplied by 3. The other side is multiplied by $\dfrac{(x+1)^3}{x^3}$, and we must prove that this is less than 3 for all $x$ greater than 3.
$\dfrac{(x+1)^3}{x^3}<3\Rightarrow \dfrac{x+1}{x}<\sqrt[3]{3}\Rightarrow 1<(\sqrt[3]{3}-1)x$
$\dfrac{1}{\sqrt[3]{3}-1}<x$
Thus we need to prove that $\dfrac{1}{\sqrt[3]{3}-1}<4$. Simplifying, we get $5<4\sqrt[3]{3}\Rightarrow 125<64*3=192$, which is true. Working backwards, we see that all $x$ greater than 3 are less efficient than 3, so we never use anything greater than 3. Therefore we use as many 3s in groups of 2 as possible, and since the remainder when 1976 is divided by 6 is 2, we can use 658 3s and one 2.
$\dfrac{1976}{3}=658.6666$, so the greatest product is $\boxed{3^{658}\cdot 2}$.

(Non-rigorous)
We demonstrate heuristically that 3's are the most efficient.
As we know that the chosen numbers must be equal, by the AM-GM inequality, we wish to maximize
Simple logarithmic differentiation shows that $\frac{S}{x} = e$ maximizes the given function. As $e$ is approximately 2.71828, we use 2's and 3's only. 3's are optimal, but we must use one 2.

Note:  This solution uses the same strategy as Solution 1 (that having the largest possible number of three's is good), but approaches the proof in a different manner.
Let $f(S) \triangleq \prod_{x \in S} x$, and $g(S) = \sum_{x\in S} x$, where $S$ is a multiset.  We fix $g(S) = 1976$, and aim to maximize $f(S)$.  Since $3(x-3) > x$ for $x \geq 5$,  we notice that $S$ must only contain the  integers $1,2,3$ and $4$.  We can replace any occurrences of $4$ in $S$ by replacing it with a couple of $2$'s, without changing $f(S)$ or $g(S)$, so we may assume that $S$ only contains the integers $1,2$ and $3$.   We may further assume that $S$ contains at most one $1$,  since any two $1$'s can be replaced by a $2$ without changing $g(S)$, but with an increase in $f(S)$.  If $S$ contains exactly one $1$, then it must also contain at least one $2$ (since $1976 \equiv 2\;(\mathrm{mod }\;3)$).   We can then replace this pair of a $1$ and a $2$ with a $3$, thus keeping $g(S)$ constant, and increasing $f(S)$.   Now we may assume that $S$ contains only $2$'s and $3$'s.
Now, as observed in the last solution, any triplet of $2$'s can be replaced by a couple of $3$'s, with $g(S)$ constant, and an increase in $f(S)$.    Thus, after repeating this operation, we will be left with at most two $2$'s.  Since $g(S) = 1976$, and $1976 \equiv 2\;(\mathrm{mod }\;3)$,  we therefore get that $S$ must have exactly one $2$ (since we already showed it consists only of $2$'s and $3$'s).   Thus, we get $\boxed{f(S) = 2\cdot 3^{658}}$.


Let $a,b,A,B$ be given reals. We consider the function defined byProve that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$

Solution:
$f(x) = 1-\sqrt{a^2+b^2}\sin (x+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \geq 0$.
$f(x+\pi) = 1+\sqrt{a^2+b^2}\sin (x+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \geq 0$
Therefore, $\sqrt{A^2+B^2}\sin (2x+\arctan\frac{A}{B}) \leq 1$. Since this identity is true for any real $x$, let the sine term be one, $\longrightarrow A^2+B^2 \leq 1$.
To get cancellation on the rightmost terms, note $\sin (x+\pi/2) = \cos x, \sin (x-\pi/2) = -\cos x$.
$f(x+\pi/4) = 1-\sqrt{a^2+b^2}\sin (x+\pi/4+\arctan\frac{a}{b}) - \sqrt{A^2+B^2}\cos2x+\arctan\frac{A}{B}) \geq 0$.
$f(x-\pi/4) = 1-\sqrt{a^2+b^2}\sin (x-\pi/4+\arctan\frac{a}{b}) + \sqrt{A^2+B^2}\cos2x+\arctan\frac{A}{B}) \geq 0$.
Let $x+\arctan\frac{a}{b} = y$.
Then $\sqrt{a^2+b^2}(\sin (y+\pi/4) + \sin (y-\pi/4)) \leq 2$
$\sqrt{a^2+b^2} \leq \dfrac{2}{\sqrt{2}(\sin y)}$. Since it's valid for all real $x$ let $\sin y = 1$, and we are done.
The above solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [1]


In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

Solution:
Denote $a = BC, b = AB = AC$ the triangle sides, $r, R$ the triangle inradius and circumradius and $s$, $\triangle$ the triangle semiperimeter and area. Let $S, T$ be the tangency points of the incircle (I) with the sides $AB, AC$. Let $K, L$ be the midpoints of of $BC, PQ$ and $M$ the midpoint of the arc $BC$ of the circumcircle (O) opposite to the vertex $A$. The isosceles triangles $\triangle AST \sim \triangle APQ \sim \triangle ABC$ are all centrally similar with the homothety center $a$. The homothety coefficient for the triangles $\triangle AST \sim \triangle ABC$ is $h_{13} = \frac{AS}{AB} = \frac{s - a}{b}$. The homothety coefficient for the triangles $\triangle AST \sim \triangle APQ$ is $h_{12} = \frac{AS}{AP}$. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles $\triangle ASK \sim \triangle APM$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles $\triangle AST \sim \triangle APQ$: $h_{12} = \frac{AS}{AP} = \frac{AK}{AM} = \frac{h}{2R}$, where h = AK is the A-altitude of the triangle $\triangle ABC$. The homothety coefficient of the triangles $\triangle APQ \sim \triangle ABC$ is then $h_{23} = \frac{h_{12}}{h_{13}} = \frac{s - a}{b} \cdot \frac{2R}{h}$ Denote $h' = AL$ the $A$-altitude of the triangle $\triangle APQ$. Then
$\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}$
$KL = h - h' = h - 2R\ \frac{s - a}{b}$
Substituting $h = \frac{2 \triangle}{a}$, $2R = \frac{ab^2}{2 \triangle}$ and $s - a = \frac{\triangle^2}{s(s - b)^2}$, we get
$KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =$
(substituting 2s = 2b + a and $s - b = \frac a 2$ for an isosceles triangle)
$= \frac{\triangle}{s}\left(\frac{2s}{a} - \frac{ab}{2(s - b)^2}\right) = \frac{\triangle}{s} \left(\frac{2b}{a} + 1 - \frac{2b}{a}\right) = \frac{\triangle}{s} = r$
which means that the point $L \equiv I$ is identical with the incenter of the triangle $\triangle ABC$.
The above solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]


We consider a point $P$ in a plane $p$ and a point $Q \not\in p$. Determine all the points $R$ from $p$ for whichis maximum.

Solution:
Let $T$ be the orthogonal projection of the point $Q$ on the plane $p$. Then, the line $PT$ is the orthogonal projection of the line $PQ$ on the plane $p$, and thus forms the least angle with the line $PQ$ among all lines through the point $P$ which lie in the plane $p$; hence, $\measuredangle RPQ\geq\measuredangle TPQ$, and equality holds if and only if the point $R$ lies on the ray $PT$ (the only exception is when $PQ\perp p$; in this case, $P = T$, so the ray $PT$ is undefined, and equality holds for all points $R$ in the plane $p$, since we always have $\angle RPQ = \angle TPQ = 90^{\circ}$).
From $\measuredangle RPQ\geq\measuredangle TPQ$, it follows that $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$; also, since the angles $\angle RPQ$ and $\angle TPQ$ are $180^{\circ}$, their half-angles $\frac{\measuredangle RPQ}{2}$ and $\frac{\measuredangle TPQ}{2}$ are < 90°, so that from $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$ we can conclude that $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$. Equality holds, as in the above, if and only if the point $R$ lies on the ray $PT$ (and, respectively, for all points $R$ in the plane $p$ if $PQ\perp p$).
On the other hand, we obviously have $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ with equality if and only if $\frac{\measuredangle QRP-\measuredangle PQR}{2}=0^{\circ}$, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base $QR$, i. e. if and only if $PR = PQ$, i. e. if and only if the point $R$ lies on the sphere with center $P$ and radius $PQ$.
Now, applying the Mollweide theorem in triangle $QPR$, we get
$\frac{QP+PR}{QR}=\frac{\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}}{\sin\frac{\measuredangle RPQ}{2}}$
$\leq\frac{1}{\sin\frac{\measuredangle RPQ}{2}}$ (since $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$)
$\leq\frac{1}{\sin\frac{\measuredangle TPQ}{2}}$ (since $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$),
and equality holds here if and only if equality holds in both of the inequalities $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ and $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ that we have used, i. e. if and only if the point $R$ lies both on the ray $PT$ (this condition should be ignored if $PQ\perp p$) and on the sphere with center $P$ and radius $PQ$.
Hence, the point $R$ for which the ratio $\frac{QP+PR}{QR}$ is maximum is the point of intersection of the ray $PT$ with the sphere with center $P$ and radius $PQ$ (or, respectively, it can be any arbitrary point on the intersection of the plane $p$ with the sphere with center $P$ and radius $PQ$ if $PQ\perp p$).
This solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [1]


(a) For which values of $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers?
(b) For which values of $n>2$ is there exactly one set having the stated property?

Solution:
Let $k = \prod p_i^{e_i}$ be the greatest element of the set, written in its prime factorization.  Then $k$ divides the least common multiple of the other elements of the set if and only if the set has cardinality at least $\max \{ p_i^{e_i} \}$, since for any of the $p_i^{e_i}$, we must go down at least to $k-p_i^{e_i}$ to obtain another multiple of $p_i^{e_i}$.  In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.
For $n > 3$, we may let $k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2)$, since all the $p_i^{e_i}$ must clearly be less than $n$ and this product must also be greater than $n$ if $n$ is at least 4.  For $n > 4$, we may also let $k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)$, for the same reasons.  However, for $n = 4$, this does not work, and indeed no set works other than $\{ 3,4,5,6 \}$.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.
Q.E.D.


Prove that if $n$ is a positive integer such that the equation $x^3-3xy^2+y^3=n$ has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.

Solution:
Suppose the equation $x^3-3xy^2+y^3=n$ has solution in integers $(x,y)$ with $y=x+k$.  Then, completing the cube yields $(y-x)^3-3x^2y+2x^3$.  Using the substitution $y=x+k$ yields $k^3-3kx^2-x^3=n$.  Notice that equality directly implies that $(k,-x)$ is also a solution to the original equation.  Applying the transformation again yields that $(-x-k,-k)$ is also a solution.  We show that these three solutions are indeed distinct: If $(x,y)=(k,-x)$ then $x=k,x+k=-x$ which only has solution $x=k=0$ which implies that $n$ is not a positive integer, a contradiction.  Similarly, since the transformation from $(k,-x)$ to $(-x-k,-k)$ and $(-x-k,-k)$ to $(x,y)$ is the same as the transformation from $(x,y)$ to $(k,-x)$, we have that the three solutions are pairwise distinct.
For the case $n=2891$, notice that $7|n$.  Considering all solutions modulo $7$ of the equation $x^3-3xy^2+y^3\equiv0\pmod{7}$ yields only $x\equiv y\equiv 0\pmod{7}$.  But, this implies that $7^3$ divides $2891$ which is clearly not true.


Let $ABC$ be an equilateral triangle and $\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$ and $CA$ (including $A$, $B$ and $C$). Determine whether, for every partition of $\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle. Justify your answer.

Solution:
The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle.
First notice that if there are at least two points of the same color on a side, then all the points in $\mathcal{E}$ which project orthogonally onto points of that color on the respective side must have the opposite color if we are to have no monochromatic right-angled triangles.
Now assume we can find a side on which there is at most one point bearing one of the colors (blue, say; we take the two colors to be red and blue). Then, by the observation above, it's obvious that the other two sides contain three blue vertices of a right-angled triangle, and we're done (all the points of those two sides are blue, except maybe for the endpoints which they have in common with the other side and the point which projects orthogonally onto the unique blue point on the third side). We can now assume all three sides contain at least two points of each color. Take $U\in BC$ s.t. $BC=3BU$. $U$ projects orthogonally onto $V\in CA$, and $V$ projects orthogonally onto $T\in AB$. We can easily see that $T$ projects orthogonally onto $U$. Again, by the observation in the previous paragraph, the points $U,V,T$ must have different colors if we are to have no monochromatic right-angled triangle, and this is impossible (we only have two colors).
This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]


Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

Solution:
First, we prove that if $BC$ and $AD$ are parallel then the claim is true:  Let $AB$ and $CD$ intersect at $E$ (assume $E$ is closer to $AD$, the other case being analogous).  Let $M,N$ be the midpoints of $AB,CD$ respectively.  Let the length of the perpendicular from $N$ to $AB$ be $r$.  It is known that the length of the perpendicular from $M$ to $CD$ is $\frac{1}{2}AB$. Let the foot of the perpendicular from $C$ to $AB$ be $H$, and similarly define $G$ for side $CD$.  Then, since triangles $MNE$ and $BCE$ are similar, we have $\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}$.  This gives an expression for $r$:
$r=\frac{1}{2}AB\cdot\frac{CH}{BG}$
Noticing that $CH=BC\sin EBC,BG=BC\sin ECB$ simplifies the expression to
$r=\frac{1}{2}AB\cdot\frac{\sin EBC}{\sin ECB}$
By the Law of Sines, $\frac{\sin EBC}{\sin ECB}=\frac{EC}{EB}$.  Since triangles $EDA,ECB$ are similar, we have $\frac{EC}{EB}=\frac{CD}{AB}$ and thus we have
$r=\frac{1}{2}AB\cdot\frac{CD}{AB}=\frac{1}{2}CD$
and we are done.
Now to prove the converse.  Suppose we have the quadrilateral with $BC$ parallel to $AD$, and with all conditions satisfied.  We shall prove that there exists no point $T$ on $CD$ such that $T$ is a midpoint of a side $CD^\prime$ of a quadrilateral $ABCD^\prime$ which also satisfies the condition.  Suppose there was such a $T$.  Like before, define the points $E,M,N$ for quadrilateral $ABCD$.  Let $t$ be the length of the perpendicular from $T$ to $AB$.  Then, using similar triangles, $\frac{ET}{t}=\frac{EN}{\frac{1}{2}CD}$.  This gives
$t=\frac{\frac{1}{2}CD\cdot ET}{EN}$
But, we must have $t=DT$.  Thus, we have
$DT=\frac{\frac{1}{2}CD\cdot ET}{EN}$
$\Rightarrow \frac{1}{2}CD\cdot EN+NT\cdot EN=\frac{1}{2}CD\cdot (EN+NT)$
$\Rightarrow NT\cdot EN=\frac{1}{2}CD\cdot NT$
Since $\frac{1}{2}CD\ne EN$, we have $NT=0$ as desired.

Let $M$ and $N$ be midpoints of $AB$ and $CD$, respectively. Let $[X]$ denote the area of figure $X$.
Suppose that $AD // BC$ and the circle centered at $M$ with diameter $AB$ touches $CD$ at $T$. We know that midsegment $MN // AD$. It follows that $[AMN] = [DMN]$, so $\frac{1}{2} AM \cdot d = \frac{1}{2} MT \cdot DN$, where $d$ is the distance from $N$ to $AB$. But we have $MT = MA$, so $DN = d$. It follows that the circle centered at $N$ with diameter $CD$ touches $AB$.
If on the other hand we have the circle with diameter $CD$ touching $AB$ at $U$(as well as the circle with diameter $AB$ touching $CD$ at $T$), we must have $\frac{1}{2} DN \cdot MT = \frac{1}{2} AM \cdot UN$ because of the equivalence of radii in circles. Hence $[ANM] = [DMN]$, so $A$ and $D$ are equidistant from $MN$ (as $MN$ is the common base). Hence $AB // MN$. Similarly $CD // MN$, and so $AB // CD$, as desired.


Given a set $M$ of $1985$ distinct positive integers, none of which has a prime divisor greater than $23$, prove that $M$ contains a subset of $4$  elements whose product is the $4$th power of an integer.

Solution: We have that $x\in M\Rightarrow x=2^{e_1}3^{e_2}\cdots 19^{e_8}23^{e_9}$.  We need only consider the exponents.  First, we consider the number of subsets of two elements, such that their product is a perfect square.  There are $2^9=512$ different parity cases for the exponents $e_1,e_2,...,e_9$.  Thus, we have at least one pair of elements out of $1985$ elements.  Removing these two elements yields $1983$ elements.  By applying the Pigeon Hole Principle again, we find that there exists another such subset.  Continuing on like this yields at least $734$ pairs of elements of $M$ whose product is a perfect square.  Let $S$ be the set of the square roots of the products of each pair. Then, by the Pigeon Hole Principle again, there exist at least two elements whose product is a perfect square.  Let the elements be $x,y$ and let $x=\sqrt{ab},y=\sqrt{cd}$ where $a,b,c,d\in M$.  Then, we have $xy=z^2$ for some $z$ which implies $abcd=z^4$ and the claim is proved.


Let $A,B$ be adjacent vertices of a regular $n$-gon ($n\ge5$) with center $O$. A triangle $XYZ$, which is congruent to and initially coincides with $OAB$, moves in the plane in such a way that $Y$ and $Z$ each trace out the whole boundary of the polygon, with $X$ remaining inside the polygon. Find the locus of $X$.

Solution:
Let $C \not = A$ the vertex which is adjacent to $B$. While $XYZ$ moves from $OAB$ to $OBC$, it is easy to see $XYBZ$ is cyclic. Thus $X$ lies on the bisector of $\angle YBZ = \angle ABC$. Moreover, $X$ is the intersection of a circle passing through $B$ (the circumcircle of $XYBZ$) and with a fixed radius (the radius is a function of $\triangle XYZ$). Therefore $X$ varies in a line segment ended in $O$. When $Y$ and $Z$ pass through the other sides, we get as locus $n$ distinct line segments, each passing throught $O$ and contained in $OV$ (but not in $\vec{OV}$) for some vertex $V$ of the polygon. Each two of these lines are obtained one from another by a rotation with center $O$.
This solution was posted and copyrighted by feliz. The original thread for this problem can be found here: [1]


Prove that there is no function $f$ from the set of non-negative  integers into itself such that $f(f(n)) = n + 1987$ for every $n$.

We prove that if $f(f(n)) = n + k$ for all $n$, where $k$ is a fixed positive integer, then $k$ must be even. If $k = 2h$, then we may take $f(n) = n + h$.
Suppose $f(m) = n$ with $m \equiv n \mod k$. Then by an easy induction on $r$ we find $f(m + kr) = n + kr$, $f(n + kr) = m + k(r+1)$. We show this leads to a contradiction. Suppose $m < n$, so $n = m + ks$ for some $s > 0$. Then $f(n) = f(m + ks) = n + ks$. But $f(n) = m + k$, so $m = n + k(s - 1) \ge n$. Contradiction. So we must have $m \ge n$, so $m = n + ks$ for some $s \ge 0$. But now $f(m + k) = f(n + k(s+1)) = m + k(s + 2)$. But $f(m + k) = n + k$, so $n = m + k(s + 1) > n$. Contradiction.
So if $f(m) = n$, then $m$ and $n$ have different residues $\pmod k$. Suppose they have $r_1$ and $r_2$ respectively. Then the same induction shows that all sufficiently large $s \equiv r_1 \pmod k$ have $f(s) \equiv r_2 \pmod k$, and that all sufficiently large $s \equiv r_2 \pmod k$ have $f(s) \equiv r_1 \pmod k$. Hence if $m$ has a different residue $r \mod k$, then $f(m)$ cannot have residue $r_1$ or $r_2$. For if $f(m)$ had residue $r_1$, then the same argument would show that all sufficiently large numbers with residue $r_1$ had $f(m) \equiv r \pmod k$. Thus the residues form pairs, so that if a number is congruent to a particular residue, then $f$ of the number is congruent to the pair of the residue. But this is impossible for $k$ odd.

Solution by Sawa Pavlov:
Let $\mathbb{N}$ be the set of non-negative integers. Put $A = \mathbb{N} - f(\mathbb{N})$ (the set of all $n$ such that we cannot find $m$ with $f(m) = n$). Put $B = f(A)$.
Note that $f$ is injective because if $f(n) = f(m)$, then $f(f(n)) = f(f(m))$ so $m = n$. We claim that $B = f(\mathbb{N}) - f(f(\mathbb{N}))$. Obviously $B$ is a subset of $f(\mathbb{N})$ and if $k$ belongs to $B$, then it does not belong to $f(f(\mathbb{N}))$ since $f$ is injective. Similarly, a member of $f(f(\mathbb{N}))$ cannot belong to $B$.
Clearly $A$ and $B$ are disjoint. They have union $\mathbb{N} - f(f(\mathbb{N}))$ which is $\{0, 1, 2, \ldots , 1986\}$. But since $f$ is injective they have the same number of elements, which is impossible since $\{0, 1, \ldots , 1986\}$ has an odd number of elements.

Consider the function $g: \mathbb{Z}_{1987} \rightarrow \mathbb{Z}_{1987}$ defined by $g(x) = f(x\; {\rm  mod }\; 1987) \;{\rm  mod }\; 1987$.  Notice that we have $f(k) + 1987 = f(f(f(k))) = f(k + 1987)$, so that $f(x) = f(y) \;{\rm  mod }\; 1987$ whenever $x = y \;{\rm mod }\; 1987$, and hence $g$ is well defined.
Now, we observe that $g$ satisfies the identity $g(g(x)) = x$, for $x \in Z_{1987}$.   Thus, $g$ is an invertible function on a finite set of odd size, and hence must have a fixed point, say $a \in \mathbb{Z}_{1987}$.  Identifying $a$ with its canonical representative in $\mathbb{Z}$, we therefore get $f(a) = a + 1987k$ for some non-negative integer $k$.
However, we then have $f(f(a)) = a + 1987$, while $f(f(a)) = f(a + 1987k)  = 1987k + f(a) = 2\cdot1987k + a$ (where we use the identity $f(x + 1987) = f(x) + 1987$ derived above, along with $f(a) = f(a) + 1987$.  However, these two equations imply that $k = \dfrac{1}{2}$, which is a contradiction since $k$ is an integer.  Thus, such an $f$ cannot exist.
Note: The main step in the proof above is that the function $g$ can be shown to have a fixed point.  This step works even if 1987 is replaced with any other odd number larger than 1.  However, for any even number $c$, $f(x) = x + \dfrac{c}{2}$ satisfies the condition $f(f(x)) = x + c$.
--Mahamaya 21:15, 21 May 2012 (EDT)


Show that the solution set of the inequality
$\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$
is a union of disjoint intervals, the sum of whose length is $1988$.

Solution:
Consider the graph of $f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$.  On the values of $x$ between $n$ and $n+1$ for $n\in\mathbb{N}$ $1\le n\le 69$, the terms of the form $\frac{k}{x-k}$ for $k\ne n,n+1$ have a finite range.  In contrast, the term $\frac{n}{x-n}$ has an infinite range, from $+\infty$ to $n$.  Similarly, the term $\frac{n+1}{x-n-1}$ has infinite range from $-n-1$ to $-\infty$.  Thus, since the two undefined values occur at the distinct endpoints, we can deduce that $f(x)$ takes on all values between $+\infty$ and $-\infty$ for $x\in(n,n+1)$.  Thus, by the Intermediate Value Theorem, we are garunteed a $n<r_n<n+1$ such that $f(r_n)=\frac{5}{4}$.  Additionally, we have that for $x>70$, the value of $f(x)$ goes from $+\infty$ to $0$, since as $x$ increases, all the terms go to $0$.  Thus, there exists some $r_{70}>70$ such that $f(r_{70})=\frac{5}{4}$ and so $f(x)\ge\frac{5}{4}$ for $x\in(70,r_{70})$.
So, we have $70$ $r_i$ such that $f(r_i)=\frac{5}{4}$.  There are obviously no other such $r_i$ since $f(x)=\frac{5}{4}$ yields a polynomial of degree $70$ when combining fractions.  Thus, we have that the solution set to the inequality $f(x)\ge\frac{5}{4}$ is the union of the intervals $(n,r_n]$ (since if $f(x)<\frac{5}{4}$ for $x\in(n,r_n)$ then there would exist another solution to the equation $f(x)=\frac{5}{4}$.
Thus we have proven that the solution set is the union of disjoint intervals.  Now we are to prove that the sum of their lengths is $1988$.
The sum of their lengths is $r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71$.  We have that the equation $f(x)=\frac{5}{4}$ yields a polynomial with roots $r_i$.  Thus, opposite of the coeficient of $x^{69}$ divided by the leading coefficient is the sum of the $r_i$.  It is easy to see that the coefficient of $x^{69}$ is $-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71$.  Thus, since the leading coefficient is $5$ we have $r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71$.  Thus, the sum of the lengths of the intervals is $63\cdot71-35\cdot71=28\cdot71=1988$ as desired.


Let $\mathbb{Q^+}$ be the set of positive rational numbers. Construct a function $f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}$ such that $f(xf(y)) = \frac{f(x)}{y}$ for all $x, y\in{Q^+}$.

Solution:
If we let $y = x = 1$ this implies $f(f(1)) = f(1)$. Plugging $f(1) = y$ in with this new fact. We get $f(f(f(1))) = \frac{f(1)}{f(1)} = 1$. Using $f(f(1)) = f(1)$ again, we see that $f(f(f(1))) = f(f(1)) = f(1) = 1$. Now plugging in $x = 1$ we get $f(f(y)) = \frac{1}{y}$. Such a function can not be continuous as if $f(x)$ is increasing or decreasing on some observable interval, $f(f(x))$ will be increasing on that interval but $\frac{1}{x}$ is decreasing on all positive intervals. This indicates the function is discrete which means we can assign a "4-chain" of $f(a) = b$, $f(b) = \frac{1}{a}$, $f(\frac{1}{a}) = \frac{1}{b}$, and $f(\frac{1}{b}) = a$. It is obvious to see that any function where $f(f(x)) = \frac{1}{x}$. Must follow such a structure. The problem occurs that we do not know whether our current value of $x$ is a $b$ or an $a$. To make non-trivial progress on this we must split all rational numbers into two sets, $S_1$ and $S_2$, such that if $E \in S_1$ then $\frac{1}{E} \in S_1$. As long as $S_1$ and $S_2$ have the same size, so the bijection $f(\textrm{an element from }S_1)$ = $(\textrm{an element from } S_2)$ can hold, there will exist a piece-wise function (possibly with a greater than countable infinity number of pieces) that satisfies f(x).
To find one of these functions we can limit our searching by proving that $f(ab) = f(a)f(b)$. To do this simply set $x = \frac{1}{f(y)}$ yielding:
\newline
$f(1) = f(f(y) \times \frac{1}{f(y)}) = \frac{f(\frac{1}{f(y)})}{y} = 1$. This means that $f(x) = y$. Since $X,Y \in Q+$ there will always be an $x$ that suffices this and a corresponding $y$. Thus $f(ab)$ with $b = f(c)$ $\leftrightarrow$ $c = \frac{1}{f(b)}$ expands into $f(af(c)) = \frac{f(a)}{c} = f(a)f(b)$.
Utilizing $f(ab) = f(a)f(b)$ we can see that if we break-down every ratio into it's prime factorization of the numerator and denominator. Our "4-chain" is simply $f(p) = q$, $f(q) = \frac{1}{p}$, $f(\frac{1}{p}) = \frac{1}{q}$, $f(\frac{1}{q}) = p$ where $p$ and $q$ are unique primes. Thus our 2 sets $S_1$ and $S_2$ is simply 2 sets of equal size containing only primes and their inverse. A good way to do this assuming we aren't given the sequence of all infinite primes (as this would require an explicit formula that does not yet exist) is to split the primes into 2 and 1 mod 6 being in one set with 3 and 5 mod 6 in the other. These sets have equal sizes from a strong form of Dirichlet's theorem on arithmetic progressions.
Thus we have created our two sets and to show that this actually works we will let $x = P_1^{E_{1x}} \times P_2^{E_{2x}} \times P_3^{E_{3x}} \times \dots$ where $P_i$ is a prime and $E_{ix}$ are integers. Define $y$ similarly. For the sake of simplicity if $i$ is odd then $P_i \in S_1$ and if $i$ is even $P_i \in S_2$. Where our "4-chain" is $f(p_n) = p_{n+1}$, $f(p_{n+1}) = \frac{1}{p_n}$, $f(\frac{1}{p_n}) = \frac{1}{p_{n+1}}$, $f(\frac{1}{p_{n+1}}) = p_n$ with $n$ being odd. Then $f(y) = P_2^{E_{1y}} \times P_1^{-E_{2y}} \times P_4^{E_{3y}} \times P_3^{-E_{4y}}\dots$ (remember $f(ab) = f(a)f(b)$).
Then $xf(y) = P_1^{E_{1x} - E_{2y}} \times P_2^{E_{2x} + E_{1y}} \times P_3^{E_{3x} - E_{4y}} \times \dots$
$f(xf(y)) = P_2^{E_{1x} - E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}} \dots$
$f(x) = P_2^{E_{1x}} \times P_1^{-E_{2x}} \times P_4^{E_{3x}} \times P_3^{-E_{4x}}\dots$
$\frac{f(x)}{y} = P_2^{E_{1x}- E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}}\dots$
Thus $f(xf(y)) = \frac{f(x)}{y}$ QED


In the plane let $C$ be a circle, $l$ a line tangent to the circle $C$, and $M$ a point on $l$.  Find the locus of all points $P$ with the following property: there exists two points $Q$, $R$ on $l$ such that $M$ is the midpoint of $QR$ and $C$ is the inscribed circle of triangle $PQR$.

Solution:
Note: This is an alternate method to what it is shown on the video.  This alternate method is too long and too intensive in solving algebraic equations.  A lot of steps have been shortened in this solution.  The solution in the video provides a much faster solution,
Let $r$ be the radius of the circle $C$.
We define a cartesian coordinate system in two dimensions with the circle center at $(0,0)$ and circle equation to be $x^{2}+y{2}=r^{2}$
We define the line $l$ by the equation $y=-r$, with point $M$ at a distance $m$ from the tangent and cartesian coordinates $(m,-r)$
Let $d$ be the distance from point $M$ to point $R$ such that the coordinates for $R$ are $(m+d,-r)$ and thus the coordinates for $Q$ are $(m-d,-r)$
Let points $S$, $T$, and $U$ be the points where lines $PQ$, $PR$, and $l$ are tangent to circle $C$ respectively.
First we get the coordinates for points $S$ and $T$.
Since the circle is the incenter we know the following properties:
$\left| RU \right| = \left| RT \right|=(m+d)$
and
$\left| QU \right| = \left| QS \right|=(m-d)$
Therefore, to get the coordinates of point $T=(T_{x},T_{y})$, we solve the following equations:
$T_{x}^{2}+T_{y}^2=r^{2}$
$\left| RT \right|^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$
$(m+d)^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$
After a lot of algebra, this solves to:
$T_{x}=\frac{2r^{2}(m+d)}{(m+d)^{2}+r^{2}}$
$T_{y}=\frac{r\left[ (m+d)^{2}-r^{2} \right]}{(m+d)^{2}+r^{2} }$
Now we calculate the slope of the line that passes through $PR$ which is perpendicular to the line that passes from the center of the circle to point $T$ as follows:
$Slope_{PR}=\frac{-T_{x}}{T_{y}}=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}$
Then, the equation of the line that passes through $PR$ is as follows:
$Line_{PR}\colon \; y+r=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( x-(m+d) \right)$
Now we get the coordinates of point $S=(S_{x},S_{y})$, we solve the following equations:
$S_{x}^{2}+S_{y}^2=r^{2}$
$\left| QT \right|^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2$
$(m-d)^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2$
After a lot of algebra, this solves to:
$S_{x}=\frac{2r^{2}(m-d)}{(m-d)^{2}+r^{2}}$
$S_{y}=\frac{r\left[ (m-d)^{2}-r^{2} \right]}{(m-d)^{2}+r^{2} }$
Now we calculate the slope of the line that passes through $PQ$ which is perpendicular to the line that passes from the center of the circle to point $S$ as follows:
$Slope_{PQ}=\frac{-S_{x}}{S_{y}}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}$
Then, the equation of the line that passes through $PQ$ is as follows:
$Line_{PQ}\colon \; y+r=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( x-(m-d) \right)$
Now we solve for the coordinates for point $P=(P_{x},P_{y})$ by calculating the intersection of $Line_{PR}$ and $Line_{PQ}$ as follows:
$\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( P_{x}-(m+d) \right)=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( P_{x}-(m-d) \right)$
Solving for $P_{x}$ we get:
$P_{x}=\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}$
Solving for $P_{y}$ we get:
$P_{y}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}-(m-d) \right)-r$
$P_{y}=\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}$
Now we need to find the limit of $P_{x}$ and $P_{y}$ as $d$ approaches infinity:
$P_{x_{d \to \infty}}=\lim_{d \to \infty} \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}=0$
$P_{y_{d \to \infty}}=\lim_{d \to \infty} \frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}=r$
This means that the locus of $P$ starts at point $(0,r)$ on the circle $C$ but that point is not included in the locus as that is the limit.
If we assume that the locus is a ray that starts at $(0,r)$ let's calculate the slope of such ray:
$Slope_{locus}=\frac{P_{y}-r}{P_{x}}$
$Slope_{locus}=\frac{\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}-r}{\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}}$
$Slope_{locus}=\frac{-r^{2}}{2m}$
Since the calculated slope of such locus at any point $P$ is not dependent on $d$ and solely dependent on fixed $r$ and $m$, then this proves the slope is fixed and thus the locus is a ray that starts at $(0,r)$ excluding that point and with a slope of $\frac{-r^{2}}{2m}$ in the cartesian coordinate system moving upwards to infinity.
We can also write the equation of the locus as: $y=\frac{-r^{2}}{2m}x+r,\;\;\forall y>r\;$and $x<0$
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


For three points $A,B,C$ in the plane, we define $m(ABC)$ to be the smallest length of the three heights of the triangle $ABC$, where in the case $A$, $B$, $C$ are collinear, we set $m(ABC) = 0$. Let $A$, $B$, $C$ be given points in the plane. Prove that for any point $X$ in the plane,
$m(ABC) \leq m(ABX) + m(AXC) + m(XBC)$.

Solution:
First we prove the claim for all points $X$ within or on the triangle $ABC$: In this specific case, suppose without loss of generality that $a\ge b\ge c$.  Then the length of the smallest height of $ABC$ is that of the height from vertex $A$.  This has a value of $\frac{2[ABC]}{a}$.  Similarly, for triangles $ABX,BCX,CAX$ we have that the length of the smallest height is twice the area of the respective triangle divided by the longest side of that triangle.  It is clear that since $a$ is the longest side of $ABC$, no two points within $ABC$ have distance exceeding $a$.  Thus, since $X$ is within $ABC$, the longest side of any of the triangles $ABX,BCX,CAX$ does not exceed $a$.  So, we have
$m(ABX)+m(BCX)+m(CAX)\ge\frac{2[ABX]}{a}+\frac{2[BCX]}{a}+\frac{2[CAX]}{a}$
$=\frac{2[ABC]}{a}=m(ABC)$
as desired.
Now on to prove the assertion for $X$ outside the triangle $ABC$.  We shall assume without loss of generality that out of the points $A,B,C$ point $A$ is that farthest from $X$.
If $ABCX$ is concave or a degenerate quadrilateral, assume without loss of generality that $B$ inside or on triangle $ACX$.  We shall prove that $m(ACX)\ge m(ABC)$ to prove the main claim.  If the shortest height of $ACX$ was from vertex $A$ then it is clear that the height of $ABC$ from $A$ is smaller than that since ray $CB$ is closer to $A$ than ray $CX$.  The case of the height from $C$ being the smallest of triangle $ACX$ is analogous, so we move on to the case of the height from $X$ is the smallest.  If this is the case, then it is clear that the height of $ABC$ from $B$ is smaller than $m(ABC)$ since $\angle XAC\ge \angle BAC$. Thus the claim is proved.
If, instead, $ABCX$ is convex, we can assign the letter $D$ to represent to point of intersection of $AX$ and $BC$.  Before proving the main claim, we shall prove that for triangle $ABX$ we have $m(ABX)\ge m(ABD)$.  We prove this by considering each of the vertices that the shortest height of $ABX$ is on.  If the shortest height is that from $X$, then it is obvious that the height from $D$ to side $AB$ is smaller than that from $X$ since $D$ is closer to $A$ than $X$ is and so $m(ABX)\ge m(ABD)$.  If the shortest height of $ABX$ was from $B$, then since the height from $B$ to $XD$ is equal to the height from $B$ to $AD$, we have $m(ABX)\ge m(ABD)$.  If instead the shortest height of $ABX$ was from $A$, the it is clear that $\angle B<90$.  Thus, the projection of $A$ onto $BD$ is on the same side of $B$ as $C$.  Now it is obvious that the height of $ABD$ from $A$ is smaller than that of $ABX$ from $A$ since the ray $BD$ is closer to $A$ than the ray $BX$.  Thus, we have $m(ABX)\ge m(ABD)$ in all cases.  Notice that the case for triangle $ACX$ is analogous.  Therefore, we have $m(ABX)+m(ACX)\ge m(ABD)+m(ACD)$.  But, this is the case of $D$ on the triangle $ABC$, and this case was shown in the first part of this proof.
So, we have that $m(ABC) \leq m(ABX) + m(AXC) + m(XBC)$ is true in all cases, as desired.


Find all ordered pairs $(m,n)$ where $m$ and $n$ are positive integers such that $\frac {n^3 + 1}{mn - 1}$ is an integer.

Solution:
Suppose $\frac{n^3+1}{mn-1}=k$ where $k$ is a positive integer.  Then $n^3+1=(mn-1)k$ and so it is clear that $k\equiv -1\pmod{n}$.  So, let $k=jn-1$ where $j$ is a positive integer.  Then we have $n^3+1=(mn-1)(jn-1)=mjn^2-(m+j)n+1$ which by cancelling out the $1$s and dividing by $n$ yields $n^2=mjn-(m+j)\Rightarrow n^2-mjn+m+j=0$.  The equation $x^2-mjx+m+j=0$ is a quadratic.  We are given that $n$ is one of the roots.  Let $p$ be the other root.  Notice that since $n+p=mj$ we have that $p$ is an integer, and so from $np=m+j$ we have that $p$ is positive.
It is obvious that $j=m=n=p=2$ is a solution.  Now, if not, and $j,m,n,p$ are all greater than $1$, we have the inequalities $np>n+p$ and $mj>m+j$ which contradicts the equations $np=m+j, n+p=mj$.
Thus, at least one of $j,m,n,p$ is equal to $1$.
If one of $m,j$ is $1$, without loss of generality assume it is $j$.  Then we have $np=m+1, n+p=m$.  That is, $np-n-p=1\Rightarrow (n-1)(p-1)=2$ which gives positive solutions $(n,p)=(3,2),(2,3)$.  These give $m=5$ and since we assumed $j=1$, we can also have $m=1$ and $j=5$.
If one of $n,p$ is $1$, without loss of generality assume it is $p$.  Then we have $n=m+j, n+1=mj$.  That is, $mj-m-j=1\Rightarrow (m-1)(j-1)=2$ which gives positive solutions $(m,j)=(3,2),(2,3)$.  These give $n=5$ and since we assumed $p=1$, we can also have $n=1$ and $p=5$.
From these, we have all solutions $(m,n)=(2,2),(5,3),(5,2),(1,3),(1,2),(3,5),(2,5),(3,1),(2,1)$.