Let $P$ be a regular 2006-gon. A diagonal of $P$ is called good if its endpoints divide the boundary of $P$ into two parts, each composed of an odd number of sides of $P$. The sides of $P$ are also called good.
Suppose $P$ has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of $P$. Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.

Solution:
Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity.
Let ABC be an iso-odd triangle, with AB and BC odd sides. This means that there are an odd number of sides of the 2006-gon between A and B and also between B and C. We say that these sides belong to the iso-odd triangle ABC.
At least one side in each of these groups does not belong to any other iso-odd triangle.
This is so because any odd triangle whose vertices are among the points between A and B has two sides of equal length and therefore has an even number of sides belonging to it in total. Eliminating all sides belonging to any other iso-odd triangle in this area must therefore leave one side that belongs to no other iso-odd triangle. Let us assign these two sides (one in each group) to the triangle ABC.
To each iso-odd triangle we have thus assigned a pair of sides, with no two triangles sharing an assigned side. It follows that at most 1003 iso-odd triangles can appear in the dissection. This value can be attained, as shows the example from the ﬁrst solution.


Consider five points $A,B,C,D$, and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral.
Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects
line $BC$ at $G$. Suppose also that $EF=EG=EC$. Prove that $\ell$ is the bisector of $\angle DAB$.

Solution:
Since $\angle{DAF}=\angle{CGF}$, $\angle{BAF}=\angle{CFG}$, it suffices to prove $CF=CG$.
Let $\angle{FCE}=\alpha$, $\angle{GCE}=\beta$, $\angle{CDE}=\gamma$. We have:

so,

Meantime, using Law of Sines on $\triangle{DEC}$, we have,

Using Law of Sines on $\triangle{BEG}$, and notice that $\angle{CBE}=\angle{CDE}=\gamma$, we have,

so,

Since $\triangle{GFC} \sim \triangle{GAB}$, and $DC=AB$, we have, $\dfrac{DC}{BG}=\dfrac{AB}{BG}=\dfrac{CF}{CG}$. Hence,

or,

There are two possibilities: (1) $\alpha+\gamma-\beta = \beta+\gamma-\alpha$, or (2) $\alpha+\gamma-\beta = 180 - (\beta+\gamma-\alpha)$. However, (2) would mean $\gamma=90$, then $EC$ would be a diameter, and $EF < EC$ because $F$ is inside the circle, so (2) is not valid. From condition (1), we have $\alpha=\beta$, therefore $CF=CG$. $\square$
Solution by Mathdummy
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


(i) If $x$, $y$ and $z$ are three real numbers, all different from $1$, such that $xyz = 1$, then prove that
$\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1$.
(With the $\sum$ sign for cyclic summation, this inequality could be rewritten as $\sum \frac {x^{2}}{\left(x - 1\right)^{2}} \geq 1$.)
(ii) Prove that equality is achieved for infinitely many triples of rational numbers $x$, $y$ and $z$.

Solution:
Consider the transormation $f:\mathbb{R}/\{1\} \rightarrow \mathbb{R}/\{-1\}$ defined by $f(u) = \frac{u}{1-u}$ and put $\alpha = f(x), \beta = f(y), \gamma = f(z)$. Since $f$ is also one-to one from $\mathbb{Q}/\{1\}$ to $\mathbb{Q}/\{-1\}$, the problem is equivalent to showing that

subject to

and that equallity holds for infinitely many triplets of rational $\alpha,\beta,\gamma$.
Now, rewrite (2) as $\alpha\beta\gamma = (1+\alpha)(1+\beta)(1+\gamma)$ and express it as

where $p=\alpha+\beta+\gamma$ and $q = \alpha\beta+\beta\gamma+\gamma\alpha$. Notice that (1) can be written as

But from $p = -1-q$, we get

with equality holding iff $q = 0$. That proves part (i) and points us in the direction of looking for rational $\alpha,\beta,\gamma$ for which $q=0$ and (hence) $p=-1$, that is:

Expressing $\alpha$ from the first equation and substituting into the second, we get

as the sole condition we need to satisfy in rational numbers.
If $\beta = \frac{b}{m}$ and $\gamma = \frac{c}{m}$ for some integers $b$,$c$,and $m$, they would need to satisfy

For $m$ to be integer, we would like $b+c$ to divide $bc$.
Consider the example

where $b+c = t^2$ divides $bc = t(t^2-t)$ for any integer $t \ne 0$. Substituting back, that gives us

A simple check shows that $\alpha,\beta,\gamma$ are rational and well defined and that $p=-1$ and $q=0$ for any integer $t$ (even for $t=0$).
Moreover, from $\lim_{t\rightarrow +\infty} \beta = 0$ and $\beta < 0$ for large $t$, we see that infinitely many $t$ generate infinitely many different triplets of $\alpha$, $\beta$, and $\gamma$. That completes the proof of part (ii).
--Vbarzov 03:03, 5 September 2008 (UTC)


Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE< \frac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the intersection of lines $EI$ and $DG$ lies on $\Gamma$.
Authors: Tai Wai Ming and Wang Chongli, Hong Kong

Solution:
Note that it suffices to prove alternatively that if $EI$ meets the circle again at $J$ and $JD$ meets $IF$ at $G$, then $G$ is the midpoint of $IF$. Let $JD$ meet $BC$ at $K$.
Observation 1. D is the midpoint of arc $BDC$ because it lies on angle bisector $AI$.
Observation 2. $AI$ bisects $\angle{FAE}$ as well.
Key Lemma. Triangles $DKI$ and $DIJ$ are similar.
Proof. Because triangles $DKB$ and $DBJ$ are similar by AA Similarity (for $\angle{KBD}$ and $\angle{BJD}$ both intercept equally sized arcs), we have $BD^2 = BK \cdot BJ$. But we know that triangle $DBI$ is isosceles (hint: prove $\angle{BID} = \angle{IBD}$), and so $BI^2 = BK \cdot BJ$. Hence, by SAS Similarity, triangles $DKI$ and $DIJ$ are similar, as desired.
Observation 3. As a result, we have $\angle{KID} = \angle{IJD} = \angle{DAE} = \angle{FAD}$.
Observation 4. $IK // AF$.
Observation 5. If $AF$ and $JD$ intersect at $L$, then $AJLI$ is cyclic.
Observation 6. Because $\angle{ALI} = \angle{AJE} = \angle{AJC} + \angle{CJE} = \angle{B} + \angle{AEC} = \angle{B} + \angle{BAF} = \angle{AFC}$, we have $LI // FK$.
Observation 7. $LIKF$ is a parallelogram, so its diagonals bisect each other, so $G$ is the midpoint of $FI$, as desired.

Let $A'$ be A-excenter $\triangle ABC \implies$

the intersection of lines $EI$ and $DG$ lies on $\Gamma$.
After pavel kozlov vladimir.shelomovskii@gmail.com, vvsss


Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal S$ are collinear. A windmill is a process that starts with a line $\ell$ going through a single point $P \in \mathcal S$. The line rotates clockwise about the pivot $P$ until the first time that the line meets some other point belonging to $\mathcal S$. This point, $Q$, takes over as the new pivot, and the line now rotates clockwise about $Q$, until it next meets a point of $\mathcal S$. This process continues indefinitely.
Show that we can choose a point $P$ in $\mathcal S$ and a line $\ell$ going through $P$ such that the resulting windmill uses each point of $\mathcal S$ as a pivot infinitely many times.

Solution:
Choose a coordinate system so that all points in $\mathcal S$ have distinct x-coordinates. Number the points $P_i=(x_i,y_i)$ of $\mathcal S$ by increasing x-coordinates: $x_1<x_2<\ldots<x_N$.
In order to divide the set $\mathcal S$ into two halves, define $n$ so that $N=2n+1+d$ where $d=0$ for an odd number of points and $d=1$ for an even number of points.
Start the "windmill" process with the line $\ell$ going vertically through the point $P_{n+1}$.  Attach a down-up direction to this line so that we can color all points as follows: Points to the left of $\ell$ (with lower x-coordinates) are blue, the pivot point on $\ell$ is white and point to the right of $\ell$ (with higher x-coordinates) are red.  We have now $n$ blue points, one white point and $n+d$ red points.
After processing the "windmill" by 180 degrees, the line $\ell$ goes vertically up-down.  Now, points with lower x-coordinates are to the right of $\ell$ and colored red; points with higher x-coordinates are to the left of $\ell$ and colored blue.
Note that at each pivot exchange, the old pivot point enters the same side of $\ell$ where the new pivot point came from.  This means that throughout the "windmill" process, the number of blue points and the number of red points stay constant, respectively: We still have $n$ blue points, one white point and $n+d$ red points.  This means that the current pivot point is $P_{n+d}$.
Note that all blue and all red points changed their color from the start of the "windmill" process.  This implies that every point was a pivot at some stage of the rotation.
For every 180 degrees of "windmill" rotation, the same argument applies: all colored points must change their color and hence be a pivot at some stage.  Infinitely many rotations imply infinitely many color changes.  This completes the proof.


Let ${{a}_{2}}, {{a}_{3}},  \cdots, {{a}_{n}}$ be positive real numbers that satisfy ${{a}_{2}}\cdot {{a}_{3}}\cdots {{a}_{n}}=1$ . Prove that

Solution:
The inequality between arithmetic and geometric mean implies

The inequality is strict unless $a_k=\frac1{k-1}$. Multiplying analogous inequalities for $k=2,\text{ 3, }\cdots \text{, }n$ yields


A configuration of $4027$ points in the plane is called Colombian if it consists of $2013$ red points and $2014$ blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian
configuration if the following two conditions are satisfied:
Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good
arrangement of $k$ lines.

Solution:
We can start off by imagining the points in their worst configuration.
With some trials, we find $2013$ lines to be the answer to the worst cases.
We can assume the answer is $2013$. We will now prove it.
We will first prove that the sufficient number of lines required for a good arrangement for a configuration consisting of $u$ red points and $v$ blue points, where $u$ is even and $v$ is odd and $u - v = 1$, is $v$.
Notice that the condition "no three points are co-linear" implies the following:
No blue point will get in the way of the line between two red points and vice versa.
What this means, is that for any two points $A$ and $B$ of the same color, we can draw two lines parallel to, and on different sides of the line $AB$, to form a region with only the points $A$ and $B$ in it.
Now consider a configuration consisting of u red points and v blue ones ($u$ is even, $v$ is odd, $u>v$).
Let the set of points $S = \{A_1, A_2, ... A_k\}$ be the out-most points of the configuration, such that you could form a convex k-gon, $A_1 A_2 A_3 ... A_k$, that has all of the other points within it.
If the set S has at least one blue point, there can be a line that separates the plane into two regions: one only consisting of only a blue point, and one consisting of the rest. For the rest of the blue points, we can draw parallel lines as mentioned before to split them from the red points.
We end up with $v$ lines.
If the set $S$ has no blue points, there can be a line that divides the plane into two regions: one consisting of two red points, and one consisting of the rest. For the rest of the red points, we can draw parallel lines as mentioned before to split them from the blue points.
We end up with $u-1 = v$ lines.
Now we will show that there are configurations that can not be partitioned with less than $v$ lines.
Consider the arrangement of these points on a circle so that between every two blue points there are at least one red point (on the circle).
There are no less than $2v$ arcs of this circle, that has one end blue and other red (and no other colored points inside the arc) - one such arc on each side of each blue point.
For a line partitioning to be good, each of these arcs have to be crossed by at least one line, but one line can not cross more than $2$ arcs on a circle - therefore, this configuration can not be partitioned with less than $v$ lines!
Our proof is done, and we have our final answer: $2013$.


Let $n\ge2$ be an integer. Consider an $n\times n$ chessboard consisting of $n^2$ unit squares. A configuration of $n$ rooks on this board is $\textit{peaceful}$ if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that, for each peaceful configuration of $n$ rooks, there is a $k\times k$ square which does not contain a rook on any of its $k^2$ squares.

Solution:
We claim the answer is $k = \lceil \sqrt{n}\rceil  - 1$, where $\lceil n\rceil$ is the ceiling function of $n$; i.e., the least integer greater than or equal to $n$. Notice that $\lceil n\rceil  < n + 1$.
First, we shall show that each $n \times n$ chessboard with a peaceful configuration of $n$ rooks contains a valid $k \times k$ square. Consider firstly the rook $R$ on the top row of the board. Because $k < n$, there exists a set $C$ of $k$ consecutive columns, one of which contains rook $R$. Consider the $n - k + 1$ $k \times k$ squares in $C$. Of them, only one contains the rook $R$ on one of its squares. Furthermore, each of the other $k - 1$ rooks in $C$ can only make up to $k$ of the $k \times k$ squares have it on one of its squares. Therefore, because

by the Pigeonhole Principle there exists a $k \times k$ square in $C$ not covered by any rook in $C$. Clearly, that square cannot be covered by any rook outside of $C$, and so it is a valid choice.
It remains to show that there exists a chessboard without a $(k+1) \times (k+1)$ square. Indeed, such a board exists. First, place a rook at the upper-left corner of the board. Next, place a rook 1 space to the right and $(k+1)$ spaces down of the first rook. Then, place a rook 1 space to the right and $(k+1)$ spaces down of the second rook, and so on, until the placement of a new rook will be under the lower boundary of the board. In that case, place a rook in the same unoccupied column and in the first unoccupied row, and continue placement of subsequent rooks. This arrangement of rooks is clearly peaceful, and because $(k+1)^2 > n$ it has the property that any $(k+1) \times (k+1)$ square in the board will be occupied by at least one rook, completing the proof.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Determine all triples of positive integers $(a,b,c)$ such that each of the numbers

is a power of 2.
(A power of 2 is an integer of the form $2^n$ where $n$ is a non-negative integer ).

Solution:
The solutions for $(a,b,c)$ are $(2,2,2)$, $(2,2,3)$, $(2,6,11)$, $(3,5,7)$, and
permutations of these triples.
Throughout the proof, we assume $a \leq b \leq c$,
so that $ab-c = 2^m$, $ca-b = 2^n$,
$bc-a=2^p$, with $m \leq n \leq p$. Note that $a>1$ since
otherwise $b-c=2^m$, which is impossible. Hence
$2^n = ac-b \geq (a-1)c \geq 2$, i.e., $n$ and $p$ are positive.
Observe that if $a=b\geq 3$, we get $a(c-1)=2^n$, so $a$ and $c-1$ are (even and)
powers of $2$. Hence $c$ is odd and $a^2-c=2^m=1$. Hence $c+1=a^2$ is also a
power of $2$, which implies $c=3$. But $a=b=c=3$ is not a
solution; hence  $a=b\geq 3$ is infeasible. We consider the remaining cases as follows.
Case 1: $a=2$. We have

From the second equation, $b$ is even. From the third equation, if
$p=1$, then $b=c=2$; if $p>1$, then $c$ is odd, which implies that
$m=0$. Hence $3b=2^n+2$ (so $n\geq 2$),
$3c=2^{n+1}+1$, and $(2^{n-1}+1)(2^{n+1}+1)=9(2^{p-1}+1)$. Hence
$1\equiv 9 \mod 2^{n-1} \implies n \leq 4$. Hence $n$ is 2 or 4, and
$(b,c)$ equals $(2,3)$ or $(6,11)$. Thus the solutions for $(a,b,c)$
are $(2,2,2)$, $(2,2,3)$ or $(2,6,11)$.
Case 2: $3\leq a<b\leq c$.
Since $(a-1)c \leq 2^n$, we have $c\leq 2^{n-1}$. Hence

Hence $b-a$ is not divisible by $2^{n-1}$, and $b+a$ is not divisible
by $2^{n-1}$ for $a\geq 5$. Adding and subtracting $ac-b=2^n$ and $bc-a=2^p$, we get

From the latter equation, $c+1$ is divisible by $4$. Hence $c-1$ is not
divisible by $4$, which implies that $b+a < 2^n$ is a multiple of $2^{n-1}$.
Hence $a \leq 4$ and $b+a=2^{n-1}$.
Consider $a=3$, which implies $3b-c=2^m$, $3c-b=2^n$, $b=2^{n-1}-3$.
Hence $2^{n-1}-3=3.2^{m-3}+2^{n-3}$, or
$2^{n-3}=2^{m-3}+1$. Hence $m=3$, $n=4$, $b=5$ and $c=7$.
Finally, consider $a=4$, $4c-b=2^n$,
$b=2^{n-1}-4$. Hence $c=3.2^{n-3}-1$. But $b \leq c$ implies
$2^{n-3} \leq 3$ and $a<b$ implies $2^{n-3}>2$. Hence there are no
solutions with $a=4$.
We obtain $(a,b,c)=(3,5,7)$ as the only solution with
$3 \leq a < b \leq c$.


Find all integers $n$ for which each cell of $n \times n$ table can be filled with one of the letters $I,M$ and $O$ in such a way that:
Note. The rows and columns of an $n \times n$ table are each labelled $1$ to $n$ in a natural order. Thus each cell corresponds to a pair of positive integer $(i,j)$ with $1 \le i,j \le n$. For $n>1$, the table has $4n-2$ diagonals of two types. A diagonal of first type consists all cells $(i,j)$  for which $i+j$ is a constant, and the diagonal of this second type consists all cells $(i,j)$ for which $i-j$ is constant.

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


Let $\mathbb{R}$ be the set of real numbers , determine all functions
$f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any real numbers $x$ and $y$

Solution:
Looking at the equation one can deduce that the functions that will work will be linear.  That is, a polynomial of at most a degree of 1.
Thus, $f$ is in the form $f(x)=mx+b$
Therefore,
$f((mx+b)(my+b))+m(x+y)+b=mxy+b$
$f(m^2xy+mb(x+y)+b^2)+m(x+y)+b=mxy+b$
$m(m^2xy+mb(x+y)+b^2)+b+m(x+y)+b=mxy+b$
$m^3xy+m^2b(x+y)+mb^2+m(x+y)+b=mxy$
$m(m^2-1)xy+m(bm+1)(x+y)+b(bm+1)=0$
Therefore,
$m(m^2-1)=0$ [Equation 1]
$m(bm+1)=0$ [Equation 2]
$b(bm+1)$ [Equation 3]
From [Equation 1] we have, $m=0,\pm 1$
From [Equation 2] we have, $m=0, bm=-1\pm 1$
From [Equation 3] we have, $b=0, bm=-1\pm 1$
When $m=0$, $b=0$, then $f(x)=0$
When $bm=-1$, $b=\frac{-1}{m}$, then since $m=\pm 1$, then $b=\mp 1$
When $bm=-1$, $b=\frac{-1}{m}$, then since $m=\pm 1$, then $b=\mp 1$ then $f(x)=\pm x \mp1$ which gives these two functions:
$f(x)=x-1$ and $f(x)=1-x$, which with $f(x)=0$ provide all the three functions for this problem.
~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.


Find all numbers $n \ge 3$ for which there exists real numbers $a_1, a_2, ..., a_{n+2}$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and

for $i = 1, 2, ..., n.$

Solution:
We find at least one series of real numbers for $n = 3,$ for each $n = 3k$ and we prove that if $n = 3k \pm 1,$ then the series does not exist.
Case 1
Let $n = 3.$ We get system of equations
We subtract the first equation from the second and get:
So $a_2 = – 1 \implies  a_1 = 2, a_3 =  – 1.$
Case 1a
Let $n = 3k, k={1,2,...}.$
Real numbers $a_1 =a_4 =...=2, a_2 = a_3 = a_5=...=-1$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and $a_{i}a_{i+1} + 1 = a_{i+2}$.
Case 2
Let $n = 4.$ We get system of equations

We multiply each equation by the number on the right-hand side and get:

We multiply each equation by a number that precedes a pair of product numbers in a given sequence $a_1, a_2, a_3, a_4, a_1, a_2.$ So we multiply the equation with product $a_1 a_2$ by $a_4$, we multiply the equation with product $a_4 a_1$ by $a_3$ etc. We get:

We add all the equations of the first system, and all the equations of the second system. The sum of the left parts are the same! It includes the sum of all the numbers $a_i$ and the sum of the triples of consecutive numbers $a_i a_{i + 1} a_{i + 2}.$ Hence, the sums of the right parts are equal, that is,

It is known that this expression is doubled

Substituting into any of the initial equations, we obtain the equation $a_1^2 + 1 = a_1,$ which does not have real roots. Hence, there are no such real numbers.
Case 2a
Let $n = 5.$ We get system of equations

We repeat all steps of Case 2 and get: there are no such real numbers.
Case 2b
Let $n = 3k \pm  1.$ We repeat all steps of cases $2$ and $2a$ and get: there are no such real numbers.
vladimir.shelomovskii@gmail.com, vvsss


In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.
Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

Solution:
The essence of the proof is to build a circle through the points $P, Q,$ and two additional points $A_0$ and $B_0,$ then we prove that the points $P_1$ and $Q_1$ lie on the same circle.
We assume that the intersection point of $AP$ and $BQ$ lies on the segment $PA_1.$ If it lies on segment $AP,$ then the proof is the same, but some angles will be replaced with additional ones up to $180^\circ$.
Let the circumcircle of $\triangle ABC$ be $\Omega$. Let $A_0$ and $B_0$ be the points of intersection of $AP$ and $BQ$ with $\Omega$. Let $\angle BAP = \delta.$

$\angle BAP = \angle BB_0A_0 = \delta$ since they intersept the arc $BA_0$ of the circle $\Omega$.
$\angle QPA_0 = \angle QB_0A_0 \implies QPB_0A_0$ is cyclic (in circle $\omega.$)
Let $\angle BAC = \alpha, \angle AA_0B_0 = \varphi.$
$\angle PP_1C = \alpha, \angle BB_0C = \alpha$ since they intersept the arc $BC$ of the circle $\Omega.$
So $B_0P_1CB_1$ is cyclic.
$\angle ACB_0 = \angle AA_0B_0 =  \varphi$ (they intersept the arc $A_0B_0$ of the circle  $\Omega).$
$\angle B_1CB_0 = \varphi.$
$\angle B_1P_1B_0 = \angle B_1CB_0 =  \varphi$ (since they intersept the arc $B_1B_0$ of the circle $B_0P_1CB_1).$
Hence $\angle PA_0B_0 = \angle PP_1B_0 = \varphi,$  the point $P_1$ lies on $\omega.$
Similarly, point $Q_1$ lies on  $\omega.$
vladimir.shelomovskii@gmail.com, vvsss


The real numbers $a$, $b$, $c$, $d$ are such that $a \geq b \geq c \geq d > 0$ and $a + b + c + d = 1$. Prove that

Solution:
Using Weighted AM-GM we get

So,
Now notice that
So, we get
Now, for equality we must have $a=b=c=d=\frac{1}{4}$
In that case we get
~ftheftics


Show that the inequality

holds for all real numbers $x_1,x_2,\dots,x_n$.

Solution:
then, since
then,

therefore we have to prove that
 for every list [Xi],
and we can describe this to

we know that

therefore,

--Mathhyhyhy 13:29, 6 June 2023 (EST)


Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying
.

Solution:
https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]
https://youtu.be/b5OZ62vkF9Y  [Video Solution by little fermat]

Answer: The unique solution is the function \( f(x) = \frac{1}{x} \) for every \( x \in \mathbb{R}^+ \). This function clearly satisfies the required property since the expression \( xf(y) + yf(x) = \frac{x}{y} + \frac{y}{x} \) is greater than 2 for every \( y \neq x \) (directly from AM-GM) and equal to 2 (with equality) for the unique value \( y = x \).
Proof: Let's consider a solution based on some ideas we encountered in the preparation classes for the Olympiad, specifically involving auxiliary sets and functions with specific properties.
The fact that for every \( x \in \mathbb{R}^+ \), there exists a unique \( y \in \mathbb{R}^+ \) that satisfies the equation \( xf(y) + yf(x) \leq 2 \) can be equivalently expressed as follows: there exists a well-defined function \( g: \mathbb{R}^+ \to \mathbb{R}^+ \) given by \( g(x) := y \), where \( y \) is the one mentioned above. The well-definedness of this function is evident due to the existence and uniqueness, and it satisfies the equation \( P(x): \quad xf(g(x)) + f(x)g(x) \leq 2 \) while applying the same property for \( x \mapsto g(x) \) gives another unique \( y := g(g(x)) \) such that \( g(x)f(y) + yf(g(x)) \leq 2 \). Therefore, we have \( xf(y) + yf(x) > 2 \) for all \( y \neq g(x) \).
Since this inequality holds for \( y = x \) (from \( xf(y) + yf(x) > 2 \)), the uniqueness assumption implies that \( g(g(x)) = x \), making \( g \) an involution (hence bijective).
Generally, working with an involution naturally leads us to consider its fixed points, especially since we aim to show that \( g(x) = x \) identically (which holds for the solution \( f(x) = \frac{1}{x}\)). Let's define the set of fixed points of \( g \) as \( \mathcal{S} := \{ x \in \mathbb{R}^+ \mid g(x) = x \} \) and show that \( \mathcal{S} = \mathbb{R}^+ \) is the entire domain.
Assume for a contradiction that some \( x \notin \mathcal{S} \) is not a fixed point, i.e., \( x \neq g(x) \). Then, the inequality \( 2xf(x) > 2 \) (derived from \( y \mapsto x \)) holds, implying \( f(x) > \frac{1}{x} \). Similarly, \( x \notin \mathcal{S} \) implies \( g(x) \notin \mathcal{S} \) (otherwise \( g(x) \in \mathcal{S} \) implies \( x = g(g(x)) = g(x) \), a contradiction), leading to \( f(g(x)) > \frac{1}{g(x)} \).
Applying these inequalities to \( P(x) \) gives \( xf(g(x)) + f(x)g(x) < 2 \), which is clearly a contradiction as \( \frac{x}{g(x)} + \frac{g(x)}{x} \geqslant 2 \), e.g., from the AM-GM inequality. Therefore, we must have \( x \in \mathcal{S} \) for every \( x \in \mathbb{R}^+ \), i.e., \( g(x) = x \).
Substituting this relationship into the original equation, we obtain \( P(x): \quad xf(x) + f(x)x \leq 2 \implies xf(x) \leq 1 \implies f(x) \leq \frac{1}{x} \) for every \( x \in \mathbb{R}^+ \). Applying \( yf(y) \leq 1 \) to the equation \( xf(y) + yf(x) > 2 \) (since \( g(x) = x \)) yields \( f(x) > \frac{2}{y} - \frac{x}{y^2} \), and taking the limit \( y \to x \) from either side results in \( f(x) \geq \frac{1}{x} \).
Combining the results, we have \( f(x) \leq \frac{1}{x} \) and \( f(x) \geq \frac{1}{x} \), implying \( f(x) = \frac{1}{x} \) as desired. \(\blacksquare\)
Note: This solution is written more extensively and with more details than necessary for a competition, especially since I include comments at certain points to encourage understanding of the ideas and explain the solution. In practice, this idea would take up only a few lines.


Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

Solution:
Denote the point diametrically opposite to a point $S$ through $S' \implies AS'$ is the internal angle bisector of $\angle BAC$.
Denote the crosspoint of $BS$ and $AS'$ through $H, \angle ABS = \varphi.$

To finishing the solution we need only to prove that $PH = AH.$
Denote $F = SS' \cap AC \implies \angle CBS = \frac {\overset{\Large\frown} {CS}}{2}  = \frac {\overset{\Large\frown} {BS}}{2}  =  \frac {\overset{\Large\frown} {AB}}{2} +  \frac {\overset{\Large\frown} {AS}}{2} =$
$=\angle FCB + \varphi \implies \angle FBS = \angle ABS \implies H$ is incenter of $\triangle ABF.$
Denote $T = S'B \cap SA \implies SB \perp TS', S'A \perp TS \implies H$ is the orthocenter of $\triangle TSS'.$
Denote $G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,$ and $G$ are concyclic.
$\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies$
points $B, G,$ and $F$ are collinear $\implies GF$ is symmetric to $AF$ with respect $TF.$
We use the lemma and complete the proof.
Lemma 1
Let acute triangle $\triangle ABC, AB > AC$ be given.
Let $H$ be the orthocenter of $\triangle ABC, BHD$ be the height.
Let $\Omega$ be the circle $BCD. BC$ is the diameter of $\Omega.$
The point $E$ is symmetric to $D$ with respect to $AH.$
The line $BE$ meets $\Omega$ again at $F \neq B$.
Prove that $HF = HD.$
Proof
Let $\omega$ be the circle centered at $H$ with radius $HD.$
The $\omega$ meets $\Omega$ again at $F' \neq D, HD = HF'.$
Let $\omega$ meets $BF'$ again at $E' \neq F'$.
We use Reim’s theorem for $\omega, \Omega$ and lines $CDD$ and $BE'F'$ and get $E'D || BC$
(this idea was recommended by Leonid Shatunov).
$AH \perp BC \implies AH \perp E'D \implies$
The point $E'$ is symmetric to $D$ with respect to $AH \implies E' = E \implies F' = F \implies HF = HD.$
vladimir.shelomovskii@gmail.com, vvsss

Identify $\Omega$ with the unit circle, and let the internal bisector of $\angle BAC$ meet $\overleftrightarrow{BS}$ at $Q$ and $\omega$ again at $T$. We set up so that
\begin{align*}
|a|=|b|=|s|&=1 \\
c &= \frac{s^2}b \\
t &= -s \\
e = -\frac{bc}a &= -\frac{s^2}a \\
d = \frac{ae(b+s) - bs(a+e)}{ae-bs} &= \frac{a^2b+abs+as^2-bs^2}{a(b+s)} \\
q = \frac{at(b+s) - bs(a+t)}{at-bs} &= \frac{2ab+as-bs}{a+b}
\end{align*}
Now we find the coordinate of $P$. We have $|p|=1$, and

Now $\frac{d-\ell}{b-c}, \frac{b-\ell}{b-e} \in \mathbb{R}$. Thus we have

and so

It remains to show that $Q$ lies on the tangent to $\omega$ at $P$. Now let $O$ be the center of $\omega$. Define the vectors

and
\begin{align*}
d' = d-p &= \frac{(2s+b-a)(a^2b+abs+as^2-bs^2) - s(b+s)(2ab+as-bs)}{a(b+s)(2s+b-a)} \\
&= \frac{a^2bs+as^3-bs^3+a^2b^2-a^3b-a^2s^2+abs^2-ab^2s}{a(b+s)(2s+b-a)} \\
&= \frac{(a-s)(b-a)(ab+s^2)}{a(b+s)(2s+b-a)}
\end{align*}
We observe that $\overline{b'} = -\frac{b'}{bp}$ and $\overline{d'} = \frac{d'}{ap}$. Thus if we define $o' = o-p$, we have

Meanwhile, we compute

So

which is pure imaginary. $\blacksquare$

Let $O$ be the circumcenter of $\triangle ABC$. We proceed with showing that $PH=AH$. Suppose that $PD$ intersects $SS'$ and $\Omega$ at $Q$ and $F \ne A$ respectively. Note that

Since $AE \perp BC$, we have $AE \perp EF$ and hence $AF$ is a diameter of $\Omega$. By similar triangles $OQ=\frac{1}{2}AD$ and therefore

Since $AF$ is a diameter of $\Omega$, $FP \perp AP \implies OH \perp AP$ and thus $H$ lies on the perpendicular bisector of $AP$. This proves the claim.


Let $a,b,c$ be real numbers. Consider the quadratic equation in $\cos{x}$ :
Using the numbers $a,b,c$, form a quadratic equation in $\cos{2x}$, whose roots are the same as those of the original equation. Compare the equations in $\cos{x}$ and $\cos{2x}$ for $a=4, b=2, c=-1$.

Solution:
Let the original equation be satisfied only for $\cos{x}=m, \cos{x}=n$.  Then we wish to construct a quadratic with roots $2m^2 -1, 2n^2 -1$.
Clearly, the sum of the roots of this quadratic must be
and the product of its roots must be
Thus the following quadratic fulfils the conditions:
Now, when we let $a=4, b=2, c= -1$, our equations are
and
This simplifies to previous equation.
The first root of the first equation $\cos {x} = \frac{-1 + \sqrt{5}}{4}$ corresponds to $\cos {2x} = \frac{-1 - \sqrt{5}}{4}$ and the second root of the first equation $\cos {x} = \frac{-1 - \sqrt{5}}{4}$ corresponds to $\cos {2x} = \frac{-1 + \sqrt{5}}{4}$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ an odd integer). Let $\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:

Solution:
Using coordinates, let $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$. Also, let $PQ$ be the segment that contains the midpoint of the hypotenuse with $P$ closer to $B$.

Then, $P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)$, and $Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)$.
So, $\text{slope}$$(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}$, and $\text{slope}$$(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}$.
Thus, $\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}$
$= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}$.
Since $[ABC]=\frac{1}{2}bc=\frac{1}{2}ah$, $bc=ah$ and $\tan{\alpha}=\frac{4nh}{(n^2-1)a}$ as desired.

Let $P, Q, R$ be points on side $BC$ such that segment $PR$ contains midpoint $Q$, with $P$ closer to $C$ and (without loss of generality) $AC \le AB$. Then if $AD$ is an altitude, then $D$ is between $P$ and $C$. Combined with the obvious fact that $Q$ is the midpoint of $PR$ (for $n$ is odd), we have


Solve the equation
$\cos^n{x} - \sin^n{x} = 1$
where $n$ is a given positive integer.

Solution: Since $cos^2x + sin^2x = 1$, we cannot have solutions with $n\ne2$ and $0<|cos(x)|,|sin(x)|<1$. Nor can we have solutions with $n=2$, because the sign is wrong. So the only solutions have $sin (x) = 0$ or $cos (x) = 0$, and these are: $x =$ multiple of $\pi$, and $n$ even; $x$ even multiple of $\pi$ and $n$ odd; $x$ = even multiple of $\pi + \frac{3\pi}{2}$ and $n$ odd.


Consider the cube $ABCDA'B'C'D'$($ABCD$ and $A'B'C'D'$ are the upper and lower bases, respectively, and edges $AA'$, $BB'$, $CC'$, $DD'$ are parallel). The point $X$ moves at constant speed along the perimeter of the square $ABCD$ in the direction $ABCDA$, and the point $Y$ moves at the same rate along the perimeter of the square $B'C'CB$ in the direction $B'C'CBB'$. Points $X$ and $Y$ begin their motion at the same instant from the starting positions $A$ and $B'$, respectively. Determine and draw the locus of the midpoints of the segments $XY$.

Solution:
First we prove a small lemma:  If the particles $X$ and $Y$ move along straight lines at constant velocities, then the locus of the midpoint of $XY$ is also a line.  This is rather trivial, since all lines in 3D space may take the parametric form $(at+n, bt+m, ct+p)$, with $t$ being time, and the average of two such lines must also have a linear parametric form.
The locus clearly starts at the midpoint of $AB'$, or the center of face $ABB'A'$.  As $X$ moves from $A$ to $B$, and $Y$ moves from $C'$ to $C$, both $X$ and $Y$ move along straight lines, so the midpoint of $XY$ traces out a line segment, starting at the midpoint of $AB'$ and ending at the midpoint of $BC'$, or the center of face $BCC'B'$.  This concludes the first phase of motion.  A quick check reveals that the locus goes nowhere during the second and fourth phases of motion, and only moves backward on the third.  Thus the locus is just the segment connecting the centers of sides $ABB'A'$ and $BCC'B'$.
solution is actually wrong, it's not segment it's parallelogram- author mixed the movement of points (even if he clearly understands what he is doing- just a little mistake) Here is correct soluton:
Answer: the rhombus CUVW, where U is the center of ABCD, V is the center of ABB'A, and W is the center of BCC'B'.
Take rectangular coordinates with A as (0, 0, 0) and C' as (1, 1, 1). Let M be the midpoint of XY. Whilst X is on AB and Y on B'C', X is (x, 0, 0) and Y is (1, x, 1), so M is (x/2 + 1/2, x/2, 1/2) = x (1, 1/2, 1/2) + (1-x) (1/2, 0, 1/2) = x W + (1-x) V, so M traces out the line VW.
Whilst X is on BC and Y is on C'C, X is (1, x, 0) and Y is ( 1, 1, 1-x), so M is (1, x/2+1/2, 1/2 - x/2) = x (1, 1, 0) + (1-x) (1, 1/2, 1/2) = x C + (1-x) W, so M traces out the line WC.
Whilst X is on CD and Y is on CB, X is (1-x, 1, 0) and Y is (1, 1-x, 0), so M is (1-x/2, 1-x/2, 0) = x (1, 1, 0) + (1-x) (1/2, 1/2, 0) = x C + (1-x) U, so M traces out the line CU.
Whilst X is on DA and Y is on BB', X is (0, 1-x, 0) and Y is (1, 0, x), so M is (1/2, 1/2 - x/2, x/2) = x (1/2, 0, 1/2) + (1-x) (1/2, 1/2, 0) = x V + (1-x) U, so M traces out the line UV.


In an $n$-gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
Prove that $a_1=a_2=\cdots = a_n$.

Solution:
Let $a_1 = p_1p_2$, $a_2 = p_2p_3$, etc.
Plot the $n$-gon on the cartesian plane such that $p_1p_2$ is on the $x$-axis and the entire shape is above the $x$-axis. There are two cases: the number of sides is even, and the number of sides is odd:
$\textbf{Case 1: Even}$
In this case, the side with the topmost points will be $p_{\frac{n}{2}+1}p_{\frac{n}{2}+2}$. To obtain the $y$-coordinate of this top side, we can multiply the lengths of the sides $a_1$, $a_2$, ... $a_{\frac{n}{2}}$ by the sine of the angle they make with the $x$-axis:

We can obtain the $y$-coordinate of the top side in a different way by multiplying the lengths of the sides $a_{\frac{n}{2}+1}$, $a_{\frac{n}{2}+2}$, ... $a_n$ by the sine of the angle they make with the $x$-axis to get the $\emph{negated}$ $y$-coordinate of the top side:

It must be true that $\textbf{(1)} = \textbf{(2)}$. This implies that $a_k = a_{k+\frac{n}{2}}$ for all $1 \leq k \leq \frac{n}{2}$, and therefore $a_1=a_2=\cdots = a_n$.
$\textbf{Case 2: Odd}$
This case is very similar to before. We will compute the $y$-coordinate of the top point $p_{frac{n+3}{2}}$ two ways:

It must be true that $\textbf{(3)} = \textbf{(4)}$. Then, we get $a_k = a_{n-k+2}$ for all $2 \leq k \leq \frac{n+1}{2}$. Therefore, $a_2=a_3=\cdots = a_n$. It is trivial that $a_1$ is then equal to the other values, so $a_1=a_2=\cdots = a_n$. This completes the proof. $\square$
~mathboy100

Define the vector $\vec{v_i}$ to equal $\cos{\left(\frac{2\pi}{n}i\right)}\vec{i}+\sin{\left(\frac{2\pi}{n}i\right)}\vec{j}$. Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length $a_i$ is parallel to $\vec{v_i}$. We then have that

But $a_i\geq a_{n-i}$ for all $i\leq \lfloor \frac{n}{2}\rfloor$, so

for all $i\leq \lfloor \frac{n}{2}\rfloor$. This shows that $a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0$, with equality when $a_i=a_{n-i}$. Therefore

There is equality only when $a_i=a_{n-i}$ for all $i$. This implies that $a_1=a_{n-1}$ and $a_2=a_n$, so we have that $a_1=a_2=\cdots =a_n$. $\blacksquare$


A circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from $\triangle ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$).

Solution:
Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow!
$[BHM]/[BCA] = \text{(perimeter of BHM/perimeter of BCA)}^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2$
Denote $[ABC]$ the area of $\triangle ABC$ and $(ABC)$ the perimeter of $\triangle ABC$.
Then $\frac{[BHM]}{[ABC]} = \left(\frac{(BHM)}{(ABC)}\right)^{2} =\left(\frac{c+a-b}{c+a+b}\right)^{2}$.
So $[BHM]=\left(\frac{c+a-b}{c+a+b}\right)^{2}\cdot [ABC]$.
We know, $r_{1}$ is the radius of the incircle of $\triangle BHM$: $r_{1}= 2 \cdot \frac{[BHM]}{(BHM)}$.
Area of the incircle of $\triangle BHM$

Area of the incircle of $\triangle ABC$:$4\pi (\frac{[ABC]}{a+b+c})^{2}$.
Sum of the area of the 4 incircles:


Given the tetrahedron $ABCD$ whose edges $AB$ and $CD$ have lengths $a$ and $b$ respectively. The distance between the skew lines $AB$ and $CD$ is $d$, and the angle between them is $\omega$. Tetrahedron $ABCD$ is divided into two solids by plane $\varepsilon$, parallel to lines $AB$ and $CD$. The ratio of the distances of $\varepsilon$ from $AB$ and $CD$ is equal to $k$. Compute the ratio of the volumes of the two solids obtained.

Solution:
Let the plane meet AD at X, BD at Y, BC at Z and AC at W. Take plane parallel to BCD through WX and let it meet AB in P.
Since the distance of AB from WXYZ is k times the distance of CD, we have that AX = k·XD and hence that AX/AD = k/(k+1). Similarly AP/AB = AW/AC = AX/AD. XY is parallel to AB, so also AX/AD = BY/BD = BZ/BC.
vol ABWXYZ = vol APWX + vol WXPBYZ. APWX is similar to the tetrahedron ABCD. The sides are k/(k+1) times smaller, so vol APWX = k3(k+1)3 vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k2(k+1)2 times smaller. Its height is 1/(k+1) times the height of A above ABCD, so vol prism = 3 k2(k+1)3 vol ABCD. Thus vol ABWXYZ = (k3 + 3k2)/(k+1)3 vol ABCD. We get the vol of the other piece as vol ABCD - vol ABWXYZ and hence the ratio is (after a little manipulation) k2(k+3)/(3k+1).


Let $k, m, n$ be natural numbers such that $m+k+1$ is a prime greater than $n+1.$ Let $c_s=s(s+1).$ Prove that the product  is divisible by the product $c_1c_2\cdots c_n$.

Solution:
For any $m, n$, $c_m - c_n = m(m+1) - n(n+1) = m^2 + m - n^2 - n = (m+n)(m-n) + m - n = (m+n+1)(m-n)$.
We can therefore write the product in the problem as follows:

But, the product of any $n$ consecutive integers is divisible by $n!$. We can prove this as follows:

Therefore, $\prod_{a = 1}^n m+a-k$ is divisible by $n!$, and $(m+k+1)\left(\prod_{a = 1}^n m+a+k+1\right)$ is divisible by $(n+1)!$. However, we are told that $m+k+1$ is prime and therefore it is not divisible by any of the numbers $1$ through $n+1$. Therefore, $\prod_{a = 1}^n m+a+k+1$ is divisible by $(n+1)!$.
Finally, it is clear that $(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)$ is divisible by $n!(n+1)! = (1 \cdot 2)(2 \cdot 3)(3 \cdot 4) \cdots (n \cdot (n+1)) =$$c_1c_2\cdots c_n$. $\square$
~mathboy100

We have that $c_1c_2c_3...c_n=n!(n+1)!$
and we have that $c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)$
So we have that $(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)=\frac{(m+n-k)!}{(m-n)!}\frac{(m+n+k+1)!}{(m+k+1)!}$ We have to show that:
$\frac{(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)}{n!(n+1)!}=\frac{(m+n-k)!}{(m-n)!n!}\frac{(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}$ is an integer
But $\frac{(m+n-k)!}{(m-n)!n!}=\binom {m+n-k}n$ is an integer and ${(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}=\binom {m+n+k+1}{n+1}\frac 1{m+k+1}$ is an integer because $m+k+1|m+n+k+1!$ but does not divide neither $n+1!$ nor $m+n!$ because $m+k+1$ is prime and it is greater than $n+1$ (given in the hypotesis) and $m+n$.
The above solution was posted and copyrighted by Simo_the_Wolf. The original thread can be found here: [1]


Consider the system of equations

with unknowns $x_1, x_2, \cdots, x_n$ where $a, b, c$ are real and $a \neq 0$. Let $\Delta = (b - 1)^2 - 4ac$. Prove that for this system
(a) if $\Delta < 0$, there is no solution,
(b) if $\Delta = 0$, there is exactly one solution,
(c) if $\Delta > 0$, there is more than one solution.

Solution:
Adding the $n$ equations together yields

Let $s_i=ax_i^2+(b-1)x_i+c$.
(a) If $\Delta<0$, then there is no solution to the quadratic equation $ax^2+(b-1)x+c$, as the determinant is negative. This implies that either $s_i>0$ for all $i$, or $s_i<0$ for all $i$. In either case the above summation cannot be 0, which implies that there are no solutions to the given system of equations. $\blacksquare$
(b) If $\Delta=0$, then there is exactly one solution to the quadratic equation $ax^2+(b-1)x+c$ (let it be $r$), and either $s_i\geq 0$ for all $i$, or $s_i\leq 0$ for all $i$. The only way that the above summation is 0 is if $s_i=0$ for all $i$. As there is exactly one $x_i$ that makes $s_i=0$ (namely $x_i=r$), then the only possible solution to the system of equations is $(x_1,x_2,\dots , x_n)=(r, r, \dots , r)$. It's not hard to show that this works, so when $\Delta=0$ the system of equations has exactly one solution. $\blacksquare$
(c) If $\Delta>1$, then there are exactly two solutions to the quadratic equation $ax^2+(b-1)x+c$. Let the roots be $r_1$ and $r_2$. If $x_i=r_1$ for all $i$, then $ax_i^2+bx_i+c=ar_1^2+br_1+c=r_1=x_{i+1}$. This shows that $(x_1,x_2,\dots ,x_n)=(r_1,r_1,\dots, r_1)$ is a solution. We can show that $(x_1,x_2,\dots ,x_n)=(r_2,r_2,\dots, r_2)$ is another solution using the same reasoning, which shows that the equation has more than one solution. $\blacksquare$


For each of $k = 1$, $2$, $3$, $4$, $5$ find necessary and sufficient conditions on $a > 0$ such that there
exists a tetrahedron with $k$ edges length $a$ and the remainder length $1$.

Solution:
A plodding question. Take the tetrahedron to be ABCD.
Take k = 1 and AB to have length a, the other edges length 1. Then we can hinge triangles ACD and BCD about CD to vary AB. The extreme values evidently occur with A, B, C, D coplanar. The least value, 0, when A coincides with B, and the greatest value √3, when A and B are on opposite sides of CD. We rule out the extreme values on the grounds that the tetrahedron is degenerate, thus obtaining 0 < a < √3.
For k = 5, the same argument shows that 0 < 1 < √3 a, and hence a > 1/√3.
For k = 2, there are two possible configurations: the sides length a adjacent, or not. Consider first the adjacent case. Take the sides length a to be AC and AD. As before, the two extreme cases gave A, B, C, D coplanar. If A and B are on opposite sides of CD, then a = √(2 - √3). If they are on the same side, then a = √(2 + √3). So this configuration allows any a satisfying √(2 - √3) < a < √(2 + √3).
The other configuration has AB = CD = a. One extreme case has a = 0. We can increase a until we reach the other extreme case with ADBC a square side 1, giving a = √2. So this configuration allows any a satisfying 0 < a < √2. Together, the two configurations allow any a satisfying: 0 < a < √(2 + √3).
This also solves the case k = 4, and allows any a satisfying: a > 1/√(2 + √3) = √(2 - √3).
For k = 3, any value of a > 0 is allowed. For a <= 1, we may take the edges length a to form a triangle. For a ≥ 1 we may take a triangle with unit edges and the edges joining the vertices to the fourth vertex to have length a.


The real numbers $a_0, a_1, \ldots, a_n, \ldots$ satisfy the condition:
The numbers $b_{1}, b_{2}, \ldots, b_n, \ldots$ are defined by
(a) Prove that $0 \leq b_n < 2$ for all $n$.
(b) given $c$ with $0 \leq c < 2$, prove that there exist numbers $a_0, a_1, \ldots$ with the above properties such that $b_n > c$ for large enough $n$.

Solution:
$b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}} = \sum_{k=1}^{n} \frac{a_k - a_{k-1}}{a_k\sqrt{a_k}} = \sum_{k=1}^{n} (a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$
Let $X_k$ be the rectangle with the verticies: $(a_{k-1},0)$; $(a_{k},0)$; $(a_{k},a_k^{-\dfrac{3}{2}})$; $(a_{k-1},a_k^{-\dfrac{3}{2}})$.

For all $k \in \mathbb{N}$, the area of $X_k$ is $(a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$. Therefore, $b_n = \sum_{k=1}^{n} [X_k]$
For all sequences $\{ a_k \}$ and all $k \in \mathbb{N}$, $X_k$ lies above the $x$-axis, below the curve $f(x) = x^{-\dfrac{3}{2}}$, and in between the lines $x = 1$ and $x = a_n$, Also, all such rectangles are disjoint.
Thus, $b_n = \sum_{k=1}^{n} [X_k] < \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx = \left[-\dfrac{2}{\sqrt{x}}\right]_{1}^{a_n} = 2 - \dfrac{2}{\sqrt{a_n}} < 2$ as desired.
By choosing $a_k = 1 + k (\Delta x)$, where $\Delta x > 0$, $b_n$ is a Riemann sum for $\int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$. Thus, $\lim_{\Delta x \to 0^+} b_n = \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$.
Therefore, $\lim_{n \to \infty} \left[ \lim_{\Delta x \to 0^+} b_n \right] = \lim_{n \to \infty} \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$ $= \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{a_n}} = \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{1+n(\Delta x)}} = 2$.
So for any $c \in [0,2)$, we can always select a small enough $\Delta x > 0$ to form a sequence $\{ a_n \}$satisfying the above properties such that $b_n > c$ for large enough $n$ as desired.


Prove that the set of integers of the form $2^k - 3(k = 2; 3; \cdots)$ contains an infinite subset in which every two members are relatively prime.

Solution:
Wlog, assume $a_{n}\ge 3$. Then say $p_{1}, p_{2}, \ldots, p_{k}\in \mathbb{N}$ are all the (pairwise distinct) primes dividing $2^{a_{n}}-3$ and let $a_{n+1}= a_{n}\cdot \prod_{i=1}^{k}(p_{i}-1)$. Obviously $p_{i}$ is odd, for any $i \in \overline{1,k}$. So $\prod_{i=1}^{k}p_{i}$ divides $2^{a_{n+1}}-1$, by Fermat's little theorem, and $\gcd(2^{a_{n+1}}-3, 2^{a_{n}}-3) = 1$. Now, by induction, it follows $\gcd(2^{a_{m}}-3, 2^{a_{n}}-3) = 1$, for any distinct $m, n \in \mathbb{N}$.
The above solution was posted and copyrighted by s.tringali. The original thread for this problem can be found here: [1]


Let $a$ and $b$ be real numbers for which the equation
$x^4 + ax^3 + bx^2 + ax + 1 = 0$
has at least one real solution. For all such pairs $(a, b)$, find the minimum value of $a^2 + b^2$.

Solution:
Substitute $z=x+1/x$ to change the original equation into $z^2+az+b-2=0$. This equation has solutions $z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}$. We also know that $|z|=|x+1/x| \geq 2$. So,

Rearranging and squaring both sides,

So, $a^2+b^2 \geq a^2+(2-2|a|)^2 = 5a^2-8|a|+4 = 5(|a|-\frac{4}{5})^2+\frac{4}{5}$.
Therefore, the smallest possible value of $a^2+b^2$ is $\frac{4}{5}$, when $a=\pm \frac{4}{5}$ and $b=\frac{-2}{5}$.
Borrowed from [1]


Prove that the number $\sum^n_{k=0}\binom{2n+1}{2k+1}2^{3k}$ is not divisible by $5$ for any integer $n\ge0.$

Solution:
Everything that follows takes place in $\mathbb F_5(\sqrt 2)$, i.e. the field we get by adjoining a root of $x^2-2=0$ to $\mathbb F_5$, the field with $5$ elements.
We have $\sum_{k=0}^n\binom{2n+1}{2k+1}2^{3k}=\sum_{k=0}^n\binom{2n+1}{2n-2k}3^k=\sum_{k=0}^n\binom{2n+1}{2(n-k)}2^{-k}$. Now, this is zero iff it's zero when we multiply it by $2^n$, so we may as well prove that $\sum_{k=0}^n\binom{2n+1}{2(n-k)}\sqrt 2^{2(n-k)}\ne 0$. The LHS is $\alpha$ from $(1+\sqrt 2)^{2n+1}=\alpha+\beta\sqrt 2,\ \alpha,\beta\in\mathbb F_5$. We have $(1-\sqrt 2)^{2n+1}=\alpha-\beta\sqrt 2$, so by multiplying them we get $-1=\alpha^2-2\beta^2$. If we were to have $\alpha=0$, then we would get $1=2\beta^2,\ \beta\in\mathbb F_5$, and this is impossible, since it would make $3=2^{-1}$ a square $\beta^2$ in $\mathbb F_5$ (i.e. $3$ would be a quadratic residue modulo $5$, and it's not).
The above solution was posted and copyrighted by grobber. 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.


On the sides of an arbitrary triangle $ABC$, triangles $ABR, BCP, CAQ$ are constructed externally with $\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\circ$. Prove that $\angle QRP = 90^\circ$ and $QR = RP$.

Consider $X$ and $Y$ so that $\triangle CQA\sim \triangle CPX$ and $\triangle CPB\sim \triangle CQY$. Furthermore, let $AX$ and $BY$ intersect at $F$. Now, this means that $\angle PBC = 45 = \angle QAC = \angle PXC$ and $\angle PCB = 15 = \angle QCA = \angle PCX$, so $\triangle BPC\cong \triangle XPC$. Thus, $XC = BC$ and $\angle BCX = 30 + 30 = 60$, so $\triangle BCX$ is equilateral. Similarly is $\triangle ACY$. Yet, it is well-known that the intersection of $BY$ and $AX$, which is $F$, must be the Fermat Point of $\triangle ABC$ if $\triangle AYC$ and $\triangle BXC$ are equilateral. Now, $\angle PBX = 60 - 45 = 15$. Similarly, $\angle PXB = 15$, so $\triangle BRA\sim \triangle BPX$, so a spiral similarity maps $\triangle BPR\sim \triangle BXA$. This implies that $\angle BRP = \angle FAB$. Similarly, $\angle ARQ = \angle FBA$, so $\angle BRP + \angle ARQ = \angle FBA + \angle FAB = 180 - \angle AFB = 60$. Then, $\angle PRQ = (180 - 15 - 15) - 60 = 90$. We also realize that $\frac {AX}{RP} = \frac {AB}{RB} = \frac {AB}{AR} = \frac {BY}{RQ}$. Now, $Y$ is a rotation of $A$ $60$ degrees around $C$ and $B$ is the same rotation of $X$ around $C$, so $AX$ maps to $YB$ from this rotation, so $AX = YB$. It follows that $RP = RQ$.
The above solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [1]

Let $\triangle ABT$ be the equilateral triangle constructed such that $T$ and $R$ are on the same side. $\triangle RTB \sim \triangle PCB \sim \triangle QCA$. We have $\frac {AT}{AC} = \frac {AR}{AQ}$ from similarity. Also we have $\angle TAC = \angle RAQ$ . So $\triangle ACT \sim \triangle ARQ$. Then $\angle ATC = \angle ARQ = m$ and $\frac {AR}{AT} = \frac {RQ}{TC}$. Similar calculations for $B$. We will have $\angle BTC = \angle BRP = 60-m$ and $\frac {BR}{AT} = \frac {RP}{TC}$. Also from the question we have $AR = BR$. So $\angle PRQ = 180 - (60-x) - (60-x) -m -(60-m) = 2x$ and $PR = RQ$.
The above solution was posted and copyrighted by xeroxia. The original thread for this problem can be found here: [2]

Define $X$ to be a point such that $\triangle AQC$ is directly similar to $\triangle AXB$.
Then, it is trivial to show that $\angle AXR=60^{\circ}$ and that $\angle RXB=45^{\circ}$; that is, $AX=AR=XR=RB$ and $\triangle RBX$ is a right isosceles triangle. If we prove that $\triangle RPQ$ is similar to $\triangle RBX$, then we will be done.
According to the property of spiral similarity, it suffices to prove that $\triangle RBP\sim\triangle RXQ.$
Since $\triangle AQC\sim \triangle AXB$, we have $\triangle AQX\sim\triangle ACB$, and this gives $\angle RBP=60^{\circ}+\angle  CBA=60^{\circ}+\angle QXA=\angle RXQ$. It remains to prove that $\frac{BP}{RB}=\frac{XQ}{RX}$.
As $RB=RX$, we must prove that $BP=XQ$. From Law of Sines on $\triangle BPC$, we have $BP=\frac{a}{2\cos 15^{\circ}}$. Since $\frac{XQ}{AX}=\frac{a}{c}$, it remains to prove that $AX=AR=\frac{c}{2\cos 15^{\circ}}$, which is easily verified. We are done. $\blacksquare$
The above solution was posted and copyrighted by fantasylover. The original thread for this problem can be found here: [3]


A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent from the volume of the box is occupied. Determine the possible dimensions of the box.

Solution:
We name a,b,c the sides of the parallelepiped, which are positive integers. We also put

It is clear that $xyz$ is the maximal number of cubes with sides of length $\sqrt[3]{2}$ that
can be put into the parallelepiped with sides parallels to the sides of the box.
Hence the corresponding volume is $V_2=2\cdot xyz$. We need $V_2=0.4\cdot V_1=0.4\cdot abc$,
hence
We give the  values of $x$ and $a/x$ for $a=1,\dots ,8$. The same table is valid for $b,y$ and $c,z$.

By simple inspection we obtain two solutions of $(1)$: $\{a,b,c\}=\{2,5,3\}$ and $\{a,b,c\}=\{2,5,6\}$.
We now show that they are the only solutions.
We can assume $\frac ax\ge \frac by \ge \frac cz$. So necessarily $\frac ax\ge \sqrt[3]{5}$. Note that
the definition of $x$ implies
hence
If $a\ge 4$ then $x\ge 3$ and $\frac ax<\sqrt[3]2(1+\frac 1x)\le \sqrt[3]2(\frac 43)<\sqrt[3]5$
since $2\cdot \frac {4^3}{3^3}<5$. So we have only left the cases $a=2$ and $a=3$. But for $a=3$
we have $a/x=3/2<\sqrt[3]5$ and so necessarily $a=2$ and $a/x=2$.
It follows
Note that the definitions of $y,z$ imply
Moreover we have from (2) and from $b/y\ge c/z$ that
If $b=2$ then $b/y=2$ and we would have $c/z=5/4<\sqrt[3]2$, which contradicts $(3)$.
On the other hand, if $b>5$ then $y>4$ and $\frac by<\sqrt[3]2(1+\frac 1y)\le \sqrt[3]2(\frac 54)<\sqrt{5/2}$
since $2^2\cdot \frac {5^6}{4^6}<\frac{5^3}{2^3}$ as $5^3<2^7$. So we have only left the
cases $b=3,4,5$. But for $b=3$ we have $b/y=3/2<\sqrt{5/2}$ and for $b=4$ we have
$b/y=4/3<\sqrt{5/2}$ and so necessarily $b=5$ and $b/y=5/3$ ($>\sqrt{5/2}$)
So we arrive finally at $a=2,b=5$ and $c/z=3/2$. If $c\ge 8$ then $z\ge 6$ and $\frac cz<\sqrt[3]2(1+\frac 1z)\le \sqrt[3]2(\frac 76)<\frac 32$ since $2\cdot \frac {7^3}{6^3}<\frac{3^3}{2^3}$. On the other hand, for $c\le 7$ there are the only two possible values $c=3$ and $c=6$ which yield the known solutions.


Let $n$ be a given number greater than 2. We consider the set $V_n$ of all the integers of the form $1 + kn$ with $k = 1, 2, \ldots$ A number $m$ from $V_n$ is called indecomposable in $V_n$ if there are not two numbers $p$ and $q$ from $V_n$ so that $m = pq.$ Prove that there exist a number $r \in V_n$ that can be expressed as the product of elements indecomposable in $V_n$ in more than one way. (Expressions which differ only in order of the elements of $V_n$ will be considered the same.)

Solution:
Lemma: there are $\infty$ many prime numbers $p \not\equiv 1 \mod n$.
Proof:
Assume that there are only finitely many of them and let their product be $k$.
Then all prime factors of $nk - 1$ must be $\equiv 1 \mod n$ since this number is coprime with $k$. But that would mean that $- 1 \equiv nk - 1 \equiv 1 \mod n$ which is impossible for $n > 2$.
Now we can tackle the problem:
There are only finetely many residue classes $\mod n$, thus we can find two primes $p,q$ with $p \equiv q \not\equiv 1 \mod n$ and coprime with $n$.
Let $s$ be the smallest positive integer with $p^s \equiv 1 \mod n$, then the same property holds for $q$ too.
Now we have that $p^s, q^s, pq^{s - 1} , p^{s - 1}q \equiv 1 \mod n$ are all indecomposable since all their nontrivial divisors are $\not\equiv 1 \mod n$. But the product $p^sq^s = (pq^{s - 1})(p^{s - 1}q)$ gives a number that is represented in two ways.
The above solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]


Let $0<f(1)<f(2)<f(3)<\ldots$ a sequence with all its terms positive$.$ The $n-th$ positive integer which doesn't belong to the sequence is $f(f(n))+1.$ Find $f(240).$

Solution:
Since the $n$-th number missing is $f(f(n))+1$ and $f(f(n))$ is a member of the sequence, it results that there are exactly $n-1$ "gaps" less than $f(f(n))$, which leads us to:
$f(f(n))=f(n)+n-1$ $(\star)$
now with a simple induction we can prove that $f(F_{n}+1) = F_{n+1}+f(1)$ , where $F_{k}$ is the Fibonacci sequence.
our next step now is to prove that $f(F_{n}+x) =F_{n+1}+f(x)$, for all $x$ with $1\leq x\leq F_{n-1}$....and agaiiiiin induction(on $n$) : :P
for $n=0$ and $n=1$ it`s trivial and now we supose that it`s true for $n-1$ and to prove that holds for $n$ , in other words $P(n-1) \Rightarrow P(n)$
$(i)$ If $x=f(y)$ for some $y$, then by the inductive assumption and $(\star)$ we have:
$f(F_{n}+x)=f(F_{n}+f(y))=f(f(F_{n-1}+y))=F_{n}+f(y)+F_{n-1}+y-1=F_{n+1}+f(x)$
$(ii)$ If $x=f(f(y))+1$ is a gap, then $f(F_{n}+x-1)+1=F_{n+1}+f(x-1)+1$ is a gap also:
$F_{n+1}+f(x)+1=F_{n+1}+f(f(f(y)))+1=f(F_{n}+f(f(y)))+1=f(f(F_{n-1}+f(y)))+1$
It follows that : $f(F_{n}+x)=F_{n+1}+f(x-1)+2=F{n+1}+f(x)$
now since we know that each positive integer $x$ is expressible as:
$x=F_{k_{1}}+F_{k_{2}}+...+F_{k_{r}}$ , where $0<k_{r}\neq 2$, $k_{i}\geq  k_{i+1}+2$
we obtain that $f(x) = F_{k_{1}+1}+...F_{k_{r}+1}$ and now that we have the general form, pe particulary calculate for $f(240)$:
we can write $240=233+5+2$, therefore $f(240)= 377+8+3=388$
Remark: it can be shown now that $f(x)=[ax]$ , where $a=\frac{1+\sqrt{5}}{2}$
The above solution was posted and copyrighted by pohoatza. The original thread for this problem can be found here: [1]


Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

Let $O_1, O_2$ be the centers of the 2 circles and let the points B, C move on the circles $(O_1), (O_2)$, recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles $\angle BO_1A = \angle CO_2A$ are equal, the isosceles triangles $\triangle AO_1B \sim \triangle AO_2C$ are similar. Consequently, the angles $\angle BAO_1 = \angle CAO_2$ are equal and the angles
$\angle BAC = BAO_1 + \angle O_1AO_2 + \angle O_2AC =$
$= BAO_1 + \angle O_1AO_2 - \angle CAO_2 = \angle O_1AO_2$
are also equal. The angles $\angle A'BA = \frac{\angle A'O_1A}{2} = \angle O_2O_1A$ are equal and the angles $\angle A'CA = \frac{\angle A'O_2A}{2} = \angle O_1O_2A$ are also equal. As a result, the point A' is on the side BC of the triangle $\triangle ABC$, which is always similar to the fixed triangle $\triangle O_1AO_2$.
Lemma (Yaglom's Geometric Transformations II, pp 68-69). Assume that a figure F' (the triangle $\triangle ABC$) moves in such a way that it remains at all times similar to a fixed figure F (the triangle $\triangle O_1AO_2$) and in addition, some point of the figure F' does not move at all (the point A). If a point B of the figure F' describes a curve $\Gamma$ (the circle $(O_1)$), then any other point M of the figure F' describes a curve $\Gamma'$ similar to the curve $\Gamma$.
Consequently, the midpoint M of the segment BC decribes a circle (Q). When the points B, C, become identical with the point A, the midpoint M of BC also becomes identical with A. Hence, the circle (Q) passes through the point A. The line BC always passes through A'. Just after the start of motion counter-clockwise, the points B, C, A' follow on the line BC in this order. The points B, C both pass through the point A' one after another and just before the end of one period of motion, the points A', B, C follow on the line BC In this order. From the continuity principle, the midpoint M of BC also passes through the point A' and consequently, the circle (Q), the locus of the midpoints M, also passes through the point A'. Thus the circle (Q) belongs to the same pencil with the circles $(O_1), (O_2)$ and it is centered on their center line $O_1O_2$. Let p be the perpendicular bisector of the segment BC intersectong the circle (Q) at a point P different from the point M. The point A' and the circle (Q) passing through this point are fixed and the angle $\angle A'MP = 90^\circ$ is right. Therefore, A'P is a fixed diameter of the circle (Q), which means that the point P is also fixed. Q.E.D.
When the points B, C are both diametrally opposite to the point A on their respective circles, the segment $O_1O_2$ is the midline of the triangle $\triangle ABC$, i.e., $BC \parallel O_1O_2$, $BC \perp AA'$. The cyclic quadrilateral AA'MP inscribed in the circle (Q) is then a rectangle. Its circumcenter Q is the intersection of its diagonals AM, A'P intersecting at their midpoint and the midpoint of AM is identical with the midpoint of $O_1O_2$. Thus the fixed point P is the reflection of the intersection A' of the circles $(O_1), (O_2)$ in the midpoint Q of the segment $O_1O_2$.
This solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]

Let $B$ and $C$ be the antipodes of $A$ on the two circles and let $D$ be the foot of the altitude from $A$, which is the other intersection point of the two circles. Also, let $M$ be the midpoint of $\overline{BC}$, and construct rectangle $ADMZ$. Our claim is that $Z$ is the fixed point.
We let $X$ and $Y$ be the two points; by the condition the angles are the same. So we have a spiral similarity
Now let $N$ be the midpoint of $\overline{XY}$. By spiral similarity, since $N$ maps to $M$, it follows $M$, $N$, $A$, and $D$ are cyclic too. So actually $N$ lies on the circumcircle of rectangle $ADMZ$, meaning $\overline{ZN} \perp \overline{XY}$, hence $ZX = ZY$ as needed.
Remark: The special point $Z$ can be identified by selecting the special case $X \to A$, $Y \to A$ and $X=B$, $Y=C$.
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]

Let the points on the circles be $E$ and $F$. Then, let $B$ the the other intersection of the two circles, and $C$ and $D$ be the centers of the circles containing $E$ and $F$, respectively. WLOG, let us name $E$ and $F$ such that $CA>DA$. Let $O$ be the midpoint of $CD$, and the antipode of $B$ with respect to the circle centered at $O$ with radius $OA$ be $P$. Let $M$ be where this such circle hits $EF$. We first note thatand alsomeaning that since $AD=DF$, $AO=OM$, and $AC=CE$, then SAS similarity gets us that $\triangle ADF \sim \triangle AOM \sim \triangle ACE$. This means we have $\frac{AO}{AM} = \frac{AC}{AE}$ so $\frac{AO}{AC} = \frac{AM}{AE}$ and because $\angle MAE = \angle OAC$, then $\triangle AOC \sim \triangle AME$ and similarly, we also have $\triangle AOD \sim \triangle AMF$. This means thatmeaning that $M$ is the midpoint of $EF$. Now, since $P$ is the antipode of $B$, then $\angle BMP = 90^{\circ} = \angle EMP = \angle FMP$ so $EP=FP$ for all $E$ and $F$ for some fixed $P$ and we are done.
This solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]


Determine the maximum value of $m^2 + n^2$, where $m$ and $n$ are integers satisfying $m, n \in \{ 1,2, \ldots , 1981 \}$ and $( n^2 - mn - m^2 )^2 = 1$.

Solution:
We first observe that since $\gcd(m,n)=1$, $m$ and $n$ are relatively prime.  In addition, we note that $n \ge m$, since if we had $n < m$, then $n^2 -nm -m^2 = n(n-m) - m^2$ would be the sum of two negative integers and therefore less than $-1$.  We now observe
,
i.e., $(m,n) = (a,b)$ is a solution iff. $(b, a+b)$ is also a solution.  Therefore, for a solution $(m, n)$, we can perform the Euclidean algorithm to reduce it eventually to a solution $(1,n)$.  It is easy to verify that if $n$ is a positive integer, it must be either 2 or 1.  Thus by trivial induction, all the positive integer solutions are of the form $(F_{n}, F_{n+1})$, where the $F_i$ are the Fibonacci numbers.  Simple calculation reveals $987$ and $1597$ to be the greatest Fibonacci numbers less than $1981$, giving $987^2 + 1597^2=3524578$ as the maximal value.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Consider infinite sequences $\{x_n\}$ of positive reals such that $x_0=1$ and $x_0\ge x_1\ge x_2\ge\ldots$.
a) Prove that for every such sequence there is an $n\ge1$ such that:
b) Find such a sequence such that for all $n$:

By Cauchy, the LHS is at least:
$\frac{(x_{0}+...+x_{n-1})^{2}}{x_{1}+...+x_{n}}.$
It is clear that $x_{0}=1$ and $x_{n}\le \frac{x_{1}+...+x_{n-1}}{n-1}$
We thus have that the LHS is at least:
$(\frac{1+2(x_{1}+...+x_{n-1})+(x_{1}+...+x_{n-1})^{2}}{x_{1}+...+x_{n-1}})(\frac{n-1}{n})$
Applying 2-variable AM-GM to $1$ and $(x_{1}+...+x_{n-1})^{2}$ the LHS is at least $4(\frac{n-1}{n})$
We then simply choose $n$ such that $4(\frac{n-1}{n}) \ge 3.999$.
Part b) is likely to occur only if both the AM-GM and Cauchy equality conditions are close to met, or if
$1=x_{1}+x_{2}+...$ and $\frac{x_{0}}{x_{1}}=\frac{x_{1}}{x_{2}}=...$, respectively. This points to the geometric series with common ratio $\frac{1}{2}$ as the equality case.
This solution was posted and copyrighted by n^4 4. The original thread for this problem can be found here: [1]

b) is immediate by taking $x_i = 2^{-i}$.
For a), we prove $LHS \ge 2^{2^0 + 2^{-1}+...+ 2^{2-n}}$, and taking $n\to \infty$ will yield the result. (Here $n=1$ gives the empty sum, so $LHS \ge 2^0=1$, which is trivial).
The proof is not hard. Let $s_i$ be the sum of the first $i$ terms, so $s_n = s_{n-2} + \dfrac{x_{n-2}^2}{x_{n-1}} + \dfrac{x_{n-1}^2}{x_n} \ge s_{n-2} + 2^{2^0} x_{n-2} \sqrt{\dfrac{x_{n-1}}{x_n}} \ge s_{n-2} + 2^{2^0} x_{n-2}$. This in turn is $s_{n-3} + \dfrac{x_{n-3}^2}{x_{n-2}} + 2^{2^0} x_{n-2}\ge s_{n-3} + 2^{2^0 +  2^{-1}} x_{n-3}$. By repeatedly applying AM-GM to the last two terms of the sum, collecting powers of two, and cancelling variables, we eventually obtain
$s_0 + \dfrac{x_0^2}{x_1} + 2^{2^0 + 2^1 + ... + 2^{3-n}}x_1 \ge 0 + 2^{2^0 + 2^1 + ... + 2^{2-n}}x_0$, as desired.
This solution was posted and copyrighted by tastymath75025. The original thread for this problem can be found here: [2]


Let $a$, $b$ and $c$ be positive integers, no two of which have a common divisor greater than $1$. Show that $2abc - ab - bc- ca$ is the largest integer which cannot be expressed in the form $xbc + yca + zab$, where $x$, $y$ and $z$ are non-negative integers.

First off, I prove $2abc-ab-bc-ca$ is un-achievable. Also, assume WLOG $a\ge b\ge c$.
Assume $2abc-bc-ca=xbc+yca+zab$, then take the equation $\pmod{ab}$ to give us $(x+1)bc+(y+1)ac\equiv 0\pmod{ab}$. By CRT and $\gcd(a,b)=1$, we take this equation mod a and mod b to give us $y+1\equiv 0\pmod{b}$, and $x+1\equiv 0\pmod{a}$ (and using all numbers are relatively prime pairwise). Substituting this back into the original equation gives us $z\le -1$, contradiction (this is where the $2abc$ part comes in).
Now, let $2abc-bc-ca-ab+k=xbc+yca+zab$, and if a solution exists where $bc-1\ge k\ge 1$ then we are done because we can add $1$ to $x$ always.
Consider the set $x=a-m$, where $m\in \{1,2,\cdots, a-1, a\}$ and $y=b-n$ where $n\in \{1,2,\cdots, b-1, b\}$.
Therefore, we get $2abc-bc-ca-ab+k=(a-m)bc+(b-n)ca+zab$. This is the same as $bc(m-1)+ca(n-1)+k=(z+1)ab$ after doing some simplifying. By CRT, this must hold mod a and mod b and because $\gcd(a,b)=1$. Mod $a$ gives us $bc(m-1)+k\equiv 0\pmod{a}$, for which a value of $m-1$ obviously exists mod $a$, which can be chosen from the set of values we have assigned for $m$. Similar method shows a value of $n-1$ exists mod $b$ from the set we have given to $n$.
Now, we already know that $x,y\ge 0$. Also, the LHS of the equation $bc(m-1)+ca(n-1)+k$ is at least $1$, therefore $z\ge 0$ and we are done.
This solution was posted and copyrighted by binomial-theorem. The original thread for this problem can be found here: [1]

First we will prove $2abc-ab-bc-ca$ is unattainable, as such:
Suppose $xbc+yca+zab=2abc-ab-bc-ca$. Then, taking this mod $a$, we have that $xbc\equiv 0\pmod{a}$, so $x\equiv 0\pmod{a}$, and $a|x$. Similarly, $b|y$, and $c|z$, so $x\ge a$, $y\ge b$, and $z\ge c$. Thus, $xbc+yca+zac\ge 3abc$, so $2abc\ge 3abc$, which is a contradiction.
Now we will prove all $n>2abc-ab-bc-ca$ is attainable, as such: consider the integer $r$ such that $r\equiv \frac{n}{bc} \pmod{a}$ and $0\le r\le a-1$. Rearranging the equation $xbc+yca+zab=n$ yields $\frac{n-xbc}{a}=yc+zb$, so set $x=r$. We see that $xbc-n=rbc-n\equiv 0 \pmod{a}$, so $\frac{n-xbc}{a}$ is a positive integer (obviously $n-xbc>0)$. Now, note that since $x\le a-1$, we have that $2abc-ab-ac-bc=(a-1)bc+abc-ab-ac\ge xbc+abc-ab-ac$, so $n>xbc+abc-ab-ac$. Thus, $n-xbc>abc-ab-ac$ so $\frac{n-xbc}{a}>bc-b-c$, so by Chicken Mc-Nugget theorem, there exist $b, c$ that satisfy the equation and are now done. $\blacksquare$
This solution was posted and copyrighted by Stormersyle. The original thread for this problem can be found here: [2]


Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\angle AOX\over OX}$, where $\angle AOX$ is measured in radians in the range $[0,2\pi)$. Prove that we can find a point $X$, not on $OA$, such that its color appears on the circumference of the circle $C(X)$.

Solution:
Let $a_{n},\ n\ge 1$ be a sequence of positive reals such that $\left(\sum_{n\ge 1}a_{n}\right)^{2}<2\pi\ (*)$. For each $n\ge 1$, let $\mathcal C_{n}$ be the circle centered at $O$ with radius $\sum_{i=1}^{n}a_{i}$.
Because of $(*)$, we can find points $x_{i,j}\in\mathcal C_{i},\ i,j\ge 1$ such that for all $i,j\ge 1$ we have $\mathcal C(x_{i,j})=\mathcal C_{i+j}$. We now forget about all the other points, and work only with the matrix $M=(x_{i,j})$.
Suppose we use $n$ colors. There must be one, $c_{1}$, which appears infinitely many times on the first row of $M$, in, say, points $x_{1,j_{1}},x_{1,j_{2}},\ldots$. Then $c_{1}$ cannot appear on the lines $j_{k}+1$, $k\ge 1$. Next, there is a color $c_{2}$ which appears infinitely often among the points $x_{j_{1}+1,j_{k}-j_{1}}$, $k\ge 2$. But then $c_{2}$ cannot appear on the lines $j_{k}+1$ for such $k$. Repeating this procedure, we reach a stage where we have a row of $M$ (infinitely many actually) on which none of our $n$ colors $c_{1},\ldots,c_{n}$ can appear. This is a contradiction.
This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]


For any polynomial $P(x) = a_0 + a_1 x + \cdots + a_k x^k$ with integer coefficients, the number of coefficients which are odd is denoted by $w(P)$.  For $i = 0, 1, \ldots$, let $Q_i (x) = (1+x)^i$.  Prove that if $i_1, i_2, \ldots , i_n$ are integers such that $0 \leq i_1 < i_2 < \cdots < i_n$, then

Solution:
We first observe that $(1+x)^{2^m} \equiv 1 + x^{2^m} \pmod{2}$, so for any polynomial $P$ of degree less than $2^m$, $w(P\cdot Q_{2^m}) = 2w (P)$.
Let $k$ be the largest power of 2 that is less than or equal to $i_n$.  We proceed by induction on $k$.
For $k = 0$, the problem is trivial.
Now assume that the problem holds for $k-1$.  We now have two cases: $i_1 \ge k$, and $i_1 < k$.
In the first case, we note that $w \left( \sum_{j=1}^{n}Q_{i_j} \right) = w \left( (1+x)^k \sum_{j=1}^{n}Q_{i_j - k} \right) = 2 w \left( \sum_{j=1}^{n}Q_{i_j - k} \right)$, which is greater than or equal to $w( Q_{i_1} )$ by assumption.
In the second case, we use the division algorithm to note that
But by assumption, $w \left( \sum_{j=0}^{k-1}a_j x^j \right) \ge w( Q_{i_1})$, and for each odd $a_j$, at least one of $(a_j + b_j )$ and $b_j$ must be odd, Q.E.D.
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To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y<0$, then the following operation is allowed: $x,y,z$ are replaced by $x+y,-y,z+y$ respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.

Solution:
The algorithm always stops. Indeed, consider the five numbers on the vertices of the pentagon as forming a $5$-tuple $\mathbf{x} = (x_1, x_2, x_3, x_4, x_5) \in \mathbb{Z}^5$. The sum of the five entries of this $5$-tuple is positive (by assumption), and stays unchanged through the entire process (since our operation does not change the sum); thus, it is always positive. Now, define the function $f : \mathbb{Z}^5 \to \mathbb{Z}$ by

Clearly, $f \geq 0$ always. Now, let $\mathbf{x}_{\text{old}} = (x_1, x_2, x_3, x_4, x_5)$ and $\mathbf{x}_{\text{new}}$ be two consecutive $5$-tuples obtained in this process. Thus, $\mathbf{x}_{\text{new}}$ is obtained from $\mathbf{x}_{\text{old}}$ by applying our operation once. Suppose, WLOG, that $y=x_4<0$, and that the last three entries $x_3, x_4, x_5$ of $\mathbf{x}_{\text{old}}$ are replaced by $x_3 + x_4, -x_4, x_5 + x_4$ in $\mathbf{x}_{\text{new}}$. Let $S = x_1 + x_2 + x_3 + x_4 + x_5$ be the sum of all entries of $\mathbf{x}_{\text{old}}$ (or $\mathbf{x}_{\text{new}}$). Then, a straightforward computation shows that $f(\mathbf{x}_{\text{new}}) - f(\mathbf{x}_{\text{old}}) = 2Sx_4 < 0$ (since $S>0$ and $x_4 < 0$). Thus, if the algorithm does not stop, we can find an infinite decreasing sequence of nonnegative integers $f_0>f_1>f_2>\cdots$ (by applying $f$ to each of the $5$-tuples). This is impossible, so the algorithm must stop. $\Box$
This solution was posted and copyrighted by tc1729. 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.


Let $x_1 , x_2 , \ldots , x_n$ be real numbers satisfying $x_1^2 + x_2^2 + \cdots + x_n^2 = 1$.  Prove that for every integer $k \ge 2$ there are integers $a_1, a_2, \ldots a_n$, not all 0, such that $| a_i | \le k-1$ for all $i$ and

We first note that by the Power Mean Inequality, $\sum_{i=1}^{n} x_i \le \sqrt{n}$.  Therefore all sums of the form $\sum_{i=1}^{n} b_i x_i$, where the $b_i$ is a non-negative integer less than $k$, fall in the interval $[ 0 , (k-1)\sqrt{n} ]$.  We may partition this interval into $k^n - 1$ subintervals of length $\frac{ (k-1)\sqrt{n} }{k^n - 1}$.  But since there are $k^n$ such sums, by the pigeonhole principle, two must fall into the same subinterval.  It is easy to see that their difference will form a sum with the desired properties.

This solution is very similar to Solution 1 but uses a slightly different approach for the first part. It suffices to find ${a_i}$ where $a_i$ is positive. Let $f(a_1, a_2, ..., a_n) = a_1| x_1|+a_2 |x_2|+\cdots+a_n |x_n|\geq0$.
By the Cauchy-Schwarz Inequality,

This implies that $\left(a_1 x_1+a_2 x_2+...+a_n x_n\right)^2\leq \sqrt{n} (k-1)$, and hence the codomain of $f(a_1, a_2, ..., a_n)$ is $\left[0, \sqrt{n} (k-1)\right]$. The rest of the proof is similar to Solution 1.
~Iraevid13
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A function $f$ defined on the positive integers (and taking positive integers values) is given by:
$\begin{matrix} f(1) = 1, f(3) = 3 \\ f(2 \cdot n) = f(n) \\ f(4 \cdot n + 1) = 2 \cdot f(2 \cdot n + 1) - f(n) \\ f(4 \cdot n + 3) = 3 \cdot f(2 \cdot n + 1) - 2 \cdot f(n), \end{matrix}$
for all positive integers $n.$ Determine with proof the number of positive integers $\leq 1988$ for which $f(n) = n.$

Considering that $f(n)=f(2n)$, the two last equations give :
$f(4n + 1)-f(4n) = 2(f(2n + 1) - f(2n))$
$f(4n + 3)-f(4n+2)= 2(f(2n + 1) - f(2n))$
And so, if $n$ is even and $2^{p+1}>n\geq 2^p>1$, we have $f(n+1)-f(n)=2^p$
So, if we have an even $n=\sum_{i=1}^{k} 2^{a_i}$, where $\{a_i\}$ is a strictly increasing sequence with $a_1>0$ ($n$ even) : $f(n+1)=2^{a_k}+f(n)$
Then $f(n)=f(\sum_{i=1}^{k} 2^{a_i})$ $=f(\sum_{i=1}^{k} 2^{a_i-a_1})$ $=2^{a_k-a_1}+f(\sum_{i=2}^{k} 2^{a_i-a_1})$
And so $f((\sum_{i=1}^{k} 2^{a_i})+1)=2^{a_k}+2^{a_k-a_1}+f(\sum_{i=2}^{k} 2^{a_i-a_1})$
And it is easy to conclude that $f(\sum_{i=1}^{k} 2^{a_i})=\sum_{i=1}^{k} 2^{a_k-a_i}$ and that applying $f(n)$ means reversing the order of binary representation of n (and this could be also easily shown with induction).
So $f(n)=n$ occurs if and only if the binary representation of n is symetrical.
It remains to count these "symetric" numbers. We have exactly $2^{\lceil\frac{m-1}{2}\rceil}$ such numbers in $[2^m,2^{m+1})$. So :
We have exactly 1 such numbers in $[1,2)$
We have exactly 1 such numbers in $[2,4)$
We have exactly 2 such numbers in $[4,8)$
We have exactly 2 such numbers in $[8,16)$
We have exactly 4 such numbers in $[16,32)$
We have exactly 4 such numbers in $[32,64)$
We have exactly 8 such numbers in $[64,128)$
We have exactly 8 such numbers in $[128,256)$
We have exactly 16 such numbers in $[256,512)$
We have exactly 16 such numbers in $[512,1024)$
Since $1988=B11111000100$, positions 2 to 6 may be any between $00000$ and $11101$, and so :
We have exactly 30 such numbers in $[1024,1988]$
And so the requested number is $1+1+2+2+4+4+8+8+16+16+30=92$
This solution was posted and copyrighted by pco. The original thread for this problem can be found here: [1]

The main claim is following.
Claim: $f(n)$ is equal to the result when $n$ is written in binary and its digits are reversed.
Proof. Follows directly by induction. $\blacksquare$
So the question asks for the number of binary palindromes which are at most $1988 = 11111000100_2$.
For $k = 1, 2, \dots, 10$ there are $2^{\left\lceil k/2 \right\rceil-1}$ binary palindromes with $k$ bits (note the first bit must be $1$). For $k=11$, the number of binary palindromes which are also less than $1988$ is $2^5 - 2$ (only $11111011111$ and $11111111111$ are missing).
Hence the final count is
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]

We claim that $f(x)$ is the reversal of the digits of $x$ in binary. We proceed by induction, because given $f(1)=f(2)=1$ and $f(3)=3$, we can determine all other values, and this property holds for $f(1)$, $f(2)$, and $f(3)$.
We prove that all operations conserve this property. For $f(2n)=f(n)$, we are adding a zero at the end, thus obviously preserving this property. For $f(4n+1) = 2f(2n+1)-f(n)$, we see that letting $4n+1=\overline{a_1a_2\ldots a_k01}_2$ gets that assuming this holds true for $f(2n+1)$ and $f(n)$, thenwhich keeps the property since $\overline{10a_ka_{k-1}\ldots a_2a_1}_2$ is the reversal of $\overline{a_1a_2\ldots a_k01}_2$ in binary. Similarly, letting $4n+1=\overline{a_1a_2\ldots a_k11}_2$, thenwhich keeps the property since $\overline{11a_ka_{k-1}\ldots a_2a_1}_2$ is the reversal of $\overline{a_1a_2\ldots a_k11}_2$. Due to this, we have proved that this property holds for all $n$ which means we need to find the number of binary palindromes below $1988$. This number is just going to be, for $n<11$-digit numbers, $2^{\lfloor \frac {n-1}2 \rfloor}$ ways, so summing up for less than $11$ digits gets $62$ ways. For $11$-digit numbers, we have $32$ ways but need to subtract a few in which are greater than $1988$; there are two numbers we don't count, namely $2015$ and $2047$. As a result, our answer is $62+30=\boxed{92}$.
This solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]


Let $n$ and $k$ be positive integers and let $S$ be a set of $n$ points in the plane such that
i.) no three points of $S$ are collinear, and
ii.) for every point $P$ of $S$ there are at least $k$ points of $S$ equidistant from $P.$
Prove that:

Let $\{A_i\}_{i=1}^n$ be our set of points. We count the pairs $(A_i,\{A_j,A_\ell\}),\ i\ne j\ne \ell\ne i$ such that $A_iA_j=A_iA_\ell$. There are $n$ choices for $A_i$, and for each $A_i$ there are $\ge\frac{k(k-1)}2$ choices for $\{A_j,A_\ell\}$, so the number of such pairs is $\ge\frac{nk(k-1)}2$. On the other hand, there are $\frac{n(n-1)}2$ choices for $\{A_j,A_\ell\}$, and for each $\{A_j,A_\ell\}$ there are at most two choices for $A_i$ (because we can't have $\ge 3$ points on the perpendicular bisector of $A_jA_\ell$). This means that the number of pairs is $\le n(n-1)$. We have thus found $n(n-1)\ge\frac{nk(k-1)}2\Rightarrow 2(n-1)\ge k(k-1)\ (*)$. If $k\ge \sqrt{2n}+\frac 12$, then $k(k-1)\ge 2n-\frac 14>2n-2$, contradicting the last inequality in $(*)$, so we're done.
This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]

So obvious simplification and the fact that $n$ is an integer gives that what we want is $n\ge\dbinom{k}{2}+1$
Now for each point, draw a circle around it that contains at least $k$ points in $S$.
For any point $P$ in $S$, let $d_P$ be the number of circles it's on.
For any point $O$ in $S$, let $f_O$ be the number of points on the defined circle with center $O$, so $f_O\ge k$ for all $O$. Since the function $\dbinom{x}{2}$ is increasing for $x\ge 1$ an integer, $\dbinom{f_O}{2}\ge \dbinom{k}{2}$ for all $O$. So $E(\dbinom{f}{2})\ge \dbinom{k}{2}$.
But notice that since any pair of two points shares at most 2 circles (otherwise we have 3 circles with centers on their perpendicular bisector, which is not allowed), $\sum_{O\in S}\dbinom{f_O}{2}\le 2\dbinom{n}{2}$, since the left side counts the number of point-pairs that share a circle (with possible circle multiplicity) in $P$ while the right side upper-bounds the number of point-pairs that can share a circle (with possible circle multiplicity). So now $\dbinom{k}{2}\le E(\dbinom{f}{2})\le (2/n)((n-1)(n)/2)=n-1$, which is what we want.
This solution was posted and copyrighted by antimonyarsenide. The original thread for this problem can be found here: [2]


Determine all integers $n > 1$ such that $\frac{2^n+1}{n^2}$ is an integer.

Let $N = \{ n\in\mathbb{N} : 2^n\equiv - 1\pmod{n^2} \}$ be the set of all solutions and $P = \{ p\text{ is prime} : \exists n\in N, p|n \}$ be the set of all prime factors of the solutions.
It is clear that the smallest element of $P$ is 3.
Assume that $P\ne\{3\}$ and let's try to determine the second smallest element $q = \min (P\setminus\{3\})$.
Let $n\in N$ be a multiple of $q$. It is important to notice that $9\not|n$ (otherwise it is easy to get that any power of $3$ divides $n$, a non-sense). Therefore, $n = 3^t n'$ where $t = 0$ or $1$ and $n'$ does not have prime divisors smaller than $q$.
Since $2^{2n}\equiv 1\pmod{q}$, the multiplicative order $r = ord_q(2)$ of 2 modulo $q$ divides $2n$. Moreover, $r$ must be even, since otherwise we would have $2^n \equiv 1\pmod{q}$, a contradiction to the required $2^n \equiv - 1\pmod{q}$.
Since $r < q$ we must have $r = 2$ or $2\cdot 3 = 6$. But the numbers $2^2 - 1 = 3$ and $2^6 - 1 = 63$ deliver only one new prime factor $7$, implying that $q = 7$. However, in this case $r = ord_7(2) = 3$, a contradiction. This contradiction proves that $P = \{3\}$ and thus $N = \{1,3\}$.
This solution was posted and copyrighted by maxal/orl. The original thread for this problem can be found here: [1]

Let $p$ be the smallest prime divisor of $n$. It is easy to check that, $n=3$ is obviously solution. Let $p^{a} || n$ . $p | 2^{2n}-1$ and $p | 2^{p-1} - 1$ (By the fermat's theorem), we obtain that $p=3$. It is also obvious that n is an odd number.
Lemma: For all $n$ positive integers, $2$ is a primitive root modulo $3^{n}$.
$3^{2a} | 2^{2n} - 1$. Using the lemma, we get that $\phi(3^{2a}) | 2n$. Using the power of three, we obtain that $3^{2a-1} | 3^{a}$. This is only possible when $a \geq 2a-1$. So $a=1$. Now $q$ be the second smallest prime divisor of $n$. $(2n,q-1) = 2 , 6$. If this equals to 2, we get $q=3$ which is a contradiction.If $(2n,q-1) = 6$ then $q|63$. We know that $q$ is different from $3$. Hence $q$ must be $7$. But this is impossible since
$2^{n} +1\equiv 2\mod 7$ when $n$ is divisible by $3$. Hence the answer is $n = 3$.
This solution was posted and copyrighted by grumpyorum. The original thread for this problem can be found here: [2]

Trivially $n=1$ is a solution. Now assume $n\neq 1$ and define $\pi(n)$ to be the smallest prime divisor of $n$. Let $\pi(n)=p\neq 2$. Then we have:
Now if $r\neq 2\mid n$ then we can't have $r\mid p-1$ because then $r\le p-1$ contradiction. Therefore $r=2$ and since $n$ is odd $\gcd(2n, p-1)=2$. HenceLet $v_3(n)=k$. By lifting the exponent we must haveLet $n=3n_1$. $n_1=1$ is a solution ($2^3+1=3^2=9$). Assume $n_1\neq 1$ and let $\pi(n_1)=q\neq 3$. By Chinese Remainder Theorem since $q\neq 3$ we have:
However $2^n+1\equiv 8^{n_1}+1\equiv 2\pmod{7}$ contradiction.
The solutions are henceforth $n=1, 3$.
This solution was posted and copyrighted by binomial-theorem. The original thread for this problem can be found here: [3]

et $n>1$ and $p$ be the smallest prime factor of $n$.Then $p|2^{2n}-1,p|2^{p-1}-1 \Rightarrow p|2^{\text{gcd}(2n,p-1)}=2^2-1=3\Rightarrow \boxed{p=3}$.
Thus $n=3^x*y$ for some positive $x$ and $y$.
Also $n^2|2^n+1 \Rightarrow 2x=v_3(n^2) \le v_3(2^n+1)=v_3(3)+v_3(n)=x+1 \Rightarrow \boxed{x=1}$.
Thus $n=3y$ for some positive integer $y$.
Now let $y>1$ and $q$ be the smallest prime divisor of $y$.Then we can deduce that $q|2^{\text{gcd}(2n,q-1)}-1=2^{2\text{gcd}(n,q-1)}-1=2^6-1=63=7*3^2 \Rightarrow \boxed{q=7}$(Note that $\text{gcd}(n,q-1)$ is $1$ or $3$).
But this gives $7|2^n+1$ which is false as $2^n+1$ leaves remainders $1,2,4$ modulo $7$.
Thus $y=1$ and $\boxed{n=3}$ if $n>1$.
Thus the solutions are $\boxed{n=1}$ and $\boxed{n=3}$
This solution was posted and copyrighted by sayantanchakraborty. The original thread for this problem can be found here: [4]
Many more solutions can be found in the thread in Contest Collections.


Let $S = \{1,2,3,\cdots ,280\}$. Find the smallest integer $n$ such that each $n$-element subset of $S$ contains five numbers which are pairwise relatively prime.

Let's look at this another way. We aim to find the least number of integers we can remove from S to leave a set which does not contain 5 pairwise coprime integers.
Let $S_p = \{p^k | p^k \leq 280\}$
And more generally $S_{p_1,p_2,...,p_l} = \{n = p_1^{q_1}p_2^{q_2}....p_l^{q_l}| n \leq 280 \}$
And finally $S_1 = \{1\}$
It is clear that these sets taken over all collections of primes and 1 forms a partition of S.
If there are representatives from five sets $S_p$, there are five mutually coprime integers, so all but four of the sets $S_p$ must be completely removed. It is clear that (with the exception of 1, which can clearly be removed first) if $p>q, |S_p| \leq |S_q|$ so it is worth removing $S_p$ before removing $S_q$ (if we are striving for a minimum).
Furthermore, if we remove two sets $S_p, S_q$, we must also (to stop there from being 5 mutually coprime integers) remove $S_{p,q}$ and so on, and these have similar ordering relations.
So once we have removed everything we need to keeping sizes of sets removed to a minimum, we are left with $S_2,S_3,S_5,S_7$ and the multi-index sets not coprime to all of these.
In other words, we have the set of multiples of 2,3,5 and 7, which can be calculated (by lots of inclusion-exclusion) to have cardinality $216$. Therefore, a subset of $S$ of size $217$ must contain 5 coprime integers.
This solution was posted and copyrighted by Ilthigore. The original thread for this problem can be found here: [1]
Firstly note that $\lfloor\sqrt{280}\rfloor = 16$. Therefore, we only need to check primes $p\le 16$ to check if a number from $1$ to $280$ is prime.

I claim the answer is $\boxed{217}$. First, let's look at a subset of $S$ that consists only of multiples of $2,3,5,7$. It is easy to calculate by PIE that the total number of elements in this set isNow, note that no matter what $5$ element subset we consider of this subset we just made, consisting of multiples of $2,3,5,7$, we will always have at least $2$ of those elements that exist in the subset that share a factor greater than $1$ by the Pigeonhole Principle since $\left\lfloor\tfrac{5}{4}\right\rfloor+1=2$. Thus, we know $n>216$ from this. We also know that there are $212$ composite numbers and $4$ prime numbers (obviously just $2,3,5,7$) in this subset of $S$. Now we do casework on the numbers $11,13$ (we don't have to check higher numbers because remember $p\le 16$!) to find the other composite numbers. We find that all of numbers, $121,143,169,187,209,221,247,253$ fills in the rest of the composite numbers from $1$ to $280$. This gives a total of $8$ numbers when counted. So we can now count up the total as $220$ composite numbers and the remaining must be prime so $59$ prime numbers ($1$ not included as prime or composite). Now suppose $X$ is a subset of our set $S$ such that $|X|=217$ ($|X|$ represents its cardinality or the number of elements in the subset $X$). Also suppose that there is no $5$ element subset of $X$ such that all $5$ elements are relatively prime to each other. This means we would have to have at most $4$ prime numbers and at least $213$ composite numbers to make this subset $X$. This means the set $S-X$ (everything that is in $S$ but not in $X$) has at most $220-213=7$ composite numbers. Now consider the following $8$ sets and notice that at least one of these must be a subset of $X$!:And obviously each of these sets has $5$ elements that are all relatively prime numbers, as desired.
This solution was posted and copyrighted by Wave-Particle. The original thread for this problem can be found here: [2]


Consider nine points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment) and each edge is either colored blue or red or left uncolored. Find the smallest value of $n$ such that whenever exactly n edges are colored, the set of colored edges necessarily contains a triangle all of whose edges have the same color.

Solution:
We show that for $n = 32$ we can find a coloring without a monochrome triangle.
Take two squares $R_1R_2R_3R_4$ and $B_1B_2B_3B_4$. Leave the diagonals of each square uncolored, color the remaining edges of $R$ red and the remaining edges of $B$
blue. Color blue all the edges from the ninth point $X$ to the red square, and red
all the edges from $X$ to the blue square. Color $R_iB_j$ red if $i$ and $j$ have the same parity and blue otherwise.
Clearly $X$ is not the vertex of a monochrome square, because if $XY$ and $XZ$ are
the same color then, $YZ$ is either uncolored or the opposite color. There is no triangle within the red square or the blue square, and hence no monochrome triangle. It remains to consider triangles of the form $R_iR_jB_k$ and $B_iB_jR_k.$ But if $i$ and $j$ have the same parity, then $R_iR_j$ is uncolored (and similarly $B_iB_j$), whereas if
they have opposite parity, then $R_iB_k$ and $R_jB_k$ have opposite colors (and similarly $B_iR_k$ and $B_jR_k$).
It remains to show that for $n = 33$ we can always find a monochrome triangle.
There are three uncolored edges. Take a point on each of the uncolored edges.
The edges between the remaining $6$ points must all be colored. Take one of
these, $X.$ At least $3$ of the $5$ edges to $X$, say $XA$, $XB$, $XC$ must be the same color
(say red). If $AB$ is also red, then $XAB$ is monochrome. Similarly, for $BC$ and $CA.$
But if $AB$, $BC$ and $CA$ are all blue, then $ABC$ is monochrome.


On an infinite chessboard, a game is played as follows. At the start, $n^2$ pieces are arranged on the chessboard in an $n$ by $n$ block of adjoining squares, one piece in each square. A move in the game is a jump in a horizontal or vertical direction over an adjacent occupied square to an unoccupied square immediately beyond. The piece which has been jumped over is removed. Find those values of $n$ for which the game can end with only one piece remaining on the board.

Solution:
The solution as described in the video is that all values of $n$ are valid except when $n$ is divisible by 3.  He describes why those values are not valid using modular math and why the other ones are by creating an algorithmic pattern of reducing the matrices by taking out pieces in subarrays of 3 by m.
In the video he made a mistake when reducing the 5x5 to a 2x2 as someone pointed out in the comments on the video.  I made sure my animated gif for when $n=5$ was correct.
I was going to write a solution to this but since Mr. Math in the video does a really beautiful solution of it I decided to write a code that will output these animated gifs with the cases for n=1 through 10 showing them whether they're valid or not depending on how many pieces are left.  I used the algorithm described by Mr. Math in the video along and some recursive functions and some visualization functions.  My code can generate the animation for any value of $n$ as long as there is enough memory for the file size and to handle the recursive function.  I had to reduce the animated gif file sizes significantly.  The original animated give for when $n=13$ was 30MB. I wrote the code in Matlab.
I don't know how to upload it to this page.  If anyone wants the code, you can send me an email and I can gladly share it.  Enjoy the animated gifs.

~ 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.