Given a finite set of points in the plane, each with integer coordinates, is it always possible to color the points red or white so that for any straight line $L$ parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on $L$ is not greater than $1$?

I'll use a well-known result: if a connected graph has $2k>0$ vertices of odd degree, then its edge set can be partitioned into $k$ paths, and if all its vertices have even degree, then it has an eulerian circuit.
We have a bipartite graph with bipartition $A,B$ here, constructed as follows: the horizontal lines which contain points from our set represent the vertices in $A$, the vertical lines represent the vertices in $B$, and the point represent the edges between the vertices corresponding to the rows and columns which contain them. What we must prove is that we can color the edge set of a bipartite graph in two colors s.t. the difference between the number of red edges and white edges adjacent to each vertex is $\le 1$ in absolute value. It suffices to prove this for connected bipartite graphs, since then we can apply the result to each connected component of the graph.
If all the vertices have even degree, then we can find an eulerian circuit. The graph is bipartite, so this circuit has an even number of edges, and we can thus color the edges in the circuit alternatively red and white so that two edges which are consecutive in the circuit have different colors. It's clear that this coloration satisfies the requirements.
If, on the other hand, the graph has $2k>0$ vertices with odd degre, then we partition the edge set into $k$ paths, and in each path we color the edges alternatively red and white. Again, it's easy to verify the required properties of the coloration.
This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]

We induct on number of points. The small cases are easily checked. Let there exist such a function for $n$ points. We will show there is a function for $n+1$ points.
If there exists a line $\ell$, parallel to any of the coordinate axes (from the next time, any line will be parallel to either of the coordinate axes, unless otherwise mentioned ), containing odd number of points, then choose a point $P_x \in \ell$, and consider $S \setminus P_x$. By inductive hypothesis there exists such a function $f : P \to \left \{ -1, 1 \right \}$ for $S \setminus P_x$. Since $P \in \ell, P \ne P_x$ contain even number of points, we have $\sum_{\substack{P \in \ell  \\ P \ne P_x}} f(P) = 0$. Let $\ell '$ be the line $\perp$ to $\ell$, passing through $P_x$. Let $\sum _{\substack{P \in \ell ' \\  P \ne P_x}} f(P) = t$, where $t \in \left \{ -1,0 , 1 \right \}$. Now for $S$ (with $n+1$ points) define $g$ as $g(P) = f(P)$ for $P \in S \setminus P_x$, and $g(P_x) = -t$, if $t \ne 0$, or $1$ or $-1$ if $t=0$. It indeed works as such a function for $n+1$ points.
If the above is not the case, i.e. if all the lines contain even number (greater than zero) of points, pick up an arbitrary point $P_y \in S$. Let $\ell$ and $\ell '$ be the two lines containing $P_y$. Also $\ell \perp \ell '$. Consider $S \setminus P_y$, for this set , by the inductive hypothesis, there exists such a function $f$. In $S \setminus P_y$ , $\ell$ and $\ell '$ contains odd number of points and if $L$ is a line different from $\ell$ and $\ell '$, then it contains even number of points. So $\sum_{\substack{P \in L \\ L \ne \ell, \ell '}} f(P) = 0$. Therefore it is seen that
$\sum_{\substack{P \in \ell \\ P \ne P_y}} f(P)$ $=  - \sum _{\substack{P \in S \\ P \notin \ell , \ell ' }} f(P)$ $= \sum_{\substack{P \in \ell ' \\ P \ne P_y}}  f(P) = t$.
Since $\ell \setminus P_y$ contain odd number of points, $t$ is either $-1$ or $1$. Wlog $t = 1$. Now for $S$, (containing $n+1$ points) define $g$ as : $g(P) = f(P)$, for $P \in S \setminus P_y$ and $g(P_y) = -t$.
So the induction is complete, and the statement is established.
This solution was posted and copyrighted by Learner94. The original thread for this problem can be found here: [2]


Let $n$ be an integer greater than or equal to 2.  Prove that if $k^2 + k + n$ is prime for all integers $k$ such that $0 \leq k \leq \sqrt{n/3}$, then $k^2 + k + n$ is prime for all integers $k$ such that $0 \leq k \leq n - 2$.

Solution:
First observe that if $m$ is relatively prime to $b+1$, $b+2$, $\cdots$, $2b$, then $m$ is relatively prime to any number less than $2b$. Since if $c\leq b$, then we can choose some $i$ to make $2^ic$ lies in range $b+1,b+2,\cdots,2b$, so $2^ic$ is relatively prime to $m$. Hence $c$ is also. If we also have $(2b+1)^2>m$, then we can conclude that $m$ is a prime. Since there must be a factor of $m$ less than $\sqrt{m}$.
Let $n=3r^2+h$ where $0\leq h<6r+3$, so $r$ is the greatest integer less than or equal to $\sqrt{n/3}$.(to see this, just let $r=\lfloor\sqrt{n/3}\rfloor$, then we can write $n=3(r+\epsilon)^2(0\leq\epsilon< 1)$, so $h=6r\epsilon+3\epsilon^2\leq 6r+3$).
Assume that $n+k(k+1)$ is prime for $k=1,2,3\cdots,r$. We show that $N=n+(r+s)(r+s+1)$ is prime for $s=0,1,2,\cdots,n-r-2$. By our observation above, it is sufficient to show that $(2s+2r+t)^2>N$ and $N$ is relatively prime to all of $r+s+1,r+s+2,\cdots,2r+2s$. We have $(2r+2s+1)^2=4r^2+4s^2+8rs+4r+4s+1$. Since $s,t\ge1$, we have $4s+1>s+2$, $4s^2>s^2$ and $6rs>3r$. Hence
Now if $N$ has a factor which divides $2r-i$ in the range $-2s$ to $r-s-1$, then so does $N-(i+2s+1)(2r-i)=n+(r-i-s-1)(r-i-s)$ which have the form $n+s'(s'+1)$ with $s'$ in range $0$ to $r$.


Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^{2} + b^{2}$. Show that
$\frac {a^{2} + b^{2}}{ab + 1}$
is the square of an integer.

Choose integers $a,b,k$ such that $a^2+b^2=k(ab+1)$
Now, for fixed $k$, out of all pairs $(a,b)$ choose the one with the lowest value of $\min(a,b)$. Label $b'=\min(a,b), a'=\max(a,b)$. Thus, $a'^2-kb'a'+b'^2-k=0$ is a quadratic in $a'$. Should there be another root, $c'$, the root would satisfy: $b'c'\leq a'c'=b'^2-k<b'^2\implies c'<b'$
Thus, $c'$ isn't a positive integer (if it were, it would contradict the minimality condition). But $c'=kb'-a'$, so $c'$ is an integer; hence, $c'\leq 0$. In addition, $(a'+1)(c'+1)=a'c'+a'+c'+1=b'^2-k+b'k+1=b'^2+(b'-1)k+1\geq 1$ so that $c'>-1$. We conclude that $c'=0$ so that $b'^2=k$.
This construction works whenever there exists a solution $(a,b)$ for a fixed $k$, hence $k$ is always a perfect square.

We proceed by way of contradiction.
WLOG, let $a\geq{b}$ and fix $c$ to be the nonsquare positive integer such that such that $\frac{a^2+b^2}{ab+1}=c,$ or $a^2+b^2=c(ab+1).$ Choose a pair $(a, b)$ out of all valid pairs such that $a+b$ is minimized. Expanding and rearranging,

This quadratic has two roots, $r_1$ and $r_2$, such that

WLOG, let $r_1=a$. By Vieta's,
$\textbf{(1) } r_2=bc-a,$ and
$\textbf{(2) } r_2=\frac{b^2-c}{a}.$
From $\textbf{(1)}$, $r_2$ is an integer, because both $b$ and $c$ are integers.
From $\textbf{(2)},$ $r_2$ is nonzero since $c$ is not square, from our assumption.
We can plug in $r_2$ for $a$ in the original expression, because $P(r_2)=P(a)=0,$ yielding $c=\frac{r^2_2+b^2}{r_2b+1}$. If $c>0,$ then $r_2b+1>0,$ and $r_2b+1\neq{0},$ and because $b>0, r_2$ is a positive integer.
We construct the following inequalities: $r_2=\frac{b^2-c}{a}<a,$ since $c$ is positive. Adding $b$, $r_2+b<a+b,$ contradicting the minimality of $a+b.$
-Benedict T (countmath1)


A permutation $\{x_1, \ldots, x_{2n}\}$ of the set$\{1,2, \ldots, 2n\}$ where $\mbox{n}$ is a positive integer, is said to have property $T$ if $|x_i - x_{i + 1}| = n$ for at least one $i \in \{1,2, \ldots, 2n - 1\}$. Show that, for each $\mbox{n}$, there are more permutations with property $T$ than without.

Let $F$ be the number of permutations with property $T$, and $F_i$ be the set of permutations $\{ x_1, x_2, \ldots, x_{2n} \}$  such that $|x_i - x_{i+1}| = n$. By the inclusion-exclusion principle,

for some $E \geq 0$. Let’s calculate the first two sums on the far right.
For any $i$, $|F_i| = 2n \cdot (2n - 2)!$, since there are are $2n$ choices for $x_i$, which fixes $x_{i+1}$, and $(2n-2)!$ choices for the remaining elements. Thus

Now let’s compute $|F_i \cap F_j|$, where $1 \leq i < j \leq 2n - 1$. It’s not possible that $|x_i - x_{i+1}| = |x_{i+1}- x_{i+2}| = n$, since that would imply $x_i = x_{i+2}$. So $|F_i \cap F_j| = 0$ if $j = i + 1$. If $j > i + 1$, then $|F_i \cap F_j| = 2n \cdot (2n - 2) \cdot (2n - 4)!$ (there are $2n$ choices for $x_i, x_{i+1}$, $2n - 2$ for $x_j, x_{j+1}$, and $(2n - 4)!$ for everything else.)
How often is $|F_i \cap F_j| \neq 0$? We know $i$ is at least $1$ and at most $2n - 3$, and for any value of $i$, there are $2n - 2 - i$ possible values for $j$. So the number of non-empty intersections $|F_i \cap F_j|$ is

Now we can compute
Let’s plug this into $(1)$:

So more than half of the $(2n)!$ permutations of $\{1, \ldots, 2n\}$ have property $T$.

Let $A_n$ be the set of permutations of $1,2, \ldots,2n$ not having property $T$ and let $B_n$ be the set of permutations of $1,2, \ldots,2n$ having exactly one value of $k\geq2$ such that $|x_k-x_{k+1}|=n$. We will prove a 1-1 correspondence between $A_n$ and $B_n$.
Consider any permutation in $A_n$. Now take $x_1$ and for the unique value $k$ such that $|x_1-x_k|=n$, put $x_1$ in the spot right before $x_k$. This will give us a permutation that belongs in the set $B_n$. Now consider a permutation in $B_n$. If you take $x_k$ and put it as the first term, you'll get a permutation that belongs in set $A_n$.Therefore, we've proved a 1-1 correspondence between the two.
Because $|A_n|$ is clearly less than the number of permutations with property $T$, we finally have that the number of permutations with property $T$ is greater than the number of permutations without $T$.
~BennettHuang


Prove that there exists a convex $1990\text{-gon}$ with the following two properties:
(a) All angles are equal.
(b) The lengths of the $1990$ sides are the numbers $1^2, 2^2,3^2,\dots, 1990^2$ in some order.

Let $\{a_1,a_2,\cdots,a_{1990}\}=\{1,2,\cdots,1990\}$ and $\theta=\frac{\pi}{995}$.
Then the problem is equivalent to that there exists a way to assign $a_1,a_2,\cdots,a_{1990}$ such that $\sum_{i=1}^{1990}a_i^2e^{i\theta}=0$.
Note that $e^{i\theta}+e^{i\theta+\pi}=0$, then if we can find a sequence $\{b_n\}$ such that $b_i=a_p^2-a_q^2, (p\not=q)$ and $\sum_{i=1}^{995}b_ie^{i\theta}=0$, the problem will be solved.
Note that $2^2-1^2=3,4^2-3^2=7,\cdots,1990^2-1989^2=3979$, then $3,7,\cdots,3979$ can be written as a sum from two elements in sets $A=\{3,3+4\times199,3+4\times398,3+4\times597,3+4\times796\}$ and $B=\{0,4\times1,4\times2,\cdots,4\times 198\}$.
If we assign the elements in $A$ in the way that $l_{a1}=3,l_{a2}=3+4\times199,l_{a3}=3+4\times398,l_{a4}=3+4\times597,l_{a5}=3+4\times796$, then clearly $\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}=0, (j\in\{1,2,3,4,5\})$
Similarly, we could assign elements in $B$ in that way ($l_{b1}=0,l_{b2}=4,\cdots$) to $e^{i\frac{2\pi}{5}}$.
Then we make $b_i$ according to the previous steps. Let $i\equiv j$ (mod 5), $i\equiv k$ (mod 199), and $b_i=l_{aj}+l_{bk}$, then each $b_i$ will be some $a_ p^2-a_q^2$ and $\sum_{i=1}^{995}b_ie^{i\theta}=\sum_{j=1}^5\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}+\sum_{k=1}^{199}\sum_{i=1}^{5}l_{bk}e^{i\frac{2\pi}{5}}=0$
And we are done.
This solution was posted and copyrighted by YCHU. The original thread for this problem can be found here: [1]

Throughout this solution, $\omega$ denotes a primitive $995$th root of unity.
We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent tobeing true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$.
Define $z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$. Then notice that and so summing yields the desired conclusion, as the left-hand side becomesand the right-hand side is the desired expression.
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]


An infinite sequence $x_0, x_1, x_2,\ldots$ of real numbers is said to be bounded if there is a constant $C$ such that $|x_i| \leq  C$ for every $i \geq 0$.
Given any real number $a > 1$, construct a bounded infinite sequence $x_0,x_1,x_2,\ldots$ such that $|x_i-x_j|\cdot |i-j|^a\geq 1$ for every pair of distinct nonnegative integers $i,j$.

Since $a>1$, the series $\sum_{k=1}^\infty\frac{1}{k^a}$ is convergent; let $L$ be the sum of this convergent series. Let $I\subset \mathbb{R}$ be the interval $[-L,L]$ (or any bounded subset of measure $\geq 2L$).
Suppose that we have chosen points $x_0,x_1,\ldots,x_{m-1}$ satisfying
$\qquad(\ast)\quad |x_i-x_j|\cdot |i-j|^a\geq 1$
for all distinct $i,j<m$. We show that we can choose $x_m\in I$ such that $(\ast)$ holds for all distinct $i,j\leq m$. The only new cases are when one number (WLOG $i$) is equal to $m$, so we must guarantee that $|x_m-x_j|\cdot |m-j|^a\geq 1$ for all $0\leq j<m$.
Let $U_j$ be the interval $(x_j-\frac{1}{|m-j|^a},x_j+\frac{1}{|m-j|^a})$, of length $\frac{2}{|m-j|^a}$. The points that are valid choices for $x_m$ are precisely the points of $I\setminus(U_0\cup \cdots \cup U_{m-1})$, so we must show that this set is nonempty. The total length $\mu(U_0\cup \cdot\cup U_{m-1})$ is at most the sum of the lengths $\mu(U_0)+\cdots+\mu(U_{m-1})=\frac{2}{m^a}+\frac{2}{(m-1)^a}+\cdots+\frac{2}{2^a}+\frac{2}{1^a}$. This is $2\sum_{k=1}^m\frac{1}{k^a}<2\sum_{k=1}^\infty \frac{1}{k^a}=2L$.
Therefore the total measure of $U_0\cup \cdots \cup U_{m-1}$ is $<2L=\mu(I)$, so $I\setminus(U_0\cup \cdots \cup U_{m-1})$ has positive measure and thus is nonempty. Choosing any $x_m\in I\setminus(U_0\cup \cdots \cup U_{m-1})$ and continuing by induction constructs the desired sequence.

The argument above would not work for $a=1$, since ${\textstyle\sum \frac{1}{k^a}}$ only converges for $a>1$. But Osmun Nal argues in this video that $x_k=C\cdot\left(k\sqrt{2}-\lfloor k\sqrt{2}\rfloor\right), C=2\sqrt2+1$ satisfies the stronger inequality $|x_i-x_j|\cdot |i-j|\geq 1$ for all distinct $i,j$; in other words, this sequence simultaneously solves the problem for all $a\geq 1$ simultaneously.

Let $n = \max \left( \left\lceil \dfrac{1}{a - 1} \right\rceil, 3 \right)$, then we have $\dfrac{\ln \left( n + 1 \right)}{\ln n} = \dfrac{\ln n + \ln \left( 1 + 1/n \right)}{\ln n} < \dfrac{\ln n + 1/n}{\ln n} = 1 + \dfrac{1}{n \ln n} < 1 + \dfrac{1}{n} \le a$
Thus $n^a > n + 1$
For any non-negative integer $i$, represent it as base-$n$ positional notation:
$i = i_m n^m + i_{m-1} n^{m-1} + \cdots + i_2 n^2 + i_1 n + i_0$, where $i_m, i_{m-1}, \cdots, i_2, i_1, i_0 \in \left\{ 0, 1, 2, \cdots, n - 1 \right\}$
We directly construct $x_i = n \left( i_0 + i_1 \left( n + 1 \right)^{-1} + i_2 \left( n + 1 \right)^{-2} + \cdots + i_m \left( n + 1 \right)^{-m} \right)$
We allow leading zeros for representing $i$ due to the leading zeros do not effect the value of $x_i$
Then $0 \le x_i \le n \left( \left( n - 1 \right) + \left( n - 1 \right) \left( n + 1 \right)^{-1} + \cdots + \left( n - 1 \right) \left( n + 1 \right)^{-m} \right) < n \left( n - 1 \right) \dfrac{1}{1 - \left( n + 1 \right)^{-1}} = n^2 - 1$
Thus $\left| x_i \right| < n^2 - 1$, which is bounded
For any two distinct non-negative integers $i \neq j$, represent them as base-$n$ positional notation:
$i = i_m n^m + i_{m-1} n^{m-1} + \cdots + i_2 n^2 + i_1 n + i_0$
$j = j_m n^m + j_{m-1} n^{m-1} + \cdots + j_2 n^2 + j_1 n + j_0$
where $i_m, i_{m-1}, \cdots, i_2, i_1, i_0, j_m, j_{m-1}, \cdots, j_2, j_1, j_0 \in \left\{ 0, 1, 2, \cdots, n - 1 \right\}$
Let $\ell$ be the minimum number such that $i_\ell \neq j_\ell$, that is, $i_0 = j_0, i_1 = j_1, \cdots, i_{\ell - 1} = j_{\ell - 1}, i_\ell \neq j_\ell$
Then we have $\left| i - j \right| \ge n^\ell$
Without loss of generality, assume that $i_\ell > j_\ell$, then we have $i_\ell - j_\ell \ge 1, j_{\ell + 1} - i_{\ell + 1}, j_{\ell + 2} - i_{\ell + 2}, \cdots, j_m - i_m \le n - 1$
$\left| x_i - x_j \right| = n \left( \left( i_\ell - j_\ell \right) \left( n + 1 \right)^{-\ell} - \left( j_{\ell+1} - i_{\ell+1} \right) \left( n + 1 \right)^{-\ell-1} - \left( j_{\ell+2} - i_{\ell+2} \right) \left( n + 1 \right)^{-\ell-2} - \cdots - \left( j_m - i_m \right) \left( n + 1 \right)^{-m} \right) \\~\\ \ge n \left( n + 1 \right)^{-\ell} - n \left( n - 1 \right) \left( \left( n + 1 \right)^{-\ell-1} + \left( n + 1 \right)^{-\ell-2} + \cdots + \left( n + 1 \right)^{-m} \right) \\~\\ > n \left( n + 1 \right)^{-\ell} - n \left( n - 1 \right) \left( n + 1 \right)^{-\ell-1} \dfrac{1}{1 - \left( n + 1 \right)^{-1}} = \left( n + 1 \right)^{-\ell}$
Thus we have $\left| x_i - x_j \right| \left| i - j \right|^a > \left( n + 1 \right)^{-\ell} n^{a \ell} > \left( n + 1 \right)^{-\ell} \left( n + 1 \right)^\ell = 1$
~Joseph Tsai, mgtsai@gmail.com


For each positive integer $n$, $S(n)$ is defined to be the greatest integer such that, for every positive integer $k \le S(n)$, $n^{2}$ can be written as the sum of $k$ positive squares.
(a) Prove that $S(n) \le n^{2}-14$ for each $n \ge 4$.
(b) Find an integer $n$ such that $S(n)=n^{2}-14$.
(c) Prove that there are infinitely many integers $n$ such that $S(n)=n^{2}-14$.

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


There are $n$ lamps $L_0, \ldots , L_{n-1}$ in a circle ($n > 1$), where we denote $L_{n+k} = L_k$. (A lamp at all times is either on or off.) Perform steps $s_0, s_1, \ldots$ as follows: at step $s_i$, if $L_{i-1}$ is lit, switch $L_i$ from on to off or vice versa, otherwise do nothing. Initially all lamps are on. Show that:
(a) There is a positive integer $M(n)$ such that after $M(n)$ steps all the lamps are on again;
(b) If $n = 2^k$, we can take $M(n) = n^2 - 1$;
(c) If $n = 2^k + 1$, we can take $M(n) = n^2 - n + 1.$


Show that there exists a set $A$ of positive integers with the following property: For
any infinite set $S$ of primes there exist two positive integers $m \in A$ and $n \not\in A$ each of which is a product of $k$ distinct elements of $S$ for some $k \ge 2$.


Let $p, q, n$ be three positive integers with $p+q<n$. Let $(x_0,x_1,\cdots ,x_n)$ be an $(n+1)$-tuple of integers satisfying the following conditions:
(i) $x_0=x_n=0$;
(ii) For each $i$ with $1 \le i \le n$, either $x_i-x_{i-1}=p$ or $x_i-x_{i-1}=-q$.
Show that there exists indices $i<j$ with $(i,j) \ne (0,n)$, such that $x_i=x_j$.


For each positive integer $n$, let $f(n)$ denote the number of ways of representing $n$ as a sum of powers of $2$ with nonnegative integer exponents.  Representations which differ only in the ordering of their summands are considered to be the same.  For instance, $f(4)=4$, because the number 4 can be represented in the following four ways:
$4;2+2;2+1+1;1+1+1+1$
Prove that, for any integer $n \ge 3$,
$2^{n^{2}/4}<f(2^{n})<2^{n^{2}/2}$.


Determine the least possible value of $f(1998),$ where $f:\Bbb{N}\to \Bbb{N}$ is a function such that for all $m,n\in {\Bbb N}$,


Determine all functions $f:\Bbb{R}\to \Bbb{R}$ such that

for all real numbers $x,y$.


Let $\overline{AH_1}$, $\overline{BH_2}$, and $\overline{CH_3}$ be the altitudes of an acute triangle $ABC$.  The incircle $\omega$ of triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ at $T_1$, $T_2$, and $T_3$, respectively.  Consider the reflections of the lines $H_1H_2$, $H_2H_3$, and $H_3H_1$ with respect to the lines $T_1T_2$, $T_2T_3$, and $T_3T_1$.  Prove that these images form a triangle whose vertices line on $\omega$.


$K > L > M > N$ are positive integers such that $KM + LN = (K + L - M + N)(-K + L + M + N)$. Prove that $KL + MN$ is not prime.

Solution:
First, $(KL+MN)-(KM+LN)=(K-N)(L-M)>0$ as $K>N$ and $L>M$.  Thus, $KL+MN>KM+LN$.
Similarly, $(KM+LN)-(KN+LM)=(K-L)(M-N)>0$ since $K>L$ and $M>N$.  Thus, $KM+LN>KN+LM$.
Putting the two together, we have
Now, we have:

So, we have:

Thus, it follows that
Now, since $KL+MN>KM+LN$ if $KL+MN$ is prime, then there are no common factors between the two.  So, in order to have  we would have to have  This is impossible as $KM+LN>KN+LM$.  Thus, $KL+MN$ must be composite.


Let $n \ge 3$ be a positive integer.  Let $C_1,C_2,...,C_n$ be unit circles in the plane, with centers $O_1,O_2,...,O_n$ respectively.  If no line meets more than two of the circles, prove that


p is a prime number. Prove that for every p there exists a q for every positive integer n, so that $n^p-p$ can't be divided by q.

Solution:
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Let N be $1 + p + p^2 + ... + p^{p-1}$ which equals $\frac{p^p-1}{p-1}$
$N\equiv{p+1}\pmod{p^2}$
Which means there exists q which is a prime factor of n that doesn't satisfy $q\equiv{1}\pmod{p^2}$.
\\unfinished


We call a positive integer alternating if every two consecutive digits in its decimal
representation are of different parity. Find all positive integers $n$ which have an
alternating multiple.

Solution:
We claim that all positive integers $n$ except multiples of $20$ have a multiple that is alternating. If $n$ is a multiple of $20$, then the units digit is $0$ and the tens digit is a multiple of $2$, so both digits are even. Now, we will prove that if $20\nmid n$, then there is a multiple of $n$ that is alternating.
We claim that there exists a $2k$ digit alternating integer that is a multiple of $2^{2k+1}$ and there exists an alternating integer with at most $k$ digits that is a multiple of $2\cdot5^k$. We will prove this by induction.
Base Case: $k=1$
We can let the number be $16$.
Inductive Step:
Let $a$ be a $2k-2$ digit alternating number such that $2^{2k-1}\mid a$. Since $a$ is even, this means that the first digit of $a$ is odd. Let $b\cdot2^{2k-1}\equiv a\pmod{2^{2k+1}}$, where $b\in\{0,1,2,3\}$. Let $x=16-2b$. We see that $x$ is alternating, so the number $x\cdot10^{2k-2}+a$ is also alternating. We also have $x\equiv-2b\pmod8$. Since $a\equiv b\cdot2^{2k-1}\pmod{2^{2k+1}}$, this means that we have
\begin{align*}
x\cdot10^{2k-2}+a&\equiv x\cdot10^{2k-2}+b\cdot2^{2k-1}\pmod{2^{2k+1}}\\
&\equiv2^{2k-2}\left(x\cdot5^{2k-2}+2b\right)\pmod{2^{2k+1}}\\
&\equiv2^{2k-2}\left(-2b\cdot5^{2k-2}+2b\right)\pmod{2^{2k+1}}\\
&\equiv -b\cdot2^{2k-1}\left(5^{2k-2}-1\right)\pmod{2^{2k+1}}.\end{align*}
Since $4\mid5^{2k-2}-1$, this means that $x\cdot10^{2k-2}+a\equiv0\pmod{2^{2k+1}}$, so $x\cdot10^{2k-2}+a$ is a $2k$ digit alternating number that is a multiple of $2^{2k+1}$.
Now, we will prove that there exists an alternating integer with at most $k$ digits that is a multiple of $2\cdot5^k$.
Base Case: $k=1$
We can let the number be $0$.
Inductive Step:
Let $a$ be a $k-1$ digit alternating number such that $2\cdot5^{k-1}\mid a$. Let $S$ be the set of digits such that $x\cdot10^{k-1}+a$ is alternating. We see that either $S=\{0,2,4,6,8\}$ or $S=\{1,3,5,7,9\}$. In each possible set, each $\mod5$ residue appears exactly once. Let $a=2\cdot5^{k-1}\cdot b$. Then, $x\cdot10^{k-1}+a=2\cdot5^{k-1}\left(x\cdot2^{k-2}+b\right)$. Therefore, there exists an $x$ such that $2\cdot5^k\mid x\cdot10^{k-1}+a$, so there exists an alternating integer with at most $k$ digits that is a multiple of $2\cdot5^k$. If $k$ is even, then $0\not\in S$, so it has exactly $k$ digits.
Now, let $n=2^a\cdot5^b\cdot m$. Then, if $20\nmid n$, then $a<2$ or $b=0$. If $b=0$, then there exists a $2a$ digit alternating integer $x$ that is a multiple of $2^{2a+1}$. Consider the sequence $a_i=x\cdot\frac{10^{2ai}-1}{10^{2a}-1}$. Then, the decimal representation of $a_i$ is the result when $x$ is written $i$ times. For any prime $p$ that divides $n$, if $p\mid10^{2a}-1$, then $\nu_p\left(10^{2ai}-1\right)=\nu_p\left(10^{2a}-1\right)+\nu_p(i)$, so $\nu_p(a_i)=\nu_p(x)+\nu_p(i)\geq\nu_p(i)$. Therefore, there exists an $i$ such that $p^{\nu_p(n)}\mid a_{ci}$ for all $c$. If $p\nmid10^{2a}-1$, then if $i=\phi(n)$, then $p^{\nu_p(n)}\mid10^{2ai}-1$ if $p\neq2,5$. Since $5\nmid n$, this means that we only need to make sure $2^a\mid x\cdot\frac{10^{2ai}-1}{10^{2a}-1}$. Since $2^{2a+1}\mid x$, this means that this is true, so if $5\nmid n$, then there exists a multiple of $n$ that is alternating.
If $a<2$, then there exists an alternating integer $x$ with exactly $2b$ digits that is a multiple of $2\cdot5^{2b}$. Then, the first digit of $x$ is odd. Let $a_i=x\cdot\frac{10^{2bi}-1}{10^{2b}-1}$. Then, if $p\mid10^{2b}-1$, then $\nu_p(a_i)\geq\nu_p(i)$, so there exists an $i$ such that $p^{\nu_p(n)}\mid a_i$. If $p\nmid10^{2a}-1$, then we can let $i=\phi(n)$, so $p^{\nu_p(n)}\mid a_i$. If $p=5$, then $5^b\mid5^{2b}\mid x$, and if $p=2$, then $a=0,1$, so $p^a\mid x$. Therefore, $n\mid a_i$ for some $i$.
Therefore, all positive integers except multiples of $20$ have an alternating multiple.


In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5  of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 problems each.


Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.


Let $n$ be a positive integer. Consider

as a set of $(n+1)^3-1$ points in three-dimensional space.
Determine the smallest possible number of planes, the union of which contain $S$ but does not include $(0,0,0)$.

Solution:
We will prove the result using the following Lemma, which has an easy proof by induction.
Lemma Let $S_1 = \{0, 1, \ldots, n_1\}$, $S_2 = \{0, 1, \ldots, n_2\}$ and $S_3 = \{0, 1, \ldots, n_3\}$. If $P$ is a polynomial in $\mathbb{R}[x, y, z]$ that vanishes on all points of the grid $S = S_1 \times S_2 \times S_3$ except at the origin, then
Proof.  We will prove this by induction on $n = n_1 + n_2 + n_3$. If $n = 0$, then the result follows trivially. Say $n > 0$.
WLOG, we can assume that $n_1 > 0$.
By polynomial division over $\mathbb{R}[y, z][x]$ we can write
Since $x-n_1$ is a monomial, the remainder $R$ must be a constant in $\mathbb{R}[y, z]$, i.e., $R$ is a polynomial in two variables $y$, $z$.
Pick an element of $S$ of the form $(n_1, y, z)$ and substitute it in the equation. Since $P$ vanishes on all such points, we get that $R(y, z) = 0$ for all $(y, z) \in S_2 \times S_3$.
Let $S_1' = \{0, 1, \ldots, n_1 - 1\}$ and $S' = S_1' \times S_2 \times S_3$.
For every point $(x, y, z)$ in $S'$ we have  where $x - n_1 \neq 0$.
Therefore, the polynomial $Q$ vanishes on all points of $S'$ except the origin. By induction hypothesis, we must have $\deg Q \geq n_1 - 1 + n_2 + n_3$. But, $\deg P \geq \deg Q + 1$ and hence we have $\deg P \geq n_1 + n_2 + n_3$.
Now, to solve the problem let $H_1, \ldots, H_m$ be $m$ planes that cover all points of $S$ except the origin. Since these planes don't pass through origin, each $H_i$ can be written as $a_i x + b_i y + c_i z = 1$. Define $P$ to be the polynomial $\prod (a_ix + b_iy + c_iz - 1)$. Then $P$ vanishes at all points of $S$ except at the origin, and hence $\deg P = m \geq 3n$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


Let $ABCD$ be a convex quadrilateral with $BA \ne BC$. Denote the incircles of triangles $ABC$ and $ADC$ by $\omega _1$ and $\omega _2$ respectively. Suppose that there exists a circle $\omega$ tangent to ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$, which is also tangent to the lines $AD$ and $CD$. Prove that the common external tangents to $\omega _1$ and $\omega _2$ intersect on $\omega$

Solution:
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Here are some hints:
Let B be the top vertex of triangle ABC, O and K are the centers of the incircles of triangles ABC and ADC with radii R and r, respectively. S is the center of the circumcircle tangential to the extensions of AB, AC and DC. And let
E be the foot of the projection of O to AB.
U be the foot of the projection of O to AC.
V be the foot of the projection of K to AC.
M be the foot of the projection of S to DC.
L be the foot of the projection of K to DC.
L be the intercept of DC and AB.
We have
/_EOB = /_LSC
/_AOU = /_SOC
/_ASO = /_KSC
/_ASK = /_OSC
/_LSB = /_KCO
UV = BC – AB
AU = VC
OA. AK. cos(/_OAK) = OC. KC. cos(/_OCK)
OK**2 = (R + r)**2 + UV**2
SK**2 = (R' + r)**2 + ML**2
Use sin (90-x) = cos x and  cos(90-x) = sinx
and characteristic of triangle a**2 = b**2 + c**2 - 2.b.c.cosine(angle) to solve.
Vo Duc Dien


Let $a_1,a_2,\ldots,a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s=a_1+a_2+\ldots+a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1,a_2,\ldots,a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.
Author: Dmitry Khramtsov, Russia

Solution:
We will use strong induction on $n$. When $n = 1$, there are no elements in $M$, so the one jump can be made without landing on a point in $M$. When $n = 2$, we consider two cases. If $a_1$ is not in $M$, then the order $a_1, a_2$ will work. If $a_1$ is in $M$, then $a_2$ is not in $M$ and the order $a_2, a_1$ will work. We will assume that the order can be chosen in such a way for all integers $n < k$ for $k \ge 3$.
When $n = k$, we can assume WLOG that $a_1<a_2< \ldots <a_k$. We can also assume that the elements of $M$ are distinct, since if two elements are identical, we can treat them as one and use induction on $n = k-1$. We now consider four cases:
Case 1: $a_k$ is in $M$ but $a_1, a_2, \ldots, a_{k-1}$ are not.
There are at most $k-2$ elements in $M$ greater than $a_k$. Then, from our assumption, there exists an order of jumps starting from $a_k$ of lengths $a_1, a_2, \ldots, a_{k-1}$ where the first jump is $a_i$ such that the grasshopper will not land in any point in $M$. We can then switch the jumps $a_k$ and $a_i$. Since $a_i$ is not in $M$ and no subsequent jumps result in the grasshopper landing in a point in $M$, this sequence is valid.
Case 2: $a_k$ is not in $M$ but at least one of $a_1, a_2, \ldots, a_{k-1}$ is.
There are at most $k-2$ elements in $M$ greater than $a_k$. Then, from our assumption, there exists an order of jumps starting from $a_k$ of lengths $a_1, a_2, \ldots, a_{k-1}$ such that the grasshopper will not land in any point in $M$. Adding $a_k$ at the beginning of this sequence then results in a valid sequence.
Case 3: $a_k$ is in $M$ and at least one of $a_1, a_2, \ldots, a_{k-1}$ is.
Consider the following pairs of distinct integers $(a_1, a_1+a_k), (a_2, a_2+a_k), \ldots, (a_{k-1}, a_{k-1}+a_k)$. Since at most $k-2$ of these points are in $M$, by the pigeonhole principle, there exists an integer $i$ such that $a_i, a_i + a_k$ are both not in $M$. Since there are at most $k-3$ elements greater than $a_k$, at most $k-3$ elements are greater than $a_i + a_k$. Then, from our assumption, there exists an order of jumps starting from $a_i + a_k$ of lengths $a_1, a_2, \ldots, a_{i-1}, a_{i+1}, \ldots, a_{k-1}$ such that the grasshopper will not land in any point in $M$. Adding $a_i, a_k$ at the beginning of this sequence then results in a valid sequence.
Case 4: None of $a_1, a_2, \ldots, a_k$ are in $M$.
Assume that there exists no valid sequence - that any order of jumps will result in the grasshopper landing on at least one element in $M$. We can select any element $m_i$ in $M$ and, from our assumption, construct a sequence beginning at some $a_j$ containing $n-1$ jumps $a_1, a_2, \ldots, a_{j-1}, a_{j+1}, \ldots, a_k$ that will not land on any of the other $n-2$ elements in $M$. Since no valid sequence exists, during this sequence of jumps, the grasshopper must land on $m_i$.
WLOG, we let $m_1 < m_2 < \ldots < m_{k-1}$. We let $i = 1$ and $j = k$ and use the corresponding sequence that only lands on $m_1$ and no other element in $M$. Let the jump immediately after landing on $m_1$ be $a_x$, where $x < n$. We swap jumps $a_x$ and $a_n$. Now, since $a_n > a_x$, all jumps before $a_n$ will land on a integer less than $m_1$. Since $m_1$ is the smallest element in $M$, none of the jumps before $a_n$ will land on a point in $M$. Since no jumps after $a_n$ will land on an element in $M$ by the construction, the grasshopper must then land on an element $m_y > m_1$ in $M$ after making the jump $a_n$. However, this would imply that the original construction would land on another element $m_y$ different from $m_1$, but this is a contradiction. So, a valid sequence must exist.
Since we have an induction on $n$, the statement must be true for all $n$, and we are done.


Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that

Prove there exist positive integers $\ell \leq s$ and $N$, such that

Solution: So for solving This Problem, we need to take a assumption that,


Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \ldots, a_n$ such that


Let $n \ge 3$ be an integer, and consider a circle with $n + 1$ equally spaced points marked on it. Consider all labellings of these points with the numbers $0, 1, ... , n$ such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels $a < b < c < d$ with $a + d = b + c$, the chord joining the points labelled $a$ and $d$ does not intersect the chord joining the points labelled $b$ and $c$.
Let $M$ be the number of beautiful labelings, and let N be the number of ordered pairs $(x, y)$ of positive integers such that $x + y \le n$ and $\gcd(x, y) = 1$. Prove that


A set of lines in the plane is in $\textit{general position}$ if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its $\textit{finite regions}$. Prove that for all sufficiently large $n$, in any set of $n$ lines in general position it is possible to colour at least $\sqrt{n}$ of the lines blue in such a way that none of its finite regions has a completely blue boundary.

Solution:
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Call the $\binom{n}{2}$ intersections, well, points. Then each line will have $n-1$ points. We call 2 points (on a line) neighbors if there are no other points on the line segment joining those 2. Then each finite region has to be a convex polygon whose any pair of neighboring vertices are, well, neighbors. Now start by coloring any line blue, and all points are uncolored (neither red nor blue).
Suppose there is an uncolored line which does not pass through any red points. Color that line blue. Now consider each of the intersections that line makes with the other blue lines. For each intersection point $X$ of the 2 blue lines $l_1,l_2$, color it blue (it was originally uncolored). Now consider the neighbors of $X$ on $l_1,l_2$, call them $A_1,A_2$ on $l_1$ and $B_1,B_2$ on $l_2$ (if there is only 1 neighbor, let the other point be an uncolored dummy). If neither $A_1$ nor $A_2$ is blue, we colour them both red. Else if neither $B_1$ nor $B_2$ is blue, we colour them both red. Otherwise, we colour the uncolored ones (out of the 4 points) red. Anyway, we would have colored at most 2 points red. Once we have done this for all the intersection points, proceed back to the start of this paragraph until no such line exists.
Now to show it works. Suppose there is a finite region polygon $P_1...P_m$ with all blue boundaries (with $P_{m+1}=P_1$). Then all the points must be blue. Consider the first point colored blue, WLOG let it be $P_2$. Suppose the previous line to be colored blue is $P_1P_2$. Then $P_2P_3$ has to be colored blue before that, and $P_3P_4$ has to be uncolored (otherwise $P_3$ will be colored blue first). So $P_3$ is uncolored. Then if $P_1$ is blue, $P_3$ will be colored red by our coloring algorithm. $P_1$ has to be uncolored and by our coloring algorithm at least one of $P_1,P_3$ will be colored red. Note that no blue line will pass through a red point. Thus there are no finite region with blue boundaries.
Now suppose $k$ lines are blue. By our algorithm, each blue point intersection will introduce at most 2 red points. Since we can't color any more lines blue, those red points will cover the remaining lines. Thus $2\binom{k}{2}\geq n-k\implies k\geq \sqrt{n}$ exact.
Note: this works for all ${n}\geq{2}$, provided there is no mistake.


The sequence $a_1,a_2,\dots$ of integers satisfies the conditions:
(i) $1\le a_j\le2015$ for all $j\ge1$,
(ii) $k+a_k\neq \ell+a_\ell$ for all $1\le k<\ell$.
Prove that there exist two positive integers $b$ and $N$ for whichfor all integers $m$ and $n$ such that $n>m\ge N$.
Proposed by Ivan Guo and Ross Atkins, Australia.

Solution:
We can prove the more general statement.
Theorem Let $T$ be a non-negative integer parameter. If given a sequence $a_1,a_2,\dots$ that satisfies the conditions:
(i) $1 \le a_j \le T+1$ for all $j \le 1;$
(ii) $k+a_k \ne \ell+a_\ell$ for all $1 \le k<\ell,$
then there exist two integers $v$ and $N$, $0 \le v \le T$ and $N>0$, for which for all integers $m$ and $n$ such that $n>m\ge N$.
Proof Consider the set of points on a plane: $\{(x,y) \mid x,y \in \mathbb N,\,y \le T+1\}$ (let's call it the base strip). The sequence is represented on it by the subset $\{(i,a_i) \mid i \in \mathbb N\}$ which is painted red. Each line $L_n$ of kind $x+y=n$, where $n \ge 2$ is an integer, will be called a carrier line. Note that the condition (ii) states that each carrier line is occupied by at most one red point. Every such line with exactly one red point on it will be called occupied while the rest carriers will be called free.
First we prove that the set of free carriers is finite. Indeed, all the carriers $L_n$ with $2 \le n \le T+m+1$, where $m \in \mathbb N$, cover the part of the base strip having $x \le m$, so there are at least $m$ red points lying on them. Thus, the number of free carriers among them is at most $T+m-m=T$. Since $m$ is arbitrary, the total number of free cariers doesn't exceed $T$. Take $N$ such that all the carriers $L_n$ with $n \ge N+2$ are occupied.
Next, take any red point and consider the points on the base strip which are lying on the point's carrier below it (i.e. having strictly greater $x$). Call it the point's trace and paint black. Finally, the set $P_x=\{\,y \mid (x,y) \text { is black}\}$ will be called a pattern, the number elements $|P_x|$ in it will be called its volume and the sum $w_x$ of all elements in $P_x$ will be called its weight.
Now let $j>N$. Lets track how $P_j$, its volume and its weight change when $j$ increases by 1. Obviously $P_{j+1}$ is $P_j$ plus added $a_j$ (red point) and shifted down by $1$; a point that goes to $0$ vanishes. This means that the volume doesn't change at all. Indeed, if $1 \notin P_j$, then $a_j=1$ or the carrier $L_{j+1}$ will remain unoccupied, so one point always vanishes going to $0$, and exactly one point is added before the shift (it may be the same point). Let $v$ be this constant volume. As a pattern never contains $T+1$, $v$ ranges from $0$ to $T$. Now it is clearly that $w_{j+1}$ is $w_j$ plus $a_j$ (red point added) and minus $v+1$ (all elements are shifted by $1$), i.e. $a_j-(v+1)=w_{j+1}-w_j$. Thus, when $n>m\ge N$: To finish, it remains to note that the weight ranges from $1+2+\dots+v$ to $(T-v+1)+(T-v+2)+\dots+T$, so its swing is:


There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.
(a) Prove that Geoff can always fulfill his wish if $n$ is odd.
(b) Prove that Geoff can never fulfill his wish if $n$ is even.


An ordered pair $(x, y)$ of integers is a primitive point if the greatest common divisor of $x$ and $y$ is $1$. Given a finite set $S$ of primitive points, prove that there exist a positive integer $n$ and integers $a_0, a_1, \ldots , a_n$ such that, for each $(x, y)$ in $S$, we have:


A convex quadrilateral $ABCD$ satisfies $AB\cdot CD=BC \cdot DA.$ Point $X$ lies inside
$ABCD$ so that
$\angle XAB = \angle XCD$ and $\angle XBC = \angle XDA.$
Prove that $\angle BXA + \angle DXC = 180^{\circ}$

Solution:
We want to find the point $X.$ Let $E$ and $F$ be the intersection points of $AB$ and $CD,$ and $BC$ and $DA,$ respectively.
The poinx $X$ is inside $ABCD,$ so points $E,A,X,C$ follow in this order.
$\angle XAB =  \angle XCD \implies  \angle XAE +  \angle XCE = 180^\circ$
$\implies AXCE$ is cyclic $\implies X$ lie on circle $ACE.$
Similarly, $X$ lie on circle $BDF.$
Point $X$ is the point of intersection of circles  $ACE$ and $\Omega = BDF.$
Special case
Let $AD = CD$ and $AB = BC \implies  AB \cdot CD = BC \cdot DA.$
The points $B$ and $D$ are symmetric with respect to the circle $\theta = ACEF$ (Claim 1).
The circle $BDF$ is orthogonal to the circle $\theta$ (Claim 2).
$\hspace{10mm} \angle FCX =  \angle BCX  =  \frac {\overset{\Large\frown} {XAF}}{2}$ of $\theta.$ $\hspace{10mm} \angle CBX = \angle XDA =  \frac {\overset{\Large\frown} {XBF}}{2}$ of $\Omega.$
$\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ$ (Claim 3) $\implies$
$\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.$
Similarly, $\angle AXD =  90^\circ \implies$
Common case
Denote by $O$ the intersection point of $BD$ and the perpendicular bisector of $AC.$ Let $\omega$ be a circle (red) with center $O$ and radius $OA = R.$
We will prove $\sin\angle BXA =\sin \angle DXC$ using point $Y$ symmetric to $X$ with respect to $\omega.$
The points $B$ and $D$ are symmetric with respect to $\omega$ (Claim 1).
The circles $BDF$ and $BDE$ are orthogonal to the circle $\omega$ (Claim 2).
Circles $ACF$ and $ACE$ are symmetric with respect to the circle $\omega$ (Lemma).
Denote by $Y$ the point of intersection of circles $BDF$ and $ACF.$
Quadrangle $BYDF$ is cyclic $\implies  \angle CBY =  \angle ADY.$
Quadrangle $AYCF$ is cyclic $\implies  \angle YAD = \angle YCB.$
The triangles $\triangle YAD \sim \triangle YCB$ by two angles, so
The points $X$ and $Y$ are symmetric with respect to the circle $\omega$, since they lie on the intersection of the circles $ACF$ and $ACE$ symmetric with respect to $\omega$ and the  circle $BDF$ orthogonal to $\omega.$
The point $B$ is symmetric to $D$ with respect to $\omega \implies$

The point $B$ is symmetric to $D$ and the point $X$ is symmetric to $Y$ with respect to $\omega,$ hence
Denote  $\angle XAB =  \angle XCD = \alpha,  \angle BXA = \varphi,   \angle DXC = \psi.$
By the law of sines for $\triangle ABX,$ we obtain $\frac {AB}{\sin \varphi} = \frac{BX}{\sin \alpha}.$
By the law of sines for $\triangle CDX,$ we obtain $\frac {CD}{\sin \psi} = \frac {DX}{\sin \alpha}.$
Hence we get $\frac{\sin \psi} {\sin \varphi}= \frac {CD}{DX}  \cdot \frac{BX}{AB} = 1.$
If $\varphi = \psi,$ then $\triangle XAB \sim  \triangle XCD \implies \frac {CD}{AB} = \frac {BX}{DX} = \frac{AX}{CX} = \frac {AD}{BC}.$
$CD \cdot BC = AB \cdot AD \implies AD = CD, AB = BC.$ This is a special case.
In all other cases, the equality of the sines follows $\varphi + \psi = 180^\circ .$
Claim 1 Let $A, C,$ and $E$ be arbitrary points on a circle $\omega, l$ be the perpendicular bisector to the segment $AC.$ Then the straight lines $AE$ and $CE$ intersect $l$ at the points $B$ and $D,$ symmetric with respect to $\omega.$
Claim 2 Let points $B$ and $D$ be symmetric with respect to the circle $\omega.$ Then any circle $\Omega$ passing through these points is orthogonal to $\omega.$
Claim 3 The sum of the arcs between the points of intersection of two perpendicular circles is $180^\circ.$
In the figure they are a blue and red arcs $\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.$
Lemma The opposite sides of the quadrilateral $ABCD$ intersect at points $E$ and $F$ ($E$ lies on $AB$). The circle  $\omega$ centered at the point $O$ contains the ends of the diagonal $AC.$ The points $B$ and $D$ are symmetric with respect to the circle $\omega$ (in other words, the inversion with respect to $\omega$ maps $B$ into $D).$ Then the circles $ACE$ and $ACF$ are symmetric with respect to $\omega.$
Proof  We will prove that the point $G,$ symmetric to the point $E$ with respect to $\omega,$ belongs to the circle $ACF$ becouse $\angle AGC = \angle AFC.$
A circle $BDE$ containing points $B$ and $D$ symmetric with respect to $\omega,$ is orthogonal to $\omega$ (Claim 2) and maps into itself under inversion with respect to the circle $\omega.$ Hence, the point $E$ under this inversion passes to some point $G,$ of the same circle $BDE.$
A straight line $ABE$ containing the point $A$ of the circle $\omega,$ under inversion with respect to $\omega,$ maps into the circle $OADG.$ Hence, the inscribed angles of this circle are equal  $\angle ADB =  \angle AGE.$
$\angle OCE =  \angle CGE (CE$ maps into $CG)$ and  $\angle OCE =  \angle BCD (BC$ maps into $DC).$
Consequently, the angles   $\angle AFC =  \angle ADB –  \angle FBD = \angle AGE -  \angle CGE =  \angle AGC.$
These angles subtend the $\overset{\Large\frown} {AC}$ of the $ACF$ circle, that is, the point $G,$ symmetric to the point $E$ with respect to $\omega,$ belongs to the circle $ACF.$
vladimir.shelomovskii@gmail.com, vvsss


Let $I$ be the incenter of acute triangle $ABC$ with $AB \neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ again at $R$. Line $AR$ meets ω again at $P$. The circumcircles of triangles $PCE$ and $PBF$ meet again at $Q$.
Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$.

Solution:
Step 1
We find an auxiliary point $S.$
Let $G$ be the antipode of $D$ on $\omega, GD = 2R,$ where $R$ is radius $\omega.$
We define $A' = PG \cap AI.$
$RD||AI, PRGD$ is cyclic $\implies \angle IAP = \angle DRP = \angle DGP.$
$RD||AI, RD \perp RG, RI=GI \implies \angle AIR = \angle AIG  \implies$

An inversion with respect $\omega$ swap $A$ and $A' \implies A'$ is the midpoint $EF.$
Let $DA'$ meets $\omega$ again at $S.$ We define $T = PS \cap DI.$
Opposite sides of any quadrilateral inscribed in the circle $\omega$ meet on the polar line of the intersection of the diagonals with respect to $\omega \implies DI$ and $PS$ meet on the line through $A$ perpendicular to $AI.$
The problem is reduced to proving that $Q \in PST.$
Step 2
We find a simplified way to define the point $Q.$
We define $\angle BAC = 2 \alpha \implies \angle AFE = \angle AEF = 90^\circ – \alpha \implies$
$\angle BFE = \angle CEF = 180^\circ – (90^\circ – \alpha) = 90^\circ + \alpha =  \angle BIC$
$(AI, BI,$ and $CI$ are bisectrices).
We use the Tangent-Chord Theorem and get
$\angle BQC =  \angle BQP +  \angle PQC = \angle BFP + \angle CEP =$
$=\angle BFE – \angle EFP + \angle CEF – \angle FEP =$
$= 90^\circ + \alpha + 90^\circ + \alpha – (90^\circ + \alpha) =$
$90^\circ + \alpha = \angle BIC \implies$
Points $Q, B, I,$ and $C$ are concyclic.
Step 3
We perform inversion around $\omega.$ The straight line $PST$ maps onto circle $PITS.$ We denote this circle $\Omega.$ We prove that the midpoint of $AD$ lies on the circle $\Omega.$
In the diagram, the configuration under study is transformed using inversion with respect to $\omega.$ The images of the points are labeled in the same way as the points themselves. Points $D,E,F,P,S,$ and $G$ have saved their position. Vertices $A, B,$ and $C$ have moved to the midpoints of the segments $EF, FD,$ and $DE,$ respectively.
Let $M$ be the midpoint $AD.$
We define $\angle MID = \beta, \angle MDI = \gamma \implies$
$\angle IMA = \angle MID + \angle MDI = \beta + \gamma = \varphi.$
$DI = IS \implies \angle ISD = \gamma.$
$MI$ is triangle $DAG$ midline $\implies MI || AG \implies$

$PI = IS \implies \angle PIS = 180^\circ – 2 \varphi =\angle DPA \implies$ point $M$ lies on $\Omega.$
$ABDC$ is parallelogram $\implies M$ is midpoint $BC.$
Step 4
We prove that image of $Q$ lies on $\Omega.$
In the inversion plane the image of point $Q$ lies on straight line $BC$ (It is image of circle $BIC)$ and on circle $PCE.$
point $Q$ lies on $\Omega$.
vladimir.shelomovskii@gmail.com, vvsss


Prove that there exists a positive constant $c$ such that the following statement is true:
Consider an integer $n > 1$, and a set $S$ of n points in the plane such that the distance between any two different points in $S$ is at least $1$. It follows that there is a line $\ell$ separating $S$ such that the distance from any point of $S$ to $\ell$ is at least $cn^{- \frac{1}{3}}$.
(A line $\ell$ separates a set of points $S$ if some segment joining two points in $S$ crosses $\ell$.)
Note. Weaker results with $cn^{- \frac{1}{3}}$ replaced by $cn^{- \alpha}$ may be awarded points depending on the value
of the constant $\alpha > \frac{1}{3}$.

Solution:
For any unit vector $v$, let $a_v=\min_{p\in S} p \cdot v$ and $b_v = \max_{p\in S} p\cdot v$. If $b_v - a_v\geq n^{2/3}$ then we can find a line $\ell$ perpendicular to $v$ such that $\ell$ separates $S$, and any point in $S$ is at least $\Omega(n^{2/3}/n) = \Omega(n^{-1/3})$ away from $\ell$.
Suppose there is no such direction $v$, then $S$ is contained in a box with side length $n^{2/3}$ by considering the direction of $(1, 0)$ and $(0, 1)$, respectively. Hence, $S$ is contained in a disk with radius $n^{2/3}$. Now suppose that $D$ is the disk with the minimum radius, say $r$, which contains $S$. Then, $r=O(n^{2/3})$. Since the distance between any two points in $S$ is at least $1$, $r=\Omega(\sqrt{n})$ too.
Let $p$ be any point in $S$ on the boundary of $D$. Let $\ell_1$ be the line tangent to $D$ at $p$, and $\ell_2$ the line obtained by translating $\ell_1$ by distance $1$ towards the inside of $D$. Let $H$ be the region sandwiched by $\ell_1$ and $\ell_2$. It is easy to show that both the area and the perimeter of $H\cap D$ is bounded by $O(\sqrt{r})$ (since $r=\Omega(\sqrt{n})$). Hence, there can only be $O(\sqrt{r})=O(n^{1/3})$ points in $H\cap S$, by that any two points in $S$ are distance $1$ apart. Since the width of $H$ is $1$, there must exist a line $\ell$ parallel to $\ell_1$ such that $\ell$ separates $S$, and any point in $S$ is at least $1/O(n^{1/3}) = \Omega(n^{-1/3})$ away from $\ell$.  Q.E.D.
Note. One can also show that $\Omega(n^{-1/3})$ is best possible.


Let $n$ be a positive integer. A Nordic square is an $n \times n$ board containing all the integers from $1$ to $n^2$ so that each cell contains exactly one number. Two different cells are considered adjacent if they share an edge. Every cell that is adjacent only to cells containing larger numbers is called a valley. An uphill path is a sequence of one or more cells such that:
(i) the first cell in the sequence is a valley,
(ii) each subsequent cell in the sequence is adjacent to the previous cell, and
(iii) the numbers written in the cells in the sequence are in increasing order.
Find, as a function of $n$, the smallest possible total number of uphill paths in a Nordic square.

Solution:
The minimum total number of uphill paths in a Nordic square is $2n(n-1)+1$
Proof:
For every pair of adjacent cells in the grid, there is at least one pathway that begins at the higher valued cell, steps to the lower valued cell, and then continues stepping to lower adjacent cells until (by finite descent) a valley is reached. When reversed, this forms an uphill pathway that terminates at that exactly that pair of adjacent cells.
There are $2n(n-1)$ such adjacent pairs (each with at least one unique uphill path). In addition, the cell containing the value 1 will always be a valley which, by itself, forms an additional (singleton) uphill path. Therefore the minimum total number of uphill paths is at least $2n(n-1)+1$.
The diagram below shows a general family of constructions of Nordic squares that achieve this lower bound, thus proving the claim:
In these constructions, numbers are progressively inserted into the white spaces in ascending order emanating outwards from the "1" in the corner in the directions of the red arrows, and the remaining j largest numbers are then inserted into the cells labelled as "mountains". In this manner, the pathway of finite descent from each pair of adjacent cells is unique, and all terminate at the "1" valley, therefore the above-derived minimum is achieved. $\blacksquare$
- Griphomaniac (talk) 13:16, 21 July 2022 (EDT)


Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and
Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$
Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$ and $CC_1C_2$ all pass through two common points.
(Note: a scalene triangle is one where no two sides have equal length.)

Solution: https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]


An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.
a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;
b) Calculate the distance of $P$ from either base;
c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.

Solution:
(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.
(b) Let $x$ be the distance from $P$ to one of the bases; then $h - x$ must be the distance from $P$ to the other base. Similar triangles give $\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}$, so $x^2 - hx + \frac{ac}{4} = 0$ and so $x = \frac{h \pm \sqrt{h^2 - ac}}{2}.$
(c) When $h^2 \ge ac$.
In our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'.  AF and BE are perpendicular to CD such that AF= BE= h.  XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0).
Let our point P be on the axis of symmetry at z distance from the origin O.
The coordinates of the points A,B,C,D,E,F and P are given in the figure.
Now,
Slope of the line $PC= (z-0)/(0-c/2) =  -2z/c$
Slope of the line $PB= (z-h)/(0-a/2) = -2(z-h)/a$
Since the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1.
i.e
$4z(z-h)=-ac$
or $z^2 - zh + ac/4= 0$
Now, solving for z, we get,  $z=  [(h + ( h^2 - ac ) ^1/2 ]/2$  and   $[(h - ( h^2 - ac ) ^1/2 ]/2$
So, z is the distance of the points from the base  CD..
Also the points are possible only when , $h^2 - ac >= 0$.. and doesn't exist for $h^2 -ac <0$


The tetrahedron $SABC$ has the following property: there exist five spheres, each tangent to the edges $SA, SB, SC, BC, CA, AB$, or to their extensions.
(a) Prove that the tetrahedron $SABC$ is regular.
(b) Prove conversely that for every regular tetrahedron five such spheres exist.

Solution:
Part (a)
Let points $P_{SA}, P_{SB}, P_{SC}, P_{BC}, P_{CA}, P_{AB}$ be the points where the smallest sphere is tangent to the edges $SA,  SB,  SC,  BC,  CA,  AB$ respectively.
For each face of the tetrahedron there is a circular cross section of the smallest sphere.  Since that circle also needs to be tangent to the edges, then that circle is the incircle of each triangular face.
From the properties of an incircle we know that:
$|AP_{AB}|=|AP_{AC}|$, $|BP_{BC}|=|BP_{BA}|$, and $|CP_{CA}|=|CP_{CB}|$ in $\Delta ABC$
There is a larger circle that is tangent to $AB$ and the extensions of $CA$ and $CB$.
This larger circle is a cross section of the sphere that's also tangent to the extension of $CS$ away from $S$ and the edges of $\Delta ABC$
Therefore, this larger circle and the incircle of $\Delta ABC$ are part of that same sphere.
In order for these two circles to be part of the same sphere and also tangent to line $AB$, then the point of the tangent of this larger circle needs to be the same as point $P_{AB}$
The only way these to circles can share the same tangent point on edge $AB$ is if $|AP_{AB}|=|BP_{AB}|$
Using the same argument with the larder circle of edge $BC$ then $|BP_{BC}|=|CP_{BC}|$
and with the larger circle of edge $AC$ then $|CP_{AC}|=|AP_{AC}|$
This results in:
$|AP_{AB}|=|AP_{AC}|=|BP_{BC}|=|BP_{BA}|=|CP_{CA}|=|CP_{CB}|$ in $\Delta ABC$
which means that
$|AB|=|BC|=|AC|$ in $\Delta ABC$, thus $\Delta ABC$ is an equilteral triangle.
Likewise,
$|AB|=|BS|=|AS|$ in $\Delta ABS$, thus $\Delta ABS$ is an equilteral triangle.
$|AS|=|SC|=|AC|$ in $\Delta ASC$, thus $\Delta ASC$ is an equilteral triangle.
$|SB|=|BC|=|SC|$ in $\Delta SBC$, thus $\Delta SBC$ is an equilteral triangle.
Since all four faces are equilateral triangles, then tetrahedron $ABCS$ is a regular tetrahedron.

Part (b)
Let's consider a regular tetrahedron in the cartesian space with center at $(0,0,0)$ with $r_c$ as the circumradius.
We can write the coordinates for the four vertices as:
$V_1=\left( \frac{2\sqrt{2}}{3}r_c ,\;0,\;-\frac{r_c}{3}\right)$
$V_2=\left( -\frac{\sqrt{2}}{3}r_c ,\;\frac{\sqrt{6}}{3}r_c,\;\frac{-r_c}{3}\right)$
$V_3=\left( -\frac{\sqrt{2}}{3}r_c ,\;-\frac{\sqrt{6}}{3}r_c,\;\frac{-r_c}{3}\right)$
$V_4=\left(0 ,\;0,\;r_c\right)$
And the midpoints of all edges as:
$M_{12}=\frac{V_1-V_2}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}{3}\right)$
$M_{13}=\frac{V_1-V_3}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;-\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}{3}\right)$
$M_{23}=\frac{V_2-V_3}{2}=\left( -\frac{\sqrt{2}}{3}r_c ,\;0,\;-\frac{r_c}{3}\right)$
$M_{14}=\frac{V_1-V_4}{2}=\left( \frac{\sqrt{2}}{3}r_c ,\;0,\;\frac{r_c}{3}\right)$
$M_{24}=\frac{V_2-V_4}{2}=\left( -\frac{\sqrt{2}}{6}r_c ,\;\frac{\sqrt{6}}{6}r_c,\;\frac{r_c}{3}\right)$
$M_{34}=\frac{V_3-V_4}{2}=\left( -\frac{\sqrt{2}}{6}r_c ,\;-\frac{\sqrt{6}}{6}r_c,\;\frac{r_c}{3}\right)$
Now we calculate the following six dot products:
$(V_1-V_2) \bullet M_{12}=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left(  \frac{\sqrt{6}}{3}r_c\right)\left( \frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$
$(V_1-V_3) \bullet M_{13}=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left(-  \frac{\sqrt{6}}{3}r_c\right)\left( -\frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$
$(V_2-V_3) \bullet M_{23}=0+\left(-  \frac{\sqrt{6}}{6}r_c\right)\left( 0\right)+ 0=0$
$(V_1-V_4) \bullet M_{13}=\left( 2\frac{\sqrt{2}}{3}r_c \right)\left( \frac{\sqrt{2}}{3}r_c \right)+0+ \left(-  \frac{4}{3}r_c\right)\left( \frac{1}{3}r_c \right)=\frac{4}{9}r_c^2-\frac{4}{9}r_c^2=0$
$(V_2-V_4) \bullet M_{13}=\left( -\frac{\sqrt{2}}{3}r_c \right)\left( -\frac{\sqrt{2}}{6}r_c \right)+\left( \frac{\sqrt{6}}{3}r_c \right)\left( \frac{\sqrt{6}}{6}r_c \right)+ \left(-  \frac{4}{3}r_c\right)\left( \frac{1}{3}r_c \right)=\frac{1}{9}r_c^2+\frac{1}{3}r_c^2-\frac{4}{9}r_c^2=0$
$(V_3-V_4) \bullet M_{13}=\left( -\frac{\sqrt{2}}{3}r_c \right)\left( -\frac{\sqrt{2}}{6}r_c \right)+\left( -\frac{\sqrt{6}}{3}r_c \right)\left( -\frac{\sqrt{6}}{6}r_c \right)+ \left(-  \frac{4}{3}r_c\right)\left( \frac{1}{3}r_c \right)=\frac{1}{9}r_c^2+\frac{1}{3}r_c^2-\frac{4}{9}r_c^2=0$
Since all these dot producs equal to $0$ that means that the lines from the center to each  of the midpoints are all perpendicular.
Now we calculate the distances from $(0,0,0)$ to all the midpoints:
$|M_{12}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$
$|M_{13}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(-\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$
$|M_{23}|=\sqrt{\left( -\frac{\sqrt{2}}{3}r_c\right)^2+\left(0\right)^2+\left(-\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$
$|M_{14}|=\sqrt{\left( \frac{\sqrt{2}}{3}r_c\right)^2+\left(0\right)^2+\left(\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$
$|M_{24}|=\sqrt{\left( -\frac{\sqrt{2}}{6}r_c\right)^2+\left(\frac{\sqrt{6}}{6}r_c\right)^2+\left(\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$
$|M_{34}|=\sqrt{\left( -\frac{\sqrt{2}}{6}r_c\right)^2+\left(-\frac{\sqrt{6}}{6}r_c\right)^2+\left(\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$
Since all distances are all the same and all dot products are $0$, then we have our first sphere at $(0,0,0)$ with a radius of $\frac{\sqrt{3}}{3}r_c$
Now we will look at the other 4 spheres.
Let $E_{ij}$ be the point of tangent of the larger sphere on the extension of line $V{i}V{j}$ in the direction of $V{i}$ to $V{j}$ and beyond $V{j}$ for $i=1,2,3,4$; $j=1,2,3,4$; and $i \ne j$
Since $|V_j E_{ij}|=|V_j M_{ij}|=\frac{1}{2} |V_j V_i|$, then $\frac{1}{2}(V_j-V_i)=E_{ij}-V_j$, thus $E_{ij}=\frac{3V_j-V_i}{2}$
Using this formula we calculate the following:
$E_{41}=\left( \sqrt{2} r_c ,\;0,\;-r_c \right)$
$E_{42}=\left( -\frac{\sqrt{2}}{2}r_c ,\;\frac{\sqrt{6}}{2}r_c,\;-r_c \right)$
$E_{43}=\left( -\frac{\sqrt{2}}{2}r_c ,\;-\frac{\sqrt{6}}{2}r_c,\;-r_c \right)$
We will start with the sphere below the base of the tetrahedron opposite of vertex $V_{4}$ below $\Delta V{1}V{2}V{3}$
The center of this larger sphere is at $C_1=(0,0,-2r_c)$ and it is tangent at points $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$
We calculate the following dot products:
$(V_1-V_2) \bullet (M_{12}-C_1)=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left(  \frac{\sqrt{6}}{3}r_c\right)\left( \frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$
$(V_1-V_3) \bullet (M_{13}-C_1)=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left(-  \frac{\sqrt{6}}{3}r_c\right)\left( -\frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$
$(V_2-V_3) \bullet (M_{23}-C_1)=0+\left(-  \frac{\sqrt{6}}{6}r_c\right)\left( 0\right)+ 0=0$
$(E_{41}-V_4) \bullet (E_{41}-C_1) = \left( \sqrt{2}r_c \right)^2+0+\left( -2r_c \right)\left( -r_c+2r_c \right)=2r_c^2-2r_c^2=0$
$(E_{42}-V_4) \bullet (E_{42}-C_1) = \left( -\frac{\sqrt{2}}{2}r_c \right)^2+\left( \frac{\sqrt{6}}{2}r_c \right)^2+\left( -2r_c \right)\left( -r_c+2r_c \right)=2r_c^2-2r_c^2=0$
$(E_{43}-V_4) \bullet (E_{43}-C_1) = \left( -\frac{\sqrt{2}}{2}r_c \right)^2+\left( -\frac{\sqrt{6}}{2}r_c \right)^2+\left( -2r_c \right)\left( -r_c+2r_c \right)=2r_c^2-2r_c^2=0$
Since all these dot producs equal to $0$ that means that the lines from the center $C_1$ to each of the point $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$ are all perpendicular.
Now we calculate the distances from $C_1=(0,0,-2r_c)$ to points $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$
$|C_1M_{12}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{5}{3}r_c\right)^2}=\sqrt{3}r_c$
$|C_1M_{13}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(-\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{5}{3}r_c\right)^2}=\sqrt{3}r_c$
$|C_1M_{23}|=\sqrt{\left( -\frac{\sqrt{2}}{3}r_c\right)^2+\left(0\right)^2+\left(-\frac{5}{3}r_c\right)^2}=\sqrt{3}r_c$
$|C_1E_{41}|=\sqrt{\left( \sqrt{2}r_c\right)^2+\left(0\right)^2+\left(-r_c\right)^2}=\sqrt{3}r_c$
$|C_1E_{42}|=\sqrt{\left( -\frac{\sqrt{2}}{2}r_c\right)^2+\left(\frac{\sqrt{6}}{2}r_c\right)^2+\left(-r_c\right)^2}=\sqrt{3}r_c$
$|C_1E_{43}|=\sqrt{\left( -\frac{\sqrt{2}}{2}r_c\right)^2+\left(-\frac{\sqrt{6}}{2}r_c\right)^2+\left(-r_c\right)^2}=\sqrt{3}r_c$
Since all distances are all the same and all dot products are $0$, then we have one of the larger spheres at $(0,0,-2r_c)$ with a radius of $\sqrt{3}r_c$
Then, the other three larger spheres which are the same size as the sphere with center at $C_1$ are congruent and tangent to their respective sides near the other faces of the tetrahedron.
and this proves that a tetrahedron with any circumradius $r_c$ will have these 5 spheres, one with radius $\frac{\sqrt{3}}{3}r_c$, at the center of the tetrahedron and the other 4 with radius $\sqrt{3}r_c$ at centers that are at a distance of $3r_c$ away from any of the vertices of the tetrahedron in the direction from that vertex to the center of its opposite face.
Thus these five spheres exist for any regular tetrahedron.
~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.