Let $K$ be the set of all positive integers that do not contain the digit $7$ in their base-$10$ representation. Find all polynomials $f$ with nonnegative integer coefficients such that $f(n)\in K$ whenever $n\in K$.

Solution:
I claim the only such polynomials are of the form $f(n)=k$ where $k\in K$, or $f(n)=an+b$ where $a$ is a power of 10, $b\in K$, and $b<a$. Obviously, these polynomials satisfy the conditions. We now prove that no other polynomial works. That is, we prove that for any other polynomial $f$ with nonnegative coefficients, there is some $n\in K$ such that $f(n)\notin K$.
We first prove the result for monomials, polynomials in which only one coefficient is nonzero. This is obvious for constant polynomials $f(n)=k\notin K$. The next simplest case is $f(n)=an$ with $a$ not a power of 10, and hence $\lg a$ is irrational. By Dirichlet's approximation theorem, the set of multiples of $\lg a$ is dense $\bmod\ 1$, and thus contains an element with fractional part in the interval $[\lg 7,\lg 8)$. In other words, there is a power of $a$ whose decimal expansion starts with a 7. Let $a^x$ be the smallest power of $a$ that is not in $K$. Obviously, $x>0$, so letting $n=a^{x-1}$ completes the proof of this part.
We have now proven that for any $a$ that is not a power of 10, there is some $n\in K$ such that $an\notin K$. We proceed to the case where $f(n)=an^x$ for $x>1$. This splits into 2 cases. If $ax$ is not a power of 10, then pick $m\in K$ such that $axm\notin K$. For any $y$, we have

If we choose $y$ to be large enough, then the terms in the expression above will not interfere with each other, and the resulting number will contain a 7 in the decimal expansion, and thus not be in $K$.
On the other hand, if $ax$ is a power of 10, then $a$ and $x$ are both powers of 10, and $x\ge10$. Obviously, $\frac12ax(x-1)$ is not a power of 10. By the previous step, which establishes the result for $x=2$, we can pick $m\in K$ such that $\frac12ax(x-1)m^2\notin K$. Then, for any $y$,

Similarly, picking a sufficient large $y$ settles this case.
Now, we extend our proof to general polynomials. If a polynomial $f(n)=a_0+a_1n+a_2n^2+...+a_xn^x$ satisfies the conditions of the problem, then for any $m,y>0$:

Similarly, choosing $y$ to be sufficiently large results in the terms not interfering with each other. If $f$ contains any monomials that do not satisfy the conditions of the problem, then picking suitable $m$ and sufficiently large $y$ causes $f(m*10^y)$ to not be in $K$. Thus, $f$ is a linear polynomial of the form $ax+b$ where $a$ is 0 or a power of 10, and $b\in K$. It suffices to rule out those polynomials where $a>0$ and $b>a$. If this is the case, then since the digit of $b$ corresponding to $a$ is not 7, there must be a single-digit number $n$ such that the digit of $f(n)$ corresponding to $a$ is 7. Therefore, we are done.
-wzs26843545602
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Let $n$ be a nonnegative integer. Determine the number of ways that one can choose $(n+1)^2$ sets $S_{i,j}\subseteq\{1,2,\ldots,2n\}$, for integers $i,j$ with $0\leq i,j\leq n$, such that:
$\bullet$ for all $0\leq i,j\leq n$, the set $S_{i,j}$ has $i+j$ elements; and
$\bullet$ $S_{i,j}\subseteq S_{k,l}$ whenever $0\leq i\leq k\leq n$ and $0\leq j\leq l\leq n$.

Note that there are $(2n)!$ ways to choose $S_{1, 0}, S_{2, 0}... S_{n, 0}, S_{n, 1}, S_{n, 2}... S_{n, n}$, because there are $2n$ ways to choose which number $S_{1, 0}$ is, $2n-1$ ways to choose which number to append to make $S_{2, 0}$, $2n-2$ ways to choose which number to append to make $S_{3, 0}$, etc. After that, note that $S_{n-1, 1}$ contains the $n-1$ in $S_{n-1, 0}$ and 1 other element chosen from the 2 elements in $S_{n, 1}$ not in $S_{n-1, 0}$ so there are 2 ways for $S_{n-1, 1}$. By the same logic there are 2 ways for $S_{n-1, 2}$ as well so $2^n$ total ways for all $S_{n-1, j}$, so doing the same thing $n-1$ more times yields a final answer of $(2n)!\cdot 2^{\left(n^2\right)}$.
-Stormersyle

There are $\frac{(2n)!}{2^n}$ ways to choose $S_{0,0}, S_{1,1} .... S_{n,n}$. Since, there are $\binom{2n}{2}$ ways to choose $S_{1,1}$, and after that, to generate $S_{2,2}$, you take $S_{1,1}$ and add 2 new elements, getting you $\binom{2n-2}{2}$ ways to generate $S_{2,2}$. And you can keep going down the line, and you get that there are $\frac{(2n)!}{2^n}$ ways to pick $S_{0,0}, S_{1,1} ... S_{n,n}$ Then we can fill out the rest of the gird. First, let’s prove a lemma.

Let $C_{j}$ represent the set of sets of the form $S_{ij}$ for $1 \le i \le n$, $a_{ij}$ denote $S_{(i+1)j} \backslash S_{ij}$, and $b_{ij} = S_{i(j+1)} \backslash S_{ij}$. Begin by considering $C_0$ and $S_{00} = \emptyset$. Then  given $S_{i0}$ we can create $S_{(i+1)0}$ by adding one element ($a_{i0}$). Using this, the number of ways to form the sequence of $S_{00}, S_{10}, \dots, S_{n0}$ are $(2n)(2n-1) \cdots (n+1)$ where we successively add one of the remaining elements of $[2n]$ to get consecutive terms in the sequence.
Now consider when we are given $C_{j}$ and we need to find $C_{j+1}$. So far, there have been $n + j$ chosen distinct elements (via $S_{nj}$). After finding $C_{j+1}$ we will have $n + j + 1$ distinct elements and so in this process we only add one unique element to sets among $C_{j+1}$. There are $2n - (n+j) = n-j$ ways to chose such a new element called $x$.
Now notice that $b_{0j}, a_{0(j+1)}, a_{1(j+1)}, \dots, a_{(n-1)(j+1)}$ is a permutation of $x,  a_{0j}, \dots, a_{(n-1)j}$ by noting $b_{nj} = x$ and,

Furthermore,

Therefore $b_{ij}, a_{i(j+1)}$ is a permutation of $a_{ij}, b_{(i+1)j}$ for $i < n$. Now let $k$ be the first $i$ such that $b_{ij} = x$. By definition, $b_{(k+1)j}, \dots = x$. Then the number of ways to order $x,  a_{0j}, \dots, a_{(n-1)j}$ is $2^{k-1}$ as there are 2 permutations for each pair before $\{b_{ij}, a_{i(j+1)}\} = \{a_{ij}, b_{(i+1)j}\}$ and each pair after is determined by $b_{ij}= b_{(i+1)j} = k, a_{i(j+1)} = a_{ij}$. For $k  = 0$, the permutation is completely determined so there is one way.
Overall the number of ways to add the $j+1$'th row is,
In total, there are $(2n)(2n-1)\cdots (n+1)$ ways to find $C_0$ and for each $C_{j}$ there are $2^n(n-j+1)$ ways for $1 \le j \le n$. So the answer is,
~Aaryabhatta1
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.


Two rational numbers $\frac{m}{n}$ and $\frac{n}{m}$ are written on a blackboard, where $m$ and $n$ are relatively prime positive integers. At any point, Evan may pick two of the numbers $x$ and $y$ written on the board and write either their arithmetic mean $\frac{x+y}{2}$ or their harmonic mean $\frac{2xy}{x+y}$ on the board as well. Find all pairs $(m,n)$ such that Evan can write $1$ on the board in finitely many steps.
Proposed by Yannick Yao

Solution:
We claim that all odd $m, n$ work if $m+n$ is a positive power of 2.
Proof:
We first prove that $m+n=2^k$ works. By weighted averages we have that $\frac{n(\frac{m}{n})+(2^k-n)\frac{n}{m}}{2^k}=\frac{m+n}{2^k}=1$ can be written, so the solution set does indeed work. We will now prove these are the only solutions.
Assume that $m+n\ne 2^k$, so then $m+n\equiv 0\pmod{p}$ for some odd prime $p$. Then $m\equiv -n\pmod{p}$, so $\frac{m}{n}\equiv \frac{n}{m}\equiv -1\pmod{p}$. We see that the arithmetic mean is $\frac{-1+(-1)}{2}\equiv -1\pmod{p}$ and the harmonic mean is $\frac{2(-1)(-1)}{-1+(-1)}\equiv -1\pmod{p}$, so if 1 can be written then $1\equiv -1\pmod{p}$ and $2\equiv 0\pmod{p}$ which is obviously impossible, and we are done.
-Stormersyle
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.


Find all polynomials $P$ with real coefficients such that holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.

Solution:
If $P(x)=c$ for a constant $c,$ then $\dfrac{c(x+y+z)}{xyz}=3c$. We have $2c=3c.$ Therefore $c=0.$
Now consider the case of non-constant polynomials.
First we have  for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$. Both sides of the equality are polynomials (of $x,y,z$). They have the same values on the 2-dimensional surface $2xyz=x+y+z$, except for some 1-dimensional curves in it. By continuity, the equality holds for all points on the surface, including those with $z=0.$ Let $z=0,$ we have $y=-x$ and $x(P(x)-P(-x))=0.$ Therefore $P$ is an even function.
(Here is a sketch of an elementary proof. Let $z=\dfrac{x+y}{2xy-1}.$ We have

This is an equality of rational expressions. By multiplying $(2xy-1)^N$ on both sides for a sufficiently large $N$, they become polynomials, say $A(x,y)=B(x,y)$ for all real $x, y$ with $x\ne 0, y\ne 0, x+y\ne 0$ and $2xy-1\ne 0.$ For a fixed $x,$ we have two polynomials (of $y$) having same values for infinitely many $y$. They must be identical. Let $y=0,$ we have $x^{N+1}(P(x)-P(-x))=0.$ )
Notice that if $P(x)$ is a solution,  then is $cP(x)$ for any constant $c.$ For simplicity, we assume the leading coefficient of $P$ is $1$:  where $n$ is a positive even number.
Let $y=\dfrac{1}{x}$, $z=x+\dfrac{1}{x}.$ we have
Simplify using $P(x)=P(-x),$
Expand and combine like terms, both sides are of the form
They have the same values for infinitely many $x.$ They must be identical. We just compare their leading terms. On the left hand side it is $2nx^{n-1}$. There are two cases for the right hand sides: If $n>2$, it is $x^{n-1}$; If $n=2$, it is $(1+a_0)x.$ It does not work for $n>2.$ When $n=2,$ we have $4=1+a_0.$ therefore $a_0=3.$
The solution: $P(x)=c(x^2+3)$ for any constant $c.$
-JZ


Let $ABC$ be a fixed acute triangle inscribed in a circle $\omega$ with center $O$. A variable point $X$ is chosen on minor arc $AB$ of $\omega$, and segments $CX$ and $AB$ meet at $D$. Denote by $O_1$ and $O_2$ the circumcenters of triangles $ADX$ and $BDX$, respectively. Determine all points $X$ for which the area of triangle $OO_1O_2$ is minimized.

Solution:
Let $E$ be midpoint $AD.$ Let $F$ be midpoint $BD \implies$

$E$ and $F$ are the bases of perpendiculars dropped from $O_1$ and $O_2,$ respectively.
Therefore  $O_1O_2 \ge EF =  \frac {AB}{2}.$
$\angle AXC = \angle ABC (AXBC$ is cyclic) $\implies \angle O O_1O_2 = \angle ABC.$
Similarly $\angle BAC = \angle O O_2 O_1 \implies  \triangle ABC \sim \triangle O_2 O_1O.$
The area of $\triangle OO_1O_2$ is minimized if $CX \perp AB$ because

vladimir.shelomovskii@gmail.com, vvsss


An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:
What is the smallest positive number of beams that can be placed to satisfy these conditions?

Solution:
Take one vertex of the cube as origin and establish 3D coordinates along the cube's edges.
Define a beam as $x-dir$ if its long edge is parallel to x-axis. Similarly for $y-dir$ and $z-dir$.
Define a beam's location as (direction, ($1 \times 1$ face's location in 2D coordinate).
For example, (y, 2, 4) indicates the beam with vertex (1, 0, 3) and (2, 2020, 4)
Apparently $x$ beam needs the other $x$ or $y$ beams to touch its $xy$ faces, $x$ or $z$ beams to touch its $xz$ faces. Similarly for $y$ and $z$ beam.
If there are only 1-dir or 2-dirs beams, it is easy to approve that $2020 \times 2020$ is the minimal number beams.
(for example, if only use $x-dir$ and $y-dir$ beams, all the $x-dir$ beams's xz faces can only be touched by $x-dir$ beams. In the other word, $2020 x-dir$ beams will be stacked to meet xz faces touch requirements in that xz layer)
Consider cases with all $3-dirs$ beams.
WLOG there is a $x-dir$ beam and it needs $x-dir$ or $y-dir$ beams to touch its $xy$ faces (unless it touches the cube surface).
And this $y-dir$ beam also needs a $x-dir$ or $y-dir$ to touch it's $xy$ faces. And so on until one which touches cube surface. So from $xy$ face perspective, it needs $2020$ beams.
Similarly from $xz$ and $yz$ face perspective, it also needs $2020$ and $2020$ beams.
Consider one beam has four $1 \times 2020$ faces and it can be counted twice. So there should be at least $2020 \times 3 \div 2=3030$ beams.
Here is one solution with 3030 beams.
$(x, 1, 1),\ (y, 1, 2),\ (z, 2, 2),$
$\cdots ,$
$(x, (2n+1), (2n+1)),\ (y, (2n+1), (2n+2)),\ (z, (2n+2), (2n+2)),$
$\cdots ,$
$(x, (2019, 2019)),\ (y, 2019, 2020),\ (z, 2020, 2020)$
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.


Let $p$ be an odd prime. An integer $x$ is called a quadratic non-residue if $p$ does not divide $x - t^2$ for any integer $t$.
Denote by $A$ the set of all integers $a$ such that $1 \le a < p$, and both $a$ and $4 - a$ are quadratic non-residues. Calculate the remainder when the product of the elements of $A$ is divided by $p$.

Solution:
This problem needs a solution. If you have a solution for it, please help us out by adding it.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.


Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$. Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs.

Solution:
Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$:
We know this must be true:
So
We require the maximum conditions for $(a_3, b_3)$
Then one case can be:
We try to do some stuff such as solving for $a_3$ with manipulations:
We showed that 3 pairs are a complete graph; however, 4 pairs are not a complete graph. We will now show that:
This is clearly impossible because $1$ is not even and also $|a_2b_1-a_1b_2| = 1$.
The answer is as follows:

$a_1$ has $0$ subtractions that follow condition while $a_2$ has $1$ and then the rest has $2$.
There are $n$ terms, so our answer be $2n-3$ and in case of $n=100$ that means
~Lopkiloinm

We claim the answer is $197$.
Study the points $(0, 0), (a_i, b_i), (a_j, b_j)$. If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\frac{1}{2}|0\times{b_i}+{a_i}\times{b_j}+{b_i}\times{0}-0\times{a_i}-{b_i}\times{a_j}-{b_j}\times{0} = \frac{1}{2}|a_ib_j - a_j b_i| = \frac{1}{2}$. Therefore, the triangle formed by the points $(0, 0), (a_i, b_i), (a_j, b_j)$ must have an area of $\frac{1}{2}$.
Two cases follow.
Case 1: Both $(a_i, b_i), (a_j, b_j)$ have exactly one coordinate equal to $0$.
Here, one point must be on the $x$ axis and the other on the $y$ axis in order for the triangle to have a positive area. For the area of the triangle to be $\frac{1}{2}$, it follows that the points must be $(1, 0), (0, 1)$ in some order.
Case 2: At least one of $(a_i, b_i), (a_j, b_j)$ does not have exactly one coordinate equal to $0$. Define $S[l]$ to be a list of lines such that each line in the list has some two lattice points that, with $(0, 0)$, form a triangle with area $\frac{1}{2}$. Note that for any such line that passes through such two lattice points, we may trivially generate infinite lattice points on the line that have nonnegative coordinates.
Note that lines $y=1$ and $x=1$ are included in $S[l]$, because the points $(1, 1), (2, 1)$ serve as examples for $y=1$ and $(1, 1), (1, 2)$ serve as examples for $x=1$. For the optimal construction, include the points $(1, 0)$ and all the points $(0, 1), (0, 2), (0, 3), ... , (0, 99)$, in that order. In this case, every adjacent pair of points would count ($98$), as well as picking $(0, 1)$ and a nonadjacent point ($99$), so this would be $98+99=197$.
To prove that this is the maximum, consider the case where some $n$ number of points were neither on $x=1$ nor on $y=1$. In this case, we would be removing $n$ adjacent pairs and $n$ options to choose from after choosing $(0, 0)$, resulting in a net loss of $2n$. By having $n$ points on some other combination of lines in $S[l]$, we would trivially have a maximum gain of $n-1$ pairs of points on the lines such that there are no lattice points between those pairs. Because these points are not on $x=1$ or $y=1$, the altitude from a given point to the line formed by $(0, 0)$ and $(0, 1)$ and $(1, 0)$ is not $1$, and so the area of the triangle cannot be $\frac{1}{2}$. Thus, by not having all points on lines $x=1$ and $y=1$, we cannot exceed the maximum of $197$. Thus, $\boxed{197}$ is our answer.
~SigmaPiE


A finite set $S$ of points in the coordinate plane is called overdetermined if $|S| \ge 2$ and there exists a nonzero polynomial $P(t)$, with real coefficients and of degree at most $|S| - 2$, satisfying $P(x) = y$ for every point $(x, y) \in S$.
For each integer $n \ge 2$, find the largest integer $k$ (in terms of $n$) such that there exists a set of $n$ distinct points that is not overdetermined, but has $k$ overdetermined subsets.


Let $ABCDEF$ be a convex hexagon satisfying $\overline{AB} \parallel \overline{DE}$, $\overline{BC} \parallel \overline{EF}$, $\overline{CD} \parallel \overline{FA}$, andLet $X$, $Y$, and $Z$ be the midpoints of $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$. Prove that the circumcenter of $\triangle ACE$, the circumcenter of $\triangle BDF$, and the orthocenter of $\triangle XYZ$ are collinear.

Let $M_1$, $M_2$, and $M_3$ be the midpoints of $CE$, $AE$, $AC$ and $N_1$, $N_2$, and $N_3$ be the midpoints of $DF$, $BF$, and $BD$. Also, let $H$ be the orthocenter of $XYZ$. Note that we can use parallel sides to see that $X$, $Z$, and $M_3$ are collinear. Thus we have  by midlines. Applying this argument cyclically, and noting the condition $AB \cdot DE = BC \cdot EF = CD \cdot FA$, $M_1$, $M_2$, $M_3$, $N_1$, $N_2$, $N_3$ all lie on a circle concentric with $(XYZ)$.
Next, realize that basic orthocenter properties imply that the circumcenter $O_1$ of $(ACE)$ is the orthocenter of $\triangle M_1M_2M_3$, and likewise the circumcenter $O_2$ of $(BDF)$ is the orthocenter of $\triangle N_1N_2N_3$.
The rest is just complex numbers; toss on the complex plane so that the circumcenter of $\triangle XYZ$ is the origin. Then we have
Note that from the above we have $h=\frac{o_1+o_2}{2}$, so $H$ is the midpoint of segment $O_1O_2$. In particular, $H$, $O_1$, and $O_2$ are collinear, as required.
~ Leo.Euler

We construct two equal triangles, prove that triangle $XYZ$ is the same as medial triangle of both this triangles. We use property of medial triangle and prove that circumcenters of constructed triangles coincide with given circumcenters.
Denote $A' =  C + E – D, B' = D + F – E, C' =  A+ E – F,$
$D' =  F+ B – A, E' =  A + C – B, F' =  B+ D – C.$
Then $A' – D'  =  C + E – D –  ( F+ B – A) = (A + C + E ) – (B+ D + F).$
Denote  $D' – A' = 2\vec V.$
Similarly we get $B' – E' = F' – C' =  D' – A'  \implies$
$\triangle A'C'E' = \triangle D'F'B'.$
The translation vector maps $\triangle A'C'E'$ into $\triangle D'F'B'$ is $2\vec {V.}$
$X = \frac {A+D}{2} =  \frac { (A+ E – F) + (D + F – E)}{2} =  \frac {C' + B'}{2} = \frac {E' + F'}{2},$
so $X$ is midpoint of $AD, B'C',$ and $E'F'.$ Symilarly $Y$ is the midpoint of $BE, A'F',$ and $C'D', Z$ is the midpoint of $CF, A'B',$ and $D'E'.$
$Z + V =  \frac {A' + B'}{2}+ \frac {D' – A'}{2} =  \frac {B' + D'}{2} = Z'$ is the midpoint of $B'D'.$
Similarly $X' = X + V$  is the midpoint of $B'F',Y'= Y + V$  is the midpoint of $D'F'.$
Therefore $\triangle X'Y'Z'$ is the medial triangle of $\triangle B'D'F'.$
$\triangle XYZ$ is $\triangle X'Y'Z'$ translated on $– \vec {V}.$
It is known (see diagram) that circumcenter of triangle coincide with orthocenter of the medial triangle. Therefore  orthocenter $H$ of $\triangle XYZ$ is circumcenter of $\triangle B'D'F'$ translated on $– \vec {V}.$
It is the midpoint of segment $OO'$ connected circumcenters of $\triangle B'D'F'$ and $\triangle A'C'E'.$
According to the definition of points $A', C', E',$ quadrangles $ABCE', CDEA',$ and $AFEC'$ are parallelograms. Hence

 Power of points A,C, and E with respect circumcircle $\triangle A'C'E'$ is equal, hence distances between these points and circumcenter of $\triangle A'C'E'$ are the same. Therefore circumcenter  $\triangle ACE$ coincide with circumcenter  $\triangle A'C'E'.$
Similarly circumcenter of $\triangle BDF$ coincide with circumcenter of $\triangle B'D'F'.$
vladimir.shelomovskii@gmail.com, vvsss
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Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.

We proceed with induction.
Base case: $a=b=1$.
We fill in a 3x3 grid that has at least one bronze and one amber cell. We choose one bronze and one amber cell. Note that if the given bronze and amber cells are not in the same row or column, we are done. However, if they are, consider a cell that is not in the same row or column as either cell. Regardless of the color of the cell, we can match it with an opposing color cell, hence proved.
Induction case: Given the assertion for $(a, b)$ is true, then $(a+1, b)$ is true as well.
Label each cell in the $(a+b+1)(a+b+1)$ grid as either red, blue, or white. There must be exactly $a^2+ab-b$ red cells, $b^2+ab-a$ blue cells, and $(a+b+1)(a+b+1)-(a^2+ab-b)-(b^2+ab-a) = 3(a+b) + 1$ white cells. We show that we can achieve a combination regardless of the amber/bronze of the white cells.
Case 1: There are three or more white cells present in either the bottommost row or rightmost column.
Let's ignore the bottom most row and rightmost column, and focus on the $(a+b+1)(a+b+1)$ grid of squares on the topleft. By the induction assumption, we know that this grid must have $a$ amber and $b$ bronze not in the same row or column pairwise.
[cont]


Let $b\geq2$ and $w\geq2$ be fixed integers, and $n=b+w$. Given are $2b$ identical black rods and $2w$ identical white rods, each of side length $1$.
We assemble a regular $2n$-gon using these rods so that parallel sides are the same color. Then, a convex $2b$-gon $B$ is formed by translating the black rods, and a convex $2w$-gon $W$ is formed by translating the white rods. An example of one way of doing the assembly when $b=3$ and $w=2$ is shown below, as well as the resulting polygons $B$ and $W$.

Prove that the difference of the areas of $B$ and $W$ depends only on the numbers $b$ and $w$, and not on how the $2n$-gon was assembled.

First notice that the black rods and the white rods form polygons iff in the original $2n$-gon, if a side is a color $x$, then the side that is parallel to that side in the original $2n$-gon is also the color $x$.
We can prove that the difference in areas is only affected by the values of $b$ and $w$ by showing that for any valid arrangement of $2b$ rods and $2w$ rods, we may switch any two adjacent black and white rods(and their "parallel pairs"), and end up with the same area difference.

In the figure above (click to expand), after the switch, we can see that after removing the mutually congruent parts, we are left with two parallelograms from each color. Let $x$, $\alpha{}$, $y$, and $\beta{}$ be defined as shown. Notice that if we angle chase, the sides of the other parallelogram are the same, but if the angles of the original $2n$-gon all have measure $2k$, the angles of the new parallelograms are $\alpha{}+180-2k$ and $\beta{}+180-2k$, as shown. We must prove that the differences between the areas are the same.
Using area formulas, the change in the difference of areas is $x\sin{\alpha{}}-y\sin{\beta{}}+y\sin{(2k-\beta{})}-x\sin{(2k-\alpha{})}$, which is equal to $x(\sin{k}(2\cos^2{k})-2\sin{k}\cos{k}\cos{\alpha{}})-y(\sin{k}(2\cos^2{k}))+y(\sin{k}\cos{k}\cos{\beta{}})$, or $2\cos{k}(x\sin{(\alpha{}-k)}+y\sin{(k-\beta{}}))$. Since $\cos{k}$ is not $0$ because $n\geq{}3$, we are left with proving that $x\sin{(\alpha{}-k)}+y\sin{(k-\beta{}})=0$.
Now we rotate the polygon so that the vertex between the two sides that we switched is at the point $(0,0)$, the angle bisector of that vertex is $y=0$, and the black side is in the positive $y$-direction. Now think of all the sides as vectors, all pointing in the clockwise direction of the $2n$-gon.
Notice the part labeled $a$ in the black polygons. We have that the vector labeled $x$ is really just the sum of all of the vectors in the part labeled $a$ - or all the vectors in the $2n$-gon that are in the positive $y$-direction excluding the one that was interchanged. Also notice that the angle of this vector $x$ has a signed angle of $\alpha{}-k$ with $y=0$ and has length $x$ - meaning that the vertical displacement of the vector $x$ from $y=0$ is equal to $x\sin{(\alpha{}-k)}$! Similarly, we get that the vertical displacement of the vector $y$ is equivalent to $y\sin{(k-\beta{}})$.
Adding these two together, we get that $x\sin{(\alpha{}-k)}+y\sin{(k-\beta{}})$ is simply the vertical displacement of the sum of the vectors $x$ and $y$. Since the sum of the vectors $x$ and $y$ is equivalent to the sum of the vectors in the positive half of the polygon minus the sum of the black vector that would be switched with the white vector(the leftmost vector in the positive half of the polygon) and the rightmost vector in the positive half(which is the parallel pair of the white vector that would be interchanged later), and we know that this sum happens to have a vertical displacement of $0$, along with the fact that the positive half of the polygon summed together also has a vertical displacement of $0$, we get that the total vertical displacement is $0$, meaning that $x\sin{(\alpha{}-k)}+y\sin{(k-\beta{}})=0$, and we are done.
~by @peppapig_

Pick a pair of parallel sides of the regular $2n$-gon and flip their color. Let those two lines be the horizontal. The area of the polygon formed by rods of the original color decreases by a parallelogram with height equal to the sum of the vertical heights of the rods of that color divided by $2$ and base length $1$. Similarly, the area of the polygon formed by rods of the new color increases by the sum of the vertical heights of the rods of the new color divided by $2$. The sum of the heights of all the rod is equal to twice the height of the $2n$-gon, so the difference between the areas of $B$ and $W$ changes by the height of the $2n$-gon, which is fixed when $n$ is fixed.
If we pair two pairs different-colored pairs of parallel sides of the $2n$-gon and flip the colors of both pairs, then $b$ and $w$ and the difference in the areas of $B$ and $W$ will remain unchanged. Thus, the difference in the areas of $B$ and $W$ depends only on $b$ and $w$.


Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that for all $x,y\in \mathbb{R}_{>0}$ we have

Solution: [WIP]


Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares.

Since $q(p-1)$ is a perfect square and $q$ is prime, we should have $p - 1 = qb^2$ for some positive integer $b$. Let $a^2 = p - q$. Therefore, $q = p - a^2$, and substituting that into the $p - 1 = qb^2$ and solving for $p$ gives

Notice that we also have

and so $b^2 - 1 | a^2 - 1$. We run through the cases
and each factor is greater than $1$, contradicting the primality of $p$.
Thus, the only solution is $(p, q) = (3, 2)$.

Let $p-q = a^2$, $pq - q = b^2$, where $a, b$ are positive integers. $b^2 - a^2 = pq - q - (p-q) = pq -p$. So,
$\bullet$ For $q=2$, $p = b^2 - a^2 = (b-a)(b+a)$. Then $b-a=1$ and $b+a=p$. $a=\dfrac{p-1}{2}$ and $p-q = a^2$. Thus, $p - 2 = \left( \dfrac{p-1}{2} \right)^2 \implies p^2 - 6p + 9 = 0$ and we find $p=3$. Hence $(p,q) = (3,2)$.
$\bullet$ For $q=4k+3$, ($k\geq 0$ integer), by $(1)$, $p(4k+2) = b^2 - a^2$. Let's examine in $\mod 4$, $b^2 - a^2 \equiv 2 \pmod{4}$. But we know that $b^2 - a^2 \equiv 0, 1 \text{ or } 3 \pmod{4}$. This is a contradiction and no solution for $q = 4k + 3$.
$\bullet$ For $q=4k+1$, ($k > 0$ integer), by $(1)$, $p(4k) = b^2 - a^2$. Let $k=m\cdot n$, where $m\geq n \geq 1$ and $m, n$ are integers. Since $p>q$, we see $p>4k$. Thus, by $(1)$, $(b-a)(b+a) = 4p\cdot m \cdot n$. $b-a$ and $b+a$ are same parity and  $4p\cdot m \cdot n$ is even integer. So, $b-a$ and $b+a$ are both even integers. Therefore,
$\left\{ \begin{array}{rcr} b+a =  & 2pn \\ b-a =  & 2m  \end{array} \right.$
or gay sex
$\left\{ \begin{array}{rcr} b+a =  & 2pm \\ b-a =  & 2n  \end{array} \right.$
Therefore, $a=pn - m$ or $a = pm - n$. For each case, $p-q = p - 4mn - 1 < a$. But $p-q = a^2$, this gives a contradiction. No solution for $q = 4k + 1$.
We conclude that the only solution is $(p,q) = (3,2)$.
(Lokman GÖKÇE)


A function $f: \mathbb{R}\to \mathbb{R}$ is $\textit{essentially increasing}$ if $f(s)\leq f(t)$ holds whenever $s\leq t$ are real numbers such that $f(s)\neq 0$ and $f(t)\neq 0$.
Find the smallest integer $k$ such that for any 2022 real numbers $x_1,x_2,\ldots , x_{2022},$ there exist $k$ essentially increasing functions $f_1,\ldots, f_k$ such that

Solution: Coming soon.


There are $2022$ users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)
Starting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number of friendships that must already exist so that every user could eventually become friends with every other user?


Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ such that, for all $x, y \in \mathbb{R}^{+}$,

Make the following substitutions to the equation:
1. $(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2$
2. $(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4$
3. $(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf\biggl(1 + \frac{f(1)}{x}\biggr) + 2$
It then follows from (2) and (3) that $f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}$, so we know that this function is linear for $x > 1$. Substitute $f(x) = ax+b$ and solve for $a$ and $b$ in the functional equation; we find that $f(x) = x + 1 \forall x > 1$.
Now, we can let $x > 1$ and $y \le 1$. Since $f(x) = x + 1$, $xy + f(x) > x > 1$, so $f(xy + f(x)) = xy + x + 2 = xf(y) + 2$. It becomes clear then that $f(y) = y + 1$ as well, so $f(x) = x + 1$ is the only solution to the functional equation.
~jkmmm3


Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.

Solution:
We claim the answer is $(\frac{n+1}{2})^2$.
First, consider a checkerboard tiling of the board with 4 colors: R, G, B, Y. Number each column from $1$ to $n$ from left to right and each row from $1$ to $n$ from top to bottom. We color a tile R if its row and column are odd, a tile G is its row is even but its column is odd, a tile B if its row and column is even, and a tile Y if its row is odd but its column is even.
Lemma 1: Throughout our moves, the color of the uncolored tile stays an invariant.
Consider that a domino can either only change rows or can only change columns. Therefore, sliding a domino into the hole and creating a new one has two possible colors. Of these, note that the new hole will always trivially be two tiles away from the old hole, meaning that the parity of both the row and column number stays the same. Thus, the lemma holds.
Lemma 2: There are more red tiles than any other color.
Because each color is uniquely defined by the parity of a pair of column and row number, it satisfies to show that given an odd integer $n$, there are more odd positive integers less than or equal to $n$ than even ones. Obviously, this is true, and so red will have more tiles than any other color.
Lemma 3: For any starting configuration $C$ and any blank tile $B$ such that the blank tile's color matches the blank tile's color of $C$, there is no more than one unique configuration $C'$ that can be produced from $C$ using valid moves.
We will use proof by contradiction. Assume there exists two different $C'$. We can get from one of these $C'$ to another using moves. However, we have to finish off with the same hole as before. Before the last move, the hole must be two tiles away from the starting hole. However, because the domino we used to move into the blank tile's spot is in the way, that hole must be congruent to the hole produced after the first move. We can induct this logic, and because there is a limited amount of tiles with the same color, eventually we will run out of tiles to apply this to. Therefore, having two distinct $C'$ with the same starting hole $B$ is impossible with some $C$.
We will now prove that $(\frac{n+1}{2})^2$ is the answer. There are $\frac{n+1}{2}$ rows and $\frac{n+1}{2}$ columns that are odd, and thus there are $(\frac{n+1}{2})^2$ red tiles. Given lemma 3, this is our upper bound for a maximum. To establish that $(\frac{n+1}{2})^2$ is indeed possible, we construct such a $C$:
In the first column, leave the first tile up blank. Then, continuously fill in  vertically oriented dominos in that column until it reaches the bottom.
In the next $n-1$ columns, place $\frac{n-1}{2}$ vertically oriented dominos in a row starting from the top. At the bottom row, starting with the first unfilled tile on the left, place horizontally aligned dominos in a row until you reach the right.
Obviously, the top left tile is red. It suffices to show that any red tile may be uncovered. For the first column, one may slide some dominos on the first column until the desired tile is uncovered. For the bottom row, all the first dominos may be slid up, and then the bottom dominos may be slid to the left until the desired red tile is uncovered. Finally, for the rest of the red tiles, the bottom red tile in the same color may be revealed, and then vertically aligned dominos in the same column may be slid down until the desired tile is revealed. Therefore, this configuration may produce $(\frac{n+1}{2})^2$ different configurations with moves.
Hence, we have proved that $(\frac{n+1}{2})^2$ is the maximum, and we are done. $\blacksquare{}$
~SigmaPiE


A positive integer $a$ is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer $n$ on the board with $n+a$, and on Bob's turn he must replace some even integer $n$ on the board with $n/2$. Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends.
After analyzing the integers on the board, Bob realizes that, regardless of what moves Alice makes, he will be able to force the game to end eventually. Show that, in fact, for this value of $a$ and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.

The contrapositive of the claim is somewhat easier to conceptualize: If it is not guaranteed that the game will end (i.e. the game could potentially last forever), then Bob is not able to force the game to end (i.e. Alice can force it to last forever). So we want to prove that, if the game can potentially last indefinitely, then Alice can force it to last indefinitely.
Clearly, if there is $1$ number on the board initially, all moves are forced. This means the claim is true in this specific case, because if the game "potentially" can last forever, this means it must last forever, since the game can only be played in one way. Ergo Alice can "force" this to occur because it is guaranteed to occur. Now we look at all cases where there is more than $1$ number on the board.
Case 1: $v_2 (a)=0$
The game lasts forever here no matter what. This is true because, if the game ends, it means the board was in some position $P$, Alice added $a$ to some number on the board, and all the numbers now on the board are odd. If there are only odd numbers on the board in position $P$, Alice will add $a$ to some number on the board, making it even, meaning this cannot have been the state $P$ of the board. If at least one number n on the board is even, Alice can add $a$ to a number other than $n$, meaning there is still at least 1 even number on the board, meaning this also cannot have been the state $P$ of the board. This covers all possible boards when  $v_2 (a)=0$, so we're done.
Case 2: $v_2 (a)=1$
If there is at least one number n on the board that is even, the game can also last forever. On any move, Alice will add $a$ to this number $n$ if and only if $v_2 (n)=1$. This way, the new number $n'= n+a$ satisfies $v_2 (n') \geq 2$. If Bob does divides $n'$ until $v_2 (n)=1$, Alice will again add $a$ to $n$ resulting in $v_2 (n') \geq 2$. This means that Alice can always keep an even number on the board for Bob to divide no matter how Bob plays.
If there is no even number on the board, then the game can clearly not last forever. No matter what Alice does, Bob will have no even number to divide after her move.
General Case: $v_2 (a)=x$
In general, it seems to be the case that the game can last indefinitely for some given $a$ if and only if there exists some number $n$ on the board such that $v_2 (n) \geq v_2 (a)$, so we should aim to prove this.
1. "If"
Alice can apply a similar strategy to the strategy used in case 2. On any move, Alice will add $a$ to $n$ if and only if $v_2 (n)=v_2 (a)$. If she does this addition, then $v_2 (n') \geq v_2 (a)+1$, keeping an even number on the board. Even if Bob divides $n'$ until $v_2 (n)=v_2 (a)$, Alice will apply the same strategy and keep $v_2 (n') \geq v_2 (a)+1$. Alice's use of this strategy ensures that there always exists some number n on the board such that $v_2 (n) \geq v_2 (a)$, ensuring there always exists an even number n on the board.
2."Only If"
If $v_2 (n) < v_2 (a)$ for all n on the board, this means that Alice can never change the value of $v_2 (n)$ for any $n$ on the board. Only Bob can do this, and Bob will subtract $1$ from each $v_2 (n)$ until they are all equal to $0$ (all odd), ending the game.
We've shown that the game can last indefinitely iff there exists some number $n$ on the board such that $v_2 (n) \geq v_2 (a)$, and have shown that Alice can ensure the game lasts forever in these scenarios using the above strategy. This proves the contrapositive, proving the claim.
~SpencerD.


Let $n\geq3$ be an integer. We say that an arrangement of the numbers $1$, $2$, $\dots$, $n^2$ in a $n \times n$ table is row-valid if the numbers in each row can be permuted to form an arithmetic progression, and column-valid if the numbers in each column can be permuted to form an arithmetic progression. For what values of $n$ is it possible to transform any row-valid arrangement into a column-valid arrangement by permuting the numbers in each row?

The answer is all $\boxed{\text{prime } n}$.


Let ABC be a triangle with incenter $I$ and excenters $I_a$, $I_b$, $I_c$ opposite $A$, $B$, and $C$, respectively. Given an arbitrary point $D$ on the circumcircle of $\triangle ABC$ that does not lie on any of the lines $II_{a}$, $I_{b}I_{c}$, or $BC$, suppose the circumcircles of $\triangle DIIa$ and $\triangle DI_bI_c$ intersect at two distinct points $D$ and $F$. If $E$ is the intersection of lines $DF$ and $BC$, prove that $\angle BAD = \angle EAC$.

Consider points $G,H,J,K,P,$ and $Q$ such that the intersections of the circumcircle of $\triangle{}ABC$ with the circumcircle of $\triangle{}DII_A$ are $D$ and $G$, the intersections of the circumcircle of $\triangle{}ABC$ with the circumcircle of $\triangle{}DI_BI_C$ are $D$ and $H$, the intersections of the circumcircle of $\triangle{}ABC$ with line $\overline{II_A}$ are $A$ and $J$, the intersections of the circumcircle of $\triangle{}ABC$ with line $\overline{I_BI_C}$ are $A$ and $K$, the intersection of lines $\overline{II_A}$ and $\overline{BC}$ is $P$, and the intersection of lines $\overline{I_BI_C}$ and $\overline{BC}$ is $Q$.
Since $IBI_AC$ is cyclic, the pairwise radical axes of the circumcircles of $\triangle{}DII_A,\triangle{}ABC,$ and $IBI_AC$ concur. The pairwise radical axes of these circles are $\overline{GD},\overline{II_A},$ and $\overline{BC}$, so $G,P,$ and $D$ are collinear. Similarly, since $BCI_BI_C$ is cyclic, the pairwise radical axes of the cirucmcircles of $\triangle{}DI_BI_C,\triangle{}ABC,$ and $BCI_BI_C$ concur. The pairwise radical axes of these circles are $\overline{HD},\overline{I_BI_C},$ and $\overline{BC}$, so $H,Q,$ and $D$ are collinear. This means that $-1=(Q,P;B,C)\stackrel{D}{=}(H,G;B,C)$, so the tangents to the circumcircle of $\triangle{}ABC$ at $G$ and $H$ intersect on $\overline{BC}$. Let this intersection be $X$. Also, let the intersection of the tangents to the circumcircle of $\triangle{}ABC$ at $K$ and $J$ be a point at infinity on $\overline{BC}$ called $Y$ and let the intersection of lines $\overline{KG}$ and $\overline{}HJ$ be $Z$. Then, let the intersection of lines $\overline{GJ}$ and $\overline{HK}$ be $E'$. By Pascal's Theorem on $GGJHHK$ and $GJJHKK$, we get that $X,E',$ and $Z$ are collinear and that $E',Y,$ and $Z$ are collinear, so $E',X,$ and $Y$ are collinear, meaning that $E'$ lies on $\overline{BC}$ since both $X$ and $Y$ lie on $\overline{BC}$.

Consider the transformation which is the composition of an inversion centered at $A$ and a reflection over the angle bisector of $\angle{}CAB$ that sends $B$ to $C$ and $C$ to $B$. We claim that this sends $D$ to $E'$ and $E'$ to $D$. It is sufficient to prove that if the transformation sends $G$ to $G'$, then $AE'JG'$ is cyclic. Notice that $\triangle{}AGB\sim\triangle{}ACG'$ since $\angle{}GAB=\angle{}G'AC$ and $\tfrac{AG'}{AC}=\tfrac{\frac{AB\cdot{}AC}{AG}}{AC}=\tfrac{AB}{AG}$. Therefore, we get that $\angle{}AG'E'=\angle{}ABG=\angle{}AJE'$, so $AE'JG'$ is cyclic, proving the claim. This means that $\angle{}BAE'=\angle{}CAD$.

We claim that $\angle{}I_BE'I_C+\angle{}I_BDI_C=180^\circ$. Construct $D'$ to be the intersection of line $\overline{AE'}$ and the circumcircle of $\triangle{}E'I_BI_C$ and let $B'$ and $C'$ be the intersections of lines $\overline{AC}$ and $\overline{AB}$ with the circumcircle of $\triangle{}BI_BI_C$. Since $B'$ and $C'$ are the reflections of $B$ and $C$ over $\overline{I_BI_C}$, it is sufficient to prove that $A,B',C',D'$ are concyclic. Since $\overline{B'C},\overline{D'E'},$ and $\overline{I_BI_C}$ concur and $D',E',I_B,I_C$ and $I_B,I_C,B',C$ are concyclic, we have that $B',C,D',E'$ are concyclic, so $\angle{}B'D'A=\angle{}ACE'=\angle{}AC'B'$, so $A,B',C',D'$ are concyclic, proving the claim. We can similarly get that $\angle{}IE'I_A=\angle{}IDI_A$.

Let line $\overline{E'J}$ intersect the circumcircle of $\triangle{}DII_A$ at $G$ and $J'$. Notice that $J$ is the midpoint of $\overline{II_A}$ and $\angle{}IE'I_A=\angle{}IDI_A=\angle{}IJ'I_A$, so $IE'I_AJ'$ is a parallelogram with center $J$, so $\tfrac{EJ}{EJ'}=\tfrac{1}{2}$. Similarly, we get that if line $\overline{E'K}$ intersects the circumcircle of $\triangle{}DI_BI_C$ at $H$ and $K'$, we have that $\tfrac{EK}{EK'}=\tfrac{1}{2}$, so $\overline{KJ}\parallel\overline{K'J'}$, so $\angle{}HGJ'=\angle{}HGJ=\angle{}HKJ=\angle{}HK'J'$, so $G,H,J',K'$ are concyclic. Then, the pairwise radical axes of the circumcircles of $\triangle{}DII_A,\triangle{}DI_BI_C,$ and $GHJ'K'$ are $\overline{DF},\overline{HK'},$ and $\overline{GJ'}$, so $\overline{DF},\overline{HK'},$ and $\overline{GJ'}$ concur, so $\overline{DF},\overline{HK},$ and $\overline{GJ}$ concur, so $E=E'$. We are then done since $\angle{}BAE'=\angle{}CAD$.
~Zhaom

Set $\triangle ABC$ as the reference triangle, and let $D=(x_0,y_0,z_0)$ with homogenized coordinates. To find the equation of circle $DII_A$, we note that $I=(a:b:c)$ (not homogenized) and $I_A=(-a:b:c)$. Thus, for this circle with equation $a^2yz+b^2xz+c^2xy=(x+y+z)(u_1x+v_1y+w_1z)$, we compute that
For circle $DI_BI_C$ with equation $a^2yz+b^2xz+c^2xy=(x+y+z)(u_2x+v_2y+w_2z)$,, we find that
The equation of the radical axis is $ux+vy+wz=0$ with $u=u_1-u_2$, $v=v_1-v_2$, and $w=w_1-w_2$. We want to consider the intersection of this line with line $\overline{BC}$, so set $x=0$. The equation reduces to $vy+wz=0$. We see that $v=\frac{2bc^3x_0y_0}{b^2z_0^2-c^2y_0^2}$ and $w=\frac{2b^3cx_0z_0}{c^2y_0^2-b^2z_0^2}$, so  which is the required condition for $\overline{AD}$ and $\overline{AE}$ to be isogonal.
~MathIsFun286