I took part in the InfinityDots MO hosted by talkon.

Q1. We call $\{x,y,z\}$ a Pythagorean triple if $x^2+y^2 = z^2$ and $z\ge 0$. It suffices to show that from any Pythagorean pair we can reach $\{0,1\}$ using the given moves.

Firstly, if $\{x,y\}$ is Pythagorean with $x,y> 0$ (since if $x=0$ then $y=1$ since $x,y$ are coprime), and futhermore we may set $z=\sqrt{x^2+y^2}$ to make $\{x,y,z\}$ a Pythagorean triple. Now we claim that the line $(x-t,y-t,z-t)$ intersects the cone again at a lattice point $(x',y',z')$. Indeed, the equation $$(x-t)^2 + (y-t)^2 = (z-t)^2$$ or equivalently $$t^2 - 2(x+y-z)t = 0$$ has a second root at $t = 2(x+y-z)$. Furthermore, this new point satisfies $0 < z' < z$, hence we can use the other move to flip the signs of $x'$ and $y'$ and keep repeating this process.

Q2. Assume otherwise that for all $m\ge N$,

$$2m\cdot a_m \le (4m-3)\cdot a_{2m-1}$$

This gives

$$ \begin{align*} a_{2^km+1}&\ge \frac{2(2^{k-1}m+1)}{2^{k+1}m+1}a_{2^{k-1}m+1}\\ &\ge … \\ &\ge \frac{2^k(2m+1)(m+1)}{(2^{k+1}m+1)(2^km+1)}a_{m+1}\\ &\ge \left(\frac{1}{2^{k}m+1}-\frac{1}{2^{k+1}{m}+1}\right)\frac{(2m+1)(m+1)}{m}a_{m+1} \end{align*}$$

Then consider

$$\begin{align*}a_m &\ge a_{2m+1} + a_{4m+1} + …\\ &\ge \frac{(m+1)}{m}a_{m+1}\end{align*}$$

Hence $a_m\ge \frac{2m-1}{m}a_{2m-1}$, contradicting our assumption

Q3. Let $\Gamma$ be the circle internally tangent to $\Omega$ (at $K'$) and tangent to segments $AB,AC$ (so that it is externally homothetic to $\omega$ through $A$). By Monge, $A,T,K'$ are collinear, so $K'$ is one of $K$ or $L$. Using lemma 1, $K'I$ bisects $\angle BK'C$, and using lemma 2 we conclude that $K'I$ bisects $\angle AK'D$.

Lemma 1. Let $ABCD$ be a cyclic quadrilateral circumscribed in circle $\gamma$. Let $X=AC\cap BD$, and let the circle tangent to segments $XC,XD$ and circle $\gamma$ touch $\gamma$ at $Y$, and let $I$ be the incenter of $\triangle XCD$. Then $YI$ bisects $\angle CYD$.

Proof. Let $I_1$ and $I_2$ be the incircles of $\triangle ACD$ and $\triangle BCD$, respectively. Using #6 in http://yufeizhao.com/olympiad/geolemmas.pdf (specifically, the lemma about $AKI'F'$ concyclic), we can angle chase a little to show that $CI_2IY$ and $DI_1IY$ are concyclic, then the fact is equivalent to $\angle DI_1I=\angle CI_2I$ which is obvious.

Lemma 2. Given a quadrilateral $ABCD$ circumscribed about an incircle centered at $I$, then $MI$ bisects $\angle AMC$ if and only if $MI$ bisects $BMD$.

Proof. For every point $X$, let $x$ denote its polar w.r.t. the incircle, and let $(xy)$ denote the intersection of lines $x,y$. Now, $MI$ bisects $\angle AMC$ iff $IM$ bisects $\angle (am)I(cm)$. However, by Desargues Involution theorem, $(am),(cm),(bm),(dm)$ and $m\cap (I)$ forms an involution, so $I(am)=I(cm)$ iff $I(bm)=I(dm)$.

Q4. We claim that if all the cells are cut, we will get at least $2n$ connected components. We can argue by lazers: if we imagine that the cuts are two-sided mirrors, then by shining a lazer perpendicularly into the grid, it will trace a path and exit the grid in another side of a cell. Since there are $4n$ such sides and they pair up in the above fashion (of course it is impossible to return to the same side), and furthermore note each path by a lazer is its own connected component, hence our claim is shown.

Now if there were $m < 2n-1$ cells that were uncut, then initially there must have been at least $2n - m > 1$ connected components.

Q6. Let’s rewrite the condition slightly. Let $h(x) = g(x) + x + \frac{1}{2}f(x)$, then the condition becomes $$h(a)f(b) + h(b)f(a)\le ab$$ For a given $f$, each $g$ gives one unique $h$ and vice versa.

Now, assume that a solution $h$ exists.

Firstly, for each $\mu\in \mathbb{R}$ such that $x + \mu f(x)\ge 0$ for all $x\in\mathbb{R}$, we get

$$ab \ge -(\mu af(b) + \mu bf(a) + \mu^2f(a)f(b))$$

so $h(x) = -\mu x - \frac{1}{2} \mu^2 f(x)$ is a solution. This implies that there can only be one unique value of $\mu$.

Suppose that $h$ is a solution. Then if $f(\alpha) = 0$ for some $\alpha\neq 0$, then:

$$\alpha (b - \frac{h(\alpha)}{\alpha} f(b))\ge 0$$

so we get $\mu = -\frac{h(\alpha)}{\alpha}$ works. Also note that $h(\alpha)\neq 0$.

Now suppose that $f(x)\ge 0$ for all 0, then if $f(\alpha) = 0$ for some $\alpha\neq 0$, then every $\mu$ between 0 and $-\frac{h(\alpha)}{\alpha}$. works, so we get infinitely many solutions. Otherwise, we can set $h(x) = -\frac{\lambda x^2}{f(x)}$ (for some fixed $\lambda > \frac{1}{2}$) when $x\neq 0$ and $h(0) =0$, then

$$h(a)f(b) + h(b)f(a) - ab\le -\frac{1}{2}(a\sqrt{\frac{f(b)}{f(a)}} + b\sqrt{\frac{f(a)}{f(b)}})^2 \le 0$$

The case $f(x)\le 0$ is similar.

Otherwise, $f$ must take both positive and negative values. Consider any $x,y$. Let $a_1=x$, $a_2=y$, then we claim that because $f$ is continuous, there exists $a_3,…,a_n$ such that $\sum f(a_i) = 0$, and unless $f(x)=cx$ for some $x$ we can also enforce that $\sum a_i \neq 0$. Then summing the given condition across all $a=a_i$,

$$f(b)\sum h(a_i)\le b\sum a_i$$

Across all possible $(a_3,…,a_n)$, we expect $(\sum h(a_i))/(\sum a_i)$ to have the same value $\mu$. Now we adjust the values slightly: instead of doing $\sum a_i$, we will do $\sum \lambda_i a_i$, where $\lambda_1 = 1 + \epsilon f(y)$ and $\lambda_2=1-\epsilon f(x)$ and $\lambda_i=1$ otherwise (where we pick $\epsilon>0$ such that the $\lambda_i$’s are all positive). Now, we sum once again to get that $(\sum \lambda_i h(a_i))/(\sum \lambda a_i) = \mu$. Expanding, this means that $f(x)h(y)-f(y)h(x)=\mu(yf(x)-xf(y))$, or $h(x) = cf(x)+\mu x$ for some $c$. Now we substitute this back to get:

$$2cf(a)f(b) + 2\mu a f(b) + 2\mu bf(a) \ge ab$$

We pick $\sum f(a_i)=0$ again, but this time:

$$(\sum a_i) (2\mu f(b) - b)\ge0$$

So either $\sum a_i$ is always 0 or $(2\mu f(b)-b)$ is always 0. In either case, we can check that $f$ is linear.