(Folklore) Consider the function sending a polygon to the polygon formed by its midpoints. Show that if we iterate this function, the limiting shape (rescaled appropriately) is an affine transform of a regular polygon.

Sketch:

• Write the explicit form out using root-of-unity filter.
• Consider every $2n$-th term (where $n$ is the number of sides of the polygon). Then, there is a clear leading order term.
• This gives us the explicit form of the “limiting shape”, which is some linear transform of the original polygon.
• Finally, if $a_i$ is the resulting vertex then we can note some relation like $a_{i-1} +a_{i+1} = c\cdot a_i$, and we can check that this $c$ is the same as if $a_i$’s formed a regular polygon.

Remarks. It’s also possible to consider the transform as a linear map on $\mathbb C^n$, then one can diagonalize that matrix to get that the limiting shape comes from the second eigenvector.

(Abyankhar-Moh) Show that if $K$ has characteristic $0$ and $K[f,g] = K[t]$ (where $f,g$ aren nonzero polynomials of $t$), then one of $\deg f, \deg g$ divides the other.

Proof. We sketch out from Richman 1984.

Let $N = [K(f,g) : K(g)]$. Then, we must have that $$1, f, f^2 , ..., f^{N-1}$$ are linearly independent over $K(f,g)$ as a $K(g)$-vector space (because the minimal polynomial of $f$ over $K(g)$ is of degree $N$). By a Gauss lemma argument, we can show that $f^N$ is a combination of lower powers with coefficients which are polynomial in $g$ (and not merely rational), so $$1, f, f^2, ..., f^{N-1}$$ is a basis for $K[f,g]$ (as a $K[g]$-module). Now, it is possible to reduce this set of polynomials so that their degrees are unique mod $\deg g$, which gives us (writing $e = \deg g / (\deg f, \deg g)$): $$1, f, f^2, ..., f^{e-1}, h_e, h_{e-1}, ..., h_{N-1}$$ Hence, every polynomial $p$ in $K[f,g]$ can be written in the form $$p = a_0(g) + a_1(g) \cdot f + a_2(g) \cdot f^2 + ... + a_e(g) h_e + ... + a_{N-1}(g) h_{N-1}$$

and furthermore, because the degrees of the basis are unique mod $\deg g$, we must have $\deg p = \deg a_i h_i$ for some unique $i$ (writing $h_i = f^i$ for $i0$ in the process), so either $\deg f | \deg g$ or $A=0$ (in which case $\deg g | \deg f$).

Remark. This was surprisingly hard, and I don’t think I could have came up with it myself. If I’m not mistaken, the original proof was geometric in nature, so I wonder whether the corresponding geometric argument is easier.

(AoPS) Show that the minimal polynomial of $\sqrt[k]{p_1}+\sqrt[k]{p_2}+…+\sqrt[k]{p_n}$ is a polynomial of $x^k$, where $p_i$ are distinct primes.

Proof. The Galois conjugates of $\sqrt[k]{p_1}+\sqrt[k]{p_2}+…+\sqrt[k]{p_n}$ are $$\zeta^{a_1}\sqrt[k]{p_1}+\zeta^{a_2}\sqrt[k]{p_2}+…+\zeta^{a_n}\sqrt[k]{p_n}$$ where $\zeta$ is a primitive $k$-th root of unity, for $a_i=0,1,...,k-1$. We can group them together if they happen to be a rotation (root-of-unity multiple) of each other. Then, the minimal polynomial is $$ \begin{align*} P(x) &= \prod_{a_1,...,a_n} (x-\zeta^{a_1}\sqrt[k]{p_1}+\zeta^{a_2}\sqrt[k]{p_2}+…+\zeta^{a_n}\sqrt[k]{p_n})\\ &= \prod_{a_2',...,a_n'} \prod_{a_1=0}^{k-1} (x-\zeta^{a_1}\sqrt[k]{p_1}+\zeta^{a_1+a_2'}\sqrt[k]{p_2}+…+\zeta^{a_1+a_n'}\sqrt[k]{p_n}) \\ &= \prod_{a_2',...,a_n'} \prod_{a_1=0}^{k-1} (x-\zeta^{a_1}(\sqrt[k]{p_1}+\zeta^{a_2'}\sqrt[k]{p_2}+…+\zeta^{a_n'}\sqrt[k]{p_n})) \\ &= \prod_{a_2',...,a_n'} (x^k-(\sqrt[k]{p_1}+\zeta^{a_2'}\sqrt[k]{p_2}+…+\zeta^{a_n'}\sqrt[k]{p_n})^k).\\ \end{align*} $$

(AoPS) For eight points in general position, let $\mathcal{K}$ be a cubic passing through the eight points. Let $P$ be a different point. Let the six tangents drawn from $P$ to $\mathcal{K}$ touch it at $T_1, T_2, T_3, T_4, T_5, T_6$. Prove all $T_i$, over all choices of $\mathcal{K}$, lie on a fixed degree-5 plane curve.

Proof.

Let $K$ be the equation of $\mathcal K$. Since $\mathcal K$ is part of a pencil, $K$ must be the linear combination of two fixed cubics $K_1, K_2$ (so we can essentially write $K = \lambda K_1 + \mu K_2$).

If $P=(P_x, P_y)$, the condition becomes $$\frac{y-P_y}{x-P_x} = -\frac{\partial_x K}{\partial_y K}$$ Therefore: $$(x\partial_x + y\partial_y - P_x\partial_x - P_y \partial_y) K = 0$$ for $(x,y)$ being a tangency point. Furthermore, $(x,y)$ also satisfies $K=0$. Write $D = x\partial_x + y\partial_y - P_x\partial_x - P_y \partial_y$. Thus we have the system $$ \begin{align*} \lambda K_1 + \mu K_2 &= 0\\ \lambda D K_1 + \mu DK_2 & = 0. \end{align*} $$ We can eliminate $\lambda$ and $\mu$ by Gaussian elimination (or just by looking at the determinant) to get a single equation $$K_1 (DK_2) - K_2(DK_1) = 0.$$ However, this is degree 6 (and not 5 as desired) so it remains to check that the degree 6 terms cancel out. For this purpose it is sufficient to assume that $K_1,K_2$ are homogeneous cubics and $D = x\partial_x + y\partial_y$. However, it is easy to check that we must have $DK_1 = 3K_1$ and $DK_2=3K_1$ so the conclusion follows.

Remark. In general, the tangency point from $P$ to a degree $n$ curve which is part of a linear family should belong on a curve of degree $3n-1$. If this linear family happens to be coaxal circles, then we have that the tangency points from $P$ to any circle lies on the same conic. I’m not sure if this is an obvious fact.