SearcharxivSearch

arXiv · 2509.05844

The $3\times 3$ Solution to the Game of Tak

Abstract

We provide answers to natural combinatorial questions which arise from learning the rules to Tak. We solve the discarded cases of the $3\times 3$ game not previously considered by Joshua Achiam, showing that Player $1$ has a forced win from any starting position, including random ones. Diagrams for both his and our solutions are provided in the hopes of making the results more accessible.

Explore related subjects

Keep this discovery

Explore connections, maps & timelines

BibTeXRIS

Joseph E. Marrow. 2025-09-06. The $3\times 3$ Solution to the Game of Tak. https://arxiv.org/abs/2509.05844

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM