Find all sets of six points in the plane, no three collinear, such that if we partition the set into two sets, then the obtained triangles are congruent.
Problem
Source: Sharygin 2018 Final Round, Grade 8 P4
Tags: geometry, combinatorics, congruent triangles
Wizard_32
31.07.2018 17:45
The official solution uses the Barycentric Leibniz theorem and some weird lemma. The answer is a hexagon with odd numbered sides and even numbered sides respectively equal.
RIP P4
62861
02.08.2018 06:00
First we show by contradiction that the points form a convex hexagon.
Let the convex hull be $\Gamma$.
Lemma. There exists a triangle of maximum area inscribed in $\Gamma$ using vertices of $\Gamma$. The proof is easy: first show that a maximal triangle $\mathcal{T}$ must have vertices on $\Gamma$, then by linearity one may push each vertex of $\mathcal{T}$ to a vertex of $\Gamma$.
Suppose for a contradiction that $\Gamma$ is not a hexagon; then let $PQR$ be the triangle of maximum area. Then the other three points $UVW$ form a triangle inside $\Gamma$ with equal area, impossible.
Label the points $ABCDEF$ in order. Then $[BCE] = [ADF]$ and $[BEF] = [ACD]$ whence $[BCDE] = [ACDF]$, and thus \[[ABF] + [CDE] = [ABC] + [DEF].\]But the two terms on each side are equal, and accordingly we obtain the areas of all six ears are equal. This forces $\overline{BC} \parallel \overline{EF} \parallel \overline{AD}$, etc. Now triangles $BCE$ and $ADF$ are congruent with $\angle CBE = \angle DAF$, so $CE = DF$. Then $CDEF$ is an isosceles trapezoid so $CD = EF$. In a similar way all six short diagonals of $ABCDEF$ are equal, $AB = CD = EF$, and $BC = DE = FA$. Then all ears are congruent, so $ABCDEF$ is equiangular. This is the final answer: any equiangular hexagon with alternating sides equal is valid.