Show that there exists a convex hexagon in the plane such that (a) all its interior angles are equal, (b) all its sides are 1, 2, 3, 4, 5, 6 in some order.
it is the 9th question inmo 1993. i cant even start this question..
$\endgroup$ 04 Answers
$\begingroup$A really simple solution makes the observation that any convex equiangular hexagon must have interior angle measure of $2\pi/3$, and thus can be constructed by considering truncation of the vertices of a suitable equilateral triangle. Suppose this triangle has side length $x$, and the vertices are truncated by lines parallel to the opposite side (so that the truncated pieces are themselves equilateral triangles). Let these pieces have side length $a$, $b$, $c$, respectively. Then the resulting equiangular hexagon has side lengths $$\{a, x - a - b, b, x - b - c, c, x - c - a\}.$$ It then suffices to find $a, b, c, x$ such that the above set is a permutation of $\{1, 2, 3, 4, 5, 6\}$. This can be done by trial and error; e.g., $a = 1$, $b = 2$, $c = 3$, $x = 9$; or by simply checking at most $\binom{6}{3} = 20$ possibilities for $a, b, c$.
$\endgroup$ 0 $\begingroup$Draw an equilateral triangle with side length 9.
Chop off equilateral triangles at each corner of side length 1, 2 and 3.
The resulting hexagon satisfies the properties
First, the vertices have to fit in a regular triangular lattice: Just check that no move of the (imaginary) drawing turtle would leave that lattice, provided that you have aligned the initial point and the initial move with the lattice. Next, you may profitably go topological and decide that you can deform the triangular lattice to a square lattice with each square divided by a northwest-to-southeast diagonal, whose length you staunchly declare to be $1$, like the squares' sides. Then it is easy to figure out something like the following (left figure). Finally, un-deform that figure back to the regular triangular lattice (right figure).
$\begin{array}{c}\Longrightarrow\\[1in]~\end{array}$
This is a proof of all possible solutions. All other solutions are congruent to these ones.
