2008 USAMO Problems/Problem 4

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Problem

(Gregory Galparin) Let $\mathcal{P}$ be a convex polygon with $n$ sides, $n\ge3$ . Any set of $n - 3$ diagonals of $\mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $\mathcal{P}$ into $n - 2$ triangles. If $\mathcal{P}$ is regular and there is a triangulation of $\mathcal{P}$ consisting of only isosceles triangles, find all the possible values of $n$ .

Solution

We label the vertices of $\mathcal{P}$ as $P_0, P_1, P_2, \ldots, P_n$ . Consider a diagonal $d = \overline{P_a\,P_{a+k}},\,k \le n/2$ in the triangulation. We show that $k$ must have the form $2^{m}$ for some nonnegative integer $m$ .

This diagonal partitions $\mathcal{P}$ into two regions $\mathcal{Q},\, \mathcal{R}$ , and is the side of an isosceles triangle in both regions. Without loss of generality suppose the area of $Q$ is less than the area of $R$ (so the center of $P$ does not lie in the interior of $Q$ ); it follows that the lengths of the edges and diagonals in $Q$ are all smaller than $d$ . Thus $d$ must the be the base of the isosceles triangle in $Q$ , from which it follows that the isosceles triangle is $\triangle P_aP_{a+k/2}\,P_{a+k}$ , and so $2|k$ . Repeating this process on the legs of isosceles triangle ( $\overline{P_aP_{a+k/2}},\,\overline{P_{a+k}P_{a+k/2}}$ ), it follows that $k = 2^m$ for some positive integer $m$ (if we allow degeneracy, then we can also let $m=0$ ).


An example for $n=17$ , $k=8$	An isosceles triangle containing the center for $n=20$ , $(x,y,z) = (0,8,16)$

Now take the isosceles triangle $P_xP_yP_z,\,0 \le x < y < z < n$ in the triangulation that contains the center of $\mathcal{P}$ in its interior; if a diagonal passes through the center, select either of the isosceles triangles with that diagonal as an edge. Without loss of generality, suppose $P_xP_y = P_yP_z$ . From our previous result, it follows that there are $2^a$ edges of $P$ on the minor arcs of $P_xP_y,\, P_yP_z$ and $2^b$ edges of $P$ on the minor arc of $P_zP_x$ , for positive integers $a,\,b$ . Therefore, we can write $n = 2 \cdot 2^a + 2^b = 2^{a+1} + 2^{b},$ so $n$ must be the sum of two powers of $2$ .

We now claim that this condition is sufficient. Suppose without loss of generality that $a+1 \ge b$ ; then we rewrite this as $n = 2^{b}(2^{a-b+1}+1).$

Lemma 1: All regular polygons with $n = 2^k + 1$ or $n=4$ have triangulations that meet the conditions.
Lemma 2: If a regular polygon with $n$ sides has a working triangulation, then the regular polygon with $2n$ sides also has a triangulation that meets the conditions.

By induction, it follows that we can cover all the desired $n$ .

Proof 1: For $n = 3,4$ , this is trivial. For $k>1$ , we construct the diagonals of equal length $\overline{P_0P_{2^{k-1}}}$ and $\overline{P_{2^{k-1}+1}P_0}$ . This partitions $\mathcal{P}$ into $3$ regions: an isosceles $\triangle P_0P_{2^{k-1}}P_{2^{k-1}+1}$ , and two other regions. For these two regions, we can recursively construct the isosceles triangles defined above in the second paragraph. It follows that we have constructed $2(2^{k-1}-1) + (1) = 2^k-1 = n-2$ isosceles triangles with non-intersecting diagonals, as desired. $\ \square$

defaultpen(linewidth(0.7)+fontsize(10));int n = 17; real r = 1; real rad = pi/2;pair pt(real k=0) { return (r*expi(rad-2*pi*k...

An example for $n=17 = 2^{4}+1$

Proof 2: We construct the diagonals $\overline{P_0P_2},\ \overline{P_2P_4},\ \ldots \overline{P_{2n-2}P_0}$ . This partitions $\mathcal{P}$ into $n$ isosceles triangles of the form $\triangle P_{2k}P_{2k+1}P_{2k+2}$ , as well as a central regular polygon with $n$ sides. However, we know that there exists a triangulation for the $n$ -sided polygon that yields $n-2$ isosceles triangles. Thus, we have created $(n) + (n-2) = 2n-2$ isosceles triangles with non-intersecting diagonals, as desired. $\ \square$

defaultpen(linewidth(0.7)+fontsize(10));int n = 10; real r = 1; real rad = pi/2;pair pt(real k=0) { return (r*expi(rad-2*pi*k...

An example for $n=10,\, n/2 = 5$

In summary, the answer is all $n$ that can be written in the form $2^{a+1} + 2^{b},\, a,b \ge 0$ . Alternatively, this condition can be expressed as either $n=2^{k},\, k \ge 2$ (this is the case when $a+1 = b$ ) or $n$ is the sum of two distinct powers of $2$ , where $1= 2^0$ is considered a power of $2$ . $\ \blacksquare$