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Difference between revisions of "2008 USAMO Problems/Problem 4"

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 $n$.

Solutions

Solution 1

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 $m=0$). $[asy] 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/n)); } for(int i=0; i $[asy] defaultpen(linewidth(0.7)+fontsize(10)); int n = 20; real r = 1; real rad = pi/2; pair pt(real k=0) { return (r*expi(rad-2*pi*k/n)); } for(int i=0; i 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 $n$.

Solution 2

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. $[asy] 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/n)); } for(int i=0; i
An example for $n=17 = 2^{4}+1$

Solution 3

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. $[asy] 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/n)); } for(int i=0; i
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 $2$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 