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

Latest revision as of 11:37, 20 April 2022

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 $m=0$ ).

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); int n = 17; real r = 1; real rad = pi/2; pair pt(real k=0) { return (r*expi(rad-2*pi*k/n)); } for(int i=0; i<n; ++i){ dot(pt(i)); draw(pt(i)--pt(i+1)); } draw(pt()--pt(8)); draw(pt()--pt(4)--pt(8),linewidth(0.7)+linetype("4 4")); draw(pt()--pt(2)--pt(4),linewidth(0.7)+linetype("4 4")); label("$d$",(pt()+pt(8))/2,WNW); label("$\mathcal{Q}$",(pt()+pt(6))/2,SE); label("$\mathcal{R}$",(pt()+pt(10))/2,W); label("$P_0$",pt(),N); label("$P_1$",pt(1),NNE); label("$P_8$",pt(8),S); label("$P_{16}$",pt(-1),NNW); label("$\cdots$",pt(2),NE); [/asy]$

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); int n = 20; real r = 1; real rad = pi/2; pair pt(real k=0) { return (r*expi(rad-2*pi*k/n)); } for(int i=0; i<n; ++i){ dot(pt(i)); draw(pt(i)--pt(i+1)); } draw(pt()--pt(8)--pt(16)--cycle); label("$O$",(0,0),W); dot((0,0)); draw(pt()--pt(4)--pt(8)--pt(6)--pt(4)--pt(2)--pt()--pt(18)--pt(16)--pt(12)--pt(8)--pt(10)--pt(12)--pt(14)--pt(16),linewidth(0.3)+linetype("4 4")); label("$P_0$",pt(),N); label("$P_1$",pt(1),NNE); label("$P_{19}$",pt(-1),NNW); label("$P_{8}$",pt(8),SE); label("$P_{16}$",pt(16),W); label("$\cdots$",pt(2),NE); [/asy]$

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.

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

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.

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); int n = 17; real r = 1; real rad = pi/2; pair pt(real k=0) { return (r*expi(rad-2*pi*k/n)); } for(int i=0; i<n; ++i){ dot(pt(i)); draw(pt(i)--pt(i+1)); } /* could rewrite recursively, if someone wants to do .. */ draw(pt(8)--pt()--pt(9)); draw(pt()--pt(4)--pt(8)); draw(pt()--pt(2)--pt(4)); draw(pt()--pt(1)--pt(2)); draw(pt(2)--pt(3)--pt(4)); draw(pt(4)--pt(6)--pt(8)); draw(pt(4)--pt(5)--pt(6)); draw(pt(6)--pt(7)--pt(8)); draw(pt(9)--pt(13)--pt(17)); draw(pt(9)--pt(11)--pt(13)); draw(pt(9)--pt(10)--pt(11)); draw(pt(11)--pt(12)--pt(13)); draw(pt(13)--pt(15)--pt(17)); draw(pt(13)--pt(14)--pt(15)); draw(pt(15)--pt(16)--pt(17)); label("$P_0$",pt(),N); label("$P_1$",pt(1),NNE); label("$P_{16}$",pt(-1),NNW); label("$\cdots$",pt(2),NE); [/asy]$
An example for $n=17 = 2^{4}+1$

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.

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.

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); int n = 10; real r = 1; real rad = pi/2; pair pt(real k=0) { return (r*expi(rad-2*pi*k/n)); } for(int i=0; i<n; ++i){ dot(pt(i)); draw(pt(i)--pt(i+1)); } draw(pt()--pt(2)--pt(4)--pt(6)--pt(8)--cycle); draw(pt()--pt(4)--pt(6)--cycle,linewidth(0.5)+linetype("4 4")); label("$P_0$",pt(),N); label("$P_1$",pt(1),NNE); label("$P_{2}$",pt(2),NE); label("$P_{3}$",pt(3),E); label("$P_{4}$",pt(4),SE); label("$P_{5}$",pt(5),S); label("$P_{6}$",pt(6),SW); label("$P_{7}$",pt(7),W); label("$P_{8}$",pt(8),NW); label("$P_{9}$",pt(9),NNW); [/asy]$
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 $2$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-(''Gregory Galparin'') Let <math>\mathcal{P}</math> be a convex polygon with <math>n</math> sides, <math>n\ge3</math>. Any set of <math>n - 3</math> diagonals of <math>\mathcal{P}</math> that do not intersect in the interior of the polygon determine a ''triangulation'' of <math>\mathcal{P}</math> into <math>n - 2</math> triangles. If <math>\mathcal{P}</math> is regular and there is a triangulation of <math>\mathcal{P}</math> consisting of only isosceles triangles, find all the possible values of <math>n</math>.
+(''Gregory Galparin'') Let <math>\mathcal{P}</math> be a [[convex polygon]] with <math>n</math> sides, <math>n\ge3</math>. Any set of <math>n - 3</math> diagonals of <math>\mathcal{P}</math> that do not intersect in the interior of the polygon determine a ''triangulation'' of <math>\mathcal{P}</math> into <math>n - 2</math> triangles. If <math>\mathcal{P}</math> is regular and there is a triangulation of <math>\mathcal{P}</math> consisting of only isosceles triangles, find all the possible values of <math>n</math>.
-__TOC__
 == Solution ==
-We label the vertices of <math>\mathcal{P}</math> as <math>P_0, P_1, P_2, \ldots, P_n</math>. Consider a diagonal <math>d = \overline{P_a,P_{a+k}},\,k \le n/2</math> in the triangulation. This diagonal partitions <math>\mathcal{P}</math> into two regions <math>\mathcal{Q},\, \mathcal{R}</math>, and is the side of an isosceles triangle in both regions. [[Without loss of generality]] suppose the area of <math>Q</math> is less than the area of <math>R</math> (so the [[circumcenter|center]] of <math>P</math> does not lie in the interior of <math>Q</math>); it follows that the lengths of the edges and diagonals in <math>Q</math> are smaller than <math>d</math>. Thus <math>d</math> must the be the base of the isosceles triangle in <math>Q</math>, from which it follows that the isosceles triangle is <math>\triangle P_aP_{a+k/2}P_{a+k}</math>, and so <math>2|k</math>. Repeating this process on the legs of isosceles triangle (<math>\overline{P_aP_{a+k/2}},\,\overline{P_{a+k}P_{a+k/2}}</math>), it follows that <math>k = 2^m</math> for some positive integer <math>m</math>.
-Now take the isosceles triangle <math>P_xP_yP_z</math> in the triangulation that contains the center of <math>\mathcal{P}</math> in its interior; if a diagonal passes through the center, selecting either of the isosceles triangles with that diagonal is an edge will suffice. Without loss of generality, suppose <math>P_xP_y = P_yP_z</math>. From our previous result, it follows that (sum of powers of <math>2</math>), eg
+We label the vertices of <math>\mathcal{P}</math> as <math>P_0, P_1, P_2, \ldots, P_n</math>. Consider a diagonal <math>d = \overline{P_a\,P_{a+k}},\,k \le n/2</math> in the triangulation. We show that <math>k</math> must have the form <math>2^{m}</math> for some nonnegative integer <math>m</math>.
-<cmath>n = 2^{a+1} + 2^{b}</cmath>
-We now claim that this condition is sufficient. Suppose WLOG that <math>a+1 \ge b</math>; then we rewrite this as
+This diagonal [[partition]]s <math>\mathcal{P}</math> into two regions <math>\mathcal{Q},\, \mathcal{R}</math>, and is the side of an isosceles triangle in both regions. [[Without loss of generality]] suppose the area of <math>Q</math> is less than the area of <math>R</math> (so the [[circumcenter|center]] of <math>P</math> does not lie in the interior of <math>Q</math>); it follows that the lengths of the edges and diagonals in <math>Q</math> are all smaller than <math>d</math>. Thus <math>d</math> must the be the base of the isosceles triangle in <math>Q</math>, from which it follows that the isosceles triangle is <math>\triangle P_aP_{a+k/2}\,P_{a+k}</math>, and so <math>2|k</math>. Repeating this process on the legs of isosceles triangle (<math>\overline{P_aP_{a+k/2}},\,\overline{P_{a+k}P_{a+k/2}}</math>), it follows that <math>k = 2^m</math> for some positive integer <math>m</math> (if we allow [[degenerate|degeneracy]], then we can also let <math>m=0</math>).
+<center><table><tr><td><asy>
+size(200); 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<n; ++i){
+ dot(pt(i));
+ draw(pt(i)--pt(i+1));
+}
+draw(pt()--pt(8));
+draw(pt()--pt(4)--pt(8),linewidth(0.7)+linetype("4 4"));
+draw(pt()--pt(2)--pt(4),linewidth(0.7)+linetype("4 4"));
+label("\(d\)",(pt()+pt(8))/2,WNW);
+label("\(\mathcal{Q}\)",(pt()+pt(6))/2,SE);
+label("\(\mathcal{R}\)",(pt()+pt(10))/2,W);
+label("\(P_0\)",pt(),N);
+label("\(P_1\)",pt(1),NNE);
+label("\(P_8\)",pt(8),S);
+label("\(P_{16}\)",pt(-1),NNW);
+label("\(\cdots\)",pt(2),NE);
+</asy>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td><td><asy>
+size(200); 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<n; ++i){
+ dot(pt(i));
+ draw(pt(i)--pt(i+1));
+}
+draw(pt()--pt(8)--pt(16)--cycle);
+label("\(O\)",(0,0),W); dot((0,0));
+draw(pt()--pt(4)--pt(8)--pt(6)--pt(4)--pt(2)--pt()--pt(18)--pt(16)--pt(12)--pt(8)--pt(10)--pt(12)--pt(14)--pt(16),linewidth(0.3)+linetype("4 4"));
+label("\(P_0\)",pt(),N);
+label("\(P_1\)",pt(1),NNE);
+label("\(P_{19}\)",pt(-1),NNW);
+label("\(P_{8}\)",pt(8),SE);
+label("\(P_{16}\)",pt(16),W);
+label("\(\cdots\)",pt(2),NE);
+</asy></td></tr><tr><td><span style="font-size:85%">An example for <math>n=17</math>,<br /><math>k=8</math></span></td><td><span style="font-size:85%">An isosceles triangle containing<br /> the center for <br /> <math>n=20</math>, <math>(x,y,z) = (0,8,16)</math></span></td></tr></table></center>
+Now take the isosceles triangle <math>P_xP_yP_z,\,0 \le x < y < z < n</math> in the triangulation that contains the center of <math>\mathcal{P}</math> 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 <math>P_xP_y = P_yP_z</math>. From our previous result, it follows that there are <math>2^a</math> edges of <math>P</math> on the minor arcs of <math>P_xP_y,\, P_yP_z</math> and <math>2^b</math> edges of <math>P</math> on the minor arc of <math>P_zP_x</math>, for positive integers <math>a,\,b</math>. Therefore, we can write
+<cmath>n = 2 \cdot 2^a + 2^b = 2^{a+1} + 2^{b},</cmath> so <math>n</math> must be the sum of two powers of <math>2</math>.
+We now claim that this condition is sufficient. Suppose without loss of generality that <math>a+1 \ge b</math>; then we rewrite this as
 <cmath>n = 2^{b}(2^{a-b+1}+1).</cmath>
-*''Lemma 1'': All <math>n = 2^k + 1</math> and <math>n=4</math> have triangulations that work.
+*''Lemma 1'': All regular polygons with <math>n = 2^k + 1</math> or <math>n=4</math> have triangulations that meet the conditions.
-*''Lemma 2'': If <math>n</math>-sided polygon works, then <math>2n</math>-sided polygon works.
-By induction, it follows that we can cover all the <math>n</math>.
+By [[induction]], it follows that we can cover all the desired <math>n</math>.
+For <math>n = 3,4</math>, this is trivial. For <math>k>1</math>, we construct the diagonals of equal length <math>\overline{P_0P_{2^{k-1}}}</math> and <math>\overline{P_{2^{k-1}+1}P_0}</math>. This partitions <math>\mathcal{P}</math> into <math>3</math> regions: an isosceles <math>\triangle P_0P_{2^{k-1}}P_{2^{k-1}+1}</math>, 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 <math>2(2^{k-1}-1) + (1) = 2^k-1 = n-2</math> isosceles triangles with non-intersecting diagonals, as desired.
+<center><asy>
+size(200); 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<n; ++i){
+ dot(pt(i));
+ draw(pt(i)--pt(i+1));
+}
+/* could rewrite recursively, if someone wants to do .. */
+draw(pt(8)--pt()--pt(9));
+ draw(pt()--pt(4)--pt(8));
+  draw(pt()--pt(2)--pt(4));
+   draw(pt()--pt(1)--pt(2));
+   draw(pt(2)--pt(3)--pt(4));
+  draw(pt(4)--pt(6)--pt(8));
+   draw(pt(4)--pt(5)--pt(6));
+   draw(pt(6)--pt(7)--pt(8));
+ draw(pt(9)--pt(13)--pt(17));
+  draw(pt(9)--pt(11)--pt(13));
+   draw(pt(9)--pt(10)--pt(11));
+   draw(pt(11)--pt(12)--pt(13));
+  draw(pt(13)--pt(15)--pt(17));
+   draw(pt(13)--pt(14)--pt(15));
+   draw(pt(15)--pt(16)--pt(17));
+label("\(P_0\)",pt(),N);
+label("\(P_1\)",pt(1),NNE);
+label("\(P_{16}\)",pt(-1),NNW);
+label("\(\cdots\)",pt(2),NE);
+</asy><br /><span style="font-size:85%">An example for <math>n=17 = 2^{4}+1</math></span></center>
+*''Lemma 2'': If a regular polygon with <math>n</math> sides has a working triangulation, then the regular polygon with <math>2n</math> sides also has a triangulation that meets the conditions.
+We construct the diagonals <math>\overline{P_0P_2},\ \overline{P_2P_4},\ \ldots \overline{P_{2n-2}P_0}</math>. This partitions <math>\mathcal{P}</math> into <math>n</math> isosceles triangles of the form <math>\triangle P_{2k}P_{2k+1}P_{2k+2}</math>, as well as a central regular polygon with <math>n</math> sides. However, we know that there exists a triangulation for the <math>n</math>-sided polygon that yields <math>n-2</math> isosceles triangles. Thus, we have created <math>(n) + (n-2) = 2n-2</math> isosceles triangles with non-intersecting diagonals, as desired.
+<center><asy>
+size(200); 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));
+}
-''Proof 1'': For <math>n = 3,4</math>, this is trivial. For <math>k>1</math>, construct <math>P_0P_{2^k}</math> and <math>P_{2^k+1}P_O</math>, then make midpoint isosceles triangles, repeat.
+for(int i=0; i<n; ++i){
+ dot(pt(i));
+ draw(pt(i)--pt(i+1));
+}
+draw(pt()--pt(2)--pt(4)--pt(6)--pt(8)--cycle);
+draw(pt()--pt(4)--pt(6)--cycle,linewidth(0.5)+linetype("4 4"));
+label("\(P_0\)",pt(),N);
+label("\(P_1\)",pt(1),NNE);
+label("\(P_{2}\)",pt(2),NE);
+label("\(P_{3}\)",pt(3),E);
+label("\(P_{4}\)",pt(4),SE);
+label("\(P_{5}\)",pt(5),S);
+label("\(P_{6}\)",pt(6),SW);
+label("\(P_{7}\)",pt(7),W);
+label("\(P_{8}\)",pt(8),NW);
+label("\(P_{9}\)",pt(9),NNW);
-''Proof 2'': Construct every other edge, then reduce etc.
+</asy><br /><span style="font-size:85%">An example for <math>n=10,\, n/2 = 5</math></span></center>
- {{incomplete|solution}}
+In summary, the answer is all <math>n</math> that can be written in the form <math>2^{a+1} + 2^{b},\, a,b \ge 0</math>. Alternatively, this condition can be expressed as either <math>n=2^{k},\, k \ge 2</math> (this is the case when <math>a+1 = b</math>) or <math>n</math> is the sum of two distinct powers of <math>2</math>, where <math>1= 2^0</math> is considered a power of <math>2</math>.
-{{alternate solutions}}
+== See also ==
+* <url>viewtopic.php?t=202905 Discussion on AoPS/MathLinks<\url>
-== Resources ==
 {{USAMO newbox|year=2008|num-b=3|num-a=5}}
-* <url>viewtopic.php?t=202905 Discussion on AoPS/MathLinks</url>
+[[Category:Olympiad Combinatorics Problems]]
+{{MAA Notice}}

2008 USAMO (Problems • Resources)
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Difference between revisions of "2008 USAMO Problems/Problem 4"

Latest revision as of 11:37, 20 April 2022

Problem

Solution

See also