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Difference between revisions of "1963 IMO Problems/Problem 3"

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for all <math>i\leq \lfloor \frac{n}{2}\rfloor</math>. This shows that <math>a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0</math>, with equality when <math>a_i=a_{n-i}</math>. Therefore
 
for all <math>i\leq \lfloor \frac{n}{2}\rfloor</math>. This shows that <math>a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0</math>, with equality when <math>a_i=a_{n-i}</math>. Therefore
  
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<cmath>\sum_{i=1}^{n}  a_i \sin{\left(\frac{2\pi}{n}i\right)}=\sum_{i=1}^{\lfloor \frac{n}{2}} a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq 0</cmath>
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<cmath>\sum_{i=1}^{n}  a_i \sin{\left(\frac{2\pi}{n}i\right)}=\sum_{i=1}^{\lfloor \frac{n}{2}\rfloor} a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq 0</cmath>
  
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There is equality only when <math>a_i=a_{n-i}</math> for all <math>i</math>. This implies that <math>a_1=a_{n-1}</math> and <math>a_2=a_n</math>, so we have that <math>a_1=a_2+\cdots =a_n</math>. <math>\blacksquare</math>
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There is equality only when <math>a_i=a_{n-i}</math> for all <math>i</math>. This implies that <math>a_1=a_{n-1}</math> and <math>a_2=a_n</math>, so we have that <math>a_1=a_2=\cdots =a_n</math>. <math>\blacksquare</math>
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1963|num-b=2|num-a=4}}
 
{{IMO box|year=1963|num-b=2|num-a=4}}

Revision as of 11:03, 29 March 2012

Problem

In an $n$-gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation

$a_1\ge a_2\ge \cdots \ge a_n$.

Prove that $a_1=a_2=\cdots = a_n$.

Solution

Define the vector $\vec{v_i}$ to equal $\cos{\left(\frac{2\pi}{n}i\right)}\vec{i}+\sin{\left(\frac{2\pi}{n}i\right)}\vec{j}$. Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length $a_i$ is parallel to $\vec{v_i}$. We then have that

\[\sum_{i=1}^{n} a_i\vec{v_i}=\vec{0}\Rightarrow \sum_{i=1}^{n} a_i\cos{\left(\frac{2\pi}{n}i\right)} = \sum_{i=1}^{n} a_i\sin{\left(\frac{2\pi}{n}i\right)} =0\]

But $a_i\geq a_{n-i}$ for all $i\leq \lfloor \frac{n}{2}\rfloor$, so

\[a_i \sin{\left(\frac{2\pi}{n}i\right)} = -a_i\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq -a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\]

for all $i\leq \lfloor \frac{n}{2}\rfloor$. This shows that $a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0$, with equality when $a_i=a_{n-i}$. Therefore

\[\sum_{i=1}^{n} a_i \sin{\left(\frac{2\pi}{n}i\right)}=\sum_{i=1}^{\lfloor \frac{n}{2}\rfloor} a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq 0\]

There is equality only when $a_i=a_{n-i}$ for all $i$. This implies that $a_1=a_{n-1}$ and $a_2=a_n$, so we have that $a_1=a_2=\cdots =a_n$. $\blacksquare$

See Also

1963 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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