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

Revision as of 16:31, 7 December 2022

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 1

Let $a_1 = p_1p_2$ , $a_2 = p_2p_3$ , etc.

Plot the $n$ -gon on the cartesian plane such that $p_1p_2$ is on the $x$ -axis and the entire shape is above the $x$ -axis. There are two cases: the number of sides is even, and the number of sides is odd:

$\textbf{Case 1: Even}$

In this case, the side with the topmost points will be $p_{\frac{n}{2}+1}p_{\frac{n}{2}+2}$ . We can multiply the lengths of the sides $a_1$ , $a_2$ , ... $a_{\frac{n}{2}}$ by the sine of the angle they make with the $x$ -axis:

$y\textrm{-coordinate} = \sum_{k = 0}^{\frac{n}{2}}$

Solution 2

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 $a_1=a_2=\cdots =a_n$ . $\blacksquare$

@@ Line 4: / Line 4: @@
 Prove that <math>a_1=a_2=\cdots = a_n</math>.
-==Solution==
+==Solution 1==
+Let <math>a_1 = p_1p_2</math>, <math>a_2 = p_2p_3</math>, etc.
+Plot the <math>n</math>-gon on the cartesian plane such that <math>p_1p_2</math> is on the <math>x</math>-axis and the entire shape is above the <math>x</math>-axis. There are two cases: the number of sides is even, and the number of sides is odd:
+<math>\textbf{Case 1: Even}</math>
+In this case, the side with the topmost points will be <math>p_{\frac{n}{2}+1}p_{\frac{n}{2}+2}</math>. We can multiply the lengths of the sides <math>a_1</math>, <math>a_2</math>, ... <math>a_{\frac{n}{2}}</math> by the sine of the angle they make with the <math>x</math>-axis:
+<cmath>y\textrm{-coordinate} = \sum_{k = 0}^{\frac{n}{2}}</cmath>
+==Solution 2==
 Define the vector <math>\vec{v_i}</math> to equal <math>\cos{\left(\frac{2\pi}{n}i\right)}\vec{i}+\sin{\left(\frac{2\pi}{n}i\right)}\vec{j}</math>. Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length <math>a_i</math> is parallel to <math>\vec{v_i}</math>. We then have that

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

Page

Toolbox

Search

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

Revision as of 16:31, 7 December 2022

Contents

Problem

Solution 1

Solution 2

See Also