Difference between revisions of "1963 IMO Problems/Problem 3"
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<math>\textbf{Case 1: Even}</math> | <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>. | + | In this case, the side with the topmost points will be <math>p_{\frac{n}{2}+1}p_{\frac{n}{2}+2}</math>. To obtain the <math>y</math>-coordinate of this top side, 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 = | + | <cmath>y\textrm{-coordinate} = \sum_{k = 1}^{\frac{n}{2}}a_n \cdot \sin \frac{2\pi(k-1)}{n}</cmath> |
==Solution 2== | ==Solution 2== |
Revision as of 16:32, 7 December 2022
Contents
Problem
In an -gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
Prove that .
Solution 1
Let , , etc.
Plot the -gon on the cartesian plane such that is on the -axis and the entire shape is above the -axis. There are two cases: the number of sides is even, and the number of sides is odd:
In this case, the side with the topmost points will be . To obtain the -coordinate of this top side, we can multiply the lengths of the sides , , ... by the sine of the angle they make with the -axis:
Solution 2
Define the vector to equal . Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length is parallel to . We then have that
But for all , so
for all . This shows that , with equality when . Therefore
There is equality only when for all . This implies that and , so we have that .
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 |