Difference between revisions of "1963 IMO Problems/Problem 3"
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<cmath> = \sum_{k = 1}^{\frac{n}{2}}a_{k+\frac{n}{2}} \cdot \sin \frac{2\pi(k)}{n}.\textbf{ (2)}</cmath> | <cmath> = \sum_{k = 1}^{\frac{n}{2}}a_{k+\frac{n}{2}} \cdot \sin \frac{2\pi(k)}{n}.\textbf{ (2)}</cmath> | ||
− | It must be true that <math>\textbf{(1)} = \textbf{(2)}</math>. | + | It must be true that <math>\textbf{(1)} = \textbf{(2)}</math>. This implies that <math>a_k \leq a_{k+\frac{n}{2}}</math> for all <math>1 \leq k \leq \frac{n}{2}</math>, and therefore |
==Solution 2== | ==Solution 2== |
Revision as of 16:42, 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
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:
We can obtain the -coordinate of the top side in a different way by multiplying the lengths of the sides , , ... by the sine of the angle they make with the -axis to get the -coordinate of the top side:
It must be true that . This implies that for all , and therefore
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 |