Difference between revisions of "1963 IMO Problems/Problem 3"
Mathboy100 (talk | contribs) (→Solution) |
Mathboy100 (talk | contribs) (→Solution) |
||
Line 31: | Line 31: | ||
<cmath>-y\textrm{-coordinate} = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot \sin \frac{2\pi(k-1)}{n}</cmath> | <cmath>-y\textrm{-coordinate} = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot \sin \frac{2\pi(k-1)}{n}</cmath> | ||
<cmath>y\textrm{-coordinate} = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot -\sin \frac{2\pi(k-1)}{n}</cmath> | <cmath>y\textrm{-coordinate} = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot -\sin \frac{2\pi(k-1)}{n}</cmath> | ||
− | <cmath> = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot \sin \frac{2\pi(n - k + | + | <cmath> = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot \sin \frac{2\pi(n - k + 1)}{n}</cmath> |
<cmath> = \sum_{k = 2}^{\frac{n+1}{2}}a_{n-k+2} \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (4)}</cmath> | <cmath> = \sum_{k = 2}^{\frac{n+1}{2}}a_{n-k+2} \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (4)}</cmath> | ||
Latest revision as of 16:58, 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 .
This case is very similar to before. We will compute the -coordinate of the top point two ways:
It must be true that . Then, we get for all . Therefore, . It is trivial that is then equal to the other values, so . This completes the proof.
~mathboy100
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 |