Difference between revisions of "1963 IMO Problems/Problem 3"
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<cmath>y\textrm{-coordinate} = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot -\sin \frac{2\pi(k-1)}{n}</cmath> | <cmath>y\textrm{-coordinate} = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot -\sin \frac{2\pi(k-1)}{n}</cmath> | ||
<cmath> = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot \sin \frac{2\pi(k-\frac{n}{2}-1)}{n}</cmath> | <cmath> = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot \sin \frac{2\pi(k-\frac{n}{2}-1)}{n}</cmath> | ||
− | <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-1)}{n}.\textbf{ (2)}</cmath> |
It must be true that <math>\textbf{(1)} = \textbf{(2)}</math>. This implies that <math>a_k = a_{k+\frac{n}{2}}</math> for all <math>1 \leq k \leq \frac{n}{2}</math>, and therefore <math>a_1=a_2=\cdots = a_n</math>. | It must be true that <math>\textbf{(1)} = \textbf{(2)}</math>. This implies that <math>a_k = a_{k+\frac{n}{2}}</math> for all <math>1 \leq k \leq \frac{n}{2}</math>, and therefore <math>a_1=a_2=\cdots = a_n</math>. | ||
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<math>\textbf{Case 2: Odd}</math> | <math>\textbf{Case 2: Odd}</math> | ||
− | This case is very similar to before. We will compute the <math>y</math>-coordinate of the top point <math>p_{frac{n+3}{2} | + | This case is very similar to before. We will compute the <math>y</math>-coordinate of the top point <math>p_{frac{n+3}{2}}</math> two ways: |
<cmath>y\textrm{-coordinate} = \sum_{k = 2}^{\frac{n+1}{2}}a_k \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (3)}</cmath> | <cmath>y\textrm{-coordinate} = \sum_{k = 2}^{\frac{n+1}{2}}a_k \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (3)}</cmath> | ||
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<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 + 2)}{n}</cmath> | <cmath> = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot \sin \frac{2\pi(n - k + 2)}{n}</cmath> | ||
− | <cmath> = \sum_{k = 2}^{\frac{n+1}{2}}a_{n-k+2} \cdot \sin \frac{2\pi(k)}{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> |
It must be true that <math>\textbf{(3)} = \textbf{(4)}</math>. Then, we get <math>a_k = a_{n-k+2}</math> for all <math>2 \leq k \leq \frac{n+1}{2}</math>. Therefore, <math>a_2=a_3=\cdots = a_n</math>. It is trivial that <math>a_1</math> is then equal to the other values, so <math>a_1=a_2=\cdots = a_n</math>. This completes the proof. <math>\square</math> | It must be true that <math>\textbf{(3)} = \textbf{(4)}</math>. Then, we get <math>a_k = a_{n-k+2}</math> for all <math>2 \leq k \leq \frac{n+1}{2}</math>. Therefore, <math>a_2=a_3=\cdots = a_n</math>. It is trivial that <math>a_1</math> is then equal to the other values, so <math>a_1=a_2=\cdots = a_n</math>. This completes the proof. <math>\square</math> |
Revision as of 17:57, 7 December 2022
Contents
Problem
In an -gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
![$a_1\ge a_2\ge \cdots \ge a_n$](http://latex.artofproblemsolving.com/4/8/b/48baac61191c39562168731a01e8e841b4d4756f.png)
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 |