Difference between revisions of "1963 IMO Problems/Problem 3"
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Prove that <math>a_1=a_2=\cdots = a_n</math>. | 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 | 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 | ||
Revision as of 17:31, 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 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 . 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 |