Difference between revisions of "1963 IMO Problems/Problem 3"
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for all <math>i\leq \lfloor \frac{n}{2}\rfloor</math>. This shows that <math>a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0</math>, with equality when <math>a_i=a_{n-i}</math>. Therefore | for all <math>i\leq \lfloor \frac{n}{2}\rfloor</math>. This shows that <math>a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0</math>, with equality when <math>a_i=a_{n-i}</math>. Therefore | ||
− | <cmath>\sum_{i=1}^{n} a_i \sin{\left(\frac{2\pi}{n}i\right)}=\sum_{i=1}^{\lfloor \frac{n}{2}} a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq 0</cmath> | + | <cmath>\sum_{i=1}^{n} a_i \sin{\left(\frac{2\pi}{n}i\right)}=\sum_{i=1}^{\lfloor \frac{n}{2}/rfloor} a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq 0</cmath> |
There is equality only when <math>a_i=a_{n-i}</math> for all <math>i</math>. This implies that <math>a_1=a_{n-1}</math> and <math>a_2=a_n</math>, so we have that <math>a_1=a_2+\cdots =a_n</math>. <math>\blacksquare</math> | There is equality only when <math>a_i=a_{n-i}</math> for all <math>i</math>. This implies that <math>a_1=a_{n-1}</math> and <math>a_2=a_n</math>, so we have that <math>a_1=a_2+\cdots =a_n</math>. <math>\blacksquare</math> |
Revision as of 10:02, 29 March 2012
Problem
In an -gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
Prove that .
Solution
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 |