Problem C. 1046. (October 2010)

C. 1046. Let $\alpha$ (n) denote the measure of the interior angles of a regular n-sided polygon. What is n if $\alpha$ (n+3)- $\alpha$ (n)= $\alpha$ (n)- $\alpha$ (n-2)?

(5 pont)

Deadline expired on November 10, 2010.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. $\displaystyle \alpha(n)= (n-2)\cdot \frac{180^\circ}{n}$, ezért a feltételt így írhatjuk fel ($\displaystyle n\ge 3$):

$\displaystyle (n+1)\cdot \frac{180^\circ}{n+3}-(n-2)\cdot \frac{180^\circ}{n}=(n-2)\cdot \frac{180^\circ}{n}-(n-4)\cdot \frac{180^\circ}{n-2}. $

$\displaystyle 180^\circ$-kal való egyszerűsítés és rendezés után $\displaystyle \displaystyle{\frac{n+1}{n+3}+\frac{n-4}{n-2}=\frac{2(n-2)}{n}}$, majd $\displaystyle n(n^2-n-2+n^2-n-12)=2(n-2)(n^2+n-6)$, amiből $\displaystyle -14n=-16n+24$, ahonnan $\displaystyle n=12$.

Statistics:

326 students sent a solution.

5 points: 277 students.

4 points: 11 students.

3 points: 17 students.

2 points: 4 students.

1 point: 2 students.

0 point: 8 students.

Unfair, not evaluated: 7 solutionss.

Problems in Mathematics of KöMaL, October 2010

326 students sent a solution.
5 points:	277 students.
4 points:	11 students.
3 points:	17 students.
2 points:	4 students.
1 point:	2 students.
0 point:	8 students.
Unfair, not evaluated:	7 solutionss.

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