Mathematical and Physical Journal
for High Schools
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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 \(\displaystyle n=12\).


327 students sent a solution.
5 points:278 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