Mathematical and Physical Journal
for High Schools
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Problem B. 4465. (September 2012)

B. 4465. The line segment A0A10 is divided into 10 equal parts. The dividing points are A_{1}, A_{2}, \ldots, A_{9} in this order. The third vertex of the regular triangle drawn over the line segment A8A10 is B. Show that \angleBA0A10+\angleBA2A10+\angleBA3A10+\angleBA4A10=60o.

Suggested by Sz. Miklós, Herceghalom

(4 pont)

Deadline expired on October 10, 2012.


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

Megoldás. Legyen az A0A1 szakasz hossza egységnyi, ekkor az A8A10B háromszög B-ből induló magassága A_9B=\sqrt{3}. A szóban forgó szögeket jelölje rendre \alpha,\beta,\gamma,\delta. Ekkor \tg\alpha=\sqrt{3}/9, \tg\beta=\sqrt{3}/7, \tg\gamma=\sqrt{3}/6,
\tg\delta=\sqrt{3}/5. Az összegképlet szerint

\tg(\alpha+\delta)=\frac{\tg\alpha+\tg\delta}{1-\tg\alpha\tg\delta}=
\frac{\sqrt{3}}{3}=\frac{\tg\beta+\tg\gamma}{1-\tg\beta\tg\gamma}=
\tg(\beta+\gamma),

vagyis \alpha+\delta=\beta+\gamma=30o. A négy szög összege tehát valóban 60o.


Statistics:

207 students sent a solution.
4 points:118 students.
3 points:3 students.
1 point:83 students.
0 point:3 students.

Problems in Mathematics of KöMaL, September 2012