Problem C. 1626. (October 2020)

C. 1626. Let \(\displaystyle F\) denote the midpoint of side \(\displaystyle BC\) in an acute-angled triangle \(\displaystyle ABC\), and let \(\displaystyle T\) be the foot of the altitude drawn from \(\displaystyle B\). Prove that if \(\displaystyle \angle FAC =30^{\circ}\) then \(\displaystyle AF=BT\).

Based on the idea of S. Róka, Nyíregyháza

(5 pont)

Deadline expired on November 10, 2020.

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

Megoldás.

Az \(\displaystyle F\)-ből \(\displaystyle AC\)-re állított merőleges talppontja legyen \(\displaystyle R\).

A \(\displaystyle BTC\) háromszögben \(\displaystyle FR\) középvonal (hiszen \(\displaystyle BT\) és \(\displaystyle FR\) párhuzamosak, \(\displaystyle F\) pedig felezőpont), így \(\displaystyle BT=2FR\).

Az \(\displaystyle AFR\) egy félszabályos háromszög (\(\displaystyle R\)-nél derékszög, \(\displaystyle A\)-nál \(\displaystyle 30^\circ\)-os szög van), így \(\displaystyle AF=2FR\).

Ezzel megmutattuk, hogy \(\displaystyle AF=BT(=2FR)\).

Statistics:

167 students sent a solution.
5 points:	105 students.
4 points:	14 students.
3 points:	6 students.
2 points:	2 students.
1 point:	3 students.
0 point:	14 students.
Unfair, not evaluated:	22 solutionss.
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Problems in Mathematics of KöMaL, October 2020