Problem C. 1493. (September 2018)
C. 1493. A triangle of unit area has sides \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\), such that \(\displaystyle a\ge b\ge c\). Show that \(\displaystyle b\ge \sqrt2\,\).
(5 pont)
Deadline expired on October 10, 2018.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Induljunk ki a háromszög területképletéből: \(\displaystyle bm_b=2\). Mivel \(\displaystyle c\), illetve \(\displaystyle m_b\) ugyanabban a derékszögű háromszögben átfogó, illetve befogó, ezért \(\displaystyle c\geq m_b\). Ezekből következik, hogy \(\displaystyle bc\geq bm_b=2\).
Ha \(\displaystyle c≤\sqrt2\), akkor \(\displaystyle b≥\frac2c≥\frac2{\sqrt2}=\sqrt2\), tehát \(\displaystyle b≥\sqrt2\).
Ha \(\displaystyle c>\sqrt2\), akkor \(\displaystyle b≥c>\sqrt2\), tehát \(\displaystyle b>\sqrt2\).
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Problems in Mathematics of KöMaL, September 2018