Maths Olympiad Prep

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Algebra Difficulty 7.1 National olympiad, round 2 Prove it

14. Given ABC:\triangle A B C: (1) If MM is any point in the plane, prove: AMsinABMsinB+CMsinCA M \cdot \sin A \leqslant B M \cdot \sin B + C M \cdot \sin C
(2) Let points A1,B1,C1A_{1}, B_{1}, C_{1} be on sides BC,AC,ABB C, A C, A B respectively, and the interior angles of A1B1C1\triangle A_{1} B_{1} C_{1} are α,β,γ\alpha, \beta, \gamma respectively, prove: AA1sinα+BB1sinβ+CC1sinγBCsinα+CAsinβ+ABsinγA A_{1} \cdot \sin \alpha + B B_{1} \cdot \sin \beta + C C_{1} \cdot \sin \gamma \leqslant B C \cdot \sin \alpha + C A \cdot \sin \beta + A B \cdot \sin \gamma.

Solution

14. (1) In quadrilateral ABMCA B M C, applying the generalized Ptolemy's theorem (Ptolemy's inequality) we get
AMBCBMAC+CMABA M \cdot B C \leqslant B M \cdot A C + C M \cdot A B

In ABC\triangle A B C, by the Law of Sines we have
AM2RsinABM2RsinB+CM2RsinCA M \cdot 2 R \sin A \leqslant B M \cdot 2 R \sin B + C M \cdot 2 R \sin C

which simplifies to
AMsinABMsinB+CMsinCA M \cdot \sin A \leqslant B M \cdot \sin B + C M \cdot \sin C
(2) From (1), we get
AA1sinαAB1sinβ+AC1sinγBB1sinβBA1sinα+BC1sinγCC1sinγCA1sinα+CB1sinβ\begin{array}{l} A A_{1} \cdot \sin \alpha \leqslant A B_{1} \cdot \sin \beta + A C_{1} \cdot \sin \gamma \\ B B_{1} \cdot \sin \beta \leqslant B A_{1} \cdot \sin \alpha + B C_{1} \cdot \sin \gamma \\ C C_{1} \cdot \sin \gamma \leqslant C A_{1} \cdot \sin \alpha + C B_{1} \cdot \sin \beta \end{array}

Adding the three inequalities, we get
AA1sinα+BB1sinβ+CC1sinγBCsinα+CAsinβ+ABsinγA A_{1} \cdot \sin \alpha + B B_{1} \sin \beta + C C_{1} \sin \gamma \leqslant B C \cdot \sin \alpha + C A \cdot \sin \beta + A B \cdot \sin \gamma

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.