In a triangle $ABC$, points $A_{1}$ and $A_{2}$ are chosen in the prolongations beyond $A$ of segments $AB$ and $AC$, such that $AA_{1}=AA_{2}=BC$. Define analogously points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$. If $[ABC]$ denotes the area of triangle $ABC$, show that $[A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}] \geq 13 [ABC]$.
Problem
Source: Iberoamerican Olympiad 1992, Problem 6
Tags: inequalities, geometry, trigonometry, circumcircle, inequalities proposed
gemath
15.05.2007 17:03
I think the const must be $10$ not $13$ \[[A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}]=[AB_{1}C_{2}]+[BC_{1}A_{2}]+[CA_{1}B_{2}]+[AA_{1}A_{2}]-2[ABC]\ge 10 [ABC] \] \[\Leftrightarrow \frac{1}{2}(\sum (a+b)(a+c)\sin A+\sum a^{2}\sin A) \ge 12[ABC] \] (Apply law of sin and $S=2R^{2}\prod\sin A$) \[\Leftrightarrow \sum (\sin A+\sin B)(\sin A+\sin C)\sin A+\sum\sin^{3}A \ge 12\prod\sin A \] \[\Leftrightarrow \sum\sin A(\sin A(\sum\sin A)+\sin B\sin C)+\sum\sin^{3}A\ge 12\prod\sin A \] \[\Leftrightarrow \sum\sin^{3}A+(\sum\sin A)(\sum\sin^{2}A)\ge 12\prod\sin A \] which is trivial by AM-GM.
daniel73
20.09.2007 13:45
If I am not mistaken, $ S =\left[A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}\right] =\left[A_{1}BC_{2}\right]+\left[B_{1}CA_{2}\right]+\left[C_{1}AB_{2}\right]+\left[AA_{1}A_{2}\right]+\left[BB_{1}B_{2}\right]+\left[CC_{1}C_{2}\right]-2\left[ABC\right]$. We need to prove that $ 30\left[ABC\right]\leq2(S+2\left[ABC\right]) =\sum_{cyclic}(a+b)^{2}\sin C+\sum_{cyclic}a^{2}\sin A = 12\left[ABC\right]+(a^{2}+b^{2}+c^{2})(\sin A+\sin B+\sin C)$. Calling $ R$ the circumradius of $ \Delta ABC$, all we need to prove is that $ \frac{9abc}{2R}= 18\left[ABC\right]\leq\frac{(a+b+c)(a^{2}+b^{2}+c^{2})}{2R}$, which is obviously true because of the AMGM inequality applied to $ a,b,c$ and to $ a^{2},b^{2},c^{2}$, with equalitty iff the triangle is equilateral.