In a triangle, the edges are extended past both vertices by the length of the edge opposite to the respective vertex.
Show that the area of the resulting hexagon is at least $13$ times the area of the original triangle.
if we represent the 3 sides of the triangle as $a,b,c$ then calculate the area of every individual region of the hexagon then we have to prove: $$\frac{(a+b)^2}{ab}+\frac{(b+c)^2}{bc}+\frac{(c+a)^2}{ca}+\frac{a}c+\frac{c}b+\frac{b}a\ge 15 $$which is always true.
The only 'tricky' part in this problem was to get rid of the sin's after calculating the area, which can be done by the extended law of sin's (using the circum radius of $ABC$. Then Am-Gm suffices.
minusonetwelth wrote:
The only 'tricky' part in this problem was to get rid of the sin's after calculating the area, which can be done by the extended law of sin's (using the circum radius of $ABC$. Then Am-Gm suffices.
It can be solved without sine law by just joining some line and making triangle pairs of same height, then you can make a ratio of area with main triangle, and adding them just use AM-GM to get $\ge 13$