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 Thread: Mathcounts Marathon View Single Post
 09-08-2007 #68 hunter34 Guest   Posts: n/a \sigh This is a long one. Time to play with graphing programs. So you have three sets of linear equations for each vertex of the triangle. Each solution for each on is $(2\frac{1}{3},-3\frac{5}{6}),(-2\frac{3}{8},\frac{7}{8}),(-5\frac{1}{5},-7\frac{3}{5})$ So take the midpoint of $(2\frac{1}{3},-3\frac{5}{6})$ and $(-2\frac{3}{8},\frac{7}{8})$ which is $(\frac{1}{48},-\frac{71}{48})$. Now take the distance between the midpoint and two endpoints, and multiply the two distances and divide by two to get the answer. I have to go soon so someone can finish it up.