Let $\mathcal{T}$ be the set of ordered triples $(x,y,z)$ of nonnegative [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_number) that lie in the [plane](https://artofproblemsolving.com/wiki/index.php/Plane) $x+y+z=1.$  Let us say that $(x,y,z)$ supports $(a,b,c)$ when exactly two of the following are true: $x\ge a, y\ge b, z\ge c.$  Let $\mathcal{S}$ consist of those triples in $\mathcal{T}$ that support $\left(\frac 12,\frac 13,\frac 16\right).$  The area of $\mathcal{S}$ divided by the area of $\mathcal{T}$ is $m/n,$  where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
This problem just requires a good diagram and strong 3D visualization.
[1999 AIME-8.png](https://artofproblemsolving.com/wiki/index.php/File:1999_AIME-8.png)
The region in $(x,y,z)$ where $x \ge \frac{1}{2}, y \ge \frac{1}{3}$ is that of a little triangle on the bottom of the above diagram, of $y \ge \frac{1}{3}, z \ge \frac{1}{6}$ is the triangle at the right, and $x \ge \frac 12, z \ge \frac 16$ the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by $x+y+z=1,\ x,y,z \ge 0$. We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third. 
[1999 AIME-8a.png](https://artofproblemsolving.com/wiki/index.php/File:1999_AIME-8a.png)
The side length of the large equilateral triangle is $\sqrt{2}$, which we can find using 45-45-90 $\triangle$ with the axes. Using the formula $A = \frac{s^2\sqrt{3}}{4}$ for [equilateral triangles](https://artofproblemsolving.com/wiki/index.php/Equilateral_triangle), the area of the large triangle is $\frac{(\sqrt{2})^2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$. Since the lines of the smaller triangles are [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) to those of the large triangle, by corresponding angles we see that all of the triangles are [similar](https://artofproblemsolving.com/wiki/index.php/Similar_triangles), so they are all equilateral triangles. We can solve for their side lengths easily by subtraction, and we get $\frac{\sqrt{2}}{6}, \frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{2}$. Calculating their areas, we get $\frac{\sqrt{3}}{8}, \frac{\sqrt{3}}{18}, \frac{\sqrt{3}}{72}$. The [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) $\frac{\mathcal{S}}{\mathcal{T}} = \frac{\frac{9\sqrt{3} + 4\sqrt{3} + \sqrt{3}}{72}}{\frac{\sqrt{3}}{2}} = \frac{14}{36} = \frac{7}{18}$, and the answer is $m + n = \boxed{025}$.
To simplify the problem, we could used the fact that the area ratios are equal to the side ratios squared, and we get $\left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{6}\right)^2 = \frac{14}{36} = \frac{7}{18}$.