AoPSWiki
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
Personal tools

1999 AIME Problems/Problem 8

From AoPSWiki

Problem

Let be the set of ordered triples of nonnegative real numbers that lie in the plane Let us say that supports when exactly two of the following are true: Let consist of those triples in that support \left(\frac 12,\frac 13,\frac 16\right). The area of divided by the area of is where and are relatively prime positive integers. Find

Solution

This problem just requires a good diagram and strong 3D visualization.

Image:1999_AIME-8.png

The region in 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 the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by . We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third.

Image:1999_AIME-8a.png

The side length of the large equilateral triangle is , which we can find using 45-45-90 with the axes. Using the formula for equilateral triangles, 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 to those of the large triangle, by corresponding angles we see that all of the triangles are similar, 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 \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 .

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}.

See also

1999 AIME (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us