2003 AMC 12A Problems/Problem 16
Problem
A point P is chosen at random in the interior of equilateral triangle . What is the probability that has a greater area than each of and ?
Solution
Solution 1
After we pick point , we realize that is symmetric for this purpose, and so the probability that is the greatest area, or or , are all the same. Since they add to , the probability that has the greatest area is
Solution 2
We will use geometric probability. Let us take point , and draw the perpendiculars to , , and , and call the feet of these perpendiculars , , and respectively. The area of is simply . Similarly we can find the area of triangles and . If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose = the height of the triangle. Setting the area of triangle greater than and , we want to be the largest of , , and . We then realize that when is the incenter of . Let us call the incenter of the triangle . If we want to be the largest of the three, by testing points we realize that must be in the interior of quadrilateral . So our probability (using geometric probability) is the area of divided by the area of . We will now show that the three quadrilaterals, , , and are congruent. As the definition of point yields, = = . Since is equilateral, is also the circumcenter of , so . By the Pythagorean Theorem, . Also, angles , and are all equal to . Angles are all equal to degrees, so it is now clear that quadrilaterals are all congruent. Summing up these areas gives us the area of . contributes to a third of that area so .
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.