Problem:
Points P and Q are chosen uniformly and independently at random on sides AB and AC, respectively, of equilateral triangle △ABC. Which of the following intervals contains the probability that the area of △APQ is less than half the area of △ABC ?
Answer Choices:
A. [83,21]
B. (21,32]
C. (32,43]
D. (43,87]
E. (87,1]
Solution:
Without loss of generality let AB=AC=BC=1; then the area of △ABC is 413. Let x=AP and y=AQ. Then the area of △APQ is
21xy⋅sin60∘=413⋅xy
The probability that the area of △APQ is less than half the area of △ABC is therefore the probability that xy<21. Graph the curve xy=21 in the unit square whose lower left corner is at the origin, as shown. Note that the curve passes through the points (21,1) and (1,21) and is concave up on the interval 21<x<1.
The probability that xy>21 is the area of the upper right "fat triangular" region with curved longest side, which is less than 41 but greater than 81. Therefore the probability that xy<21 lies between 1−41=43 and 1−81=87.
Note: The required area can be calculated to be 21+21ln2≈0.85.
