A woman's "measurements" are three numbers that measure, in order, the inches around her bust, waist and hips.
Now, a certain beauty contest winner had a 36-23-34 figure.
Although no two contestants had exactly the same measurements, the two runners-up differed by less than an inch in each measurement from the winner and the waist of each was two-thirds the hips of the other.
If the sum of the three measurements was the same for all three girls, what were the vital statistics of the two runners-up (the tape is accurate to only a quarter of an inch)?
Let us denote the bust, waist and the hips of the contestants by b, w and h. Let the first runners up and the second runners up respectively be denoted by f and s,
The absolute difference between two corresponding measurements of a contestant is <= 3/4. Since the hips of f is precisely 1.5 times that of the waist of g, and vice versa, it follows that four times the measurement of hips of each of f and s must be a multiple of 3.
Since the hip measurement of the winner is 34, we must have:
|4*34 - t| <=3, where t is a multiple of 3.
Or, |136 - t|<=3
The only possible values of t satisfying the above inequality occurs at t = 135, 138.
Accordingly, h = 135/4, 138/4, so that:
w = (2/3)*(138/4), (2/3)*(135/4) = 23, 90/4
Now, the sum of measurements for each of the three girls = 36+23+34
= 93
So, (w, h) = (23, 135/4) gives: b = 93 - 23 - 135/4 = 145/4
Also, (w, h) = (90/4, 138/4) gives: b = 93 - 90/4 - 138/4 = 36
Since the bust determines the place of a given contestant in the
competition, the measurements of f and s are now as follows:
Place...................... Measurements
First Runners Up....... 36 1/4- 23 - 33 3/4
Second Runners Up.... 36 - 22 1/2 - 34 1/2
Puzzle Solution
