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 362334 figure.
Although no two contestants had exactly the same measurements, the two runnersup differed by less than an inch in each measurement from the winner and the waist of each was twothirds 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 runnersup (the tape is accurate to only a quarter of an inch)?
The measurements of the two runners up are B1W1H1 and B2W2H2. It is given that W1=(2/3)*H2 and W2=(2/3)*H1. Since the tape is accurate only to a quarter inch, H1 and H2 must be multiples of 3/4. The winners hips are 34=136/4. The only measurements that are multiples of 3/4 and are within one inch of this value are 135/4 (33 3/4) and 138/4 (34 1/2). Since we are not differentiating between the two runners up, let H1=135/4 and H2=138/4. Then
W1=(2/3)*138/4=92/4=23 and
W2=(2/3)*135/4=90/4=22 1/2.
Since B1+H1+W1=B2+W2+H2=36+23+34=93
B1=932333 3/4=36 1/4
B2=9322 1/234 1/2=36
To sum up, the measurements of the two runners up are 36 1/42333 3/4 and 3622 1/234 1/2.

Posted by Bryan
on 20030528 09:51:38 