A not so innocent game
Three kids were playing in the park. Ann was 4m from Bert who was 3m from Cecilie who was 2m from Ann. Suddenly, Cecilie started to wonder how far she was from the midpoint between Ann and Bert.
Problem source: Pims.Math.
Three kids were playing in the park. Ann was 4m from Bert who was 3m from Cecilie who was 2m from Ann. Suddenly, Cecilie started to wonder how far she was from the midpoint between Ann and Bert.
Problem source: Pims.Math.
April 3rd, 2009 at 11:51 am
Half the battle is defining the problem. The three kids form an acute triangle. I define a cartesian coordinate system such that Ann is at (-a, 0), Bert is at (b, 0) and Cecilie is at (0, c); such that a, b, and c are all positive real numbers.
From the problem definition and Pythagoras,
a+b = 4
b^2+c^2=3^2
a^2+c^2=2^2
Working the Ann-Bert equation:
b=4-a
b^2=16-8a+a^2
b^2-a^2=16-8a
Subtracting the two Cecilie equations:
(b^2+c^2)-(a^2+c^2)=9-4
b^2-a^2 = 5
Combining the two values for b^2-a^2
16-8a=5
8a=11
a=11/8
Plug a into Ann-Bert equation
11/8 + b = 4
b=21/8
Plug a into the Ann-Cecilie equation:
(11/8)^2+c^2 = 2^2
c^2=256/64-121/64 = 135/64
c=sqrt(135)/8
as a check, 1.5m<c<<1.6m
as a check, plug b and c into the Bert-Cecilie equation:
(21/8)^2+(sqrt(135)/8)^2 =?= 3^2
441/64 + 135/64 =?= 576/64
The numbers check.
But Cecilie wants to know how far she is from the midpoint. The midpoint is:
(b-2, 0) = (2-a, 0)
(5/8, 0)
From this, we can compute the final distance
=sqrt ( (5/8)^2 + 135/64)
=sqrt(160/64)
=sqrt(10/4)
=sqrt(10)/2
or just under 1.6m