A geometry problem by Sahil Bansal
A jogging park has two identical circular tracks touching each other and a rectangular track enclosing the two circles. The edges of the rectangle are tangential to the circles. Two friends, A & B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track, A jogs along the rectangular track while B jogs along the two circular tracks in a figure of eight. Approximately, how much faster than A does B have to run, so that they take the same time to return to their starting point? (Take pi = 3.1416)
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1 solution
We know that, time = distance/speed
Here, the time taken by A to cover the rectangular park = time taken by B to cover the two circles.
Let the radius of each circle = r units
Since the edges of rectangle are tangential to the circle, length of rectangle = 2r + 2r = 4r & Breadth of the rectangle = r + r = 2r
Perimeter of rectangle (distance travelled by A) = 2(l + b) = 2(4r + 2r) = 2(6r) = 12r
Perimeter of two circles(distance travelled by B) = 2(2 pi r) = 4 pi r = 4 X 3.1416 r = 12.5664 r
Let speed of A = v (a) Let speed of B = v (b)
Hence, 12r/v(a) = 12.5664r/v(b)
That implies, v(a) / v(b) = 12r/12.5664r
Both r(s) get cancelled, so
v(a) / v(b) = 12 / 12.5664
Now, v (b) = v(a) x 12.5664 /12
Let us consider v(a) = 1
That implies, v(b) = 12.5664 / 12
Now, v(b) - v(a) = 0.5664/12
So, speed of b is more than that of a by -
[v(b) - v(a)] / v(a) x 100 % = 0.5664/12 x 100 = 56.64/12 = 4.72 %