Hindustan Times ST (Mumbai) - Live

Pretty rings all in a row

- By Dilip D’Souza

It’s the Olympics, so how about a puzzle that gets you to look more closely at the famous symbol of the Games, the five circles?

As you see, the circles are linked, pairwise. Each link produces a small area that’s common to the two linked circles. For example, the blue and yellow circles share a common area between points A and B on their circumfere­nces (ie, A and B are the two points where the blue and yellow circles intersect). There are four such shared spaces: blue and yellow, yellow and black, black and green, green and red.

Assume each circle has a diameter of 2 cm. Assume that if you draw a straight line from A to B, its length is 1 cm. Assume the shared spaces are all equal in area.

Question: What is the total area of the four shared spaces? (All you need is elementary geometry and arithmetic).

Hint: Note that the area of each circle is πr2, or π times the square of its radius. Since its diameter is 2, the radius is 1 cm; thus the area of each circle is πcm2.

Call the centre of the blue circle O.

Draw lines from O to A (OA), O to B (OB), and A to B (AB). OA and OB are both radii of the blue circle, thus they are 1 cm long. So is AB. Thus the triangle OAB is an equilatera­l triangle, so each of its angles is 60º; in particular, the angle at point O is 60º. Since a circle is 360º all the way around, you can draw six such equilatera­l triangles in the blue circle, all equal, all radiating from point O.

That divides the circle into six equal slices: the one defined by points O, A and B, and five more. Thus the area of slice OAB is one sixth of the area of the circle, thus π/6 cm2.

What’s the area of the triangle OAB? Draw a line from the midpoint of AB, call it C, to O. AC = 1/2 cm, OA = 1 cm, and the angle at C (angle OCA) is 90º. Thus by Pythagoras, OC = √3/2 cm.

Nearly done! You now have all you need to calculate the area of triangle OAB; then the outer part of slice OAB, ie, outside the line AB but within the blue circle. The rest is simple.

For the answer, see bottom of the page

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