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March 16, 2020

Area of Circle - Singapore Mathematics Olympiad - 2013

Try this beautiful problem from Singapore Mathematics Olympiad based on Area of Circle.

Problem - Area of Circle


On the xy - plane , let S denote the region consisting of all points

(x,y ) for which \(|x+ \frac {1}{2} y | \leq 10 \) and \(|x|\leq 10 \) and \(|y|\leq 10 \). The largest circle centered at ( 0,0 ) that can be fitted in the region S has area \(k\pi\). Find the value of k.

  • 81
  • 80
  • 79
  • 84

Key Concepts


Area of Circle

2 D Geometry

Check the Answer


Answer: 80.

Singapore Mathematics Olympiad - 2013 - Senior Section - Problem No. 14

Challenges and Thrills - Pre - College Mathematics

Try with Hints


We can start the sum from here :

The region S is the hexagon enclosed by lines

\(x = \pm 10 \) ; \( y = \pm 10 \) and \( x + \frac {1}{2} y = \pm 10\).

So the largest circle contained in S must be tangent to \(x + \frac {1}{2} y = \pm 10\)

Try to find the radius from this ....................

We can start this hint by using a diagram:

Area of circle - figure

So following the last hint we can understand that its radius is the distance from the origin (0,0) to \( x + \frac {1}{2} y = 10 \)

r = \( \frac {10}{ \sqrt {1+ ({\frac {1}{2}})^2}} = 4\sqrt {5} \).

Now it very easy to find the rest of the solutiion.........

So the area of a circle is \(\pi {r}^2\)

= \(\pi {(4\sqrt 5)}^2 \)

= \(80 \pi\) (Answer).

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