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AMC 10 USA Math Olympiad

Area of Circle – Singapore Mathematics Olympiad – 2013

Try this beautiful problem from Singapore Mathematics Olympiad based on Area of Circle. You may use sequential hints to solve the problem.

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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