And Now, the Other Way

Quarter Wit, Quarter WisdomToday we will work with circles inscribed in regular polygons.

We begin by considering an equilateral triangle whose each side is of length ‘a’. Recall that every triangle has an incircle i.e. a circle can be inscribed in every triangle. The diagram given below shows the circle of radius ‘r’ inscribed in an equilateral triangle.

How can we find the relation between ‘r’ and ‘a’? Every angle of an equilateral triangle is 60 degrees. Since it is an equilateral triangle, due to the symmetry, angle OBD = angle OBA = 30 degrees. So we see that triangle BOD is a 30-60-90 triangle. So the ratio of the sides OD:BD = 1: sqrt(3) = r : a/2.

Therefore, a = 2 * sqrt(3) * r

Side of the triangle = 2 * sqrt(3) * Radius of the circle

As discussed last week, there are many other methods of getting this result. We can use the altitude method.

Altitude of an equilateral triangle is given by (sqrt(3)/2)*a. The incenter is at a distance 2/3rd of the altitude so OD (radius) = (1/3)* (sqrt(3)/2)*a = a/2*sqrt(3)

Or Side of the triangle = 2*sqrt(3) * Radius of the circle

Now we will look at a square.

The figure itself shows us that r = a/2

Side of the square = 2 * Radius of the circle

There is no need to delve deeper into it. Though, here is something for you to think about: Can you have a circle inscribed in a rectangle?

Now let’s consider a circle inscribed in a regular hexagon.

We know that the interior angle of a regular hexagon is 120 degrees. OA will bisect that angle making angle OAD = 60 degrees. Since AB is tangent to the circle, OD will be perpendicular to AB. Hence OAD is a 30-60-90 triangle. Therefore, a/2 : r = 1: sqrt(3)

Hence, a = 2r/sqrt(3)

Side of the hexagon = (2/sqrt(3)) * Radius of the circle

Again, remember, you are not expected to ‘know’ these results so don’t try to learn them up. You can always derive any relation you want once you know some basic tricks. The intent of these posts is to familiarize you with those tricks.

Next week, we will look at some interesting Geometry questions based on these concepts!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!