10.010 Each side of an equilateral triangle is divided into three equal parts

Problem 10.010

Problem Statement:

Each side of an equilateral triangle is divided into three equal parts, and the corresponding division points, counting in the same direction, are connected. An inscribed circle with a radius of r = 6 cm is drawn in the resulting equilateral triangle. Determine the side lengths of the triangles.

What does this mean?

We have a large equilateral triangle. Imagine we’ve taken each of its sides and divided it into three equal segments. Then, we connected these points to form a new, smaller triangle inside. This smaller triangle also turns out to be equilateral. A circle is inscribed within this smaller triangle, and we know its radius is 6 cm. Our task is to find the lengths of the sides of both triangles: the large one and the inner one.

Let’s refer to Figure 10.10 to better understand how these triangles are formed.

Solution

Step 1: Finding the side length of the inner (smaller) triangle.

• Why are we doing this? We know the radius of the circle inscribed specifically in this smaller triangle. There’s a special formula that relates the inradius to the side length of an equilateral triangle. By using this formula, we can easily find the side length of the smaller triangle.
• How are we doing this?

  We know that for an equilateral triangle with an inscribed circle, its area can be expressed in two ways:

  1. Using the triangle’s side length (a): Area = \frac{a^2\sqrt{3}}{4}
  2. Using its perimeter (P) and the inradius (r): Area = \frac{P \cdot r}{2}. Since the perimeter of an equilateral triangle is 3a, this becomes Area = \frac{3a \cdot r}{2}.

  Let’s equate these two area formulas to solve for the side length a (which is represented as FB, BM, or FM on our diagram – these are all sides of the inner triangle).

  \frac{a^2\sqrt{3}}{4} = \frac{3a \cdot r}{2}

  Now, substitute the known radius value r = 6 cm:

  \frac{a^2\sqrt{3}}{4} = \frac{3a \cdot 6}{2}
  \frac{a^2\sqrt{3}}{4} = 9a

  Since a is a side length, it cannot be zero. Therefore, we can divide both sides of the equation by a:

  \frac{a\sqrt{3}}{4} = 9

  Now, let’s solve for a:

  a\sqrt{3} = 9 \cdot 4
  a\sqrt{3} = 36
  a = \frac{36}{\sqrt{3}}

  To rationalize the denominator, we multiply both the numerator and the denominator by \sqrt{3}:

  a = \frac{36 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{36\sqrt{3}}{3} = 12\sqrt{3}

  So, the side length of the inner triangle (e.g., FB) is 12\sqrt{3} cm.

Step 2: Finding the side length of the outer (larger) triangle.

• Why are we doing this? We’ve found the side length of the smaller triangle. Now we need to find the side length of the larger triangle. To do this, we’ll use the relationship between the sides of the large and small triangles, which is given in the problem statement (the division of sides into three equal parts).
• How are we doing this?

  Let’s consider triangle AFM (see diagram).

  1. Angle: Since the large triangle ASC is equilateral, all its interior angles are 60^\circ. Therefore, angle \angle FAM = 60^\circ.
  2. Sides AF and AM: According to the problem statement, side AC (and AS, SC) is divided into three equal parts.
     • AF is one of these three parts, so AF = \frac{1}{3} AS.
     • AM consists of two such parts (from A to K and from K to M), so AM = \frac{2}{3} AC. Since AC = AS (sides of the large equilateral triangle), then AM = \frac{2}{3} AS.
  3. Side FM: Side FM is a side of the inner triangle, which we’ve already found: FM = FB = 12\sqrt{3} cm.

  Now, having two sides (AF and AM) and the included angle (\angle FAM = 60^\circ) in triangle AFM, we can use the Law of Cosines to find the third side FM.

  The Law of Cosines states: c^2 = a^2 + b^2 - 2ab \cos C.
  In our case:

  FM^2 = AF^2 + AM^2 - 2 \cdot AF \cdot AM \cdot \cos \angle FAM

  Substitute the expressions for AF and AM in terms of AS:

  FM^2 = \left(\frac{AS}{3}\right)^2 + \left(\frac{2AS}{3}\right)^2 - 2 \cdot \left(\frac{AS}{3}\right) \cdot \left(\frac{2AS}{3}\right) \cdot \cos 60^\circ

  Recall that \cos 60^\circ = \frac{1}{2}:

  FM^2 = \frac{AS^2}{9} + \frac{4AS^2}{9} - 2 \cdot \frac{2AS^2}{9} \cdot \frac{1}{2}
  FM^2 = \frac{AS^2}{9} + \frac{4AS^2}{9} - \frac{2AS^2}{9}
  FM^2 = \frac{AS^2 + 4AS^2 - 2AS^2}{9}
  FM^2 = \frac{3AS^2}{9}
  FM^2 = \frac{AS^2}{3}

  Now, take the square root of both sides to find FM:

  FM = \sqrt{\frac{AS^2}{3}} = \frac{AS}{\sqrt{3}}

  We know that FM is a side of the smaller triangle, and we’ve already found its value: FM = 12\sqrt{3} cm.
  Let’s equate these two expressions for FM:

  \frac{AS}{\sqrt{3}} = 12\sqrt{3}

  To solve for AS, multiply both sides by \sqrt{3}:

  AS = 12\sqrt{3} \cdot \sqrt{3}
  AS = 12 \cdot 3
  AS = 36 cm

  Therefore, the side length of the outer triangle is 36 cm.

Answer

The side length of the inner triangle is 12\sqrt{3} cm, and the side length of the outer triangle is 36 cm.