Problem of the week #66 - July 1st, 2013

[Jameson]

The diagram shows a circle with an equilateral triangle inside of it and one equilateral triangle outside of it. Calculate the ratio of the larger triangle to the smaller triangle.
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[Jameson]

Congratulations to the following members for their correct solutions:

1) caffeinemachine
2) MarkFL
3) soroban
4) Sudharaka

Solution (from soroban):

Spoiler

The diagram shows a circle with an inscribed equilateral triangle
and a circumscribed equilateral triangle.
Calculate the ratio of the areas of the large triangle to the small triangle.

                    *
                   * *
                  *   *
                 *     *
                *       *
               *         *
              *   o * o   *
             *o    *:*    o*
            o     *:::*     o
           o     *:::::*     o
          *     *:::::::*     *
         *o    *:::::::::*    o*
        * o   *:::::::::::*   o *
       *  o  *:::::::::::::*  o  *
      *     *:::::::::::::::*     *
     *     *  *  *  *  *  *  *     *
    *       o               o       *
   *          o           o          *
  *  *  *  *  *   o o o   *  *  *  *  *

Rotate the small triangle 60^o.

                    *
                   * *
                  *   *
                 *     *
                *       *
               *         *
              *           *
             *             *
            *               *
           *::*::*::*::*::*::*
          * *:::::::::::::::* *
         *   *:::::::::::::*   *
        *     *:::::::::::*     *
       *       *:::::::::*       *
      *         *:::::::*         *
     *           *:::::*           *
    *             *:::*             *
   *               *:*               *
  *  *  *  *  *  *  *  *  *  *  *  *  *

We have four congruent equilateral triangles.

The ratio is [tex]4:1[/tex]

Alternate solution (from MarkFL):

Spoiler

I will let the are of the larger triangle be $A_L$ and the area of the smaller triangle be $A_S$. We may find both areas in therm of $r$, the radius of the circle.

If we draw line segments from the center of the circle to the vertices of the smaller triangle, we obtain 3 isosceles triangle, where two of the sides of the triangles is $r$, and the angle subtended by the two known sides is \(\displaystyle \frac{360^{\circ}}{3}=120^{\circ}\)

Using the formula for the area of a triangle \(\displaystyle A=\frac{1}{2}ab\sin(C)\) and summing the areas 3 triangles, we find:

\(\displaystyle A_S=3\left(\frac{1}{2}r^2\sin\left(120^{\circ} \right) \right)=\frac{3\sqrt{3}}{4}r^2\)

If we draw line segments from the center of the circle to the mid-point of each side of the larger triangle, and to each vertex, we will have divided the larger triangle into 6 $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Thus, we may add the areas of these triangles to find:

\(\displaystyle A_L=6\left(\frac{1}{2}2r^2\sin\left(60^{\circ} \right) \right)=3\sqrt{3}r^2\)

And so the requested ratio is:

\(\displaystyle \frac{A_L}{A_S}=\frac{3\sqrt{3}r^2}{\frac{3\sqrt{3}}{4}r^2}=4\)