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

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#### Jameson

Staff member

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

Staff member
Congratulations to the following members for their correct solutions:

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

Solution (from soroban):
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.
Code:
                    *
* *
*   *
*     *
*       *
*         *
*   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 60o.

Code:
                    *
* *
*   *
*     *
*       *
*         *
*           *
*             *
*               *
*::*::*::*::*::*::*
* *:::::::::::::::* *
*   *:::::::::::::*   *
*     *:::::::::::*     *
*       *:::::::::*       *
*         *:::::::*         *
*           *:::::*           *
*             *:::*             *
*               *:*               *
*  *  *  *  *  *  *  *  *  *  *  *  *
We have four congruent equilateral triangles.

The ratio is $$4:1$$

Alternate solution (from MarkFL):
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:

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

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