Three Circles

In the figure below, 3 equal circles have been drawn so that each one passes through the centers of the other two. Is the area of overlap of the three triangles more or less than a quarter of the area of a circle?
Source: Julia Robinson Mathematics Festival

SOLUTION

The figure above shows the overlap of the three triangles in close-up. Let x denote the area between a circle and a chord, for example between a circle and chord \overline{AC} and let y denote the area of triangle ABC. Then,
Area of overlap = 3x+y

Area of triangle ABC
Triangle ABC is equilateral because its three sides are radii of congruent circles. Let r be the radius. \overline{BD} is an altitude and equals \frac{r\sqrt 3}{2} because triangle BDC is a 30^\circ-60^\circ-90^\circ triangle with a hypotenuse of length r.
Area of \triangle ABC=\frac{1}{2}\left (AC\right )\left (BD\right )
=\frac{1}{2}\left (r\right )\left (\frac{r\sqrt 3}{2}\right )
=\frac{r^2\sqrt 3}{4}

Calculate x
x = area of sector ABC- area of \triangle ABC
=\frac{60}{360}\pi r^2-\frac{r^2\sqrt 3}{4}

=\frac{\pi r^2}{6}-\frac{r^2\sqrt 3}{4}

Area of overlap
Area of overlap = 3x+y
=3\left (\frac{\pi r^2}{6}-\frac{r^2\sqrt 3}{4}\right )+\frac{r^2\sqrt 3}{4}

=\frac{\pi r^2}{2}-\frac{3r^2\sqrt 3}{4}+\frac{r^2\sqrt 3}{4}

=\frac{\pi r^2}{2}-\frac{r^2\sqrt 3}{2}

=\left (\frac{\pi-\sqrt 3}{2}\right )r^2

Compare areas
A quarter of the area of a circle = \left (\frac{\pi}{4}\right )r^2.
\frac{\pi-\sqrt 3}{2}=.7048
\frac{\pi}{4}=.7854
Since .7048<.7854 we conclude that the area of overlap is less than a quarter of the area of a circle.

Answer: Area of overlap is less than a quarter of the area of a circle.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

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