inverse star in a rectangle solution in Notes

# Solution to the Inverse Star in a Rectangle Puzzle

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[[InverseStarinaRectangle.png:pic]]

> All four triangles are equilateral.  What fraction of the rectangle do they cover?
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## Solution by Properties of an [[Equilateral Triangle]]

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[[InverseStarinaRectangleLabelled.png:pic]]
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In the above diagram, let $x$ be the length of the side $A B$.  The "height" of the yellow triangle, which is the [[perpendicular distance]] of $I$ to the side $D C$ is half that, so from [[lengths in an equilateral triangle]], the length of $D C$ is $\frac{x}{\sqrt{3}}$.  The height of $I$ above $A B$ is $\frac{x \sqrt{3}}{2}$ so the full height of the rectangle is:

$$
\frac{x \sqrt{3}}{2} + \frac{x}{2\sqrt{3} = \frac{2 x}{\sqrt{3}}
$$

The area of the rectangle is therefore $\frac{2 x^2}{\sqrt{3}}$.

The height of the red triangle, being the [[perpendicular distance]] of $I$ to $F E$, is half the length of $D C$ meaning that $F E$ has length $\frac{x}{\sqrt{3}} \div \sqrt{3} = \frac{x}{3}$.  The [[area of an equilateral triangle]] is $\frac{\sqrt{3}}{4}$ times the square of its side length, so the total area of the triangles is:

$$
\frac{\sqrt{3}}{4} \left( 1 + 2 \times \frac{1}{3} + \frac{1}{9} \right) x^2 = \frac{4 x^2}{3\sqrt{3}}
$$

This means that the equilateral triangles cover $\frac{2}{3}$rds of the rectangle.

## Solution by [[Isosceles Triangles]]

With the points labelled as in the above diagram, triangle $I C B$ is [[isosceles]] because angle $I \hat{C} B$ is $120^\circ$ and angle $C \hat{B} I$ is $30^\circ$.  This means that $I C$ is the same length as $C B$, so triangle $I D C$ has the same "base" length as $I C B$ (along $D B$) and they both have $I$ as apex above this length, hence they have the same area.  By the same argument, triangles $I E F$ and $I D E$ have the same area.

The four triangles therefore have the same area as the [[trapezium]] $A E D B$.  Moreover, $E D$ is a third of the length of $G D$.  Comparing the trapezium area with the rectangle shows that the trapezium is $\frac{2}{3}$rds of the area of the rectangle.