Ratio of area covered by four equilateral triangles in a rectangle

Question

The following puzzle is taken from social media (NuBay Science communication group).

It asks to calculate the fraction (ratio) of colored area in the schematic figure below where the four colored triangles are supposed to be equilateral. The sides of the rectangle are not mentioned.

At first one might think that the problem is not well-posed. However, it turns out that the fact that such a configuration exists for the rectangle at hand (note that for example for a square it is clearly impossible to have such a configuration) yields a condition on the proportions of the rectangle. This condition in turn allows to determine the ratio.

The question here is to determine the condition on the proportions of the rectangle and the fraction (ratio) of the colored area.

Both turn out to be unique and the problem is thus well-posed.

Note: The question is self-answered, see this answer.

Answer 1

Immediately obvious is the fact that half the base of the green triangle equals the altitudes of each of the yellow and orange triangles, thus the green to yellow and orange triangle similarity ratio is $\sqrt{3}$, and by the same reasoning, the yellow and orange triangles to the red triangle have similar ratio $\sqrt{3}$. If the altitude of the red triangle is $1$, then the width of the rectangle is $2\sqrt{3}$ and the height is $1 + 3 = 4$, for an aspect ratio of $2 : \sqrt{3}$.

If we look at the rectangle that encloses half the red and yellow triangles, the white triangle is equal in area to the full red triangle, and the half yellow triangle is half the area of the rectangle. Therefore, the colored areas in that rectangle comprise $4/6 = 2/3$ of the area of that rectangle. Since this relationship is the same regardless of the scale, the whole figure is shaded by $2/3$.

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It is worth entertaining a generalization of the given figure to acute isosceles triangles. Suppose the half-angle of the apex of the yellow triangle is $\theta$; then for $0 < \theta \le \pi/4$, the triangles are in similarity ratio $1 : \cot \theta : \cot^2 \theta$ from smallest to largest, and the rectangle has aspect ratio $$\frac{1 + \cot^2 \theta}{2 \cot \theta} = \csc 2\theta.$$ The ratio of the shaded area to the rectangle's area is simply $$\frac{1}{2}\sec^2 \theta.$$ For the equilateral case, $\theta = \pi/6$.

Answer 2

Let's call the side of the green triangle $l$, and the side of the yellow/orange triangles $x$. The height of the green triangle is $l\frac{\sqrt 3}2$, and the height of the orange triangle is $x\frac{\sqrt 3}2$. But from the figure, this is also $l/2$. So $$x\frac{\sqrt 3}2=\frac l2$$ or $$x=\frac l{\sqrt 3}$$ Then the height of the rectangle is $$l\frac{\sqrt 3}2+\frac x2=l\left(\frac{\sqrt 3}2+\frac 1{2\sqrt 3}\right)$$ So the ratio of the long length in the rectangle to the short one is $$r=\frac{\sqrt 3}2+\frac 1{2\sqrt 3}\approx1.1547$$ For the red triangle, of side $y$, the height is $$y\frac{\sqrt 3}2=\frac x2$$ so $$y=\frac x{\sqrt 3}$$ You can express then $x, y$ in terms of $l$ and then calculate all areas for the triangles and the area of the rectangle. Everything will contain a factor of $l^2$.

Alternative for the last part If one draws the perpendiculars from the intersection point of the triangles to the sides, you have 4 similar rectangles. The width of the bottom left rectangle is $\frac l2$ and the height is $\frac {l\sqrt 3}2$. Then the area of the rectangle is $$\frac {l^2\sqrt 3}4$$ Half of that is green. The area of the orange part is $$\frac 12\frac x2\frac l2=\frac{l^2}{8\sqrt 3}$$ Then the colored region to the rectangle region is $$R=\frac{\frac 12\frac {l^2\sqrt 3}4+\frac{l^2}{8\sqrt 3}}{\frac {l^2\sqrt 3}4}=\frac 12+\frac 16=\frac 23$$

Answer 3

The 4 write triangles are isosceles with their base angles all $30^\circ$, i.e. A and B tri- and bi-sects the sides of the rectangle respectively. So, X is a quarter from the top vertically and midway horizontally. Assuming unit rectangle area

$$\text{I}= \frac14\cdot\frac16 \cdot 1 = \frac1{24},\>\>\>\>\> \text{II }= \frac12\cdot\frac14 \cdot 1 = \frac1{8}$$

with the sum $2(\frac1{24}+\frac18 ) = \frac13$. Thus, the fraction of the colored areas is $\frac23$.

Answer 4

If $S$ is the side and $H$ is the height of an equilateral triable then its area $A=\frac{\sqrt{3}}{4}S^2=\frac{H^2}{\sqrt{3}}.$

Let the base of rectangle to be $k$. Then the area of the biggest equilateral triangle is $A_1= \frac{\sqrt{3}~k^2}{4}$ Let the height of the rectangle be $h $ then the height of the automatic equilateral triangle above the first one is $y=(h-\frac{\sqrt{3}~k}{2})$, then the area $A_2$ of this second one is $A_2=\frac{1}{\sqrt{3}}(h-\frac{\sqrt{3}~k}{2})^2$ The third and the fourth identical left and right triangle of height $k/2$ will be critically equilateral if $$y=\frac{k}{2} \tan (\pi/6) \implies h=\frac{2k}{\sqrt{3}}\implies A_2= \frac{k^2}{12\sqrt{3}}$$ Their area $A_3=A_4=\frac{k^2}{4\sqrt{3}}$ Finally, the ratio ($R$) of the area of all four equilateral triangles $A=A_1+A_2+2A_3$ which are colored to the area of the rectangle $hk$, is given as $$R=\frac{\sqrt{3}}{2k^2}\left(\frac{\sqrt{3}~k^2}{4}+\frac{k^2}{12\sqrt{3}}+\frac{k^2}{2\sqrt{3}}\right)=\frac{2}{3},~\text{iff}~\frac{h}{k}=\frac{2}{\sqrt{3}}$$