Ratio of area covered by four equilateral triangles in a rectangle

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 $$R=\frac{\frac 12\frac {l^2\sqrt 3}4+\frac{l^2}{8\sqrt 3}}{\frac {l^2\sqrt 3}4}=\frac 12+\frac 16=\frac 23$$


enter image description here

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 $2/3$.

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, $\theta = \pi/6$.

enter image description here


enter image description here

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 $\frac23$.

Tags:

Puzzle

Area

Euclidean Geometry

Triangles

Quadrilateral

Related

A problem concerning a parallelogram and a circle Evaluation of double summation Is the sequence $(B_n)_{n \in \Bbb{N}}$ unbounded, where $B_n := \sum_{k=1}^n\mathrm{sgn}(\sin(k))$? A summation similar to Vandermonde identity Alternative approaches to showing that $\Gamma'(1/2)=-\sqrt\pi\left(\gamma+\log(4)\right)$ Does there exist an operation that could turn the set of all negative real numbers into an abelian group? Prove or Disprove : There exists a continuous bijection from $\mathbb{ R}^2$ to $\mathbb{R} $ If for invertible matrices $A$ and $X$, $XAX^{-1}=A^2$ then eigenvalues of $A$ are $n^{th}$ roots of unity. $\det(I+A)=1+\operatorname{Tr}(A)$ if $\operatorname{rank}(A)=1$ An Interesting Question I Posed to Myself About $\pi$ as an Average. When writing $\int_{1}^{x}x^2dx$, is $x$ a variable or a constant? $\int_0^1f(x) dx =0$, $\int_0^1xf(x) dx =0$. How to show that f has at least two zeros?

Recent Posts