# Solution to the Stacked Triangles Puzzle +-- {.image} [[StackedTriangles.png:pic]] > Three of the triangles in this stack are equilateral. Find the sum of the pink and green areas. =-- ## Solution by [[Pythagoras' Theorem]] and [[Equilateral Triangles]] +-- {.image} [[StackedTrianglesLabelled.png:pic]] =-- With the points labelled as above, let $a$ be the length of $B C$, $b$ of $A C$, and $c$ of $A B$. Then these are also the lengths of the blue, purple, and yellow triangles respectively. The area of an [[equilateral triangle]] is $\frac{\sqrt{3}}{4}$ times the square of its side length, so considering the areas of the blue and yellow triangles gives: $$ \frac{\sqrt{3}}{4} (a^2 + c^2) = 100 $$ Line segments $A E$ and $B C$ are [[parallel]] so the "height" of $A$ above $B C$ is the same as the "height" of $B$ above $A E$, which is $\frac{\sqrt{3}}{2} a$. The green area is therefore $$ \frac{1}{2} \times \frac{\sqrt{3}}{2} a \times c = \frac{\sqrt{3}}{4} a c $$ Considering the [[right-angled triangle]] $A D C$, the length of $A D$ is $\frac{a + c}{2}$ and the length of $C D$ is $\frac{\sqrt{3}}{2}c - \frac{\sqrt{3}}{2}a$. [[Pythagoras' theorem]] therefore gives the following formula for $b^2$: $$ \begin{aligned} b^2 &= \left(\frac{\sqrt{3}}{2}c - \frac{\sqrt{3}}{2}a\right)^2 + \left(\frac{a + c}{2}\right)^2 \\ &= \frac{3}{4} c^2 + \frac{3}{4} a^2 - \frac{3}{2} a c + \frac{1}{4} a^2 + \frac{1}{4} c^2 + \frac{1}{2} a c \\ &= a^2 + c^2 - a c \end{aligned} $$ The area of the purple triangle is $\frac{\sqrt{3}}{4} b^2$ so the combined purple and green regions have total area: $$ \frac{\sqrt{3}}{4} b^2 + \frac{\sqrt{3}}{4} a c = \frac{\sqrt{3}}{4} a^2 + \frac{\sqrt{3}}{4} c^2 = 100 $$ ## Solution by [[Cosine Rule]] and [[Area of a Triangle]] With $a$, $b$, and $c$ as above, the [[cosine rule]] applied to the green triangle gives: $$ b^2 = a^2 + c^2 - 2 a c \cos(60^\circ) = a^2 + c^2 - a c $$ So $b^2 + a c = a^2 + c^2$. The sine rule for the [[area of a triangle]] shows that the areas of the three triangles are: $$ \frac{1}{2} a^2 \sin(60^\circ), \quad \frac{1}{2} b^2 \sin(60^\circ), \quad \frac{1}{2} c^2 \sin(60^\circ), \quad \frac{1}{2} a c \sin(60^\circ) $$ So the purple and green areas are $$ \frac{1}{2} (b^2 + a c) \sin(60^\circ) = \frac{1}{2} (a^2 + c^2) \sin(60^\circ) = 100 $$ ## Solution by [[Invariance Principle]] If the yellow and blue triangles are the same size, the green triangle becomes equilateral and all four are in fact [[congruent]], whereupon the area of the green and purple is the same as that of the yellow and blue.