# Find side of an equilateral triangle inscribed in a rhombus

The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer diagonal. Determine the length of a side of this triangle. Express your answer in the form $k\left(4 \sqrt{3} − 3\right)$ where k is a vulgar fraction.

• In what form should the answer be expressed? $43-3=40$, so this does not make sense. – chaosflaws Sep 21 '14 at 12:20
• Sorry that should read k(4 root of 3 minus 3) where k is a vulgar fraction.Thanks – megan Sep 21 '14 at 12:32
• Have a look at my edited question to get a feeling for how LaTeX works. ;) (as soon as my edit becomes peer-reviewed) – chaosflaws Sep 21 '14 at 12:35
• Gosh, Thanks so much to Jack D'Aurizio and chaosflaws. I am happy and quite depressed at what I don't know!But thanks Guys ;) – megan Sep 21 '14 at 14:32
• Have another look at @N. F. Taussig's answer. It probably is the most straightforward answer. – chaosflaws Sep 21 '14 at 23:17

Given the picture, let $x$ be the side of the equilateral triangle. We have: $$6 = \frac{x}{2}\cot\arctan\frac{4}{3}+\frac{x}{2}\cot\frac{\pi}{6},$$ or: $$6 = \frac{3x}{8}+\frac{\sqrt{3}\,x}{2},$$ so: $$48 = x(4\sqrt{3}+3)$$ and: $$48(4\sqrt{3}-3) = 39 x,$$ so $k=\color{red}{\frac{16}{13}}$.

• A way to describe the second equality without explicit trig is to note that 1) the second term is the altitude of the equilateral triangle, 2) the first term is the width of the adjacent acute triangle, which has proportions $6:8$. – Semiclassical Sep 21 '14 at 14:23

Hopefully one can see anything in this drawing...

You know that all the green and all the blue lines are equal. Additionally, you know that the top green line is parallel to one of the diagonals and that the diagonals are perpendicular. Therefore, the top green line is perpendicular to the short diagonal as well.

From the Pythagorean theorem follows $$b^2-\left(\frac{b}{2}\right)^2=\left(n-x\right)^2$$ where $n$ is the short diagonal, $a$ is the length we are looking for and $x$ is the orange bit. This equation still has an $x$ in it; we need it in terms of $n$ and $m$.

Using the Intercept theorem, one can infer $$\frac{x}{\frac{b}{2}}=\frac{\frac{n}{2}}{\frac{m}{2}} \implies x=\frac{nb}{2m}$$ where $m$ denotes the long diagonal.

Algebra: \begin{align*}\frac{3}{4}b^2=\left(n-x\right)^2&=\left(n-\frac{nb}{2m}\right)^2\\ \frac{\sqrt{3}}{2}b&=n-\frac{nb}{2m}\\ \frac{\sqrt{3}}{2}b+\frac{nb}{2m}&=n\\ b\left(\frac{\sqrt{3}}{2}+\frac{n}{2m}\right)&=n\\ b=\frac{n}{\frac{\sqrt{3}}{2}+\frac{n}{2m}}&=\frac{2mn}{\sqrt{3}m+n}\end{align*}

Substituting the given values: $$b=\frac{96}{8\sqrt{3}+6}=\frac{48}{4\sqrt{3}+3}=\frac{48\left(4\sqrt{3}-3\right)}{\left(4\sqrt{3}+3\right)\left(4\sqrt{3}-3\right)}=\frac{48\left(4\sqrt{3}-3\right)}{\left(4\sqrt{3}\right)^2-3^2}=\frac{16}{13}\left(4\sqrt{3}-3\right)$$

• It should be $\frac{16}{13}$ in the very last line, since $(4\sqrt{3})^2-3^2=39.$ – Jack D'Aurizio Sep 21 '14 at 14:11
• Thanks. Neat solution over there, although I am not sure whether the OP can use trig ;) – chaosflaws Sep 21 '14 at 14:20

Here is a coordinate geometry solution:

The diagonals of a rhombus are perpendicular bisectors of each other. Suppose that the diagonals intersect at the origin of the coordinate plane, that the vertices at the ends of the short diagonal are $A(-3, 0)$ and $C(3, 0)$, that the vertices at the ends of the long diagonal are $B(0, 4)$ and $D(0, -4)$, and that one vertex of the inscribed equilateral triangle is the vertex $C$ of the rhombus.

The other vertices of the equilateral triangle must lie on sides $\overline{AB}$ and $\overline{AD}$ of the rhombus. Let $E$ be the vertex of the equilateral triangle on side $\overline{AD}$ and $F$ be the vertex of the equilateral triangle that lies on side $\overline{AB}$. Since the line containing an altitude of an equilateral triangle is the perpendicular bisector of the base, points $E$ and $F$ are equidistant from the $x$-axis. Thus, if the coordinates of point $F$ are $(a, b)$, then the coordinates of point $E$ are $(a, -b)$. If $s$ is the length of a side of the equilateral triangle, then $s = b - (-b) = 2b$.

In an equilateral triangle of side length $s$, the length of an altitude is $$\frac{s\sqrt{3}}{2}$$

Since $a < 0$, the length of the altitude is

$$3 - a = \frac{s\sqrt{3}}{2}$$

Since $s = 2b$, we obtain

$$3 - a = b\sqrt{3}$$

Solving for $a$ yields

$$3 - b\sqrt{3} = a$$

The equation of $\overleftrightarrow{AB}$ is

$$y = \frac{4}{3}x + 4$$

Hence, at point $F(a, b)$,

\begin{align*} b & = \frac{4}{3}(3 - b\sqrt{3}) + 4\\ 3b & = 4(3 - b\sqrt{3}) + 12\\ 3b & = 12 - 4b\sqrt{3} + 12\\ 3b + 4b\sqrt{3} & = 24\\ b(3 + 4\sqrt{3}) & = 24\\ b & = \frac{24}{3 + 4\sqrt{3}}\\ b & = \frac{24}{3 + 4\sqrt{3}} \cdot \frac{3 - 4\sqrt{3}}{3 - 4\sqrt{3}}\\ b & = \frac{24(3 - 4\sqrt{3})}{9 - 48}\\ b & = -\frac{24}{39}(3 - 4\sqrt{3})\\ b & = -\frac{8}{13}(3 - 4\sqrt{3})\\ b & = \frac{8}{13}(4\sqrt{3} - 3) \end{align*}

Hence,

$$s = 2b = \frac{16}{13}(4\sqrt{3} - 3)$$