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If two vertices of an equilateral triangle are (0,0) and (3,√3) then find the third vertex. The first thing I did was calculated the distance of the given points and tried to make an equation including the third vertex but instead of being simplified, the term went on being complex.

Ger Wyn
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  • a straightforward geometric approach should give you one of the solutions very easily. the other requires a little more work – David Holden Dec 29 '15 at 16:09
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    Draw a picture. Let $O$ be the origin, and let $P=(3,\sqrt{3})$. Note that if $\theta$ is the angle that $OP$ makes with the positive $x$-axis, then $\tan\theta=\frac{1}{\sqrt{3}}$, so $\theta$ is $30$ degrees. – André Nicolas Dec 29 '15 at 16:33

3 Answers3

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Use complex numbers. Multiply $3+i\sqrt{3}$ by $e^{i \pi /3}$ to get the third vertex. This is because the origin is already one of the vertices.

The other possible vertex you can get by rotating counterclockwise, i.e. multilying by $e^{-i\pi /3}$

GaussTheBauss
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Using matrices, going anticlockwise the other vertex is $$\left(\begin {matrix}\cos 60 &-\sin 60 \\ \sin 60 & \cos 60\end{matrix}\right)\left (\begin{matrix}3\\ \sqrt{3}\end{matrix}\right)$$

For the clockwise possibility, multiply by the inverse matrix instead i.e. replace $60$ with $-60$ in the above.

David Quinn
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If you notice that the angle from the $x$-axis to $(3,\sqrt3)$ is $30°$, then you will see that the third vertex must be either on the $y$-axis or the reflection of $(3,\sqrt3)$ over the $x$-axis. (Note that $(3,\sqrt3)$ is a scalar multiple of $(\frac{\sqrt3}2,\frac12)$.)

If you didn't see that, then here's a general geometric solution. Let $A=(0,0)$ and $B=(3,\sqrt3)$. We can find the third vertex by looking at the perpendicular bisector of $AB$ and locating the points on it at the right distance from $A$. The midpoint $M$ of $AB$ is $(\frac32,\frac{\sqrt3}2)$. If $P$ is the third vertex, then the distances $\overline{MP}$ and $\overline{AM}$ are in a ratio of $\sqrt3:1$. Therefore, since $\overline{AM} = \sqrt3$, we have $\overline{MP}=\sqrt3\cdot\sqrt3=3$.

The unit vector perpendicular to $AB$ is $\frac1{2\sqrt3}(-\sqrt3,3) = \left(-\frac12,\frac{\sqrt3}2\right)$. Therefore, the two choices for $P$ are:

$$\left(\frac32,\frac{\sqrt3}2\right) \pm 3\left(-\frac12,\frac{\sqrt3}2\right)$$

giving $(0,2\sqrt3)$ and $(3,-\sqrt3)$.

Théophile
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