Let the distance between two parallel lines be 5 units and a point P lie between the lines…

Question

Let the distance between two parallel lines be 5 units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle $P Q R$ is formed such that $Q$ lies on one of the parallel lines, while R lies on the...

Answer: 28

Accepted Answer

We set up a coordinate system so that the two parallel lines are given by

$y=0\, \, y=5,$

since their distance is 5 units. Choose point

$P=(0,1)$

so that the distance from $P$ to the line $y=0$ is 1 unit (and its distance to the line $y=5$ is 4 units).

Let point

$Q=(a,0)$

be on the line $y=0$, and let point

$R=(b,5)$

be on the line $y=5$. Since triangle $PQR$ is equilateral with side length $s$, we require:

$PQ=PR=QR=s.$

A convenient method is to ``rotate'' $Q$ about $P$ by an angle of ${60}^{\circ }$ to obtain $R$. In complex-number (or vector) terms, if we translate so that $P$ is at the origin, then the rotation is given by

${e}^{i{60}^{\circ }}=\cos {60}^{\circ }+i\sin {60}^{\circ }=\frac{1}{2}+i\frac{\sqrt{3}}{2}.$

Thus, writing $Q$ in vector form relative to $P$, we have

$Q-P=(a,-1).$

Rotating this by ${60}^{\circ }$ gives

$R-P=(a\cos {60}^{\circ }-(-1)\sin {60}^{\circ },\, a\sin {60}^{\circ }+(-1)\cos {60}^{\circ }).$

Substituting the values $\cos {60}^{\circ }=\frac{1}{2}$ and $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$, we obtain

$\begin{gathered} R-P =(\frac{a}{2}+\frac{\sqrt{3}}{2},\, \frac{a\sqrt{3}}{2}-\frac{1}{2}), \ \, R =(\frac{a+\sqrt{3}}{2},\, 1+\frac{a\sqrt{3}}{2}-\frac{1}{2})=(\frac{a+\sqrt{3}}{2},\, \frac{a\sqrt{3}+1}{2}). $

Since $R$ lies on $y=5$, its $y$-coordinate must equal 5:

$\frac{a\sqrt{3}+1}{2}=5.$

Solve for $a$:

$ a\sqrt{3}+1 =10, \ a\sqrt{3} =9, \ a =\frac{9}{\sqrt{3}}=3\sqrt{3}. $

Now, the side length $s$ (which is the distance $PQ$) is given by

$ {s}^{2} =P{Q}^{2}={(3\sqrt{3}-0)}^{2}+{(0-1)}^{2} \ =(3\sqrt{3}{)}^{2}+{1}^{2} \ =27+1 \ =28. \end{gathered}$

Thus, the square of side $QR$ is

$(QR{)}^{2}={s}^{2}=28.$

Solutions · 1