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 $PQR$ is formed such that $Q$ lies on one of the parallel lines, while $R$ lies on the other. Then $(QR)^2$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 28

Accepted Answer

Set the parallel lines as $y=0$ and $y=5$, with $P=(0,1)$ (distance 1 from $y=0$, distance 4 from $y=5$). Let $Q=(a,0)$ on $y=0$.

Rotating $Q-P=(a,-1)$ by $60^\circ$ about $P$ gives $R-P$:
$$R-P = \left(a\cos60^\circ-(-1)\sin60^\circ,\ a\sin60^\circ+(-1)\cos60^\circ\right) = \left(\frac a2+\frac{\sqrt3}2,\ \frac{a\sqrt3}2-\frac12\right)$$
$$R = \left(\frac{a+\sqrt3}2,\ \frac{a\sqrt3+1}2\right)$$

Since $R$ lies on $y=5$:
$$\frac{a\sqrt3+1}{2}=5 \implies a\sqrt3=9 \implies a = 3\sqrt3$$

$$s^2 = PQ^2 = (3\sqrt3)^2+1^2 = 27+1=28$$

$(QR)^2 = s^2 = 28$

Solutions · 1