Question

Two parallel lines - are at a distance 1 apart and a point P between them is at a distance p from one of them. Q and R are chosen on these two parallel, lines such that $\Delta$ PQR is an equilateral triangle. Prove that the length of the side of this equilateral triangle is$\sqrt{\frac{p^2-p+1}{3}}$

Answer 1

y = 0 and y = 1 are two parellel lines , distance QL = 1apart . P is point between then such that PM = p and $\Delta$ PQR is equilateral
Let Q lie on y = 1 and R on y = 0 may be chosen as origin (0,0) if RP be inclined at $\theta$ , then
p = a sin $\theta$ ........(1)
QL = 1 = QR sin $(\theta +{60}^{\circ })=asin(\theta +{60}^{\circ })$
or 1 = a $(sin\theta cos{60}^{\circ }+cos\theta sin{60}^{\circ })$ .... (2)
We have to eliminate $\theta$ between (1) and (2) and the find the length of side a in terms of known quantity p.
${(1-\frac{asin\theta }{2})}^2=\frac{3a^{2}co{s}^2\theta }{4}=\frac{3a^2}{4}(1-si{n}^2\theta )$
Now put sin $\theta$ = p/a from (1)
${(1-\frac{p}{2})}^2+\frac{3a^2}{4}[1-\frac{p^2}{a^2}]=\frac{3a^2}{4}-\frac{3p^2}{4}$
or 4 - 4p = $p^2+3p^2=3a^2$
or a = 2$\sqrt{\frac{p^2-p+1}{3}}$