1. Set up the parallel lines

Let the two parallel lines be $y=0 \quad \text{and} \quad y=5$ since the distance between them is $5$ units.

The point $P$ lies between them and is at a unit distance from one of them. So $P$ must be at distance $1$ from one line and hence distance $4$ from the other.

Without loss of generality, take $P=(0,1)$ so that $P$ is $1$ unit from the line $y=0$ and $4$ units from the line $y=5$.

2. Use the condition of equilateral triangle

Let the side length of the equilateral triangle be $QR=PQ=PR=s$ We need to find $s^2=(QR)^2$.

Now $Q$ lies on one parallel line and $R$ on the other. There are two possible cases:

• $Q$ on $y=0$, $R$ on $y=5$
• $Q$ on $y=5$, $R$ on $y=0$

We test the feasible one.

3. Take the feasible case

Suppose $Q=(x_1,0), \qquad R=(x_2,5).$ Then $PQ=s \Rightarrow x_1^2+(1-0)^2=s^2 \Rightarrow x_1^2+1=s^2$ $PR=s \Rightarrow x_2^2+(1-5)^2=s^2 \Rightarrow x_2^2+16=s^2$ Subtracting, $x_1^2-x_2^2=15.$

Also, $QR=s \Rightarrow (x_1-x_2)^2+(0-5)^2=s^2$ $\Rightarrow (x_1-x_2)^2+25=s^2.$

But from above, $s^2=x_1^2+1$, so $(x_1-x_2)^2+25=x_1^2+1$ $x_1^2-2x_1x_2+x_2^2+25=x_1^2+1$ $x_2^2-2x_1x_2+24=0.$

This approach is a bit messy. A cleaner way is to use the altitude property of an equilateral triangle.

4. Use altitude formula of an equilateral triangle

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

Since $Q$ and $R$ lie on the two parallel lines, the segment $QR$ joins the two lines. The point $P$ is at perpendicular distances $1$ and $4$ from the two lines. For the triangle to be equilateral, the perpendicular distance from $P$ to side $QR$ must be the altitude.

A more direct coordinate construction is easier:

Take $Q=(a,0), \qquad R=(b,5).$ Then $PQ^2=a^2+1, \qquad PR^2=b^2+16.$ Since triangle is equilateral, PQ=PR \Rightarrow a^2+1=b^2+16 \Rightarrow a^2-b^2=15. \tag{1} Also, $QR^2=(a-b)^2+25.$ And since $PQ=QR$, $a^2+1=(a-b)^2+25$ $a^2+1=a^2-2ab+b^2+25$ 2ab-b^2=24. \tag{2}

From (1), $a^2=b^2+15.$ Substitute into $(a-b)^2=a^2-2ab+b^2=(b^2+15)-2ab+b^2=2b^2-2ab+15.$ Using (2), $2ab-b^2=24 \Rightarrow -2ab=-24-b^2.$ So $(a-b)^2=(b^2+15)+(-24-b^2)+b^2=b^2-9.$ This is still not the neatest route.

Let us instead use the standard fact for an equilateral triangle on base $QR$:

If $M$ is midpoint of $QR$, then $PM \perp QR, \qquad PM=\frac{\sqrt{3}}{2}s.$ Also, $PQ^2=PM^2+QM^2=\left(\frac{\sqrt{3}}{2}s\right)^2+\left(\frac{s}{2}\right)^2=s^2.$ So this is consistent.

Now observe the key geometric fact: as $Q$ and $R$ lie on two parallel lines $5$ units apart, and $P$ is at distances $1$ and $4$ from them, the only possible placement is when the side $QR$ makes an angle $\theta$ with the parallels such that the perpendicular distances from $P$ to the endpoints match equal radii. This gives the standard relation $\text{distance between parallels} = \sqrt{s^2-1^2}+\sqrt{s^2-4^2}$ which is cumbersome.

A much simpler and correct method is to use coordinates with circles.

5. Intersection of circles approach

Since $PQ=PR=s$, points $Q$ and $R$ lie on circles centered at $P=(0,1)$.

• If $Q$ is on $y=0$, then from $PQ=s$: $x_Q^2+1=s^2 \Rightarrow x_Q=\pm\sqrt{s^2-1}.$
• If $R$ is on $y=5$, then from $PR=s$: $x_R^2+16=s^2 \Rightarrow x_R=\pm\sqrt{s^2-16}.$

Now $QR^2=(x_Q-x_R)^2+25.$ But since triangle is equilateral, $QR^2=s^2.$ Hence (x_Q-x_R)^2+25=s^2. \tag{3}

To make the distance as required, $Q$ and $R$ must lie on opposite sides of the vertical through $P$, so $|x_Q-x_R|=\sqrt{s^2-1}+\sqrt{s^2-16}.$ Thus from (3), $\left(\sqrt{s^2-1}+\sqrt{s^2-16}\right)^2+25=s^2.$ This is impossible since the left side exceeds $25$.

So the correct assignment must be: $Q \text{ on } y=5, \quad R \text{ on } y=0.$ Then $x_Q^2+16=s^2, \qquad x_R^2+1=s^2.$ Similarly, $QR^2=(x_Q-x_R)^2+25=s^2.$ Now the feasible configuration uses points on the same side of the vertical through $P$, so $|x_Q-x_R|=\sqrt{s^2-1}-\sqrt{s^2-16}.$ Therefore $\left(\sqrt{s^2-1}-\sqrt{s^2-16}\right)^2+25=s^2.$ Expand: $s^2-1+s^2-16-2\sqrt{(s^2-1)(s^2-16)}+25=s^2$ $s^2+8-2\sqrt{(s^2-1)(s^2-16)}=0$ $2\sqrt{(s^2-1)(s^2-16)}=s^2+8$ Square both sides: $4(s^2-1)(s^2-16)=(s^2+8)^2.$ Let $t=s^2.$ Then $4(t-1)(t-16)=(t+8)^2$ $4(t^2-17t+16)=t^2+16t+64$ $4t^2-68t+64=t^2+16t+64$ $3t^2-84t=0$ $3t(t-28)=0.$ Since $t>0$, $t=28.$ Thus $QR^2=s^2=28.$

6. Final answer

$\boxed{28}$