Key Concepts and Formulas

• Rotation of a Point: Rotating a point $(x, y)$ about the origin by an angle $\theta$ results in a new point $(x', y')$ where $x' = x \cos \theta - y \sin \theta$ and $y' = x \sin \theta + y \cos \theta$. For rotation about a point $(x_0, y_0)$, translate to the origin, rotate, and translate back.
• Equilateral Triangle: All sides are equal in length, and all interior angles are 60 degrees.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. The square of the distance is $(x_2 - x_1)^2 + (y_2 - y_1)^2$.

Step-by-Step Solution

Step 1: Setting up the Coordinate System

We establish a convenient coordinate system to represent the given geometric configuration.

• Let the two parallel lines be $L_1: y = 0$ and $L_2: y = 5$. This simplifies calculations as the lines are parallel to the x-axis.
• The point $P$ lies between the lines at a unit distance from $L_1$. Thus, let $P = (0, 1)$.
  • Why this choice? Choosing the lines as $y=0$ and $y=5$, and placing P on the y-axis simplifies the coordinates and leverages symmetry, without loss of generality.

Step 2: Defining the Coordinates of Points Q and R

We define the coordinates of points $Q$ and $R$ based on the problem statement.

• $Q$ lies on $L_1: y = 0$, so $Q = (a, 0)$ for some unknown $a$.
• $R$ lies on $L_2: y = 5$, so $R = (b, 5)$ for some unknown $b$.
• We aim to find $(QR)^2$, which is the square of the side length of the equilateral triangle $PQR$.

Step 3: Applying the Rotation Principle

We use the rotation formula to find the coordinates of $R$ by rotating $Q$ about $P$ by $60^\circ$.

• Vector $PQ$ relative to $P$: The coordinates of $Q$ relative to $P$ are $(x_Q - x_P, y_Q - y_P) = (a - 0, 0 - 1) = (a, -1)$.
• Rotating this vector by $60^\circ$: Let the rotated vector be $(x'_R, y'_R)$. Using the rotation formulas with $\theta = 60^\circ$: $x'_R = a \cos 60^\circ - (-1) \sin 60^\circ$ $y'_R = a \sin 60^\circ + (-1) \cos 60^\circ$ Substitute $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$: $x'_R = a \left(\frac{1}{2}\right) + \frac{\sqrt{3}}{2} = \frac{a + \sqrt{3}}{2}$ $y'_R = a \left(\frac{\sqrt{3}}{2}\right) - \frac{1}{2} = \frac{a\sqrt{3} - 1}{2}$
• Finding the absolute coordinates of $R$: Translate back by adding the coordinates of $P(0,1)$: $R = (x_P + x'_R, y_P + y'_R) = \left(0 + \frac{a+\sqrt{3}}{2}, 1 + \frac{a\sqrt{3}-1}{2}\right)$ $R = \left(\frac{a+\sqrt{3}}{2}, \frac{a\sqrt{3}+1}{2}\right)$
  • Why this step? The rotation formula gives the coordinates relative to the center of rotation ($P$). To get the actual coordinates of $R$ in our established coordinate system, we must add $P$'s coordinates back.

Step 4: Using the Condition for R

We use the fact that $R$ lies on the line $y=5$ to solve for $a$.

• Since $R$ lies on $y = 5$, its y-coordinate must be 5: $\frac{a\sqrt{3} + 1}{2} = 5$ Solving for $a$: $a\sqrt{3} + 1 = 10$ $a\sqrt{3} = 9$ $a = \frac{9}{\sqrt{3}} = 3\sqrt{3}$
  • Why solve for 'a'? The variable 'a' defines the position of point Q. Once 'a' is known, we have the full coordinates of Q, which allows us to calculate the side length of the equilateral triangle.

Step 5: Calculating $(QR)^2$

We calculate $(QR)^2$ using the fact that $QR = PQ$ in an equilateral triangle.

• Since $PQR$ is equilateral, $QR = PQ$, so $(QR)^2 = (PQ)^2$.
• We have $P = (0, 1)$ and $Q = (a, 0) = (3\sqrt{3}, 0)$.
• Using the distance formula: $(PQ)^2 = (3\sqrt{3} - 0)^2 + (0 - 1)^2$ $(PQ)^2 = (3\sqrt{3})^2 + (-1)^2$ $(PQ)^2 = 27 + 1$ $(PQ)^2 = 28$ Therefore, $(QR)^2 = 28$.

Step 6: Checking for the other rotation direction

Let us consider rotating by $-60^\circ$.

• Rotating this vector by $-60^\circ$: Let the rotated vector be $(x'_R, y'_R)$. Using the rotation formulas with $\theta = -60^\circ$: $x'_R = a \cos (-60^\circ) - (-1) \sin (-60^\circ)$ $y'_R = a \sin (-60^\circ) + (-1) \cos (-60^\circ)$ Substitute $\cos (-60^\circ) = \frac{1}{2}$ and $\sin (-60^\circ) = -\frac{\sqrt{3}}{2}$: $x'_R = a \left(\frac{1}{2}\right) - (-1) \left(-\frac{\sqrt{3}}{2}\right) = \frac{a}{2} - \frac{\sqrt{3}}{2} = \frac{a-\sqrt{3}}{2}$ $y'_R = a \left(-\frac{\sqrt{3}}{2}\right) + (-1) \left(\frac{1}{2}\right) = -\frac{a\sqrt{3}}{2} - \frac{1}{2} = \frac{-a\sqrt{3}-1}{2}$
• Finding the absolute coordinates of $R$: Translate back by adding the coordinates of $P(0,1)$: $R = (x_P + x'_R, y_P + y'_R) = \left(0 + \frac{a-\sqrt{3}}{2}, 1 + \frac{-a\sqrt{3}-1}{2}\right)$ $R = \left(\frac{a-\sqrt{3}}{2}, \frac{-a\sqrt{3}+1}{2}\right)$
• Since $R$ lies on $y = 5$, its y-coordinate must be 5: $\frac{-a\sqrt{3} + 1}{2} = 5$ Solving for $a$: $-a\sqrt{3} + 1 = 10$ $-a\sqrt{3} = 9$ $a = \frac{-9}{\sqrt{3}} = -3\sqrt{3}$
• We have $P = (0, 1)$ and $Q = (a, 0) = (-3\sqrt{3}, 0)$.
• Using the distance formula: $(PQ)^2 = (-3\sqrt{3} - 0)^2 + (0 - 1)^2$ $(PQ)^2 = (-3\sqrt{3})^2 + (-1)^2$ $(PQ)^2 = 27 + 1$ $(PQ)^2 = 28$ Therefore, $(QR)^2 = 28$.

Common Mistakes & Tips:

• Sign Errors: Be careful with signs when applying the rotation formulas, especially when rotating about a point other than the origin.
• Coordinate System Choice: A well-chosen coordinate system can simplify calculations. Placing one line along the x-axis is often a good starting point.
• Alternative Approaches: While the rotation method is elegant, the problem can also be solved using the distance formula and setting up a system of equations. However, this is generally more computationally intensive.

Summary

By setting up a convenient coordinate system and utilizing the rotation formula to represent the coordinates of point $R$ in terms of point $Q$, we were able to use the condition that $R$ lies on the line $y=5$ to solve for the unknown coordinate $a$. This allowed us to find the coordinates of $Q$ and subsequently calculate the side length of the equilateral triangle, $(QR)^2$, which is equal to 28.

Final Answer

The final answer is \boxed{28}.