If p and q are real numbers between 0 and 1 such that the points (p,1), (1,q) and (0,0) form an equilateral triangle, then what is $(p+q)$ equal to?

Calculation:

Given points A(0,0), B(p,1), and C(1,q), which form an equilateral triangle. We are told (0 < p,q < 1).

Compute the squared lengths of the sides:

\(AB^{2} = (p - 0)^{2} + (1 - 0)^{2} = p^{2} + 1\)

\(AC^{2} = (1 - 0)^{2} + (q - 0)^{2} = 1 + q^{2}\)

\(BC^{2} = (1 - p)^{2} + (\,q - 1\,)^{2} = (1 - p)^{2} + (q - 1)^{2} = 2\,(1 - p)^{2}\)

Because the triangle is equilateral, all three squared lengths are equal:

⇒ \(AB^{2} = AC^{2} \)

⇒ \(p^{2} + 1 \;=\; 1 + q^{2} \;\Longrightarrow\; p^{2} = q^{2} \;\Longrightarrow\; p = q \)

both p and q are positive in (0,1)

Let p = q = t Then

⇒ \(AB^{2} = t^{2} + 1 \)

⇒ \(BC^{2} = 2\,(1 - t)^{2} \)

Equate AB² and BC²:

⇒ \(t^{2} + 1 \;=\; 2\,(1 - t)^{2} \;=\; 2\bigl(1 - 2t + t^{2}\bigr) \;=\; 2 - 4t + 2t^{2}. \)

Simplify:

⇒ \(t^{2} + 1 = 2 - 4t + 2t^{2}\;\Longrightarrow\;0 = 2 - 4t + 2t^{2} - (t^{2} + 1) = t^{2} - 4t + 1. \)

So t satisfies:

⇒ \(t^{2} - 4t + 1 = 0 \quad\Longrightarrow\quad t = \frac{4 \pm √{16 - 4}}{2} = \frac{4 \pm 2√{3}}{2} = 2 \pm √{3}. \)

Since 0 < t < 1, we take t = 2 - √3 (note 2 + √3> 1) is not allowed. Hence,

⇒ \(p = q = 2 - √{3}.\)

Therefore:

 \(p + q \;=\; (2 - √{3}) + (2 - √{3}) \;=\; 4 - 2√{3}.\)

Hence, the correct answer is Option 4.