Q201 / 269 · NDA 2025 (Phase 1) · Mathematics
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?
- A$\sqrt{2}$
- B$\sqrt{2} - 1$
- C$2 - \sqrt{3}$
- Correct answerD$4 - 2\sqrt{3}$
Solution
Let the points be $O(0, 0)$, $A(p, 1)$, and $B(1, q)$.
Since $\triangle OAB$ is equilateral, $OA^2 = OB^2 = AB^2$.
$OA^2 = p^2 + 1$
$OB^2 = 1 + q^2$
$AB^2 = (p - 1)^2 + (1 - q)^2$
From $OA^2 = OB^2$, we get $p^2 + 1 = 1 + q^2 \Rightarrow p^2 = q^2$.
Since $p$ and $q$ are between $0$ and $1$, $p = q$.
From $OA^2 = AB^2$, we get $p^2 + 1 = (p - 1)^2 + (1 - p)^2$.
$p^2 + 1 = 2(p^2 - 2p + 1)$
$p^2 + 1 = 2p^2 - 4p + 2$
$p^2 - 4p + 1 = 0$
Solving for $p$, we get $p = \dfrac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.
Since $0 < p < 1$, we must have $p = 2 - \sqrt{3}$.
Therefore, $p + q = 2p = 2(2 - \sqrt{3}) = 4 - 2\sqrt{3}$.
Answer: $4 - 2\sqrt{3}$
Since $\triangle OAB$ is equilateral, $OA^2 = OB^2 = AB^2$.
$OA^2 = p^2 + 1$
$OB^2 = 1 + q^2$
$AB^2 = (p - 1)^2 + (1 - q)^2$
From $OA^2 = OB^2$, we get $p^2 + 1 = 1 + q^2 \Rightarrow p^2 = q^2$.
Since $p$ and $q$ are between $0$ and $1$, $p = q$.
From $OA^2 = AB^2$, we get $p^2 + 1 = (p - 1)^2 + (1 - p)^2$.
$p^2 + 1 = 2(p^2 - 2p + 1)$
$p^2 + 1 = 2p^2 - 4p + 2$
$p^2 - 4p + 1 = 0$
Solving for $p$, we get $p = \dfrac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.
Since $0 < p < 1$, we must have $p = 2 - \sqrt{3}$.
Therefore, $p + q = 2p = 2(2 - \sqrt{3}) = 4 - 2\sqrt{3}$.
Answer: $4 - 2\sqrt{3}$
Hence, the correct answer is option: 'D'.