UPSC NDA I 2025 Mathematics Paper

Concept:For an equilateral triangle, all sides are equal. Use the distance formula to equate squared side lengths.Explanation:Let A(0,0), B(p,1), C(1,q). Compute squared distances:$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$Since triangle is equilateral, $AB^2 = AC^2$ gives $p^2+1 = 1+q^2$, so $p^2=q^2$. As p,q in (0,1), we have $p=q$. Let $p=q=t$.Now equate $AB^2$ and $BC^2$: $t^2+1 = 2(1-t)^2$. Expand right side: $2(1-2t+t^2)=2-4t+2t^2$. Thus $t^2+1 = 2-4t+2t^2$. Rearranging gives $0 = t^2-4t+1$. Solve: $t = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$. Since $0 < t < 1$, choose $t = 2 - \sqrt{3}$. Hence $p = q = 2 - \sqrt{3}$.Therefore $p+q = 2(2-\sqrt{3}) = 4 - 2\sqrt{3}$.Answer:$p+q = 4-2\sqrt{3}$ which corresponds to option D.