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: ‌AB2=(p−0)2+(1−0)2=p2+1 ‌AC2=(1−0)2+(q−0)2=1+q2 ‌BC2=(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: ‌⇒AB2=AC2 ‌⇒p2+1=1+q2⇒p2=q2⇒p=q both p and q are positive in (0,1) Let p=q=t Then ‌⇒AB2=t2+1 ‌⇒BC2=2(1−t)2 Equate AB2 and BC2 : ⇒t2+1=2(1−t)2=2(1−2t+t2)=2−4t+2t2 Simplify: ⇒t2+1=2−4t+2t2⇒0=2−4t+2t2−(t2+1)=t2−4t+1. So t satisfies: ⇒t2−4t+1=0‌‌⟹‌‌t=‌
4±√‌16−4
2
=‌
4±2√‌3
2
=2±√‌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.