Math olympiad 1988 problem 6: canonical solution (2) without Vieta jumping

Geometric solution to Q6: Consider a rectangle with sides 𝑎, 𝑏 Its diagonal has length $$ \sqrt{a^2+b^2} $$

This diagonal is a side of square A

Area A $$= a^2+b^2$$ Area B = $$ 𝑐. \sqrt{a^2+b^2}$$ Geometrically Q6 becomes equation (1) , $$ 𝑘 = Area A / Area B $$

Length 𝑎 is projected onto the side square A to yield a length, 𝑐

Now, $$ 𝑐 = 𝑎/cosϴ $$ And $$cosϴ = 𝑏/√{a^2+b^2}$$ So $$𝑐 = 𝑎/𝑏. √{a^2+b^2}$$ Area B $$= 𝑎/𝑏. {a^2+b^2}$$ Then from (1) $$𝑘 = 𝑏/𝑎 $$ (2)

Now if Area B = $$𝑎𝑏 + 1$$ as in Q6 then $$𝑎𝑏 + 1 = a/b.{a^2+b^2}$$ $$b^2 + 𝑏/𝑎 = {a^2+b^2}$$ $$𝑏 = a^3$$ From (2) $$𝑘 = a^2,$$ a perfect square