$$𝑀𝑇 𝐴𝐼$$ is a parallelogram of area $$\frac{40}{41}$$ square units such that $$𝑀𝐼 = 1/𝑀𝑇$$. If $$𝑑$$ is the least possible length of the diagonal $$MA$$, and $$d^{2} = \frac{a}{b}$$, where $$𝑎, 𝑏$$ are positive integers with $$gcd(a,b) = 1$$, find $$|a - b|$$.

Let the adjacent sides of the parallelogram be represented by the vectors
$$\vec{u} \;=\; \overrightarrow{MT},\qquad \vec{v} \;=\; \overrightarrow{MI}.$$

Denote their lengths by
$$|\vec{u}| = MT = x,\qquad |\vec{v}| = MI = y.$$

The question gives the relation $$MI = \dfrac1{MT},$$ hence
$$y = \dfrac1x.$$

Let $$\theta$$ be the angle between the two adjacent sides (between $$\vec{u}$$ and $$\vec{v}$$). The area of a parallelogram is $$|\vec{u}\times\vec{v}| = xy\sin\theta$$, so

$$x\Bigl(\dfrac1x\Bigr)\sin\theta = \dfrac{40}{41}\;\;\Longrightarrow\;\;\sin\theta = \dfrac{40}{41}.$$

Because $$\sin\theta = \dfrac{40}{41},$$ the cosine may be either
$$\cos\theta = \pm\dfrac{9}{41}.$$

The diagonal whose length we need is $$\overrightarrow{MA} = \vec{u} + \vec{v},$$ so

$$d^2 = |\vec{u}+\vec{v}|^{\,2} = |\vec{u}|^{2} + |\vec{v}|^{2} + 2\,\vec{u}\!\cdot\!\vec{v} = x^{2} + \dfrac1{x^{2}} + 2\!\left(x\cdot\dfrac1x\right)\cos\theta = x^{2} + \dfrac1{x^{2}} + 2\cos\theta.$$

To find the least possible value of $$d^2$$ we must minimise both terms: 1. $$x^{2} + \dfrac1{x^{2}}$$, and 2. $$2\cos\theta$$ (choose the smaller of the two possible cosines).

$$d^{2}_{(1)} = x^{2} + \dfrac1{x^{2}} + \dfrac{18}{41}.$$ The minimum of $$x^{2} + \dfrac1{x^{2}}$$ occurs at $$x = 1$$ (by differentiating or AM ≥ GM), giving
$$d^{2}_{(1)\,\min} = 2 + \dfrac{18}{41} = \dfrac{100}{41}.$$

$$d^{2}_{(2)} = x^{2} + \dfrac1{x^{2}} - \dfrac{18}{41}.$$ Again the minimum of $$x^{2} + \dfrac1{x^{2}}$$ is attained at $$x = 1$$, so
$$d^{2}_{(2)\,\min} = 2 - \dfrac{18}{41} = \dfrac{64}{41}.$$

Since $$\dfrac{64}{41} \lt \dfrac{100}{41},$$ the least possible value of $$d^{2}$$ is

$$d^{2}_{\min} = \dfrac{64}{41}.$$

This fraction is already in its lowest terms, so $$a = 64,\; b = 41.$$ Therefore
$$|a - b| \;=\; |64 - 41| \;=\; 23.$$

Answer: 23