Side Squared

Relevant wiki: Solving Triangles - Problem Solving - Medium

Firstly note that $BD = \sqrt{l^2 + l^4}$ .

Using the cosine rule on $\triangle ABD$ , we get $\large l^2 +l^4 = 2l^2-2l^2 \cos \color{#3D99F6}{\alpha} \, \Rightarrow \, l^2 = 1-2 \cos \color{#3D99F6}{\alpha} = \color{#03d300}{X} \quad (*)$ The maximum value of $\color{#03d300}{X}$ is $\text{max} (1-2 \cos \color{#3D99F6}{\alpha})=3$ when $\color{#3D99F6}{\alpha}=\pi$ . This occurs as the quadrilateral tends to a right triangle with hypotenuse $DAB$ side $2l$ .

As $\color{#03d300}{X}=l^2 \ge 0$ , its minimum value is $0$ . This occurs as the quadrilateral tends to an equilateral triangle of side $l$ . We can substitute $\color{#3D99F6}{\alpha} = \dfrac{\pi}{3}$ into $(*)$ to check.

Thus $0<\color{#03d300}{X}<3 \, \Rightarrow \, a+b = \large \color{gree}{\boxed{3}}$ .

A simpler solution:

We know that $BD=\sqrt{l^2+l^4}$ .

In any given triangle, the sum of two sides is always greater than the remaining side.

Thus, $l+l>\sqrt{l^2+l^4} \implies (2l)^2>l^2 +l^4 \implies 3l^2>l^4 \implies 3>l^2$ .

Quite clearly, for smaller values of $l$ we have $l^2$ approaching 0.

Since $X=l^2$ , $0<X<3\implies a+b=3$ .

Maximum value is at $\alpha = \pi$ :

$l^2 + l^4 = 4l^2 \rightarrow X_{max} = l^2 = 3$ (Pythagorean theorem)

Minimum value is at $\alpha = 0$ :

$l = 0 \rightarrow X_{min} = 0$

$3-0 = \boxed3$