Can the angle take on any value - 2

An isosceles trapezoid can be transformed into a parallelogram while keeping the same area and perimeter:

This problem is now the same as the inspiration problem , where it was found that $\sin \theta \geq \cfrac{16A}{P^2} = \cfrac{16 \cdot 250}{80^2} = \cfrac{5}{8}$ .

Therefore, $a = 5$ , $b = 8$ , and $a + b = \boxed{13}$ .

Label the two parallel sides as $a$ and $b$ respectively; also label the height as $h$ . The hypotenuse of the right angled triangle is $d = \frac{h}{\sin(\theta)}$ . The area of a trapezium is given by $A = \frac{a+b}{2}h$ , while the perimeter for the isosceles trapezium is $P = 2d+a+b$ .

Using the area equation, we find that $a + b = \frac{2A}{h}$ . Substituting this into the perimeter equation along with the expression for $d$ yields $\frac{2h}{\sin(\theta)} + \frac{2A}{h} = P$ $\frac{h}{\sin(\theta)} = \frac{P}{2} - \frac{A}{h} = \frac{Ph - 2A}{2h}$ $\sin(\theta) = f(h) = \frac{2h^2}{Ph-2A}$ Differentiate using the quotient rule: $f'(h) = \frac{4h(Ph-2A) - 2Ph^2}{(Ph-2A)^2} = \frac{2Ph^2 -8Ah}{(Ph-2A)^2}$ Set the numerator to zero to find when $f'(h) = 0$ . $2Ph^2 - 8Ah = 2h(Ph - 4A) = 0 \longrightarrow h = 0, \frac{4A}{P}$ Since $h = 0$ results in $A=0$ , this is not a valid solution so $h = \frac{4A}{P}$ must be correct. Thus, the minimum of $\sin(\theta)$ is: $f\left( \frac{4A}{P}\right) = \frac{2\left( \frac{4A}{P}\right)^2}{P\left( \frac{4A}{P}\right) - 2A} = \frac{\frac{32A^2}{P^2}}{2A} = \boxed{\frac{16A}{P^2}}$ Substituting in $P = 80$ and $A = 250$ gives the final answer. $\frac{16\times250}{80^2} = \frac{16\times250}{16\times5\times80} = \frac{50}{80} = \frac{5}{8}$ Therefore, $a+ b = 5 + 8 = \boxed{13}$ .

Any isosceles trapezoid can be associated with a parallelogram with the same perimeter, area, and angles. Just reflect half of the trapezoid across the midsegment that is parallel to the two bases. And the relationship works in reverse. So the answer to this problem is the same as the minimum value of $\sin \theta$ in a parallelogram that meets the same criteria. That was the question addressed in the Inspiration problem , with general solution $\boxed{ \sin \theta \geq \dfrac{16A}{P^2} }$ .

For the values given here, we get $\sin \theta \geq \dfrac{16(250)}{80^2} = \dfrac{5}{8}$ . So $a + b = 5 + 8 = \boxed{13}$ .

Denote by $P$ the perimeter of trapezium $ABCD$ , by $A$ its area and by $\theta$ the acute angle $\angle ADC$ . Let $x=AD=BC$ and $y=AB$ , the smaller of the two parallel sides of $ABCD$ . Let $E$ , $F$ be points on $CD$ such that $AE\bot DC\bot BF$ . Then, we have the relations

$\begin{aligned} P & =AD+BC+AB+DE \\ & =2x+2y+2x\cos \theta \\ & \Rightarrow 2y=P-2x-2x\cos \theta \ \ \ \ \ (1) \\ \end{aligned}$

$\begin{aligned} & A=\dfrac{1}{2}\left( AB+CD \right)AE\Rightarrow 2A=\left( 2y+2x\cos \theta \right)x\sin \theta \\ & \overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,2A=\left( P-2x-\cancel{2x\cos \theta }+\cancel{2x\cos \theta } \right)x\sin \theta \\ & \Rightarrow \left( 2\sin \theta \right){{x}^{2}}-\left( P\sin \theta \right)x+2A=0 \\ \end{aligned}$

The latter is a quadratic equation which must have a solution for $x$ , thus its discriminant must be non-negative, i.e.

$\begin{aligned} & \Delta \ge 0\Leftrightarrow {{P}^{2}}{{\sin }^{2}}\theta -16A\sin \theta \ge 0 \\ & \Leftrightarrow \sin \theta \left( {{P}^{2}}\sin \theta -16A \right)\ge 0 \\ & \overset{0{}^\circ <\theta <90{}^\circ }{\mathop{\Leftrightarrow }}\,{{P}^{2}}\sin \theta -16A\ge 0 \\ & \Leftrightarrow \sin \theta \ge \dfrac{16A}{{{P}^{2}}} \\ \end{aligned}$

Hence, the minimum value for $\sin \theta$ is ${{\left( \sin \theta \right)}_{\min }}=\dfrac{16A}{{{P}^{2}}}$ Substituting the specific values $A=250$ and $P=80$ , we get ${{\left( \sin \theta \right)}_{\min }}=\dfrac{16\times 250}{{{80}^{2}}}=\dfrac{5}{8}$ .

Consequently, we find $AD=BC=x=20$ , $AB=y=20-\dfrac{5\sqrt{39}}{2}\approx 12.2$ and $DC\approx 27.8$ .

For the answer, $a=5$ , $b=8$ , thus, $a+b=\boxed{13}$ .

Area $A = \dfrac{1}{2}(B + b)s\sin(\theta) = 250 \implies \sin(\theta) = \dfrac{500}{(B + b)s}$

and

Perimeter $P = B + b + 2s = 80 \implies B + b = 80 - 2s \implies \sin(\theta) = \dfrac{250}{(40 - s)s} \implies \theta = \arcsin(\dfrac{250}{40s - s^2})$

$\implies \dfrac{d\theta}{ds} = \dfrac{500(s - 20)}{s(40 - s)\sqrt{(s^2 - 40s - 250)(s^2 - 40s + 250)}} = 0$

$\implies s = 20$ and $5(4 - \sqrt{6}) < s < 20 \implies \dfrac{d\theta}{ds} < 0$ and $20 < s < 5(4+ \sqrt{6}) \implies \dfrac{d\theta}{ds} > 0$

$\implies$ a min occurs at $s = 20 \implies \sin(\theta) = \dfrac{250}{400} = \dfrac{5}{8} = \dfrac{a}{b} \implies a + b = \boxed{13}.$