More Wrecked Angles

$Let~QE \perp AD ~with~ E~ on~ AD.~\\\text{Let BQ=SD=AE=X.}\\\text {BQ increases as diagonal QS decreases. But diagonal PR } \nless 50.\\ \text {But diagonals of a rectangle are equal (and bisect one another).}\\\implies ~minimum~ QS \nless 50. ~\text{So from the sketch, } ~QS^2=QE^2 + ES^2\\\implies~50^2=36^2 + (50 - 2X)^2.~~~\therefore~X=\dfrac 1 2 \{50-\sqrt{50^2-36^2} \}.\\=a-\sqrt b. ~\therefore~a+b= \Large \color{#D61F06}{326}$

The centroids of the rectangles are concurrent (Why?) - call this point O. Since PQRS is a rectangle, its diagonals are congruent, bisect eachother, and intersect at O. The length of half a diagonal OQ decreases when the length of BQ increases. Since OQ=OP and P is on AB, the shortest OP can be is 25, We investigate this case: consider the right triangle with two vertices Q and O and with one leg including Q parallel to AB. The leg parallel to BC would have length \sqrt{301}, so the length of BQ is 25 - \sqrt{301}.

Helpful way to think of this is you have a circle centered at the center of the rectangle and PQRS are its intersections with the rectangle. With a little thought BQ=25-SQRT(r^2-18^2) where r is the radius of this circle. The minimum value of r is half the length of the rectangle from A to D if it is to intersect all 4 sides. So the max value of this happens when r=25 and from that we get 25-sqrt(301)

Write $P = (0,y), Q = (x, 36), R = (50, 36-y), S = (50-x, 0)$ . For $PQRS$ to be a rectangle, we must have slope $(PQ) \cdot$ slope $(QR) = -1$ . Thus, $-1 = \frac{36-y}{x} \cdot \frac{y}{x-50} = \frac{36y-y^2}{x^2 - 50x},$ and so, $y^2 - 36y = x^2-50x$ $(y - 18)^2 = (x-25)^2 -301$ $y = 18 \pm \sqrt{(x-25)^2 - 301}.$ So, if $(x,y)\in \mathbb{R}^2$ , we must have $(x-25)^2 - 301 \geq 0$ , or $\lvert x-25\rvert \geq \sqrt{301}.$

Finally, $BQ < QC \implies x < 25$ , so $\lvert x-25\rvert = 25 - x$ , and $BQ = x \leq 25 - \sqrt{301}.$

x=25 - SQRT(301), A + B= 25 +301=326

Sorry, spent a lot of time, trying to load an image on Imgur​ and get the link to embed it here. But it just did not work. Is there some other way of doing it?

Let (f : 1-f) be the ratio in which P divides AB. This will be repeated by point R on the opposite side CD as (1-f : f). By symmetry, the midpoint of PR always remains at the center (O) of ABCD.

Now, if we use these pairs of points P & R as the diameter for various values of f, we will get a family of concentric circles that intersect BC at points Q. And smaller the circle gets further Q will be from B.

The smallest circle is had (or the closest P & R get) is when P & R are midpoints of AB and CD respectively. This gives radius = OQ = 25 and $BQ = 25 - \sqrt{25^2 - 18^2}$

Hence a = 25, b = 301 and $\textbf a + b = 326$