Work those Quadriceps

This solution discusses the method of how to find every cyclic quadrilateral that fits under the given parameters of this question. However, this does make it lengthy and you may wanna skip some of the computations if you like.

Let $\angle CDA = \theta.$ thus by cyclic properties, $\angle ABC = 180- \theta.$ Now, using circle theorems, since $AD$ is a diameter $\angle ACD = 90$ . Finally, let segment $\overline{AC}=x$ .

By Pythagorean theorem, $x^2=d^2-b^2$

By law of cosines, $x^2=2a^2-2a^2\cos(180-\theta)$ . Setting the two equal to each other (also note that $\cos(180-\theta)=-\cos(\theta)$ )

$d^2-b^2=2a^2+2a^2\cos(\theta)$

Also, remember that $b^2=d^2\cos(\theta)^2$

$d^2-d^2\cos^2(\theta)-2a^2(1+\cos(\theta))=0$

$d^2(1-\cos(\theta))(1+\cos(\theta))-2a^2(1+\cos(\theta))=0$

$(d^2-d^2\cos(\theta)-2a^2)(1+\cos(\theta))=0$

Once again, $\cos(\theta)=\frac{b}{d}$

$(d^2-db-2a^2)(1+\cos(\theta))=0~~(i)$

One of the two equates to 0, obviously $1+\cos(\theta)\neq 0$ . So using the quadratic we have

$d=\dfrac{b\pm\sqrt{b^2+8a^2}}{2}$

Note that since $d$ can't be negative, the $\pm$ is actually just a $+$ .

Using this formula on integral square roots (a little lemma I made whose properties I have still yet to fully exploit), we obtain the following equation for integral solutions for $d$ and $a$ . Note that the lead coefficient of $b$ here is 1 which is a perfect square.

$b=\dfrac{8a^2-n^2}{2n}$

Where n is an arbitrary variable we will be playing with here. Plugging in this value for $b$ yields

$d=\dfrac{\dfrac{8a^2-n^2}{2n}+\sqrt{\dfrac{64a^2-16a^2n^2+n^4}{4n^2}+8a^2}}{2}$

$d=\dfrac{\dfrac{8a^2-n^2}{2n}+\sqrt{\dfrac{(8a^2+n^2)^2}{(2n)^2}}}{2}$

$\boxed{d=\dfrac{4a^2}{n}}~(ii)$

Going back to equation (i) with our new value of d

$d^2-db-2a^2\longrightarrow \dfrac{16a^4}{n^2}-\dfrac{4a^2b}{n}-2a^2=0$

$\longrightarrow (8a^2-2bn-n^2)2a^2=0$

Since $a\neq0$

$\boxed{a=\sqrt{\dfrac{2bn+n^2}{8}}}$

Finally, substituting our value of $a$ from equation (ii) $a=\dfrac{\sqrt{dn}}{2}$

$\dfrac{\sqrt{dn}}{2}=\sqrt{\dfrac{2bn+n^2}{8}}$

$\boxed{b=\dfrac{2d-n}{2}}$

From this final equation, it can be easily observed that $n$ MUST be even for $b$ to be even

We now have all the equations we need to make our cyclic quad under the parameters of the question. They are:

$d=\dfrac{4a^2}{n}$

$b=\dfrac{2d-n}{2}$

When I finally got this result, I had to tell myself, "Sometimes, it's the small things that are beautiful. Diamonds aren't simply found in a minute" #minecraft (no I don't play this game)

From here, it's a simple plug and chug of $a$ along with even values of $n$ .

On the chart displayed below. I have a few of the values of $a,n,d,b$

The 4 smallest values are $19, 22, 38, 41$

Which have values of

Thus or answer is $19+22+38+41=\boxed{120}$

Let $O$ be the centre of the circle, $M$ be the midpoint of $AB$ , and $d = AD$ . Draw the line segments $OM$ , $OB$ and $OC$ . Let $\theta = \angle AOM$ . Then $a = AB = 2MA = 2OB \sin(\theta) = d \sin(\theta)$ .

By congruent triangles, we can see that $\angle AOB = 2\theta$ and $\angle BOC = 2\theta$ , so $\angle AOC = 4\theta$ . Now Euclid's Elements III.20 says the angle at the circumference is half the angle at the centre (in other words $\angle ADC = \dfrac{1}{2}\angle AOC$ ). Hence $\angle ADC = 2\theta$ .

Draw line segment $AC$ . By Thales' theorem, $\angle ACD = 90^{\circ}$ . Therefore, $b = CD = AD \cos(2\theta) = d \left(1 - 2\sin^2(\theta)\right) = d\left(1 - 2\left(\dfrac{a}{d}\right)^2\right) = d - \dfrac{2a^2}{d}$

Hence $2a^2$ is divisible by $d$ . Let's look at a few cases:

$a = 1 \rightarrow d | 2$ . Now $d$ can't be $1$ since $d > a$ . It also can't be $2$ because then we would have $b = a$ . Hence there are no solutions in this case.

$a = 2 \rightarrow d | 8$ . Now $d = 1,2$ is too small, and $d = 4$ leads to $a = b$ . However $d = 8$ works, giving a solution of $(a,b,d) = (2,7,8)$ .

$a = 3 \rightarrow d | 18$ . Here $d = 1,2,3$ are too small and $d = 6$ leads to $a = b$ . However $d = 9, 18$ work, giving solutions of $(3,7,9)$ and $(3,17,18)$ .

$a = 4 \rightarrow d | 32$ . Here $d = 1,2,4$ are too small and $d = 8$ leads to $a = b$ . However, $d = 16,32$ give the solutions $(4,14,16)$ and $(4, 31, 32)$ .

So far the 4 smallest perimeters are $19, 22, 38, 41$ . All that's left to do is prove that these are the smallest possible perimeters. However, this is a little tedious and 2 other people have already done it, so I'll skip this part.

Hence the answer is $19+22+38+41 = \boxed{120}$ .

$I ~have ~given~ a ~full~ list ~so~ that~ changes~ in ~variables~ can~ be~ seen.\\ For ~the ~solution~ of ~this ~problem ~there ~is~ a ~short~ cut.\\ ~~~~\\ (1)~ As~ 'a' increases,~ perimeter~ S ~decreases.\\ (2)~ When~(d-c)=a,~and~ d=2a,~ then~ S=5a.\\ (3) ~If~ there~ is~ no~ factor~ m*n,~ a<m,n<2a,\\ ~~~~ S=5a~ is~ minimum~ S_m~ for~ that~ value~ of~ a.\\ ~~~~(3a)~If~S_m >~ our~ desired~value~41, ~we~end~our~investigation.\\ ~~~~(3bi)~If~S_m \leq~our~ desired~value~41, \\ ~~~~~~~~~~~~we~go~for~value~of~(d-c),~before~it~was~a.\\ ~~~~(3bii)~if~THIS~S>41~we~end~our~investigation.\\ ~~~~(3biii)~If~S_m \leq 41,~go~for~value~of~(d-c),~before.\\ (4) ~If~ there~ is~a~ factor~ m*n,~ a<m,n<2a,\\ ~~~~~~~~~this ~S~is~same~as~one~of~the~S~we~saw,\\ ~~~~~~~~~but~ with~different~a, ~c,~ and~ d.~ Neglect.\\ ~~~~~\\$