Complex Cyclic with Surprising Solution

Name the lengths of $BD$ and $AE$ , ${ p }_{ 1 }$ and ${ p }_{ 2 }$ respectively. Name the length $AB$ , $a$ and name $AD$ and $BE$ , $b$ and $c$ respectively.

Since $AB$ is the diameter, according to Thales' theorem, the angles $ADB$ and $AEB$ are both right angles. Therefore, $ADB$ and $AEB$ are both right angled triangles . Therefore you can use Pythagoras' theorem on $ADB$ :

${ AD }^{ 2 }+{ B }E^{ 2 }{ =AB }^{ 2 }\\ { b }^{ 2 }+{ p }_{ 1 }^{ 2 }={ a }^{ 2 }\\ { p }_{ 1 }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }\\ { p }_{ 1 }^{ 2 }=(a+b)(a-b)$

The only factors of ${ p }_{ 1 }$ are ${ p }_{ 1 }$ and $1$ , because ${ p }_{ 1 }$ is a prime number. Therefore, the only factors ${ p }_{ 1 }^{ 2 }$ are $1$ , ${ p }_{ 1 }$ and ${ p }_{ 1 }^{ 2 }$ . Since $a$ and $b$ are positive integers, $a+b\ge a-b$ . Therefore, either $(a-b)={ p }_{ 1 }$ and $(a+b)={ p }_{ 1 }$ or $(a-b)=1$ and $(a+b)={ p }_{ 1 }^{2}$ .

If $(a-b)={ p }_{ 1 }$ and $(a+b)={ p }_{ 1 }$ :

${ p }_{ 1 }=a+b\\ { p }_{ 1 }=a-b\\ a+b=a-b\\ 2b=0\\ b=0$

Therefore, ${ p }_{ 1 }\neq a+b$ and ${ p }_{ 1 }\neq a-b$ . This means that $(a-b)=1$ and $(a+b)={ p }_{ 1 }^{2}$ . Since that is true, $a-b=1$ can be rearranged to get $a=b+1$ .

We can use a similar argument on the other triangle $AEB$ to get $a=c+1$ . Since $a=c+1$ and $a=b+1$ , $b+1=c+1$ so $b=c$ . Therefore, $AD=BE$ . In the right angled triangles, $ADB$ and $AEB$ , they share a hypotenuse $AB$ and they two sides of equal length ( $AD=BE$ ). Therefore, according to $RHS$ (Right angle, hypotenuse and side), $ADB$ and $AEB$ are congruent. Therefore, $AE=BD$ .

If the two pairs of opposite sides ( $AE, BD$ ) and ( $AD, BE$ ) are equal, $ADBE$ is a parallelogram. However, since are also two right angles in the quadrilateral, the quadrilateral $ADBE$ must be a rectangle. Since $ADBE$ is a rectangle the diagonals $AB$ and $DE$ are of equal length.

Now if we go back to the original equation we were given to solve we can replace some of the values now we know $AE=BD$ , $AD=BE$ , $AB=DE$ :

${ (AD-5) }^{ 2 }+{ (BD-2) }^{ 2 }+{ (AE+2) }^{ 2 }+{ (BE-1) }^{ 2 }\\-2(AB-3)(DE-3)-18$

${ (AD-5) }^{ 2 }+{ (BD-2) }^{ 2 }+{ (BD+2) }^{ 2 }+{ (AD-1) }^{ 2 }\\-2(AB-3)(AB-3)-18$

${ (AD-5) }^{ 2 }+{ (BD-2) }^{ 2 }+{ (BD+2) }^{ 2 }+{ (AD-1) }^{ 2 }\\-{ 2(AB-3) }^{ 2 }-18$

${ A }D^{ 2 }-10AD+25+{ BD }^{ 2 }-4BD+4+{ BD }^{ 2 }+4BD+4\\+{ A }D^{ 2 }-2AD+1-2{ AB }^{ 2 }+12AB-18$

$2{ A }D^{ 2 }+2{ BD }^{ 2 }-2{ AB }^{ 2 }+12AB-12AD-2$

$2({ AD }^{ 2 }+{ BD }^{ 2 })-2{ AB }^{ 2 }+12(AB-AD)-2$

Since $ADB$ is a right angled triangle, we can use Pythagoras' theorem to show ${ AD }^{ 2 }+{ BD }^{ 2 }={ AB }^{ 2 }$ . Earlier we showed $a-b=1$ , therefore, $AB-AD=1$ .

Now if we go back to the equation, we can substitute the values of $AB-AD=1$ and ${ AD }^{ 2 }+{ BD }^{ 2 }={ AB }^{ 2 }$ :

$2({ AD }^{ 2 }+{ BD }^{ 2 })-2{ AB }^{ 2 }+12(AB-AD)-2\\ 2({ AB }^{ 2 })-2{ AB }^{ 2 }+12(1)-2\\ 0+12-2\\ \boxed { 10 }$