None of My Business?

Lemme tell u a secret... $ABCE$ is actually a parallelogram . Sounds peculiar? Lemme show you...

Set the above figure onto a complex plane with $AB$ as the real axis and $A$ be the origin, the points $A$ , $B$ , $C$ , $E$ , $M$ , $N$ , $O$ and $P$ are represented by complex numbers $a$ , $b$ , $c$ , $e$ , $m$ , $n$ , $o$ and $p$ respectively (the $e$ here is not the constant $2.71828\ldots$ , don't be confused! I won't be using that constant here).

Let's set $o=1$ , then $b=2$ . Suppose to get to $P$ from $B$ we go $x$ units along the real axis and 1 unit up the imaginary axis, then $p=2+x+i$ , thus $c=2+2x+2i$ . $\therefore \overrightarrow{BC}=c-b=2x+2i \\ \therefore \overrightarrow{CO}=o-c=-(1+2x)-2i$

Since $AE\parallel BC$ and $|AB|=2$ , thus $e=k_0x+k_0i$ for some value $k_0\in \mathbb R$ .

Let's first find the value of $n$ . Since $n=c+\overrightarrow{CN}$ , suppose $n=k_1p$ and $\overrightarrow{CN}=k_2\overrightarrow{CO}=-k_2(1+2x)-2k_2i$ where $k_1,k_2\in\mathbb R$ .

We have $k_1p=c-k_2(1+2x)-2k_2i \\ k_1(2+x)+k_1i=2+2x-k_2(1+2x)+2(1-k_2)i$

Equating the coefficients of the imaginary part, we have $k_1=2(1-k_2)$ .

Equating the coefficients of the real part and substituting $k_1=2(1-k_2)$ , we have $k_1(2+x)=2(1-k_2)(2+x)=2+2x-k_2(1+2x) \\ 4+2x-4k_2-2k_2x=2+2x-k_2-2k_2x \\ 4-4k_2=2-k_2 \\ \therefore k_2=\frac23$ $\therefore k_1=2(1-\frac23)=\frac23$ $\therefore n=\frac23p$

Denote $S_\text{a shape}$ as the area of a shape, since $\frac{|AM|}{|AN|}=\frac{S_{\triangle AMO}}{S_{\triangle AMO}+S_{\triangle NOM}}=\frac35$ , thus $m=\frac35n=\frac25p=\frac45+\frac25x+\frac25i$

Now, because $o=e+\overrightarrow {EO}$ , suppose $\overrightarrow{EO}=k_3\overrightarrow{MO}$ where $k_3\in\mathbb R$ , then $\begin{aligned} o=1&=e+k_3(o-m) \\&=k_0x+k_0i+k_3(1-\frac45-\frac25x-\frac25i) \\&=k_0x+k_0i+\frac15k_3-\frac25k_3x-\frac25k_3i \\&=(k_0-\frac25k_3)x+\frac15k_3+(k_0-\frac25k_3)i \end{aligned}$

By equating the coefficients again we get $k_0=\frac25k_3$ and $(k_0-\frac25k_3)x+\frac15k_3=1$ which we can get $k_3=5$ which in turn implies $k_0=2$ .

Thus, $e=\overrightarrow{AE}=k_0x+k_0i=2x+2i=\overrightarrow{BC}$ , which implies that $EC\parallel AB$ , $ABCE$ is a parallelogram.

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Now we have proven that $ABCE$ is a parallelogram, the only circumscribed parallelograms are rectangles, so $ABCE$ must be a rectangle. Let's calculate the area of this rectangle.

$\because |AB|=2|AO| \\ \therefore S_{\triangle ABN}=2S_{\triangle AON}=2\times(10+15)=50$ $\because |n|=\frac23|p| \\ \therefore S_{\triangle ABP}=\frac32 S_{\triangle AON}=\frac32\times50=75$ $\therefore S_{ABCE}=4S_{\triangle ABP}=4\times75=300=AB\times BC$

300 can be factorised as $\begin{aligned} 1\times300 \\ 2\times150 \\ 3\times100\\4\times75\\5\times60\\6\times50\\10\times30\\12\times25\\15\times20 \end{aligned}$

Since the diameter of the circle is an integer, and the diameter of the circle is $\sqrt{AB^2+BC^2}$ , we see that only 15 and 20 satisfies the problem.

Hence, the diameter of the circle is $\sqrt{15^2+20^2}=25$ .