2010 AMC 12B Problem 22

Let $a=AB,$ $b=BC,$ $c=CD,$ $d=DA$ with $bc=ad=k;$ comparing areas gives $(ab+cd)\cdot AC=2k\cdot BD$

With Ptolemy’s $AC\cdot BD=ac+bd,$ eliminate $AC$ to get $BD^2=\tfrac12(a^2+b^2+c^2+d^2)$

Let $a=AB,$ $b=BC,$ $c=CD,$ $d=DA$ and $k=bc=ad.$ Writing each triangle’s area in terms of the circumradius and using $[ABC]+[CDA]=$ $[BCD]+[ABD]$ gives $(ab+cd)\cdot AC=2k\cdot BD.$

Ptolemy’s theorem gives $AC\cdot BD=ac+bd.$ Eliminating $AC,$ $$\begin{aligned} BD^2 &=\frac{(ac+bd)(ab+cd)}{2k} \\ &=\frac12\left(a^2+b^2+c^2+d^2\right). \end{aligned}$$

The sides are distinct integers below $15$ with $bc=ad,$ so neither $11$ nor $13$ can appear (each is prime and would need a matching factor on the other side).

If the largest side is at most $12,$ the four squares sum to at most $12^2+10^2+9^2+8^2=389,$ because $11$ is unavailable.

Now suppose the largest side is $14,$ and write the others as $s_1\gt s_2\gt s_3.$ The product condition must pair $14$ with $s_3,$ so $14s_3=s_1s_2.$ Hence one of $s_1,s_2$ equals $7.$ If $s_1=7,$ the sum of squares is less than $14^2+7^2+6^2+5^2=306.$ If $s_2=7,$ then $s_1=2s_3,$ so the largest possibility is $(s_1,s_2,s_3)=(12,7,6).$ Thus $$\begin{aligned} 2BD^2 &=14^2+12^2+7^2+6^2 \\ &=425, \end{aligned}$$ so $BD\le\sqrt{\dfrac{425}{2}}.$ Equality is attained by the cyclic quadrilateral with side order $(a,b,c,d)=(14,12,7,6),$ for which $bc=ad=84.$

Thus, the correct answer is D.