Question

Let $ABCD$ be a cyclic quadrilateral. The side lengths of $ABCD$ are distinct integers less than $15$ such that $BC\cdot CD=AB\cdot DA.$ What is the largest possible value of $BD?$

$\sqrt{\dfrac{325}{2}}$

$\sqrt{185}$

$\sqrt{\dfrac{389}{2}}$

$\sqrt{\dfrac{425}{2}}$

$\sqrt{\dfrac{533}{2}}$

Solution:

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 $AC,$ $BD^2=\frac{(ac+bd)(ab+cd)}{2k}=\frac12\left(a^2+b^2+c^2+d^2\right).$

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).

To maximize, take the largest side $14.$ Writing the others as $s_1\gt s_2\gt s_3$ with $14s_3=s_1s_2,$ the best case is $s_2=7,$ giving $s_1=2s_3$ and $(a,b,c,d)=(14,12,7,6).$ Then $2BD^2=14^2+12^2+7^2+6^2=425,$ so $BD=\sqrt{\dfrac{425}{2}}.$

Thus, the correct answer is D.

2010 AMC 12B Problem 22

Solution:

Problem 22 in Other Years