Problem:
Let ABCD be a cyclic quadrilateral. The side lengths of ABCD are distinct integers less than 15 such that BC⋅CD=AB⋅DA. What is the largest possible value of BD ?
Answer Choices:
A. 2325
B. 185
C. 2389
D. 2425
E. 2533 Solution:
Let R be the circumradius of ABCD and let a=AB,b=BC, c=CD,d=DA, and k=bc=ad. Because the areas of △ABC,△CDA, △BCD, and △ABD are
4Rab⋅AC,4Rcd⋅AC,4Rbc⋅BD, and 4Rad⋅BD
respectively, and Area(△ABC)+Area(△CDA)=Area(△BCD)+Area(△ABD) : it follows that
4RAC(ab+cd)=4RBD(bc+ad)=4RBD(2k)
that is, (ab+cd)⋅AC=2k⋅BD. By Ptolemy's Theorem ac+bd=AC⋅BD. Solving for AC and substituting into the previous equation gives
None of the sides can be equal to 11 or 13 because by assumption a,b,c, and d are pairwise distinct and less than 15 , and so it is impossible to have a factor of 11 or 13 on each side of the equation bc=ad. If the largest side length is 12 or less, then 2BD2≤122+102+92+82=389, and so BD≤2389. If the largest side is 14 and the other sides are s1>s2>s3, then 14s3=s1s2. Thus 7 divides s1s2 and because 0<s2<s1<14, it follows that either s1=7 or s2=7. If s1=7, then 2BD2<142+72+62+52=306. If s2=7, then 2s3=s1, and it follows that 2BD2≤142+72+122+62=425. Therefore BD≤2425 with equality for a cyclic quadrilateral with a=14,b=12,c=7, and d=6.