Problem:
A convex quadrilateral has area 30 and side lengths 5,6,9, and 7, in that order. Denote by θ the measure of the acute angle formed by the diagonals of the quadrilateral. Then tanθ can be written in the form nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
The quadrilateral has area 30 and the side lengths are given. Suppose that AC and BD intersect at P. Then let's label lengths, AX=a, XB=b, XC=c, XD=d.
The first way to start is to use the given side lengths, 'bashing' law of cosines, yielding these equations:
a2+b2−2abcos(180−θ)=52
b2+c2−2bccos(θ)=62
c2+d2−2cdcos(180−θ)=92
d2+a2−2adcos(θ)=72
Notice that cos(θ)=−cos(180−θ).
So we can turn some of the negatives into positives and make them into cos(θ).
Now, the other piece of information that we're given is the area, so we can use [ABC]=21absinC repeatedly:
2absin(180−θ)+2cdsin(180−θ)+2adsin(θ)+2bcsin(θ)=30
and rewriting noticing that sin(θ)=sin(180−θ).
2ab+bc+cd+dasin(θ)=30
This yields
sin(θ)=ab+bc+cd+da60
Notice that this is useful as we can rearrange the Law of Cosines equation to rewrite cos(θ) as
cos(θ)(2ab+2bc+2cd+2da)=21→cos(θ)=ab+bc+cd+da221
Now, since we have sin(θ) and cos(θ), with the same common denominator, our answer is merely
ab+bc+cd+da221ab+bc+cd+da60=21120=740→40+7=47
The problems on this page are the property of the MAA's American Mathematics Competitions