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:
Answer (047):
Label the vertices of the quadrilateral A,B,C, and D so that AB=5,BC=6,CD=9, and DA=7. Let P be the intersection of the diagonals of ABCD, and let x=PA,y=PB,z=PC, and w=PD. Let θ=∠APD=∠BPC. The area condition is
30=[ABCD]=[APB]+[BPC]+[CPD]+[DPA]=21xysinθ+21yzsinθ+21zwsinθ+21wxsinθ=21(x+z)(w+y)sinθ
To each of the four triangles formed by the sides of ABCD and the diagonals AC and BD, apply the Law of Cosines to obtain
52=x2+y2+2xycosθ,92=z2+w2+2wzcosθ,62=y2+z2−2yzcosθ72=w2+x2−2wxcosθ
Thus
21=(52+92)−(62+72)=[x2+y2+w2+z2+2(xy+zw)cosθ]−[x2+y2+w2+z2−2(yz+wx)cosθ]=2(xy+zw+yz+wx)cosθ=2(x+z)(y+w)cosθ
Dividing the first equation by the second yields
tanθ=2212⋅30=740
The requested sum is 40+7=47. Had tanθ been negative, the desired angle would have been the supplement of ∠APD.
Note: This result also follows from Bretschneider's Formula for the area of a quadrilateral given its sides and the sum of two opposite angles. Thus one could solve for the sum of opposite angles and deduce the needed angle between the diagonals.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions