Problem:
In convex quadrilateral ABCD,β Aβ
β C,AB=CD=180, and ADξ =BC. The perimeter of ABCD is 640 . Find β1000cosAβ. (The notation βxβ means the greatest integer that is less than or equal to x.)
Solution:
Let β A=β C=Ξ±,AD=x, and BC=y. Apply the Law of Cosines in triangles ABD and CDB to obtain
BD2=x2+1802β2β
180xcosΞ±=y2+1802β2β
180ycosΞ±
Because xξ =y, this yields
cosΞ±=2β
180(xβy)x2βy2β=360x+yβ=360280β=97β
Thus β1000cosAβ=777.
OR
Assume without loss of generality that AD is the greater of AD and BC. Then there is a point P on AD with AP=BC. Because β³BAPβ
β³DCB, conclude that BP=BD, and altitude BH of isosceles β³BPD bisects PD. Now cosA= AH/180, and because AH=AP+(PD/2)=ADβ(PD/2),
AH=2AP+ADβ=2BC+ADβ=2640β2β
180β=140
Thus cosA=140/180=7/9, and β1000cosAβ=777β.
The problems on this page are the property of the MAA's American Mathematics Competitions
