Problem:
Let ABCDE be a convex pentagon with AB∥CE,BC∥AD,AC∥DE, ∠ABC=120∘,AB=3,BC=5, and DE=15. Given that the ratio between the area of △ABC and the area of △EBD is m/n, where m and n are relatively prime positive integers, find m+n.
Solution:
Let AD intersect CE at F. Extend BA through A to R so that BR≅CE, and extend BC through C to P so that BP≅AD. Then create parallelogram PBRQ by drawing lines through D and E parallel to AB and BC, respectively, with Q the intersection of the two lines. Apply the Law of Cosines to triangle ABC to obtain AC=7. Now
ABRA=FCEF=ACDE=715, so ABRB=722
Similarly, CBPB=722. Thus parallelogram ABCF is similar to parallelogram RBPQ. Let K=[ABCF]. Then [RBPQ]=(22/7)2K. Also,
[EBD]=[RBPQ]−21([BCER]+[ABPD]+[DFEQ])=[RBPQ]−21([RBPQ]+[ABCF])=21([RBPQ]−[ABCF])
Thus
[EBD][ABC]=21[(722)2K−K]21K=49484−11=43549
and m+n=484.
OR
Apply the Law of Cosines to triangle ABC to obtain AC=7. Let AD intersect CE at F. Then ABCF is a parallelogram, which implies that [ABC]= [BCF]=[CFA]=[FAB]. Let ED/AC=r=15/7. Since AC∥DE, conclude EF/FC=FD/AF=r. Hence
[ABC][EBD]=[ABC][BFE]+[ABC][EFD]+[ABC][DFB]=[BCF][BFE]+[CFA][EFD]+[FAB][DFB]=r+r2+r=r2+2r=435/49
implying that m/n=49/435 and m+n=484.
OR
Apply the Law of Cosines to triangle ABC to obtain AC=7. Let AD and CE intersect at F. Then ABCF is a parallelogram, which implies that △ABC≅ △CFA. Note that triangles AFC and DFE are similar. Let the altitudes from B and F to AC each have length h. Then the length of the altitude from F to ED is 15h/7. Thus
[EBD][ABC]=21⋅15(h+h+715h)21⋅7h=15⋅297⋅7=43549
so m+n=484.
The problems on this page are the property of the MAA's American Mathematics Competitions