Problem:
Rectangle ABCD has side lengths AB=84 and AD=42. Point M is the midpoint of AD, point N is the trisection point of AB closer to A, and point O is the intersection of CM and DN. Point P lies on the quadrilateral BCON, and BP bisects the area of BCON. Find the area of △CDP.
Solution:
Let Q be the projection of O onto CD. Because △AND∼△QDO,DQOQ=2842=23, and because △QOC∼△DMC,84−DQOQ=8421=41. Solving this system of two equations yields DQ=12 and OQ=18. Then the area of △BON is 2BN(42−OQ)=672, and the area of △BCO is 2BC(84−DQ)=1512. Because △BON has a smaller area than △BCO, point P must lie on CO, and △BPC must have area 2672+1512=1092. Thus the distance from P to BC is BC2.1092=52, and the distance from P to CD is 52⋅CDDM=13. The requested area of △CDP is 213⋅CD=546.
OR
Let R be the intersection of lines MC and AB, and let Q be the intersection of lines DN and BC. Let X,Y,Z, and W be the projections of point O onto line segments AB,BC,CD, and AD, respectively. Because △AND∼△BNQ,BQ=84. Because △RAM∼△CDM,RA=84. Because △QCO∼△DMO, it follows from OWOY=MDCQ=6 that OY=72 and OW=12. Because △RNO∼△CDO, it follows from OZOX=DCRN=34 that OX=24 and OZ=18. Then [BCON]=[BON]+[BCO]=2OX⋅BN+2OY⋅BC=672+1512=2184. Because [BCO]> 21[BCON], it must be that P lies on OC. Then the result follows as above.
OR
Position the rectangle in the coordinate plane with vertices A(0,0),B(84,0),C(84,42), and D(0,42), midpoint M(0,21), and trisection point N(28,0). Then lines CM and DN have equations y=41x+21 and y=−23x+42, which intersect at O(12,24). The Shoelace Formula gives [BCON]=2184. Clearly P must lie on OC, and [BCP]=1092, from which it follows that the coordinates of P are (32,29). The Shoelace Formula then gives the area of △CDP as 546.
The problems on this page are the property of the MAA's American Mathematics Competitions