Problem:
A trapezoid has side lengths 3,5,7, and 11 . The sum of all the possible areas of the trapezoid can be written in the form of r1n1+r2n2+r3, where r1,r2, and r3 are rational numbers and n1 and n2 are positive integers not divisible by the square of a prime. What is the greatest integer less than or equal to
r1+r2+r3+n1+n2?
Answer Choices:
A. 57
B. 59
C. 61
D. 63
E. 65 Solution:
Let ABCD be a trapezoid with AB∥CD and AB<CD. Let E be the point on CD such that CE=AB. Then ABCE is a parallelogram. Set AB=a,BC=b,CD=c, and DA=d. Then the side lengths of △ADE are b,d, and c−a. If one of b or d is equal to 11 , say b=11 by symmetry, then d+(c−a)≤7+(5−3)<11=d, which contradicts the triangle inequality. Thus c=11. There are three cases to consider, namely, a=3,a=5, and a=7. If a=3, then △ADE has side lengths 5,7 , and 8 and by Heron's formula its area is
41(5+7+8)(7+8−5)(8+5−7)(5+7−8)=103
The area of △AEC is 83 of the area of △ADE, and triangles ABC and AEC have the same area. It follows that the area of the trapezoid is 21(353).
If a=5, then △ADE has side lengths 3,6 , and 7 , and area
41(3+6+7)(6+7−3)(7+3−6)(3+6−7)=45.
The area of △AEC is 65 of the area of △ADE, and triangles ABC and AEC have the same area. It follows that the area of the trapezoid is 31(325).
If a=7, then △ADE has side lengths 3,4 , and 5 . Hence this is a right trapezoid with height 3 and base lengths 7 and 11 . This trapezoid has area 21(3(7+11))=27.
The sum of the three possible areas is 2353+3325+27. Hence r1=235,r2=332, r3=27,n1=3,n2=5, and r1+r2+r3+n1+n2=235+332+27+3+5=63+61. Thus the required integer is 63 .