Problem:
Isosceles trapezoid ABCD has parallel sides AD and BC, with BC<AD and AB=CD. There is a point P in the plane such that PA=1,PB=2,PC=3, and PD=4. What is ADBC ?
Answer Choices:
A. 41
B. 31
C. 21
D. 32
E. 43
Solution:
Let the trapezoid have vertices A(−a,0),B(−b,c),C(b,c), and D(a,0), where a>0,b>0, and c>0. Let P have coordinates (p,q) for real numbers p and q. The given conditions imply
(p+a)2+q2(p+b)2+(c−q)2(p−b)2+(c−q)2(p−a)2+q2=1=4=9, and =16
Subtracting the fourth equation from the first gives pa=4−15, and subtracting the third equation from the second gives pb=−45. Hence ADBC=ab=(B)31.
OR
Let X and Y be feet of the perpendiculars from P to (parallel) lines AD and BC respectively. (In the diagram below, point Y lies to the left of segment BC and point X lies on segment AD. The solution generalizes to other configurations without issue through the use of directed lengths.)
By the Pythagorean Theorem AP2−AX2=XP2=DP2−DX2. It follows that DX2−AX2=DP2−AP2=42−12=15, so
AD⋅(DX−AX)=(DX+AX)(DX−AX)=DX2−AX2=DP2−AP2=15
Likewise, BP2−BY2=YP2=CP2−CY2, so CY2−BY2=CP2−BP2=32−22=5 and
BC⋅(CY+BY)=(CY−BY)(CY+BY)=CY2−BY2=5
By symmetry, AX+BY=DX−CY, so CY+BY=DX−AX. Therefore
ADBC=AD⋅(DX−AX)BC⋅(CY+BY)=31
Let ℓ denote the common perpendicular bisector of AD and BC, and let Q denote the reflection of P across ℓ. By symmetry, AP=DQ=1,AQ=DP=4,BP=CQ=2, and BQ=CP=3. Furthermore, APQD is an isosceles trapezoid and is therefore a cyclic quadrilateral. By Ptolemy's Theorem AD⋅PQ+AP⋅DQ=AQ⋅PD, so AD⋅PQ=42−12=15. Likewise, BC⋅PQ=32−22=5. Therefore
ADBC=AD⋅PQBC⋅PQ=(B)31
Note: Such trapezoids do in fact exist. For example, if a=2 and b=32 in the first solution, then (p,q)=(−815,837) and c=241(97+1463). A degenerate example has vertices at A(0,0), B(1,0),C(2,0), and D(3,0), with P at (−1,0).
