Problem:
Let ABCD be an isoceles trapezoid having parallel bases AB and CD with AB>CD. Line segments from a point inside ABCD to the vertices divide the trapezoid into four triangles whose areas are 2,3,4, and 5 starting with the triangle with base CD and moving clockwise as shown in the diagram below. What is the ratio CDABβ ?
Answer Choices:
A. 3
B. 2+2β
C. 1+6β
D. 23β
E. 32β Solution:
Answer (B): Let X and Y be the feet of the perpendiculars from interior point P to AB and CD, respectively. Let r=AB and s=CD.
The area of β³APB=21ββ rβ PX=4. Thus PX=r8β. Similarly, PY=s4β. Now using the formula for the area of a trapezoid yields
Thus the ratio Ο=srβ satisfies Ο+2Οβ1=4; solving yields Ο=2Β±2β. Because r>s, the ratio srβ is 2+2β. Indeed, such a trapezoid exists. Let s=1 and r=2+2β. The height from base AB to point P is 8β42β, and the height from base CD to point P is 4 . Thus the height of the trapezoid is 12β42β and the area of the trapezoid is
Also, β³ABP has area
21ββ (3+2β)β (12β42β)=14
21ββ (2+2β)(8β42β)=4
and β³CDP has area
21ββ 1β 4=2
Point P is 85β of the way from left to right along the line segment extending across the interior of the trapezoid through P parallel to the bases. Because the combined area of β³ADP and β³CPB is 14β6=8, this placement of P will ensure that the areas of β³ADP and β³CPB are 5 and 3 , respectively, as needed.
