Problem:
Let AB be a diameter of a circle and C be a point on AB with 2⋅AC=BC. Let D and E be points on the circle such that DC⊥AB and DE is a second diameter. What is the ratio of the area of △DCE to the area of △ABD?
Answer Choices:
A. 61​
B. 41​
C. 31​
D. 21​
E. 32​
Solution:
Let O be the center of the circle. Each of â–³DCE and â–³ABD has a diameter of the circle as a side. Thus the ratio of their areas is the ratio of the two altitudes to the diameters. These altitudes are DC and the altitude from C to DO in â–³DCE. Let F be the foot of this second altitude. Since â–³CFO is similar to â–³DCO,
DCCF​=DOCO​=DOAO−AC​=21​AB21​AB−31​AB​=31​
which is the desired ratio.
OR
Because AC=AB/3 and AO=AB/2, we have CO=AB/6. Triangles DCO and DAB have a common altitude to AB so the area of △DCO is 61​ the area of △ADB. Triangles DCO and ECO have equal areas since they have a common base CO and their altitudes are equal. Thus the ratio of the area of △DCE to the area of △ABD is 31​.
Answer: C​.
The problems on this page are the property of the MAA's American Mathematics Competitions
.jpg)
