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Equal Areas on Circles (Posted on 2016-11-19) Difficulty: 4 of 5
Circles P and Q have radii 1 and 4 respectively and are externally tangent at point A. Point B is on P and point C is on Q such that BC is a common external tangent of the two circles. A line L through A intersects P again at D and intersects Q again at E. Points B and C lie on the same side of the line L and the areas of triangle DBA and triangle ACE are equal. Find this common area.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Analytic solution Comment 1 of 1
Place P and the origin and Q at (5,0), A=(1,0)
Its easy to show B=(-3/5,4/5) C=(3/5,16,5)

Line L has some slope, m and so equation y=m(x-1)

Its a fair amount of work to find points D and E, but knowing A helps.
For D:
x^2+m^2(x-1)^1=1
x=(1-m^2)/(m^2+1)
y=(-2m)/(m^2+1)

For E:
(x-5)^2+m^2(x-1)^2=16
x=(m^2+9)/(m^2+1)
y=(8m)/(m^2+1)

To find an area formula for DBA, contruct K where segment BD intersects the x-axis. K=((m-2)/(m+2),0)
The add the ares of ABK and ADK
After some simplifying the area is
.8(2m-1)/(m^2+1)

To find an area formula for ACE, construct F and G on the x-axis under C and E.  Then ACE=ACF+FCEG-AEG
After a lot of simplifying the area is
6.4(2-m)/(m^2+1)

Setting the two areas equal and solving gives
m=1.5

From which each area formula gives the solution: 64/65


  Posted by Jer on 2016-11-19 09:00:48
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