Sunday, June 24, 2012

Problem 773: Intersecting Circles, Midpoint, Secant line, Bisection

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 773.

1. http://img59.imageshack.us/img59/328/problem773.png

Draw new lines per attached sketch
We have ∠BDA=1/2*∠BQA= ∠BQO
And ∠BCA=1/2*∠BOA =∠BOQ
∆BOQ similar to ∆BCD …. ( case SAS)
∆ BCD is the image ∆BOQ in the spiral similarity geometry transformation with following details.
center = point B , angle of rotation θ= ∠OBC=∠QBD , ratio of similarity= BC/BO=BD/BQ
Since M and E are midpoints of OQ and CD , so E is the image of M in this geometry transformation.
So we have ∆BME similar to ∆BOC => MB=ME and E is the point of intersection of CD and circle M

Let F be diametrically opposite to A w.r.t. cirle (M).
Join EF. So ∠AEF = 90°, E is the foot of ⊥ from F on CD.
Let O', M', Q' be the feet of the ⊥s from O, M, Q on CD.
So CO' = O'A, AM' = M'E, AQ' = Q'D
Join OF, QF.
OAQF is a parallelogram since diagonals OQ, AF bisect each other.
OA = FQ and OF = AQ
So O'A = EQ' and O'E = AQ'
Now CE = CA + AE = 2O'A + AE
= 2EQ' + AE = EQ' + (EQ' + AE)
= EQ' + AQ' = EQ' + Q'D = ED as required.

3. Ref: For figure of Problem 773, please go to the following link:

4. Let OC=r and QD=R
This is true, if and only if
EA*EC=ED*EA or
EO^2-r^2=R^2-EQ^2 or
R^2+r^2=EO^2+EQ^2 or
AO^2+AQ^2=EO^2+EQ^2 or
2(OQ^2)+(AM^2)/2=2(OQ^2)+(EM^2)2 or
AM=EM.
But this is true

5. http://www.subirimagenes.net/./pictures/57349088113a71fd591ec83ceeb19ce8.png

6. Let X,Y, and Z be the feet of the perpendiculars from O,M, and Q to CD. Then XA=(1/2)AC and AZ=(1/2)AD, so XZ=(1/2)CD, so XY=YZ=(1/4)CD, so XY=XA+AY=(1/2)AC+(1/2)AE=(1/2)CE=(1/4)CD, so CE=(1/2)CD.

7. (O ,r) ,(Q,R) The two circles
EA.EC= EO^2 - r^2, EA.ΕD=R^2-EQ^2
EA.EC- EA.ΕD= EO^2+ EQ^2-( r^2+ R^2)=2ΕΜ^2+2ΟΜ^2-2ΑΜ^2-2ΟΜ^2=0(by Theorem median in triangles ΟΕQ,OAQ, and because ΑΜ=ΕΜ ).So, EA.EC- EA.ΕD=0 ,EA.EC=EA.ΕD, EC=ED