Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 1031.
Sunday, July 20, 2014
Geometry Problem 1031: Trapezoid, Intersecting Circles, Common Chord, Tangent Line, Midpoint, Parallel Lines
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See sketch for location of S, L, N
ReplyDeleteWe have SE^2=SA.SB and SF^2=SC.SD
SB/SA=SC/SD
1. Calculate p^2= (SE.SD)^2= SA.SB.SD^2
And p1^2=(SA.SF)^2= SC.SD.SA^2
(p/p1)^2= (SB/SA).(SD/SC)= 1
So p=p1 => quadrilateral AFED is cyclic
Similarly quadrilateral BFEC is cyclic
2. We have ∠(FAE)= ∠(FDE)= ∠(BFC)= ∠ BEC) => AE//FC and FD//BE
And FNEL is a parallelogram
We also have NE.NA=NF.ND => N will be on radical line GH of circles O and Q
Similarly L will be on GH
M is the intersection of 2 diagonals of a parallelogram FNEL so M is the midpoint of EF.
To Peter Problem 1031
DeleteSketch link?
below is the sketch of problem 1031
ReplyDeletehttp://s25.postimg.org/xlwtf6fjz/pro_1031.png
Peter Tran
link is broken
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