Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 757 details.
Tuesday, May 29, 2012
Problem 757: Equilateral Triangle, Circumcircle, Chords, a Half of Harmonic Mean
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Denote S(XYZ)= area of triangle XYZ
ReplyDeleteWe have S(ADB)=S(ADE)+S(DEB)
Or ½.b.c.sin(120)=1/2.b.x.sin(60)+1/2.c.x sin(60)
Note that sin(60)=sin(120)
So b.c=b.x+c.x => x=b.c/(b+c)
Tr.ADE///Tr.CDB since /_CDB=/_ADE both=60 and /_DAE=/_DCB (same sector); hence x/b = c/CD --(1)
ReplyDeleteBy Ptolemy or by Problem 256, CD=b+c --(2)
Combine (1)&(2) to get, x/b=c/(b+c) or 1/x=1/b+1/c
By similar triangles (ADE ~ BEC)&(BDE ~ AEC),
ReplyDeletex/b = BE/BC, x/c = AE/AC
Since ABC is equiliteral, (BE + AE = AC = BC)
Then x/b + x/c = BE/BC + AE / AC = 1
The remaining could easily done by rearranging of terms
Q.E.D.
By Sine Rule,
ReplyDeletex/b + x/c
= sin DAE / sin AED + sin DBE / sin DEB
= sin DCB / sin BEC + sin DCA / sin AEC
= BE/AC + EC/AC
= (BE + EC) / AC
= BC/AC
= 1
etc
A pure geometric solution:
ReplyDeleteLet be P on ray AD such that DP=DB. As ADBC is cyclic, x=bc(b+c) QED
Video solution is coming, Go-Greetings :)
WTF, what happened with my post?!?!!?
Deletei will complete this: (...) cylic, triangles ADE and APB are similar, then (...)
Go-greetings!
Here is my solution...
ReplyDeletehttps://www.youtube.com/watch?v=FRng4sMQJCg