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

Labels:
chord,
circumcircle,
equilateral,
harmonic mean,
triangle

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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!