Geometry Problem

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

Click the figure below to see the complete problem 679.

## Saturday, October 22, 2011

### Problem 679: Triangle, Circle, Median, Proportion

Labels:
circle,
median,
proportions,
similarity,
triangle

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http://img580.imageshack.us/img580/7519/problem679.png

ReplyDeleteDraw RMS // DFE ( see picture)

Note that Area(ABM)=area(BMC) ( same altitudes and bases )

Area(ABM)= Area(BMC)=½* c*MN =1/2*a*MP

So a/c=MN/MP

Note that triangle RBS is isosceles and triangle MNR similar to tri. MPS

So MN/MP=MR/MS=FD/FE=x/y

So a/c=x/y

Peter Tran

let be r the radius,B1,B2 the angles at vertex B

ReplyDeleteusing the sine law

in DBF:sinB1/x=sinF/r

in EBF:sinB2/y=sinF/r

in ABM:sinB1/(0.5b)=sinA/BM

in CBM:sinB2/(0.5b)=sinC/BM

in ABC:sinA/a=sinC/c

hence a/c=x/y

Draw AG//DE and CH//DE, giving K and L the intersections with BM. r = radius circle.

ReplyDeletehttp://s73.photobucket.com/albums/i207/HenkieR/PuzzleSolutions/GoGeometryProblem679solution_rlg.jpg

∆MCL = ∆MAK (opposite angles, Z-angles and AM=CM)

so LC = AK.

∆BEF ~ ∆BCL (aa), so y/r = LC/a => LC = ay/r

∆BDF ~ ∆BAK (aa), so x/r = AK/c => AK = cx/r

Because LC = AK, this gives ay/r = cx/r

=> ay = cx => a/c = x/y

QED.

Draw a line CPQ parallel to EFD intersecting BM at P and AB at Q

ReplyDeleteSince BDE and BQC are similar isosceles triangles, CP = ky and PQ = kx

Applying Menelaus to triangle ACQ

=> (BQ/BA)*(AM/MC)*(CP/PQ) = 1

=> (a/c)*(1)*(ky/kx) = 1

=> a/c = x/y