Geometry Problem

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

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## Monday, January 20, 2014

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## Monday, January 20, 2014

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Geometry Problem 962. Triangle, Two Cevians, Areas, Equal Products

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Let BE=mAE, BD=nCD.

ReplyDeleteLet S(ABC) denote the area of ABC.

From Menelaus theorem,

nCF=(m+1)EF

mAF=(n+1)DF

S₁ = S(ADC) = 1/(n+1) S(ABC)

S₃ = S(AEC) = 1/(m+1) S(ABC)

S₂ = S(BEF) = n/(m+n+1) S(BEC) = mn/[(m+1)(m+n+1)] S(ABC)

S₄ = S(BDF) = m/(m+n+1) S(BDA) = mn/[(n+1)(m+n+1)] S(ABC)

Thus,

S₁×S₂ = S₃×S₄ = mn/[(m+1)(n+1)(m+n+1)] S(ABC)²

Extend BF and cut AC at G

ReplyDeleteBy Ceva, (BD/DC)*(CG/GA)*(AE/EB) = 1

So BD*CG*AE = BE*AG*DC

S1*S2

=ΔADC*ΔBEF

=(DC/BC)*ΔABC * (BE/BA)*ΔBAF

=(DC/BC)*ΔABC * (BE/BA)*(BF/BG)*ΔBAG

=(DC/BC)*ΔABC * (BE/BA)*(BF/BG)*(AG/AC)*ΔABC

=(DC/BC)*(BE/BA)*(AG/AC)* [ΔABC]^2*(BF/BG)

=(DC*BE*AG)/(BC/BA/AC) * [ΔABC]^2*(BF/BG)

=(AE*BD*GC)/(AB/BC/AC) * [ΔABC]^2*(BF/BG)

=(AE/AB)*(BD/BC)*(GC/AC) * [ΔABC]^2*(BF/BG)

=(AE/AB)*ΔABC * (BD/BC)*(BF/BG)*(GC/AC)*ΔABC

=(AE/AB)*ΔABC * (BD/BC)*(BF/BG)*ΔBGC

=(AE/AB)*ΔABC * (BD/BC)*ΔBFC

=S3*S4

http://imagizer.imageshack.us/v2/800x600q90/42/epgz.png

ReplyDeleteLet ED cut AC at K

Let L, M are projection of E and D over AC

Note that (K,E,H,D)=-1

And HE/HD=KE/KD=EL/DM….. (1)

But HE/HD= h2/h4

We have S2/S4=h2/h4 …. ( triangles have the same base)

And S3/S1=EL/DM…. (triangles have the same base)

Per (1) we have h2/h4=EL/DM => S2/S4=S3/S1 => S1.S3=S3.S4

Extend ED to AC at X. ED cuts BF at Y. S1/S3=perpendicular from D to AC/perpendicular from E to AC=XD/XE, and S4/S2=perpendicular from D to BF/perpendicular from E to BF=DY/EY.

ReplyDeleteApplying Menelaus to BED and transversal XAC, we get that

AE/AB*BC/DC*DX/EX=1

Applying Ceva to BED and point F, we obtain

AE/AB*BC/DC*YD/YE=1

From these two equations we see that S1/S3=DX/EX=YD/YE=S4/S2