Geometry Problem

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

This entry contributed by Ajit Athle.

Click the figure below to see the problem 755 details.

## Saturday, May 19, 2012

### Problem 755: Equilateral Triangle, Circumcircle, Chord, Arc, Midpoints, Perpendicular

Labels:
arc,
chord,
circumcircle,
equilateral,
midpoint,
perpendicular,
triangle

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http://img407.imageshack.us/img407/8651/problem755.png

ReplyDeleteConnect OE and FC

A,O and E are collinear …..(F is midpoint of arc BC)

OA=FC=Radius of circle

OE ⊥EA …(E is midpoint of DA)

∠DCF=∠EAO… ( face same arc DBF)

∆FGC congruence to ∆ OEA (case ASA) => GC=EA

∠DAB =∠DCB …(face same arc DB)

∆BGC congruence to ∆ BEA… ( case ASA)=> BG=BE and ∠GBC=∠EBA

So ∠EBG=∠ABC=60 => ∆EBG is equilateral

Let T be the rotation 60° clockwise about B.

ReplyDeleteThen

T(C) = A

T(F) = O

Since ∠ADC = ∠ABC = 60°,

so T(line DC) = line DA

Let T(G) = G', then G' lies on DA.

Since FG⊥DC, we have OG'⊥DA,

hence, G' = E.

As a result, T(G) = E, T(BG) = BE.

Hence, ∆BEG is equilateral.

Let O be the origin, z(P) be the complex number representing point P.

ReplyDeleteLet z(C)=1, then z(B)=ω, z(A)=ω^2, z(F)=-ω^2.

Let z(D)=ε, then z(E)=(ε+ω^2)/2.

Since FG⊥DC, let [z(D)-z(G)]/[z(F)-z(G)]=ki.

k = |z(D)-z(G)| / |z(F)-z(G)]| = √3

because ∠DFG = 60°.

Now [z(D)-z(G)]/[z(F)-z(G)] = i √3

ε - z(G) = i √3 [-ω^2 - z(G)] = -i√3 ω^2 - i√3 z(G)

[1 - i√3] z(G) = ε + i√3 ω^2

-2ω z(G) = ε + i√3 ω^2

z(G) = -1/2 ω^2 [ε + i√3 ω^2] = -1/2 ω^2 ε - i√3/2 ω

Consider

z(B) + ω z(E) + ω^2 z(G)

= ω + ω(ε+ω^2)/2 + ω^2 [-1/2 ω^2 ε - i√3/2 ω]

= ω + 1/2 ωε + 1/2 - 1/2 ωε - i√3/2

= ω - ω

= 0

As a result, ∆BEG is equilateral.

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27405&p=134056

ReplyDelete