## Saturday, May 19, 2012

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

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

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

2. Let T be the rotation 60° clockwise about B.

Then
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.

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

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