## Tuesday, January 14, 2014

### Geometry Problem 956: Two Equilateral Triangles, Center, Collinear Points, Midpoint, Right Triangle, 30, 60, 90 Degrees

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

Click the figure below to see the complete problem 956.

#### 4 comments:

1. Let z(P) be the complex number represent P.
Let z(A)=0, z(C)=1, and let ω=cos(2π/3)+i sin(2π/3)=−1/2+√3/2 i.
Let BD=k.

Then we have
z(B)=1+ω
z(D)=(1+ω)(1+k)
z(E)=1+ω(1+k)
z(O)=1/3 [1+ω+1+ω+k+ωk+1+ω+ωk]=1/3 [3+3ω+k(1+2ω)]
z(F)=1/2 [1+ω+ωk]

z(O)−z(F)=1/6 [3+3ω+2k+ωk]=√3/12 [√3(k+1)+(k+3)i]
z(C)−z(F)=1/2 [−ω−ωk]=−1/4 [−(k+3)+√3(k+1)i]

Since (−√3 i)[z(O)−z(F)]=z(C)−z(F),
thus OF⊥CF and CF=√3 OF.
Therefore, ΔCFO is 30°-90°-60° Δ.

2. http://imagizer.imageshack.us/v2/1024x768q90/833/6b5z.png

Draw O’ such that FO=FO’ ( see sketch)
EOAO’ is a parallelogram => EO=O’A=OB and OE//AO’
∠ (CBO)= ∠ (EOx)=150
But ∠ (O’AC)= ∠ (EOx) =150 … angles with sides parallels
So ∆ (CBO) congruence to ∆ (CAO’)….(case SAS)
∆ (CAO’) is the image of ∆ (CBO) in the rotation (center C, rotation angle=60)
So ∆ (OCO’) is a equilateral triangle with median CF => ∆ (CFO)= 30, 90, 60 triangle

3. FC and ED meet at point P. FC=FP because triangles FAC and FEP are congruent.
With side AC=a and ED=b, and midpoints of AC and ED denoted A' and B' respectively, OP^2=PB'^2+OB'^2=OC^2=OA'^2+CA'^2=a^2+ab+b^2/3 after working out the algebra.
With OP=OC and FC=FP, <OFC must be 90.
Furthermore PC^2=3a^2+3ab+b^2 in triangle PEC by cosine rule(Check it yourself), so FC^2=(PC^2)/4=(sqrt(3)/2)^2*OC^2.
Therefore FC=sqrt(3)/2*OC and <FCO=30.

4. Problem 956

Fetch CK perpendicular to AB,so <KCB=30 and KC/BC=√3 /2,but FK=//EB/2 then FK/OB=
√3 /2, <CKF=<CKB+<BKF=90+60=150,<CBO=<CBD+<DBO=120+30=150,then triangle
CKF similar triangle CBO.Therefore <KCF=<BCO so <FCO=30.Triangle KCB is similar
With triangle FCO (KC/BC=FC/OC) so Δ CFO is 30°-90°-60.
MANOLOUDIS APOSTOLIS
4th HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE