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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 956.
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+ωz(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.
http://imagizer.imageshack.us/v2/1024x768q90/833/6b5z.pngDraw O’ such that FO=FO’ ( see sketch)EOAO’ is a parallelogram => EO=O’A=OB and OE//AO’∠ (CBO)= ∠ (EOx)=150But ∠ (O’AC)= ∠ (EOx) =150 … angles with sides parallelsSo ∆ (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
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.
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 triangleCKF similar triangle CBO.Therefore <KCF=<BCO so <FCO=30.Triangle KCB is similarWith triangle FCO (KC/BC=FC/OC) so Δ CFO is 30°-90°-60.MANOLOUDIS APOSTOLIS 4th HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE