Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, June 23, 2018
Geometry Problem 1357: Regular Dodecagon, Diagonal, Concurrency, Collinearity, Points, Lines
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ReplyDeleteLet OA=OB= r
Connect OA, OB, AC and BF
∠ (AOB)= 360/12= 30 degrees ; ∠ (CAJ)= ½ x 180= 90; ∠ (NCA)= ½(30+30+30)= 45
NAC is a right isosceles triangle => NA= AC= r and ∠ (CNA)= 45
NAOB is a rhombus ( NA=OA=OB and NA//OB)
So ∠ (BNA)= ∠ (AOB)=30 and ∠ (CNB)= 15 and ∠ (NBC)=180-15-30=135
∠ (NBC) supplement to ∠ (CBF) so N,B,F are collinear
Is PN=PA=PB=AB=BC=> P=circumcenter of triangle NAB=> <CBN=135.But <NBC_<CBF=180.Therefore the points N,B,F are collinear.
ReplyDeleteExtend CB to meet AN at X. Drop a perpendicular XY from X to CN
ReplyDeleteIts easy to see that XAC is a right triangle and that XB = BC = AB = a say.
Now XYC is a 30-60-90 Tr. so XY = a, XN = √2.a
So XN2 = 2a2 = XB.XC, hence < XNB = < XCN = 30
But < ANC = 45 since < CAN = 90 and < ACN = 45
and so < CNB = 45-30 =15
Further < FBC = 45 = 30 +15 = < BNC + < BCN.
So we conclude N,B,F must be collinear