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

Labels:
collinear,
concurrent,
diagonal,
dodecagon,
geometry problem,
regular polygon

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https://photos.app.goo.gl/Jo6diAsRxYjufs1h9

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