Geometry Problem. Post your solution in the comment box below.

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

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## Saturday, February 11, 2017

### Geometry Problem 1313: Regular Decagon, Pentadecagon, Equilateral Triangle, Congruence

Labels:
congruence,
decagon,
equilateral,
pentadecagon,
regular polygon

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Problem 1313

ReplyDeleteIs <COB=360/10=36, then <COQ=36/2=18, so <ABC=144.But <CQB=360/15=24 and <MBC=156, so <ABM=360-144-156=60, therefore triangle ABM is equilateral.Similar

triangle CDN is equilateral (OC=OD, CN=DN) so ON is perpendicular bisector of CD

then <NOC=36/2=18 or <NOQ=18+18=36.But <CQN=24, <OQC=24/2=12 (OB=OC, QB=QC)

and <NQO=24+12=36=<NOQ.Therefore NO=QN.

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

‹ABM=360°-(144+156)=60°, AB=BM => ∆ABM equilateral

ReplyDeleteE,F on CD, BC. ‹EOF=180°-144=36°,

‹CQN=180-144=36°, ‹FQC=180-(90+78)=12° => ‹FQN=36° or ∆ONQ isoceles

Exterior angle of decagon = 360/10 =36

ReplyDeleteExterior angle of pentadecagon = 360/15 =24

Hence < ABM = 36+24 = 60.

But AB = BC = BM

So ∆ ABM is equilateral

Similarly ∆DCN is equilateral. Hence ∆ OCN ≡ ∆ ODN.

Therefore < NOC = ½ < COD = ½ X 360/10 = 18. Similarly < COQ = 18 and so <

NOQ = 36.

Also < NQO = 1½ X 360/15 = 36

Therefore ON = QN

Sumith Peiris

Moratuwa

Sri Lanka

Number of sides of regular polygons which when placed together so that one side is common and forms an angle of 60 (other than 10,15)

ReplyDelete7,42

8,24

9,18

12,12

Good info Sumith.

DeleteAny two regular polygons that satisfy the equation 6(n1+n2) = n1*n2 form an angle of 60.

Where n1 = no of sides of first polygon

n2 = No of sides of second polygon.

Yes the above pairs are the only ones that satisfy the equation 360/n + 360/m = 60 which is equivalent to the equation u have given.

DeleteThose are the only integer solutions