## Saturday, February 11, 2017

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Problem 1313
Is <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

2. ‹ABM=360°-(144+156)=60°, AB=BM => ∆ABM equilateral
E,F on CD, BC. ‹EOF=180°-144=36°,
‹CQN=180-144=36°, ‹FQC=180-(90+78)=12° => ‹FQN=36° or ∆ONQ isoceles

3. Exterior angle of decagon = 360/10 =36
Exterior 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

4. 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)
7,42
8,24
9,18
12,12

1. Good info Sumith.

Any 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.

2. 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.

Those are the only integer solutions