## Friday, September 2, 2022

### Geometry Problem 1495: Circle, Parallel Chords, 30 Degree Angle, Radius Squared

Geometry Problem 1495. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below. More Details
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1. Trigonometry Solution

If O is the centre, Tr.s AOC, COE, DOF, BOD are all equilateral

So if < AOB = 2@ then < EOF = 120-2@

AB = 2r.sin@
CD = 2r.sin(60+@) = r.(sqrt3.cos@ + sin@)
EF = 2r.sin (60-@) = r.(sqrt3.cos@ - sin@)

AB^2 + CD^2 + EF^2
= r^2.( 4sin^2 @ + sin^2 @+ 3cos^2 @ + 2sqrt3.cos@sin@
+ sin^2 @ + 3cos^2 @ - 2sqrt3.cos@sin@)
= r^2.(6sin^2 @ + 6 cos^2 @)
= 6r^2

Sumith Peiris
Moratuwa
Sri Lanka

1. Geometry Solution using Pythagoras
Let AB = a, CD = b and EF = c. Let AH be the height of trapezoid ABCD, H on CD
Now AC = CE = BD = DF = r (they all subtend 60 degrees at the centre)

So CH = (b-a)/2, HD = (a+b)/2 and hence AH = (a+b)/(2sqrt.3)

Using Pythagoras for Tr. ACH, r^2 = (b-a)^2 /4 + (a+b)^2 /12,
from which 3r^2 = a^2 + b^2 - ab...........(1)

Similarly we can prove that
3r^2 = b^2 + c^2 - bc...............................(2)

(1) - (2) gives b = a + c ........(3) upon simplification assuming a not = to c
(1) + (2) gives 6r^2 = a^2 + b^2 + c^2 + (b^2 - ab - bc) and from (3)
the expression in the parenthesis = 0

Therefore a^2 + b^2 + c^2 = 6r^2

Sumith Peiris
Moratuwa
Sri Lanka

2. 3. I think this proof only goes in the special case when CD passes through O. Otherwise AD and DE are not equidistant and hence AC is not CE

1. I have not assumed that CD passes through O. Do read my proof carefully.

Sumith Peiris

4. Geometry solution: It is not difficult to prove CD=EF+AB (1). Join B to F, from ABFE (Pt Theorem) we get CD²=3R² + AB . EF (2) From (1) and (2) we get the result