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.

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Trigonometry Solution

ReplyDeleteIf 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@)

Squaring and adding

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

Geometry Solution using Pythagoras

DeleteLet 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

nice

ReplyDeleteI 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

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

DeleteSumith Peiris

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

ReplyDelete