Challenging Geometry Puzzle: Problem 1569. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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Trigonometry Solution
ReplyDeleteLet the Incentre of Triangle BED be O and let DE be tangent to the incircle at K & let BD be tangent to the Incircle at L
So KOLD is a Square of side r
Let the Square CDEF be of side u and let AE = EG = v
Let t = tan A
From Triangle EOK,
r/ (u-r) = tan (A-45) = (t-1)/(t+1) from which upon simplification
u = r(2t/(t-1)).............. (1)
From Triangle CFG,
u/(u-v) = t so v/u = (t-1)/t............. (2)
(1) X (2)
v = 2r or EG = 2r
Sumith Peiris
Moratuwa
Sri Lanka
https://photos.app.goo.gl/LwBnYrJzUHqLL1NQ8
ReplyDeleteConnect CE and we have angle ECD=45
Triangle BED is a right triangle and triangle AEG is isosceles
We have 2r= BD+ED-BE
But BD+ED= BD+DC= BC=BA( triangle ABC is isosceles)
So 2r= BA-BE= AE=EG
Geometry Solution
ReplyDeleteLet the circle (Centre O) be tangent to AD at P, ED at Q and BE at S
Let the square be of side u and let AB = BC = a
Now OPDQ is a square of side r
So EQ = u - r = ES and
Also BP = BC - PC = a - u - r = BS
Hence AE = AB - ES - BS = a - (u - r) - (a - u - r) = 2r
Now EG // BC hence AE = EG = 2r
Sumith Peiris
Moratuwa
Sri Lanka