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