Challenging Geometry Puzzle: Problem 1597. 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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With power of point A with respect to circle C we find AD=3 (or, use of Menelaos theorem). Next, by Heron formula find area of triangle ACD, from which you can get the altitude DE, then AE is piece of cake!
ReplyDeletecorrection for my error in the calculation ( see my comment)
ReplyDeletethe corrected values are u=3 and x= CE=6.5
draw line from point C vertical to line AB,insect at F. so DF=FB=1. Conect DE. so CEDF form a circle. so AD*(AD+1)= AE*8. Also AD*(AD+2)=1*(8+7) ; so AD=3. AE=3/2, CE=8-3/2=6.5
ReplyDeleteFrom point C, draw a line perpendicular to line AB, intersecting AB at point F, so
ReplyDeleteBF=FD=1. Connect point D to point E. Then, quadrilateral CFDE lies on a circle.
so AE⋅AC=AD⋅AF . Given AC=8 and AF=AD+1, this becomes: AE⋅8=AD⋅(AD+1)
Now, let circle C intersect line AC at point G. Applying Power of a Point again:
AG⋅(AC+radius of circle C)=AD⋅AB
AG=AC-BC=1, AC=8, and the radius of circle C is 7, we get:
1⋅(8+7)=AD⋅(AD+2), that is15=AD⋅(AD+2)
Solving this gives AD=3. Substituting back, AE⋅8=3⋅(3+1)=12⇒AE= 3/2, therefore,
EC=AC−AE=8− 3/2 =6.5
F = midpoint DB, BF = 1
ReplyDeleteFC = 7^2 - 1^2 = sqrt(48)
(AD+1)^2 = 64 - 48 = 16
AD^2 + 2AD - 15 = 0
AD = -5, 3
cos (angle FAC) = 4/8 = .5 angle BAC = 60 deg
AE = 3/2 = 1.5
EC = 8 - 1.5 = 6.5