Challenging Geometry Puzzle: Problem 1588. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
Gain comprehensive insights! Click below to reveal the complete details.
Click for additional details.
Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.
Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.
Trigonometry Solution
ReplyDeleteLet < OAQ = @ so < OAC = 2@ (since Tr.s OAQ & CAQ are congruent SSS)
tan @ = 1/2 = t (say)
Triangle AOD is isosceles, so
AD = 2r.cos2@ = 2r. (1-t^2)/(1+t^2) = 2r. (1-1/4)/(1+1/4) = 6r/5
Therefore CD = AD - AC = 6r/5 - r = r/5
Sumith Peiris
Moratuwa
Sri Lanka
Geometry Solution
ReplyDeleteLet U be the mid point of isosceles triangle OAC
Let AQ & OC intersect at V
S(OAQ) = 1/2 X r X r/2 = 1/2 X AQ X OV
So OV = (r^2 / 2) / (V5 X r/2 ) = r /V5
OC = 2.r/V5 & AV = 2.r/V5
2.S(OAC) = OU.r = AV.OC
Hence OU = (2.r/V5).(2.r/V5) / r = 4r/5
So AU = 3r/5
and CU = r - 3r/5 = 2r/5
Therefore CD = 3r/5 - 2r/5 = r/5
Sumith Peiris
Moratuwa
Sri Lanka
Key steps of 2nd Geometry Solution
ReplyDeleteLet U be the mid point of AC
Let AQ & OC meet at V and OU & AQ at W
Let AU = p
AOVU is concyclic
WU = p/2 (angle bisector theorem)
Write Pythagoras for Triangle AOU
p = 3r/5
CU = 2r/5
CD = 6/5
Sumith Peiris
Moratuwa
Sri Lanka
Errata
DeleteCD = r/5 not 6/5
https://photos.app.goo.gl/xgT5B3wdSGKy41Hr8
ReplyDeleteDefine point P .
Power of C to circle O= CA.CD=r.CD= CO^2-r^2=-CB^2
In right triangle AOQ we have OC^2= 4.OP^2= r^4/AQ^2= ⅘. r^2
So CB^2= ⅕.r^2 and CD= r/5