Challenging Geometry Puzzle: Problem 1589. 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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ReplyDeleteDefine point P and Q
Due to symmetry we have
Arc(AC)=Arc(PQ ) and Arc(BC)=Arc(BP)
Angle CAD= 45=> Arc(BC)=Arc(BP)=45
And Arc(AC)=Arc(PQ)=45
AD become angle bisector of angle OAB
So BD/OD= BA/OA= sqrt(2)
Let AC = DC = a, OD = b and radius = R
ReplyDeleteSo AD = V2.R
OADC is a concyclic quadrilateral and so using Ptolemy
ab + aR = V2.aR from which
b = (V2-1).R
and so R = (V2 + 1).b (using conjugate of V2 - 1)
Now BD/OD = (R-b)/b = R/b - 1
Therefore BD/OD = (V2 + 1) - 1 = V2
(Using Trigonometry of course is a shortcut since we can see that < ACO = < ADO = 67.5 and tan 67.5 = V2+1)
Sumith Peiris
Moratuwa
Sri Lanka
2nd Geometry Solution
ReplyDeleteAC = CD so < ADC = 45
Since AODC is concyclic , < AOC = < ADC = 45
Since AO = CO, < ACO = 67.5 = < ADO
Hence < OAD = 22.5 = < BAD
So AD bisects < OAB
Therefore BD / OD = AB / AO = V2
Sumith Peiris
Moratuwa
Sri Lanka
3rd Geomtery Solution using similar Triangles
ReplyDeleteLet radius = R and AC = DC = a
Triangles OCD and BAD are similar having angles 45 and 22.5 (see my solution 2 above)
So BD / OD = V2a / a = V2R / R = V2
Sumith Peiris
Moratuwa
Sri Lanka