Challenging Geometry Puzzle: Problem 1592. 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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Rotate triangle QEF 180 degrees around point F, so that F maps to itself, E maps to D (DF=FE), and Q maps to Q' on line OD. (Q', Q, F are also collinear)
ReplyDeleteBecause DQ'=EQ=BQ, and DO=OB, so DQ'+DO=OB+BQ -> OQ'=OQ.
Next, we know that OQQ' is an isosceles right triangle, and Q', Q, F are collinear, so angle(BQF)=45 degrees. From this, angle(FQE)=45 degrees. So BQF is congruent to FQE (SAS).
So DF=FB. We also know that ODF is cong. to OBF (SSS), so angle(FOQ)=45 degrees.From this, OFQ is also an isosceles right triangle. Let BQ=FQ=x. So sqrt(2)x=1+x, from this, we deduce that [BQ=sqrt(2)+1]. QED
Trigonometry Solution
ReplyDeleteLet BQ = R
< EBQ = 45 = < DBO
So < DBE = 90
Hence FB = FE and
Triangles FBQ and FEQ are congruent SSS and so
< FQE = 45 and < FEQ = 67.5 and < FEB = 22.5 and < BDE = 67.5
Hence Tan 67.5 = V2.R / V2 = R = V2 + 1
Sumith Peiris
Moratuwa
Sri Lanka
Geometry Solution
ReplyDeleteLet BQ = R
< EBQ = < DBO = 45
So < DBE = 90 and FB = FE
Hence Triangles BFQ and EFQ are congruent SSS
So < FQO = 45
Also Triangles ODF and OBF are congruent SSS
So <FOQ = 45
Hence Triangle OFQ is Right Isosceles and
R^2 + R^2 = (R + 1)^2 from which
V2.R = R + 1 and so
R = 1/(V2 - 1) = V2 + 1
Sumith Peiris
Moratuwa
Sri Lanka