Proposed Problem
Click the figure below to see the complete problem 428 about Quadrant of a circle, Square, Angle bisector, Measurement.
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Complete Problem 428
Level: High School, SAT Prep, College geometry
Sunday, February 21, 2010
Problem 428: Quadrant of a circle, Square, Angle bisector, Measurement
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ang CFB = 90° ( draw symetrical on the left )
ReplyDeletedraw at B tang to OB
extend CF to G ( G on tangent )
=>
BCG is isoceles ( CD bisector )
at right tr CFB and FBG, FB altitude, is common for two tr
a² - x² = ( 2b )² - ( a - x )² ( BG = 2DE = 2b )
a² - x² = 4b² - a² + 2ax - x²
2ax = 2a² - 4b²
ax = a² - 2b²
x = (a² - 2b²)/a
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b² + b² = R² ( from tr ODE , R = OB = OD )
2b² = R²
b² = R²/2 (1)
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a² = R² + b² ( from tr COB )
a² = R² + R²/2 (from (1) )
a² = (3R²)/2 (2)
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substitute (1) & (2) at x =>
x = R√6/6
a/x = (R√6/2)/(R√6/6) ( a from (2) )
a/x = 3
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Let the side of the square be y. Then OC =√2 y and BC =√3 y. Draw FG perpendicular to OB. OG =x*OB/BC=x√(2/3) and FG = y+ x*OC/BC =y+x/√3. Now OF^2=OB^2=2y^2=(x√(2/3))^2+( y+x/√3)^2 or x^2 +2xy/√3 –y^2=0 which is a quadratic yielding x = y/√3 as the only admissible value. Hence x/BC = (y/√3)/√3 y = 1/3.
ReplyDeleteAjit
Erratum: Read OD=√2y (OC=y) in the proof above.
ReplyDeleteAjit
CF cuts OB at P, such that <CPB=<FCD=<DCB=<CBP making triangle PCB isosceles and P on circle. <FOB=2<FPB=<FCD+<DCB=<FCB so FCOB is cyclic. Using power of point P, we can say that
ReplyDeletea(a+x)=(PO=asqrt(2/3))*(PB=2asqrt(2/3))=2a^2*2/3. Solving for x gets desired answer.
Complete rectangle COBT and parellelogram COTU.
ReplyDeleteT is the mid point of BU and so OT which is parellel to CU bisects FB perpendicularly since Tr. OFB is isoceles and so BF is an altitude of isoceles Tr. CUB.
Writing the area if this Tr. in 2 ways ;
FB. a = OB. OC. Squaring.
FB^2 = 2a^2 . a^2/ 3 / a^2
So FB ^2 = 8a^2/3 and so applying Pythagoras to Tr. BCF we see that this is = to a^2 - x^2 from which x^2 = a^2 / 9
So x = a/3
Sumith Peiris
Moratuwa
Sri Lanka
Dear Antonio - do u have a proof for this which does not use Pythagoras?
ReplyDeleteProblem 428
ReplyDeleteThe power of point C of the circle (O, OB) is CF’.CB=CD’.CD or CF.CB=CD^2.Is CD^2=R^2/2
α^2=OC^2+R^2 (F,F’ and D,D’ are symmetrical to the ΟΑ) ,then a^2=3R^2/2.But a.x=R^2/2
so a.x=a^2/3 (a^2/3=R^2/2 ).Therefore x=a/3.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
2nd Solution
ReplyDeleteExtend BO and FC to meet at Z.
Easily Tr. BCZ is isosceles & ZO = OF = OB = R (say)
So < BFZ = 90, hence OCFB is concyclic
Hence ZC. ZF = ZO. ZB, therefore
a(a+x) = 2R^2 = 2 (2a^2/3) since upon using Pythagoras on Tr. BCO, R^2 = 2a^2/3
Hence a+x = 4a/3 and x = a/3
Sumith Peiris
Moratuwa
Sri Lanka
Let the side of the square be 1 unit, then the radius of the circle = √2 units
ReplyDelete=>CB=a=√3 units
and BF=√(3-x^2)
Extend BC to meet the circle at P
Extend BO to meet the circle at Q
m(OCB)=m(ACP)=m(ACF)=m(OCQ)
=> FC when extended intersects the circle at Q
=>m(QFB)=m(CFB)=90
=>O,C,F,B are concyclic
Applying Ptolemy's to OCFB
√2x+√(3-x^2)=√6
=>3x^2-4√3x+3=0
=>x=1/√3=a/3
Solution 3
ReplyDeleteApply Ptolemy to OCFB
R.a = R.x + (R/sqrt2).sqrt(a^2 - x^2)
(a-x)^2 = (a^2 - x^2)/2
Divide by a-x which is not = to zero
a-x = (a + x)/2 from which
x = a/3
Sumith Peiris
Moratuwa
Sri Lanka
Solution 4
ReplyDeleteLet the square be of side p so that OB = sqrt2.p and BC = sqrt3.p = a
Extend BO to meet the circle at G.
FCG is collinear and CG = BC = a
< GFB = 90 = < COB and Tr.s BFG and BCO are similar
So GF/BG = BO/BC i.e. a + x = 2.sqrt2.p X sqrt2.p / sqrt3.p = 4p/sqrt3 = 4a/3
Therefore x = a/3
Sumith Peiris
Moratuwa
Sri Lanka
Solution 5
ReplyDeleteTrigonometry Solution
Let < BOC = u, then cos u = sqrt(2/3)
x = a cos 2u = a (2.cos^2 u -1) = a(4/3 - 1) = a/3
Sumith Peiris
Moratuwa
Sri Lanka