Proposed Problem

Click the figure below to see the complete problem 428 about Quadrant of a circle, Square, Angle bisector, Measurement.

See also:

Complete Problem 428

Level: High School, SAT Prep, College geometry

## Sunday, February 21, 2010

### Problem 428: Quadrant of a circle, Square, Angle bisector, Measurement

Labels:
angle bisector,
circle,
circular sector,
inscribed,
measurement,
quadrant,
square

Subscribe to:
Post Comments (Atom)

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

---------------------

b² + b² = R² ( from tr ODE , R = OB = OD )

2b² = R²

b² = R²/2 (1)

---------------------

a² = R² + b² ( from tr COB )

a² = R² + R²/2 (from (1) )

a² = (3R²)/2 (2)

----------------------

substitute (1) & (2) at x =>

x = R√6/6

a/x = (R√6/2)/(R√6/6) ( a from (2) )

a/x = 3

----------------------

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