Geometry Problem
Click the figure below to see the complete problem 494 about Circular Sector, 90 Degrees, Semicircle, Chord, Parallel.
See also:
Complete Problem 494
Level: High School, SAT Prep, College geometry
Saturday, August 7, 2010
Problem 494: Circular Sector, 90 Degrees, Semicircle, Chord, Parallel
Subscribe to:
Post Comments (Atom)
Line AB intersect circle O’ at M. M is midpoint of AB . Triangle AO’M is isosceles right triangle
ReplyDeleteQuadrilateral ACBD is a parallelogram and diagonals CD and AB intersect at midpoint M of AB and CD.
E is midpoint of AC (corresponding points of dilation transformation centered at A with dilation factor =1/2 ).
Angle ADC=45 (Angle face 90 degrees arc AM)
Consider circle O’ and point C . We have CE.CA=CM.CD .
Since M and E are the midpoints of CD and CA so CA=CD and CM=CE
triangle ACD is isosceles right triangle.
Triangles ACM congruence with DCE ( case SAS)
So x=DE=AM =O’A*SQRT(2)=2*SQRT(2)
Just confirming Peter's excellent solution by analytical geometry: Let O be (0,0). So A:(0,4),B:(4,0). The two circles are: x^2+y^2=16 & x^2+(y-2)^2=4. Now let C be (c,d) and D be (a,b). The given data can be translated into following eqns: a^+(b-2)^2=4, c^2+d^2=16,bc=(a-4)(d-4) and c^2+(d-4)^2=(a-4)^2+b^2 which when solved give us: a=8/5,b=4/5,c=12/5 and d=16/5.
ReplyDeleteNow we can write the eqn. of AC as y=-x/3+4 and determine E as (6/5,18/5) as the intersection of the smaller circle and AC. Now x^2=(8/5-6/5)^2+(4/5-18/5)^2 or x^2=8 -->x=2√2
Ajit
ang AOC = ang AO'E (isoc tr & A common)
ReplyDeleteang COB = ang DO'O ( OD bisect)
=> AO'E + DO'O = 90° => EO'D = 90°
=> x² = 2² + 2²
x = 2√2
To c.t.o.e.:
ReplyDeleteWill u pl explain what u mean by "ang COB=ang DO'O (OD bisect)". Unfortunately, it isn't very clear how u arrive at EO'D = 90° which, of course, has to be correct.
Vihaan
To Vihaan:
ReplyDeleteOD perpendicular to AD and CD (AD // CD)
BOC isoceles
=> OD bisector
DO'O = arc OD
BOD = 1/2 arc OD ( OB tg to O' and OD chord )
BOC = 2 BOD = arc OD
=> DO'O = BOC
( nice solution of 495 )
Thanks. BTW, I'm very curious to know what "c.t.e.o" stands for!
ReplyDeletevihaanup@gmail.com
1. Angle AEO is 90 degrees since AEO is an inscribed triangle on the diameter of the smaller circle.
ReplyDelete2. Angle ACB is 135 degrees since it inscribes the 270 degree arc on the larger circle.
3. ACBD is a parallelogram by the problem definition and how it was constructed. So angle CAD is supplementary to ACB and 45 degrees.
4. Let F be the intersection of AD and OE. Triangle AEF is a right isosceles by angle chasing. AE = EF and AF = sqrt(2) AE
5. AEDO is a cyclic quadrilateral by definition on the smaller circle so AFO is similar to EDF and DE/ AO = EF / AF or DE = AO * EF / AF = 4 * AE / sqrt(2) * AE =
2 sqrt(2)
Bonus: AE = EC since ACO is isosceles. Further drawing in OC and doing some angle chasing shows CDO is congruent to OBD by SAS so CD = BD and triangle ACD and BCD are both isosceles right triangles. That means CD when extended intersects the other end of the diameter of the larger circle and in fact is 2x scaled version of AEO.
Since <AOB = 90, <ACB = 135.
ReplyDeleteHence in \\ogram ADBC, <EAD = 45
So <EO'D = 2 X 45 = 90
So in isosceles Tr.DEO', x^2 = 2^2 + 2^2 = 8 and so x = 2sqrt2
Sumith Peiris
Moratuwa
Sri Lanka
Sumith
ReplyDeleteExcellent solution !
Peter Tran
Thanx Peter
Delete