Proposed Problem

See also: Complete Problem 283, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Thursday, April 23, 2009

### Problem 283: Circular Sector 90 degrees, Semicircle, Circle inscribed, Radius

Labels:
circle,
circular sector,
radius,
semicircle

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With O as origin, the smallest circle with its centre at (r,h) is represented by (x-r)^2+(y-h)^2=r^2 while the circle with centre at C is:

ReplyDelete(x-R/2)^2+y^2=R^2/4. These two equations when solved simultaneously give: x=(2r+R)h^2 + or -2h^2(2rR-h^2)^(1/2)/(4(h^2+r^2)+R^2-4rR).

For them to be touching each other at one point (and not intersect) we can conclude that h^2 =2rR. Like wise, the two circles given by:(x-r)^2+(y-h)^2=r^2 and x^2+y^2=R^2 are touching at a point if only if -r^2h^4+2(rRh)^2-r^2R^4+4r^4R^2 =0 where h^2 = 2rR gives us, -4r^4*R^2 + 4r^4*R^2 + 4r^3*R^3 - r^2*R^4 = 0 or 4r^3*R^3 = r^2*R^4 or r = R/4

Ajit: ajitathle@gmail.com

can someone give a better answer involved with circle geometry please!!! urgent!!!

ReplyDeleteTo anonymous (Problem 283):

ReplyDeleteIn the acute triangle OCD apply Euclid's Elements Book II, Proposition 13

i need like a step by step way to complate this + reasoning.. i dont really understand the above comment... thanks :)

ReplyDeleteApplying Euclid's Book II, 13th Proposition to Tr. OCD:(R/2+r)^2 =(R/2)^2+ OD^2 - rR. Now substitute OD = (R - r) in this equation to obtain: R = 4r,QED.

ReplyDeleteIf X is the foot of the perpendicular from D to OC, DX^2 =

ReplyDelete(R-r)^2 - r^2 = (r + R/2)^2 - (R/2 - r)^2

Upon simplifying

r = R/4

Sumith Peiris

Moratuwa

Sri Lanka

See the

ReplyDeletedrawingDefine

F the tangent intersection of CirleD and CircleO

E intersection of CircleD and OA

G intersection of CircleD and CircleC

Define the line L tangent of CircleD and CircleO at F

F in CircleO => OF ⊥ L, F in CircleD => DF ⊥ L

=> O, D and F are aligned

Define H intersection of ED and CircleD

ED⊥OA => ED//OB

=> ∠FDH=∠DOC

Draw semi circle with center O and define I intersection of OB and the semi circle

F and I on CircleO => ∠FOB=2 ∠FIB

F and E on CircleD => ∠FDH=2 ∠FED

∠FDH=∠FOB => ∠FIB = ∠FED and F, E , I are aligned

D, G and C are aligned and DH//CO=> ∠DCO = ∠CDH

=> ∠GED = ∠GBC

ED//BC => E, G and B are aligned

OI=OB and OA⊥IB => ∠EIO = ∠EBO

OB//ED =>∠EBO= ∠BED

=> ∠GED=∠FED

=>∠GDH=2∠GED and ∠GCO=2∠GBO

∠GED=∠FED => ∠DCO=∠DOC

=> ODC is isosceles => DO=DC

DO=R-r and DC=r+R/2

=>R-r=r+R/2

=>R/2=2r

Therefore r=R/4