Proposed Problem

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Complete Problem 310

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Friday, June 26, 2009

### Problem 310: Circle Inscribed in a Semicircle, 45 degree Angle

Labels:
45 degrees,
angle,
circle,
semicircle

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With O as the origin we have A:(-R,0) and B:(R,0). Let D be (p,0) and inner circle have a radius r; hence C is (p,r). Since the two circles touch each other at E we can show that R^2-p^2=2rR, by solving the two equations simultaneously and setting the discriminant=0 to obtain a single point as the solution. Now if we substitute r=(R^2-p^2)/2R in the equation above we can determine E as [2pR^2/(R^2+p^2),R(R^2-p^2)/(R^2+p^2)]. Further by the distance formula, AE^2 =4pR^3/(R^2+p^2)+ 2R^2 while BE^2= 2R^2 - 4pR^3/(R^2+p^2) Hence, AE^2/BE^2 = (R+p)^2/(R-p)^2 = AD^2/DB^2 since AD=R+p and DB=R-p. In other words, AE/BE = AD/DB or D divides AB in the ratio of the two other sides of triangle ABE. Hence we can say DE biscts angle AEB which 90 deg since AB is a diameter of the larger circle. Hence angle AED=45 deg.

ReplyDeleteQED. Ajit: ajitathle@gmail.com

Forgot to mention that the equations of the two circles above are: x^2 + y^2 = R^2 and (x-p)^2 +(y-r)^2 = r^2. Also, the discriminant referred to is: -p^6 + 2(R^2)(p^4)-(R^4)(p^2) + (2rRp)^2 which when equated to 0 gives us: R^2- p^2 = 2rR.

ReplyDeleteIf there's another easier way to obtain this condition, will someone please show me the same?

Ajit

Dear Ajit,

ReplyDeleteAnother way:

1. Isosceles triangle DCE; therefore...

2. Isosceles triangle AOE; therefore...

3. Right triangle ODC; therefore...

Q.E.D.

Antonio

Yes, indeed!

ReplyDeleteO,C & E are collinear and OC=R-r while OD=p and CD=r which gives us (R-r)^2 = p^2 + r^2 from where R^2 - p^2 = 2Rr.

Thanks, Antonio.

Ajit

This problem can be done with plane geometry!

ReplyDeleteAuxiliary Construction Hint:

Draw tangent line at E; extend Ray ED and Arc EA turning counterclockwise till they intersect at F; and finally connect AF and BE.

And furthermore, connect D and the intersection of AE and circle C (call the intersection G).

ReplyDeleteDraw full circle diameter AB and a radius OF perpendicular to diameter AB ( F on extended portion of the circle)

ReplyDeleteDE intersect OF at F’ . We will prove F’ coincide with F and angle AED face 90 degrees arc.

We have CD perpendicular to AB and points O, C and E are collinear ( circle C tangent to dia. AB and circle O)

Since CD //OF’ so triangles CDE similar to triangle OF’E ( Case AA)

And triangle OF’E is isosceles

So OF’=OE=OF= radius of circle O and point F’ coincide with F and angle AED=45

Peter Tran

Let < OAE = < OEB = p. Then < EOB = 2p.

ReplyDeleteNow O, C, E are collinear, DC is perp. To OB and < DCE is isoceles.

Hence < CDE = alpha- p and < EDB = alpha + p. Hence < CDB = 2. Alpha = 90, so alpha = 45.

Sumith Peiris

Moratuwa

Sri Lanka

Problem 310

ReplyDeleteThe points O, C, E are collinear.Let <OEA=<OAE=x, <CED=<CDE=y. But CD, AE are perpendicular in BD , EB respectively.Then <BDE=2x+y=90-y or 2x+2y=90 so

x+y=45.But <AED=x+y=45.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Very Simply,

ReplyDeleteDC = CE so < OCD = 2 X < OED ....(1)

OE = OA so < EOD = 2 X < AEO ....(2)

(1) + (2) < OCD + < EOD = 90 = 2 X (< OED + < AED) = 2 Alpha

Sumith Peiris

Moratuwa

Sri Lanka