Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemClick the figure below to see the complete problem 310.See also:Complete Problem 310Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
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.QED. Ajit: firstname.lastname@example.org
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.If there's another easier way to obtain this condition, will someone please show me the same? Ajit
Dear Ajit,Another way:1. Isosceles triangle DCE; therefore...2. Isosceles triangle AOE; therefore...3. Right triangle ODC; therefore...Q.E.D.Antonio
Yes, indeed!O,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!Auxiliary 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).
Draw full circle diameter AB and a radius OF perpendicular to diameter AB ( F on extended portion of the circle)DE 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=45Peter Tran
Let < OAE = < OEB = p. Then < EOB = 2p.Now 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 PeirisMoratuwaSri Lanka
Problem 310The 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 sox+y=45.But <AED=x+y=45.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE