Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemClick the figure below to see the complete problem 311.See also:Complete Problem 311Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
Let O be (0,0) so that A:(-R,0)& B:(R,0). Also let D be(p,0)so that C is (p,r) with the radius of the inner circle being r. We therefore have: x^2+y^2=R^2 ---(1) and (x-p)^2+(y-r)^2=r^2--(2). Solve these two equations and set the discriminant equal to zero so as to obtain a unique solution for the circles to touch each other at E. You my use Quickmath to do this and obtain: R^2 - p^2 =2rR or C is (p,(R^2-p^2)/2R). Substitute this in equation (2) to obtain the co-ordinates of E as (2pR^2/(p^2+R^2),R(R^2-p^2)^2/((p^2+R^2)). We can now obtain length DE by the distance formula as DE^2=(R^2 -P^2)^2/(R^2+p^2). As depicted in the diagram DE = x while (R+p)=a and (R-p)=b. Hence, x^2 = a^2*b^2/[(a+b)^2/4 +(a-b)^2/4] or x^2 = 2(a^2)(b^2)/(a^2+b^2) or 1/a^2 + 1/b^2 = 2/x^2 QEDAjit: email@example.com
we know that the points O,C,E lie on a line hence angleOCE=180. Join C,D then CD perpendicular to AB and we have AD=a, BD=b so OA=OB=a+b/2=OE so OD=a-b/2 and we have CE=CD=r(let).let AngleCOD=P then angleECD=90+P and using cosinerule for trCDE,x^2=r^2+r^2-2.r.rCos(90+P)hence x^2=2r^2(1+ sinP) from right trOCD sin P=CD/OD and we have OE=OC+CE, OE=(CD^2+OD^2)^1/2 +CE , a+b/2 = (r^2+(a-b/2)^2)^1/2 +r, r=ab/a+b henceOC=OE-CE=OE-CD=a+b/2-ab/a+b =a^2+b^2 /2(a+b) so sin P = 2ab/a^2+b^2 hence x^2=2(ab/a+b)^2(1+ 2ab/a^2+b^2), 2/x^2 = 1/a^2 + 1/b^2
Let DC meet AE at M. Let DN be an altitude of Tr. ADM, N on AEFrom Problem 310 < AED = < BED = 45 and since BDME is cyclic DM = b and since Tr. DEN is right isoceles DN = x/sqrt2Applying Pythagoras to right Tr. ADMa^2 * b^2 = AM^2Divide both sides by a^2 b^21/a^2 * 1/b^2 = (AM/ab)^2 = 1/DN^ 2 = 2/x^2 where I have used the fact that AM. DN = ab considering the area of Tr. ADMSumith PeirisMoratuwaSri Lanka