Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 636.
http://img28.imageshack.us/img28/7780/problem636.pngLet circle F tangent to CE and circle O at M and N ( see picture)Triangles AON and MFN are similar isosceles triangles ( MF // AO and angle AON= angle MFN)So angle NAO= angle NMF and 3 points A, M and N are collinearNote that angle ANB =90 and quadrilateral MNBC is cyclicSo AM.AN =AC.ABPoint A has the same power to circle center D and circle center F so A must be on common tangent from G of these circlesPeter Tran
Let EC touch circle (F;x) at XLet GM cut EC at Y. clearly YX=YG=YC .So Y is the midpoint of XC.Let angle XYF = αt=tan α=x/XY=x/√(rx)=√(x/r)tan ∠XYG=2t/(1–t2)=2√(x/r)/[1–(x/r)]=2√(rx)/(r–x) = 2√(rx)/r–[r(R–r)/R] from Problem 635= 2R√(rx)/r2 = 2R√x/(r√r)tan ∠AYC =AC/YC=(AB–CB)/YC=2(R–r)/√(rx)=2(Rx/r)/√(rx)=2R√x/(r√r)∠XYG=∠AYC X,Y,G are collinear
Without loss of generality I asume that AO = OB = 1.Call DC = DB = R and FG = r.Drop a perpendicular from F to AB, giving X.Then: AD = 2-R, FD = R+r, XD = R-r, GD = R.From problem 635: r = R(1-R) = R-R².Call Y the meeting point of AB with the tangent GM.[To prove that Y = A.] Triangle DXF is similar with triangle DGY (both a right angle and the same angle at D).FD:XD = YD:GD -> (R+r):(R-r) = YD:RYD = R*(R+r)/(R-r)Substituting r = R-R² givesYD = R*(2R-R²)/(R²)= 2-RSo YD = ADY = AQED.
Type error:Please read last sentence of my solution as "A,Y,G are collinear"
To Peter Tran,"Point A has the same power to circle center D and circle center F so A must be on common tangent from G of these circles"The above sounds strange.
To Anonymous .GM is radical line of circles D and F.Point A has the same power to 2 circles, so it must lie on GM which is the common tangent of 2 circles
Hello Tran, thanks to your comment.AM.AN=AG1.AG1 and AC.AB=AG2.AG2(AG1 is tangent line to circle F, AG2 is tangent line to circle D)Then length AG1=AG2 ---> it is correct.But, please explain why angel BAG1 is equal to angle BAG2.
To AnonymousNo G1 or G2 is required.Power of A to circle F= AM.AN ( A,M, N collinear)Power of A to circle O= AC.ABSince MNBC is cyclic so AM.AM=AC.AB => Power of A to circle F=Power of A to circle OSo A must lie on radical line of circles F and O which is common tangent of 2 circles
See http://imageshack.us/photo/my-images/547/imageqx.png/According to your "Power theorem", length AG1=AG2=AG3, I can understand.But, see my image.
Apply inversion at C with any inversion radius.The inversion image : Line EC unchange ; arc BC becomes a line through B parrallel to EC ; arc AB appears unchange (becomes a circle with diameter B'A'); Circle F becomes a circle tangent to the parrallel lines and touching the circle AB externally. Note that the image of circle F has a diameter equals to B'C' B'G' is the external tangent of the two circles. Hence, B'G'^2 = (A'B' * B'C').[This is from the formula of the length of external tangent]So, B'G' is tangent to a circle through A'C'G'.By the property of inversion, it is a line passing through A and tangent to arc BC at G.Q.E.D.
Reference my proof in Problem 639 where I independently proved that DH = DA, Tr.s HCD and AGD are congruent SASHence < AGD = <HCD = 90 and so the tangent at G passes thro ASumith PeirisMoratuwaSri Lanka