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 680.
CD = b-aED is diameter of the small circle (Thales).So by Pythagoras: (2r)² =(b-a)² + c², giving a² + b² + c² = 4r² + 2ab (result 1)Let F be the orthogonal projection of O to AB.AF = ½(a+b).Triangle DEC is similar to triangle DOF, so OF = ½c.Pythagoras in triangle OAF:R² = (½(a+b))² + (½c)², giving a² + b² + c² = 4R² - 2ab (result 2)Adding results 1 and 2 gives:2(a² + b² + c²) = 4r² + 4R²So: a² + b² + c² = 2(r² + R²)QED
http://img24.imageshack.us/img24/8664/problem680.pngLet F and G are projections of O over CE and ABF and G are the midpoints of CE and ABWe have OF^2=r^2-(c/2)^2 and OG^2=R^2-(a+b)^2/4OF^2+OG^2=r^2=r^2-c^2/4+R^2-a^2/4-b^2/4-a.b/2or a^2+b^2+c^2=4R^2-2.a.b (1)Note that a.b= -Power of point C to Outer circle , radius RSo a.b= -(OC^2-R^2)= R^2-r^2Replace this value in equation (1) we will get a^2+b^2+c^2=2(R^2+r^2)Peter Tran
Just want to prove that a*b=R^2-r^2 in a slightly different way. Draw a line thru' A & O meeting the inner circle in P & Q resply. Now AP*AQ=AC*AD or (R-r)*(R+r)=a*b since AD=BC=b. Hence, R^2-r^2 =a*b. And (2r)^2=(b-a)^2+c^2 from rt. angled Tr. ECD. Hence etc.Ajit
Let M be the midpoint of CD (AB also). Let CD = 2d.AM = MD implies a + d = b - d, 2d = b - a.EOD is a diameter.Easy to note 4r^2 = c^2 + 4d^2 = c^2 + (b - a)^2AlsoR^2 - r^2 = AM^2 - CM^2 = (a + d)^2 - d^2 = a^2 + 2ad = ab.Hence2(R^2 + r^2) = 2(R^2 - r^2) + 4r^2 = 2ab + c^2 + (b - a)^2 = a^2 + b^2 + c^2
draw a perpendicular to AB from center intersecting AB at F therefore by theorem AF=BF=a+b/2CF=AF-CA=b-a/2draw OG perpendiculr to EC from center meeting EC at Gtherefore EG=GC=OF=c/2 (OGCF is rectangle)in tri(OCF) by phythagors theorem (Pt)4r sq = c sq + a sq -2ab + b sq .......... 1in tri(OAF) by Pt4R sq = c sq + b sq+ a sq +2ab ...............21 + 2we get the required result
Draw ON perp to AB. ON bisects AB (resp CD)AD = AN + ND = NB + NC = BC = b DE is a diameter.In triangle AED, AO is median.So AE^2 + AD^2 = 2(AO^2 + OE^2)(a^2 + c^2) + b^2 = 2(R^2 + r^2) etc