Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 672.
Friday, September 23, 2011
Problem 672: Internally tangent circles, Chord, Tangent, Geometric Mean
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Extend AD to meet circle O again at E
ReplyDeleteCircles O and O' touch each other at A
So points A, O', O are collinear
OA being a diameter(of circle O'),
ODA is a right angle
So OD bisects the chord ADE in circle O
Now a.b = AD.DB = AD.DE = AD^2
Hence AD = sqrt(a.b)
Extend AD to meet circle O at E. Since OA is a diameter of circle O',OD is perpendicular to AD or OD is perpendicular chord AE of circle O. In other words D is the midpt. of AE which means AD*DC = AD*DE = AD^2 (intersecting chords) or a*b = AD^2 or AD is the GM of a & b.
ReplyDeleteAjit : ajitathle@gmail.com
Typo Correction:
ReplyDeleteBD * DC = AD * DE = AD^2 since AD = DE
Ajit
Chinese version
ReplyDeletehttp://imgsrc.baidu.com/forum/pic/item/500fd9f9d72a6059e27a5d452834349b033bba03.jpg
Let AD extended meet circle O at P
ReplyDelete< ODA = 90 so PD = DA = x
Hence easily in circle O,
PD. DA = a.b i.e. x^2 = a.b
Sumith Peiris
Moratuwa
Sri Lanka