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

Labels:
chord,
circle,
geometric mean,
metric relations,
tangent

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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