Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 680.

## Sunday, October 23, 2011

### Problem 680: Concentric Circles, Radii, Chords, Perpendicular, Metric Relations

Labels:
chord,
concentric circles,
metric relations,
perpendicular,
radius

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CD = b-a

ReplyDeleteED 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 + 4R²

So: a² + b² + c² = 2(r² + R²)

QED

Nice proof

Deletehttp://img24.imageshack.us/img24/8664/problem680.png

ReplyDeleteLet F and G are projections of O over CE and AB

F and G are the midpoints of CE and AB

We have OF^2=r^2-(c/2)^2 and OG^2=R^2-(a+b)^2/4

OF^2+OG^2=r^2=r^2-c^2/4+R^2-a^2/4-b^2/4-a.b/2

or a^2+b^2+c^2=4R^2-2.a.b (1)

Note that a.b= -Power of point C to Outer circle , radius R

So a.b= -(OC^2-R^2)= R^2-r^2

Replace 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.

ReplyDeleteAjit

Let M be the midpoint of CD (AB also).

ReplyDeleteLet 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)^2

Also

R^2 - r^2 = AM^2 - CM^2

= (a + d)^2 - d^2 = a^2 + 2ad

= ab.

Hence

2(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

ReplyDeletetherefore by theorem AF=BF=a+b/2

CF=AF-CA=b-a/2

draw OG perpendiculr to EC from center meeting EC at G

therefore EG=GC=OF=c/2 (OGCF is rectangle)

in tri(OCF) by phythagors theorem (Pt)

4r sq = c sq + a sq -2ab + b sq .......... 1

in tri(OAF) by Pt

4R sq = c sq + b sq+ a sq +2ab ...............2

1 + 2

we get the required result

Draw ON perp to AB. ON bisects AB (resp CD)

ReplyDeleteAD = 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