Geometry Problem

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

Click the figure below to see the complete problem 636.

## Monday, July 18, 2011

### Problem 636: Semicircle, Diameter, Perpendicular, Inscribed Circle, Common Tangent

Labels:
circle,
common tangent,
diameter,
perpendicular,
semicircle

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http://img28.imageshack.us/img28/7780/problem636.png

ReplyDeleteLet 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 collinear

Note that angle ANB =90 and quadrilateral MNBC is cyclic

So AM.AN =AC.AB

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

Peter Tran

Let EC touch circle (F;x) at X

ReplyDeleteLet GM cut EC at Y. clearly YX=YG=YC .

So Y is the midpoint of XC.

Let angle XYF = α

t=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.

ReplyDeleteCall 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:R

YD = R*(R+r)/(R-r)

Substituting r = R-R² gives

YD = R*(2R-R²)/(R²)= 2-R

So YD = AD

Y = A

QED.

Type error:Please read last sentence of my solution as "A,Y,G are collinear"

ReplyDeleteTo Peter Tran,

ReplyDelete"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 .

DeleteGM 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.

DeleteAM.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 Anonymous

DeleteNo G1 or G2 is required.

Power of A to circle F= AM.AN ( A,M, N collinear)

Power of A to circle O= AC.AB

Since MNBC is cyclic so AM.AM=AC.AB => Power of A to circle F=Power of A to circle O

So 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/

DeleteAccording to your "Power theorem", length AG1=AG2=AG3, I can understand.

But, see my image.

Apply inversion at C with any inversion radius.

ReplyDeleteThe 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 SAS

ReplyDeleteHence < AGD = <HCD = 90 and so the tangent at G passes thro A

Sumith Peiris

Moratuwa

Sri Lanka