## Monday, July 18, 2011

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

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

Click the figure below to see the complete problem 636.

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

2. Let EC touch circle (F;x) at X
Let 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

3. Without loss of generality I asume that AO = OB = 1.
Call 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
Y = A
QED.

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

5. To Peter Tran,
"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.

1. To Anonymous .

GM 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

2. Hello Tran, thanks to your comment.

AM.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.

3. To Anonymous
No 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

4. See http://imageshack.us/photo/my-images/547/imageqx.png/

According to your "Power theorem", length AG1=AG2=AG3, I can understand.
But, see my image.

6. Apply inversion at C with any inversion radius.
The 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.

7. Reference my proof in Problem 639 where I independently proved that DH = DA, Tr.s HCD and AGD are congruent SAS

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

Sumith Peiris
Moratuwa
Sri Lanka