Peter Gyarmati

Posted on Monday, 24 February, 2003 - 07:45 am:

The points E and F lie on the extension of the side AB of a rhombus ABCD. The tangents drawn from E and F to the inscribed circle of the rhombus intersect the line AD at the points E' and F'. Determine the ratio DE':DF', given that BE:BF.

David Loeffler

Posted on Monday, 24 February, 2003 - 09:13 am:

I would be strongly tempted to guess that DE'/DF' = BF/BE. The reason I say this is that the mapping E |-> E' looks like it ought to be a projective involution. It thus preserves cross-ratios, and since it maps B to infinity and infinity to D, the result follows.
I'll get back to you when I have checked the definitions.

David

David Loeffler

Posted on Monday, 24 February, 2003 - 12:43 pm:

I think that we have BE.DE'=BF.DF'=

r^{2} \sec^{2} θ

where

θ =

angle OAB. This implies the result I gave above.

I need to patch a few gaps in the proof though.

David

David Loeffler

Posted on Monday, 24 February, 2003 - 02:12 pm:

Yes. I have the proof. Let G be the foot of the perpendicular from O to AB and consider triangle AOG. Then AO

= r cosec θ

. Hence AB

= r cosec θ \sec θ

Consider triangle AEE'. Let angles $ρ$ , $σ = 1 / 2$ AEE', 1/2 AE'E respectively. Then $2 ρ$ , $2 σ$ , $2 θ$ are the angles of triangle AEE' and consequently we obtain $ρ + σ + θ = π / 2$ . Thus if $u$ , $v$ , $w$ are the tangents of $ρ$ , $σ$ , $θ$ respectively, $u v + v w + w u = 1$ .

Now we have AG $= r \cos \thtea$ and GE $= r \cot ρ$ . Thus BE=AE-AB= $r \cos θ + r \cos ρ - r \sec θ cosec θ$ . It is easily verified that $\sec θ cosec θ - \cot θ = \tan θ$ .

Hence we obtain BE $= r (\cot ρ - \tan θ) = r (1 / u + w)$ . Similarly DE' $= r (1 / v + w)$ .

Hence BE.DE'=

$r^{2} (\frac{1}{u} + w) (\frac{1}{v} + w) = r^{2} \frac{(1 - u w) (1 - v w)}{u v} = r^{2} \frac{1 - u w - v w + u v w^{2}}{u v} = r^{2} \frac{u v + u v w^{2}}{u v} = r^{2} (1 + w^{2}) = r^{2} \sec^{2} θ$

as required.

Can anyone supply a more insightful proof? I am sure that there is some neat composition of perspectivities which gives the same result.

David

Peter Gyarmati

Posted on Monday, 24 February, 2003 - 08:22 pm:

Thanks David. All is clear.

Peter

David Loeffler

Posted on Tuesday, 25 February, 2003 - 02:40 pm:

So you spotted the two typos, then.

Peter Gyarmati

Posted on Thursday, 27 February, 2003 - 02:36 pm:

Yes I spotted them:

r² *(1+vw+uw+uvw² )/(uv)=r² *(1+(1-uv)+uvw² )/(uv)=r² *(2-uv+uvw² )/(uv)=r² *((2)/(uv)+(w² -1))

Here it is obvious that we have to eliminate u and v. I don't see yet how to, however if we don't solve this, the proof is not finished yet.

Peter

David Loeffler

Posted on Thursday, 27 February, 2003 - 04:23 pm:

The mistake is in the transcription from my original (hopefully correct) working. It should be BE = r( 1/u - w), DE' = r( 1/v - w). This does simplify as I said, indeed it is only these two formulae and the first line of the next that are wrong.

David

Peter Gyarmati

Posted on Thursday, 27 February, 2003 - 05:34 pm:

I don't know how couldn't I see the mistake in the formulae of BE and DE'.
Thanks for the quick answer.

Peter