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'= r² sec²q where q = 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 cosecq. Hence AB =r cosecqsecq.

Consider triangle AEE'. Let angles r, s = 1/2 AEE', 1/2 AE'E respectively. Then 2r, 2s, 2q are the angles of triangle AEE' and consequently we obtain r+s+q = p/2. Thus if u, v, w are the tangents of r, s, q respectively, u v+v w+w u=1.

Now we have AG =r cos\thtea and GE =r cotr. Thus BE=AE-AB= r cosq+r cosr-r secqcosecq. It is easily verified that sec qcosecq-cotq = tanq.

Hence we obtain BE =r(cotr-tanq)=r(1/u+w). Similarly DE' =r(1/v+w).

Hence BE.DE'=

r² æ
ç
è 1
u
+w ö
÷
ø æ
ç
è 1
v
+w ö
÷
ø =r² (1-u w)(1-v w)
u v
=r² 1-u w-v w+u v w²
u v
=r² u v+u v w²
u v
=r²(1+w²)=r² sec²q

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