Re: Geometry challenge
- From: bill <b92057@xxxxxxxxx>
- Date: Mon, 15 Oct 2007 22:03:23 -0700
On Oct 15, 5:39 am, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
bill wrote :
On Oct 11, 6:02 pm, TefJlives <gmarkow...@xxxxxxxxx> wrote:
Anyone who is interested in doing a Euclidean geometry problem, please
follow the link below, and let me know what you think.
http://www.umcs.maine.edu/~markov/GregMarkowskyShoemakerKnife.pdf
Greg
adc", "aeb", "bfc", & "gdh" are similar right
triangles!
I suspect the 4 triangles with "k" and "l" as
apexes are also similar to the four larger
triangles?
Bill J
You are right. All these triangles are similar...
We have first to proove what is just asserted in the statement :
that is gh parallel to ac /hence/ d,g,a in line and d,h,c in line.
Considering a dilation from center d, which transforms circle (gh) into
circle (ac), because these circles are tangent in d.
Hence diameter gh is transformed into diameter ac (any diameter is
transformed into a parallel diameter). Hence d,g,a in line.
But as this doesn't involve the circle with diameter gh is tangent to
/all three/ other circles, this can't be suficient to proove the
wanted properties !
Given the three circles (ab) (bc) (ac), you can choose any point d on
(ac) and any circle tangent in d to (ac), even not tangent to the two
others, this results into all the 10 right triangles in the figure
being similar !
But there is only one position of point d and one radius for the
circle (gh) where properties 1 and 2 (and the unknown property 3
corrected) are true, and all circles are tangent.
You must involve in your proof the tangency property of the circles,
for instance using points m and n, contact points of circle (gh) with
circles (ab) and (bc), points which are not drawn on the given figure.
Note that b,m,g in line, as well as b,n,h.
I have a proof for property 2, but which I feel is too long :
Property 2 is an interesting case. We know that "bedf" is a
rectangle.
If "bedf" is a square, then "be" = "bf". It is possible to
simplify the
diagram as follows:
Start with the two adjacent semicircles "aeb" and "bfc"
Draw line "ae" and extend beyond "e" indefinitely.
Draw line '"eb".
Draw line "bf" perpendicular to "eb"
Draw line '"cf" and extend to meet line "ae" at "d"
Now you have rectangle "bedf" without using semicircle "adc" and
the inscribed circle.
e
|
|
a---------- ---b-------c
Consider triangle "aec" aec is a right triangle.
Then ae / be = be / cf or be * be = ae * cf
Triangles "aeb" and bfc" are similar triangles, where ae / bf = be /
cf. Then,
be * bf = ae * cf = be * be.
This is not exactly a 'formal' proof, but I hope you understand what I
am
trying to say?
Bill J
Consider the inversion with pole b which exchanges points a and c.
This [details omitted] results into point d on the circle from which
we see segment bc under a 45° angle.
Hence angle bdf is 45° hence dfbe is a square.
[OT]
Very disturbing to name points in little letters. Usually points are
named in capital letters, this leaves the little letters vor values,
for instance an ABC triangle, with sides a,b,c.
Regards.
--
Philippe C., mail : chephip+n...@xxxxxxx
site :http://chephip.free.fr/ (recreational mathematics)
.
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