-----Original Message----- From: email@example.com [mailto:firstname.lastname@example.org] On Behalf Of Zbigniew Karno Sent: Wednesday, September 12, 2007 4:42 PM To: email@example.com; firstname.lastname@example.org Subject: Re: Obvious but hard to prove?
On 10 Wrz, 07:57, Steve Gray <stev...@roadrunner.com> wrote: >> Place three points A,B,C in the plane, not in a line. Put another >> point P strictly inside the triangle. Draw all three circles P,A,B; >> P,B,C; and P,C,A. Show that the three circles have no interior points >> in common, that is, their intersection is null (excluding the >> circumferences). I'd like a proof but don't have one.
to which we received the following reply
>Lat ABC be a triangle in the plane and let P be a point inside >ABC. Let us denote by S_a, S_b and S_c circles passing by B,P,C, >C,P,A and A,P,B, respectively. Denote by K_a, K_b and K_c circular >discs with the boundary circles S_a, S_b and S_c, respectively. >Let I_a, I_b and I_c be the centers of the circles S_a, S_b and S_c.
>To prove the above claim it is required to show that
> P = K_a \cap K_b \cap K_c.
>Consider three points D, E and F on circles S_a, S_b and S_c, >respectively, opposite to the point P with respect to I_a, I_b and >I_c, respectively. Thus PD, PE and PF are diameters of the circles >S_a, S_b and S_c, respectively.
>Now consider the triangle DEF. Observe that points A, B and C >belong to the sides EF, FD and DE, respectively, of the triangle >DEF, because the sides I_aI_b, I_bI_c and I_cI_a of I_aI_bI_c are >midlines of the intervals PC, PA and PB, respectively, and hence >the triangle DEF is similar to the triangle I_aI_bI_c. > Then ABC is contained in DEF and therefore the point P is inside >DEF. It follows that DEF is the union of three triangles DPE, EPF >and FPD, and furthermore PD = DPE \cap FPD, PE = DPE \cap EPF >and PF = FPD \cap EPF. Hence P = DPE \cap EPF \cap FPD.
>To complete observe that the lenses K_a \cap K_b, K_b \cap K_c and >K_c \cap K_a are entirely contained in the triangles DPE, EPF and >FPD, respectively, because ...... <clip>
I don?t think this last statement is correct. You are saying: Given two circles PCD, PCE then the lens like common part of the circles is entirely within the triangle PDE. Well, suppose you take two circles (say equal) very nearly one on top of the other. Then the centres of these circles are well within their common part. Thus PD (which contains the centre of one of the circles) IS NOT outside the lens. (Draw a figure to convince your self.)
Here, however, is a simple proof of the original problem:
Consider first the circles BPC, CPA. If the third circle APB has common points with the lens like common part of the two above circles, then some point of the circumference of circle APB will cut arc PC somewhere in between P and C. Call this point Q. Note then that circles APB and BPC contain Q. In other words these circles have three common points, B, P, Q. But a theorem in Euclid?s Elements (which is easy to prove) states that two circles with three common points coincide. But this is absurd because circles APB, BPC do not coinside (one contais the point A but not the other). QED.
All the best, Michael Lambrou.
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