Talk:Fermat point

Incomplete Proof?[edit]

As I was reading through the proof, I found it to be (to the best of my knowledge) incomplete. Here is why: "Because AR=AB, AC=AQ, by construction, and because angle RAC = angle BAQ, therefore triangle RAC and BAQ are congruent." Is it assumed that RC = BQ? If not, I do not see any proof that <RAC = <BAQ.

Now, I am no math wiz. If I am missing something here, please forgive my ignorance.

Angle RAC = angle RAB + angle BAC and angle BAQ = angle BAC + angle CAQ. Both angle RAB and angle CAQ are 60 degrees, because the triangles RAB and CAQ are equilateral, and hence <RAC = <BAQ. -- Jitse Niesen (talk) 02:17, 19 March 2006 (UTC)[reply]

Ah, yes, that makes sense. Thanks for the quick response! I have updated the main page accordingly.
Wolever 03:37, 19 March 2006 (UTC)[reply]

Incorrect Property?[edit]

I think the final property is wrong. The circumcenter of the Napoleon triangle is the centroid, not the Fermat point, of the original triangle .... MRFS 19:39, 11 January 2007 (UTC)[reply]

I have no idea, and I don't feel like thinking about it. So, I'm just playing it safe and I'll move the statement here:

"The triangle formed by joining the centers of the three regular triangles in the construction is also a regular triangle(Napoleon's theorem), and the circumcenter of this triangle is the fermat point of the original triangle."

Hopefully, somebody will find a reference for it, or for MRFS's statement. -- Jitse Niesen (talk) 13:59, 9 April 2007 (UTC)[reply]

See http://mathworld.wolfram.com/OuterNapoleonTriangle.html. All web statements need to be treated with caution, but if you take ABC to be an isosceles triangle with base angles of 30 degrees then its Fermat point is A and it is quite obvious that A isn't the centroid of the Napoleon! MRFS 18:11, 15 April 2007 (UTC)[reply]

Thanks, that's a nice example! I'm happy to have the corrected statement back in the article, but I'm not totally sure it belongs in the article now that the center of Napoleon's triangle does not coincide with the Fermat point. So, I leave it up to you. -- Jitse Niesen (talk) 05:34, 16 April 2007 (UTC)[reply]

You're quite right - the statement doesn't belong here so I think it is best omitted! MRFS 19:02, 18 April 2007 (UTC)[reply]

BEC??[edit]

What the hell is BEC? E is never mentioned...

I can only assume that the diagram used to be labeled sympathetically with the first proof and got changed for some reason. Here is the "proof" in my own words using that diagram. Let F be an arbitrary point, and rotate the triangle BFC sixty degrees clockwise about B to form a congruent triangle BGD. By the details of the construction, BCD and BFG are both equilateral triangles, and D's location is independent of our initial choice for F. Due to all the congruences lying around the place, it follows that BF = FG and CF = GD, and therefore, AF + BF + CF = AF + FG + GD. So if we wish to find a point F that minimizes the former sum, then our problem is reduced to that of the latter sum. The distance AF + FG + GD obviously cannot be less than that of AD. Obviously, the minimum will occur when we choose F on AD.

I place that last sentence in italics because it is not immediately true. If you consider F to be any point on AD other than the Fermat point, then G will not lie on AD and the total distance may be greater than for a point F near the Fermat point but not on AD. What does follow from the above is that if there is a point F on AD such that G is also on AD then that F is the unique minimum point (and, therefore, the unique minimum point is on AD). I'm not saying that I don't believe that; merely that it requires more work (and perhaps more advanced techniques than first-semester plane geometry) to be properly demonstrated. I'd rather see the proof fixed than removed as it is particularly accessible, but I'm not certain as to the most straightforward way to do that. MatthewDaly (talk) 22:41, 19 June 2008 (UTC)[reply]

Perhaps similarly, in Case 2 of the geometric proof, is the point F meant to have been constructed identically to in Case 1? I.e. by rotating AB by 60 degrees? I guess part of the confusion is that in the diagram for Case 2, ABP appears more or less equilateral, so it is unclear whether F is meant to be an image of B under rotation or an image of P under reflection, or maybe something else, who knows? — Preceding unsigned comment added by 128.250.5.248 (talk) 03:16, 18 February 2014 (UTC)[reply]

Suggestion[edit]

Unfortunately a lack of proper definition in this area is causing massive confusion. The Article defines the Fermat point as "the solution to the problem of finding a point F inside a triangle ABC such that the total distance from the three vertices to point F is the minimum possible." That seems reasonable as it corresponds to the question that Fermat originally asked. Similarly the first isogonic centre (aka X13) merits its name because it is defined as the point P where the angles BPC, CPA, APB are all equal (to 120^o). As far as I can see these are the two most common definitions in use today.

The big problem is that there is a widespread belief (perpetuated here) that the Fermat point and X13 are one and the same. THIS IS QUITE UNTRUE. In fact they coincide if and only if no angle of the triangle exceeds 120^o. All the machinery for correcting this mistake is already in the Article but it needs to be pieced together in a slightly different way.

As well as demonstrating that the 3 lines in Construction(1) are concurrent Proof(3) also shows (implicitly) that all their angles of intersection are 120^o. Therefore the Construction(1) actually yields X13, NOT the Fermat point.

Provided no angle of the triangle exceeds 120^o then X13 will be inside the triangle (or on its perimeter). Turning to the diagram immediately beside Proof(3) and locating F at X13 then F will lie on AD and G will be forced onto AD because the triangle BFG is regular. With F and G both on AD it follows that AF+BF+CF=AF+FG+GD=AD and since AF+FG+GD is at least AD this means F (ie X13) is also the Fermat point of the triangle. However if an angle of the triangle does exceed 120^o then the argument fails because X13 is outside the triangle. In that case, as is well known, the Fermat point lies at the obtuse angled vertex. MRFS (talk) 11:48, 29 July 2008 (UTC)[reply]

¶Under construction the article clearly states "The Fermat point has a near-identical twin called the first isogonic center or X(13) and it is important not to confuse the two." The problem you identify seems to be fixed. In clearly distinguishing the cases of the Fermat point, one can see that the Fermat point and the first isogonic center are the same only in a case when there is no angle equal to or greater than 120 degrees. Additionally, it is clearly stated that in the case in which an angle in the triangle is equal to or greater than 120 degrees, the Fermat point must lie on the obtuse angled vertex as it logically cannot lie outside of the triangle. Perhaps this proof can be elaborated on as to why the Fermat point must lay on this point even when a simple construction of the point would point to it lying outside of the triangle. — Preceding unsigned comment added by 128.148.231.12 (talk) 14:50, 18 July 2013 (UTC) 138.16.64.23 (talk) 14:59, 18 July 2013 (UTC)AR[reply]

First variation[edit]

The proof that the constructed point is the Fermat point is known in differential geometry as a "first variation argument". This establishes that all angles must be 2 $π$ /3. It may be worth noting this or linking it to the appropriate page. Tkuvho (talk) 08:08, 19 July 2013 (UTC)[reply]

Too many proofs?[edit]

Frankly this article is an awful mess ....

The geometric proof in §2 is the only one that deals satisfactorily with Case 1.
§3.4 doesn't belong here. It shows that the 3 red lines in Fig 1. are concurrent but it doesn't prove that they cut one another at 60° and it doesn't even mention the Fermat point.
§3.1 should be retained because it shows that the 3 red lines cut at 60°, in other words at X13.
§3.2 probably deserves no more than a passing mention. What it doesn't say is that it deals purely with Case 2, some of its signs are wrong, and when it talks about "some algebraic manipulations" it really means a minimum of several more pages of very detailed work.
§3.3 is a lovely proof but very long-winded and it only considers Case 2. Here's a much shorter version that covers Case 1 as well.

Let O, A, B, C, X be any five points in a plane. Denote the vectors OA, OB, OC, OX by u, v, w, x respectively, and let i, j, k be the unit vectors from O along u, v, w.
Now |u| = u⋅i = (u − x)⋅i + x⋅i ≤ |u − x| + x⋅i and similarly |v| ≤ |v − x| + x⋅j and |w| ≤ |w − x| + x⋅k.
Adding gives |u| + |v| + |w| ≤ |u − x| + |v − x| + |w − x| + x⋅(i + j + k).
If u, v, w meet at O at angles of 120° then i + j + k = 0 so |u| + |v| + |w| ≤ |u − x| + |v − x| + |w − x| for all x.
In other words XA + XB + XC ≥ OA + OB + OC and hence O is the Fermat point of ABC.
This argument fails when the triangle has an angle Ĉ > 120° because there is no point O where u, v, w meet at angles of 120°. Nevertheless it is easily fixed by redefining k = − (i + j) and placing O at C so that w = 0. Note that |k| ≤ 1 because the angle between the unit vectors i and j is Ĉ which exceeds 120°. Since |0| ≤ |0 − x| + x⋅k the third inequality still holds, the other two inequalities are unchanged. The proof now continues as above (adding the three inequalities and using i + j + k = 0) to reach the same conclusion that O (or in this case C) must be the Fermat point of ABC.
Does anyone disagree? MRFS (talk) 09:55, 21 September 2013 (UTC)[reply]

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