Fermat’s Point applies to isosceles triangles

geometry

Fermat's Point applies to equilateral triangles.

Recently as I searched isosceles triangles on Wolfram Mathworld, I learnt that the same principle applies to similar isosceles triangles.

Besides the fact that the total distance from the three vertices of the triangle to Fermat's Point is the minimum possible, how is Fermat's Point different with the points formed by the isosceles triangles?
Why is Fermat's point recorded in the Encyclopedia of Triangle Centers as X(13), while the other points formed by the isosceles triangles are omitted?

Best Answer

Because you can vary the heights of the isosceles triangles and get different points, whereas triangle centres must depend solely on the base triangle itself. In the limit of isosceles triangle heights approaching zero, the centroid is obtained.

Related Solutions

[Math] How to show that the point P (Miquel point) is in the circle formed by the centers of the other 4 circles

Hint: We recognize that the circumcircles are hard to work with directly. Instead, let's consider the points $O^*, A^*, B^*, C^*$, which are obtained by expansion from point $P$ by a factor of 2.

We then want to show that these 5 points are concyclic.We focus on showing that $O^*, B^*, P, C^*$ are con cyclic. This follows easily from slight angle chasing. Hence we are done.

Judging from the phrasing of your question, I think you have shown that the 4 circles are concurrent. If not, here is an approach:

Let $\Gamma$ be the circumcircle of $ABC$ and $\Gamma_A$ be the circumcircle of $AB'C'$ and the rest defined similarly. Show that $\Gamma, \Gamma_A, \Gamma_B$ are concurrent by slight angle chasing. Hence, the 4 circles are concurrent.

[Math] Maximal area covered by two triangles in unit circle

I whipped up a quick Matlab program that computes area stochastically (generate 20000 points; count how many are inside either triangle), and then ran it on 10,000 randomly generated triangle pairs (i.e., 6 random points on the unit circle; the first 3 define one triangle; the next three define the other. WLOG, I made the first angle be 0). Each time a new "largest area" was found, I printed out the triangles. It instantly found an area of ~1.25; then steadily improved this to 1.87 over the next couple of minutes. A previous instance of the program reached 1.97 or so...and this one just jumped to ~1.92, with a solution that's very nearly the two 45-90-45 triangles erected on a diameter.

That suggests to me that the correct answer is indeed this pair of triangles, with area 2 as the optimum.

>> testCircles
area = 1.2587; angles are [0 4.13265878886387 1.69578766384379 2.78030964099446 4.4988452659303 2.81878425776269]
area = 1.4833; angles are [0 1.03997694447012 4.97128628805005 5.53558413650189 4.31034051247746 2.03741390656017]
area = 1.5065; angles are [0 2.583714405482 3.95028117360417 3.41185810372066 4.81492678793123 1.47312279640955]
area = 1.8173; angles are [0 4.99055083947817 1.93427156493409 1.86260140259647 3.04043013930809 5.0276959720494]
area = 1.8517; angles are [0 3.01875721392096 1.30210531664363 4.12564514448203 2.38462146306761 5.8391459560257]
area = 1.87; angles are [0 1.78060248058588 4.66075033844476 2.08372315689924 5.4697587833186 3.76689571284346]
area = 1.9184; angles are [0 4.43108946197308 1.4092884007527 4.31113371053254 2.84429076671212 1.32017294630991]

Here's a picture of the most recent success: a near-optimal pair of triangles

The associated angles/area are

area = 1.9428; angles are [0 1.51926494165392 3.23093144636552 0.0773804130776986 3.34779844425289 4.92681116635565]

Best Answer

Related Solutions

[Math] How to show that the point P (Miquel point) is in the circle formed by the centers of the other 4 circles

[Math] Maximal area covered by two triangles in unit circle

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