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April 13

Euler line forms a Fixed Angle with a Given Side

We are given a fixed triangle side BC, and asked to find the locus described by a third point A, so that the Euler line of the resulting triangle ABC forms a given angle with side BC. I have solved the problem for 0° and 90° using GeoGebra, but am unable to solve for the general case. (For 90°, the locus of A is -of course- the perpendicular bisector of segment BC, and for 0° we have an ellipse whose minor axis is the segment BC, and whose major axis is BC√3 ). — 79.113.236.232 (talk) 19:50, 13 April 2017 (UTC)[reply]

Here's an idea that will certainly work: coordinatize in your favorite coordinate system, so that B and C are fixed and A has variable coordinates. Choose two points on the Euler line, say the centroid and cicumcenter; give their coordinates. This determines the equation of the Euler line, hence its slope. Solving for a fixed value of the slope gives an implicit equation for the coordinates of A. The solution set is your locus. It will always be some conic section. --JBL (talk) 20:41, 13 April 2017 (UTC)[reply]

Euler line#Slope gives the slope of the Euler line in terms of the slopes of the sides as

m_{E}=-{\frac {m_{1}m_{2}+m_{1}m_{3}+m_{2}m_{3}+3}{m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}}}.

Set the slope of side BC equal to 0, and set the slope of the Euler line at a given value (the tangent of the required angle). This gives one equation in the two unknown remaining slopes, so one of the remaining slopes can be solved for parametric on the other one. Loraof (talk) 21:53, 13 April 2017 (UTC)[reply]

The questioner says that for an angle of 0°, we have an ellipse whose minor axis is the segment BC, and whose major axis is BC√3 . I would note that this ellipse must be punctured at the two endpoints of the major axis. If B=(–1, 0) and C=(1, 0), so BC=2, the upper end of the (vertical) major axis is at A= (0, ${\sqrt {3}}$ ) (since from the origin to A is half the major axis, the latter being given by the questioner as $2{\sqrt {3}}$ ). Then AB=AC=2, and we have an equilateral triangle, which does not have an Euler line since all the points that would be on the Euler line coincide with each other. Loraof (talk) 17:51, 14 April 2017 (UTC)[reply]

Quite right. I saw no point in adding extra clutter. — 79.113.215.18 (talk) 05:43, 15 April 2017 (UTC)[reply]

For an angle α below 60° I get an ellipse whose center coincides with that of the segment BC, and whose minor axis forms an angle of α⁄2 with the same segment. However, I am unable to obtain a general formula for the lengths of its two axes. For α above 60°, the locus becomes a hyperbola, and for α = 60°, we have two parallel lines. — 79.113.215.18 (talk) 05:43, 15 April 2017 (UTC)[reply]

@79.113.215.18:, @79.113.236.232: Maybe this will work. The equilateral triangle with horizontal base BC of length 1 has m_2=0, m_1 =sqrt(3), and m_3= –sqrt(3). By the slope formula I gave above, the Euler slope is given as 0/0. I'm guessing that this means the equilateral triangle is a limiting case of any of your ellipse points as the x coordinate goes to the midpoint of the side BC. If that speculation is correct, that gives us two points on the ellipse regardless of the required Euler slope (the points above and below the BC axis giving an equilateral triangle). Next, consider a triangle with point A very near the BC axis. In the slope formula, m_2=0 and m_3 approx.=0, so by the formula m_1 = –3/m_E which is strictly positive. So as A goes to the BC axis it cannot be approaching any point other than B or C (lest the slope m_1=0). So by symmetry, the ellipse (regardless of the Euler slope) must cross the BC axis at B and at C.

Now we have four points on any ellipse: B, C, and the two points giving an equilateral triangle. Is that what you found with GeoGebra? If so, these four pieces of information along with the line of the minor axis (slope alpha/2 and going through the midpoint of BC) should be enough to let you specify the ellipse. Loraof (talk) 17:17, 19 April 2017 (UTC)[reply]

Thanks yo your observations, I've finally managed to find a fifth point, namely the ellipse's intersection with the vertical axis at (0, 4m_E). — 86.121.233.7 (talk) 00:26, 20 April 2017 (UTC)[reply]

Resolved

April 16

Poncelet's closure theorem

Illustration of Poncelet's porism for n = 3, a triangle that is inscribed in one circle and circumscribes another.

By the n=3 case of Poncelet's closure theorem for circles, if the equation of Euler's theorem in geometry is satisfied, relating the inradius, circumradius, and distance between the incenter and circumcenter, then for given incircle and circumcircle, every point A on the circumcircle is a vertex of a triangle ABC having those circles. As we walk vertex A completely around the circumcircle from a given starting point X, dragging B and C along (with slippage) so as to maintain the given incircle, the original triangle occurs a total of three times: when A=X, when B=X, and when C=X.

For each of the three portions of this walkaround, what is the nature of the closed path traced out by (1) the triangle's centroid, and (2) its orthocenter? Does it trace a locus of a particular named type? Does the locus enclose a convex set? Loraof (talk) 01:22, 16 April 2017 (UTC)[reply]

The question can be visualized by this figure from the article, in which unfortunately the animation is too fast. I'm interested especially in the locus of centroids. Loraof (talk) 14:21, 17 April 2017 (UTC)[reply]

You can see a slowed-down version of the animation here - just paste in the URL of the animated gif. AndrewWTaylor (talk) 16:33, 17 April 2017 (UTC)[reply]

This came up in Richard Schwartz's Einstein Lecture at the AMS sectional meeting at Indiana U a couple of weeks ago. He has a java applet on his webpage. I am moderately confident that he asserted that the locus of the centroid is a circle. (As with many things in Euclidean geometry, this shouldn't be particularly difficult if you just throw down a coordinate system.) --JBL (talk) 19:42, 17 April 2017 (UTC)[reply]

Thanks to both of you! Loraof (talk) 14:42, 18 April 2017 (UTC)[reply]

Having gone through the calculations, out of curiosity: if we take R = 1 and the distance between the two centers to be d, then Euler's theorem gives r = (1 - d^2)/2. The locus of the centroid is a circle whose center is between the centers of the incircle and circumcircle, at distance 2d/3 from the circumcenter; it has radius d^2/3. I did not try to work out the details for the orthocenter. -JBL (talk) 20:31, 19 April 2017 (UTC)[reply]

Thanks, JBL—this is exactly what I wanted. I tried to work this out with Cartesian coordinates, but I became hopelessly lost in an algebraic morass. Could you give me a sketch of your derivation? Loraof (talk) 21:40, 19 April 2017 (UTC)[reply]

Nothing interesting or helpful, I'm afraid -- I just bashed through the calculations with the aid of a computer algebra system. Of course I've cleverly closed the file without saving anything, but if you like I can try to run through it again tomorrow and post the salient bits. (Or e-mail it.) --JBL (talk) 22:39, 19 April 2017 (UTC)[reply]

Ternary and n-terms ratios equivalent representations

How many binary ratios are needed for an equivalent representation of a ternary ratio of numbers a:b:c? What is the general case for n-term ratios a:b:c:d:.:....:q? (Thanks.)--82.137.11.181 (talk) 13:47, 16 April 2017 (UTC)[reply]

n–1 are needed. E.g., for a:b:c, knowing a:b and b:c gives you a:b:c. In general, if you know all binary ratios of adjacent items, you can string together the whole n-item ratio. Loraof (talk) 16:39, 16 April 2017 (UTC)[reply]

And the necessary number of ratios cannot be less than n–1 because those n–1 sequential binary ratios are all independent pieces of information, and you need to have at least as many binary ratios as there are independent pieces of information to convey. Loraof (talk) 20:22, 16 April 2017 (UTC)[reply]

Hypercube/hypersphere volume question.

For number of dimensions n, take the n-hypercube of edge 2 units centered at the origin. Circumscribe the n-hypersphere around it. Divide the hypersphere's volume into n parts (Vn,V(n-1), ... V0) based on how many of coordinates of each point have absolute value<1. Is Vn (the n-hypercube) always the largest volume of the parts for each number of dimensions? For example if n=4, does the hypercube have more hypervolume than the points where only 3 of the 4 coordinates have absolute value<1? Naraht (talk) 18:03, 16 April 2017 (UTC)[reply]

An interesting question. I cannot answer it, but here is a formulation that could be helpful for an answer based on calculus. Define $V_{k,n}(r)$ the hypervolume of the part of the unit-radius hypersphere in n dimensions that verifies that at least k coordinates are smaller than r. Then the original question is whether the inequality $V_{k+1,n}(r)-V_{k,n}(r)\leq V_{n,n}(r)-V_{n-1,n}(r)$ holds for all k and n if $r={\sqrt {n}}/2$ (the side of the hypercube that fits into a sphere of radius 1). I imagine $V_{k,n}(r)$ could have an explicit formulation. Tigraan^{Click here to contact me} 18:32, 16 April 2017 (UTC)[reply]

The half-edge of the hypercube is 1, so the hypersphere's radius (which equals the distance from the hypercube's center to a vertex) is ${\sqrt {n}}.$ By n-sphere#Closed forms, when n is even the volume of a hypersphere with unit radius is ${\frac {\pi ^{n/2}}{(n/2)!}},$ so the volume of a hypersphere with radius ${\sqrt {n}}$ is the unit volume times the radius raised to the power n: ${\frac {(n\pi )^{n/2}}{(n/2)!}}.$ The volume of the hypercube with edge 2 is $2^{n}.$ So the fraction of the hypersphere's volume that is within the hypercube is ${\frac {(n/2)!}{(\pi n/4)^{n/2}}}.$ I think this declines sharply as n rises. For example, for n=12, this fraction is about .001. Since you have divided the hypersphere volume into only 13 parts, the hypercube Vn cannot be the largest. Loraof (talk) 23:54, 16 April 2017 (UTC)[reply]

That sharp decrease as n rises is sometimes called the curse of dimensionality. 50.0.136.56 (talk) 03:42, 18 April 2017 (UTC)[reply]

The example on that page is for the reverse, sphere inscribed in a cube, but I agree that a similar example could be set up for this.Naraht (talk) 15:15, 18 April 2017 (UTC)[reply]

Thanks for that link. And in fact, the section of that article Curse of dimensionality#Distance functions deals with the case of a hypersphere inscribed in a hypercube, and again the ratio of their volumes goes to 0 as n goes to infinity. Loraof (talk) 15:15, 18 April 2017 (UTC)[reply]

April 18

Bayes' Theorem

The wiki article about it has an illustration with a decision tree that I couldn't understand. Is the following an ok way to think of it and explain it?

Think of a big square with area 1 representing a probability space. Draw two filled circles A and B (maybe unequal sizes) inside the square for the events A and B, so they partly overlap like a Venn diagram. Their areas are the probabilities Pr[A] and Pr[B]. Let C be the intersection of A and B.
Pr[A|B] just means you've picked a point in B and now you want to know the likelihood that it's also in A. That is, it's in the intersection C. So this likelihood is Pr[C]/Pr[B]. Similarly Pr[B|A] is Pr[C]/Pr[A]. Rearranging terms you get Pr[A|B]=Pr[B|A]*Pr[A]/Pr[B] which is Bayes' Theorem.

If yes, is this somehow a wonderful or amazing fact? Why is it usually written as that equation, instead of thinking of intersecting events like the circles above? Thanks! 50.0.136.56 (talk) 03:52, 18 April 2017 (UTC)[reply]

Let's add a link to the article: Bayes' theorem. StuRat (talk) 04:32, 18 April 2017 (UTC)[reply]

Thanks. There's also a separate article Bayes' rule whose purpose I couldn't discern. 50.0.136.56 (talk) 04:37, 18 April 2017 (UTC)[reply]

That's just a redirect to the same article. --76.71.6.254 (talk) 10:53, 18 April 2017 (UTC)[reply]

That was not a redirect until today's edit by Trovatore (special:diff/775983852). --CiaPan (talk) 12:44, 18 April 2017 (UTC)[reply]

It's possible I jumped the gun a little on that one, but I figured it was easy to undo if someone didn't like it. Looking on the talk page, apparently there's some concern that the rule might be useful even in situations where the hypotheses of the theorem are not strictly met. It seems to me that that clarification could be made within a single article, but it may be that it's content that ought to be merged. --Trovatore (talk) 18:12, 18 April 2017 (UTC)[reply]

Oh. You might've mentioned it here, since it was being discussed. Oh well, not important. --76.71.6.254 (talk) —Preceding undated comment added 18:44, 18 April 2017 (UTC)[reply]

April 19