1) Is every transcendental number also irrational? (I would like a proof if it is so, or not)

2) I tried to make a general formula for finding a center of a circle going through 3 points (not on the same line), when having only the coordinates of the three points. I got a pretty nasty looking formula. Is there any simple formula?

3) To find the centroid (the meeting point of the medians) point of a triangle you can use:
((x₁ +x₂ +x₃ )/3,(y₁ +y₂ +y₃ )/3)
Is there a same for the incenter (the meeting point of the angle bisectors) point or the orthocenter (the meeting point of the heights)?

Thanks for anyhelp,

Yatir

Yatir,

1. Yes, every transcendental number is irrational. To see this, we have the number a/b is a root of the equation bx-a=0.

2. The best strategy for finding the circumcentre would probably be finding the intersection of perpendicular bisectors. I am not convinced that this will give anything "not nasty-looking", whatever the phrase means.

3. I know the incentre result is truly horrible if you use the angle-bisectors. Probably a nicer way is to use vectors and the result would probably look better (whatever that means). The orthocentre result would probably be messier than the circumcentre result. I would suggest a vector method here too. By the way, there necessarily exists formulae for these, just whether the formula is "nice".

Kerwin

About (1), of course somehow it slipped my mind
About (2), this is the way I thought of doing it, and the X coordinate of the center is:

Where (x₁ ,y₁ ), (x₂ ,y₂ ) and (x₃ ,y₃ ) are the 3 points.
This is what I call "Nasty-looking" :-)

And about (3), I guess you're right and the vector method is the easiest.

Thanks,
Yatir

Just to add a note to what you were saying, Kerwin,

The formulae for the orthocentre and circumcentre are both in some sense "nice" in that they can be expressed as ratios of polynomials in x1..x3, y1..y3.

However, this is impossible for the incentre (and a lot of other triangle centres) as they are only expressible if you throw in the side lengths, leading to square root terms in your function and generally increased messiness.

You may find it interesting to note that if the corners are (0,0), (1,0) and (0,1) the coordinates of the incentre are irrational, suggesting the absence of any nice formula.

PS. You may notice a similarity between the formulae for the orthocentre and the circumcentre. This is because they are closely related points - the technical term is that they are "isogonal conjugates" of each other. If a point P can be expressed as a rational polynomial in the x's and y's, so can its conjugate P'.

David

David, I'm not familiar with the term "isogonal conjugates". Would you care exlaining it?

Thanks,
Yatir

Let P, P' be points in the plane of triangle ABC. Then if the angles ABP = P'BC, BCP = P'CA, and CAP = P'AB, P and P' are isogonal conjugates.

It's not obvious from this definition that P' exists at all; I will write more some other time (i.e. not 2.45 AM British time :-).)


David

Yatir, one way of making your fomula looks nicer is to take all 3 expressions for your circumcentre and invoke the result a/b=c/d => a/b=(a+b)/(c+d)=c/d. That should introduce symmetry into your formula.

David, I wasn't sure what Yatir meant by "nasty" when I wrote my last message. All the centres are symmetric functions of the vertices, though they may not be polynomials....

Kerwin

Kerwin, I don't really understand what you mean by: a,b,c and d?
And what 3 expressions are you talking about?


Yatir

Anyway, isogonal conjugates. Essentially, suppose we have a triangle and a point P.

The lines joining P to the three vertices are called the cevians of that point (there is much debate as to how this should be pronounced!). Let D, E, F be the feet of the cevians, as in the diagram.



Now, there is a well-known theorem called Ceva's theorem that states that given three arbitrary cevians AD, BE and CF, the three of them all meet at a point P if and only if
(AF)/(FB) (BD)/(DC) (CE)/(EA) = 1

Now, let's express this in terms of the angles x,y and z as shown on the diagram. We see that

(BD)/(DC)
= (area BAD) / (area DAC) as the two triangles have the same height
= (1/2 AB AD sin x)/(1/2 AD AC sin (A-x))
= (AB sin x)/(AC sin (A-x))

Similarly AF/FB = (AC sin z)/(BC sin (C-z)) and CE/EA = (BC sin y)/(AB sin (A-y))

Hence the Ceva condition is that the lines intersect if and only if

(AB sin x)/(AC sin (A-x)) (AC sin z)/(BC sin (C-z)) (BC sin y)/(AB sin (A-y)) = 1

Now, all the side lengths suddenly cancel from this fraction, and we are left with

sin x sin y sin z = sin(A-x) sin (B-y) sin (C-z)

Now, here is the clever bit :-)

If we replace x, y and z in the above equation with A-x, B-y, C-z, then it clearly remains true.

That is, if you have a set of cevians concurring in a point P, the new set of cevians you get by reflecting each cevian about the bisector of the angle of the original triangle will also concur at a new point P'.



(Sorry for the crowded diagram - can't be helped!)

The point P' is then said to be the isogonal conjugate of the original point P.

(Every point in the plane of a triangle has an isogonal conjugate unless it lies on the circumcircle, in which case the reflected cevians will be parallel and the point P' recedes to infinity).

David

Xi=

æ è

å

fi(xl,xm,xn)

ö ø

/

æ è

å

gi(xl,xm,xn)

ö ø

where the sums are taken over l m n even permutations of 123, x_ns are position vectors of the vertices.

Now we have homogeneity with cyclic symmetry, there may be some way of simplifying the expressions to something nice.

Kerwin

Equation of a Circle, given three points

Problem: Give the general solution to this question:

Find a circle given 3 points (a,b), (c,d), and (e,f)

Your answer should be in the form of an equation of the circle centered at point (h,k) with radius r, and should look like this:

(x-h)² + (y-k)² = r²

Solve for h, k, and r in terms of a, b, c, d, e, and f.

Solution:

Start with these three equations:

(a-h)² + (b-k)² = r²
(c-h)² + (d-k)² = r²
(e-h)² + (f-k)² = r²

Multiply them out, and you get these three equivalent equations:

a² - 2ah + h² + b² - 2bk + k² = r²
c² - 2ch + h² + d² - 2dk + k² = r²
e² - 2eh + h² + f² - 2fk + k² = r²

Multiply the first by (e-c), the second by (a-e), and the third by (c-a):

(a² - 2ah + h² + b² - 2bk + k²)(e-c) = r²(e-c)
(c² - 2ch + h² + d² - 2dk + k²)(a-e) = r²(a-e)
(e² - 2eh + h² + f² - 2fk + k²)(c-a) = r²(c-a)

Multiply out these three equations, then add 'em up. Amazingly, all the squared h, k, and r terms go away, so you can easily solve for h and k

a²e - a²c - 2aeh + 2ach + h²e - h²c + b²e - b²c - 2bek + 2bck + k²e - k²c = r²e - r²c
c²a - c²e - 2cah + 2ceh + h²a - h²e + d²a - d²e - 2dak + 2dek + k²a - k²e = r²a - r²e
e²c - e²a - 2ech + 2eah + h²c - h²a + f²c - f²a - 2fck + 2fak + k²c - k²a = r²c - r²a
----------------------------------------------------------------------------------------------------------------------------
a²e - a²c + c²a - c²e + e²c - e²a + b²e - b²c + d²a - d²e + f²c - f²a -2bek + 2bck - 2dak + 2dek - 2fck + 2fak = 0

Now get all the "k" terms to one side, and solve:

a²(e-c) + c²(a-e) + e²(c-a) + b²(e-c) + d²(a-e) + f²(c-a) = 2k(b(e-c)+d(a-e)+f(c-a))
(a²+b²)(e-c) + (c²+d²)(a-e) + (e²+f²)(c-a) = 2k(b(e-c)+d(a-e)+f(c-a))
k = ((a²+b²)(e-c) + (c²+d²)(a-e) + (e²+f²)(c-a)) / (2(b(e-c)+d(a-e)+f(c-a)))

Solve for h the same way:

h = ((a²+b²)(f-d) + (c²+d²)(b-f) + (e²+f²)(d-b)) / (2(a(f-d)+c(b-f)+e(d-b)))

Now, to find r², just plug your "h" and "k" into one of the original equations:

r² = (a-h)² + (b-k)²

Graeme, This leades to something close to what I've done.
David, Thanks, I never thought of cheva's theorem that way.

Yatir