Markiyan Hirnyk

Markiyan Hirnyk
8 years, 252 days

These are Posts that have been published by Markiyan Hirnyk

Take a look at this link.

Let  us consider the general case of symbolic values C(xC,yC). I make use of the idea suggested by edgar in : no assumptions.

restart; with(geometry); point(A, 0, 0);
point(B, 1, 0);
point(C, xC, yC);
point(MA, (xC+1)*(1/2), (1/2)*yC);
point(MC, 1/2, 0);
point(MB, (1/2)*xC, (1/2)*yC);
point(E, (0+1+xC)*(1/3), (0+0+yC)*(1/3));# the center of mass
line(l1, x = 1/4, [x, y]);
The coordinates of the center of the first described circle are found as the solutions of the system of the equations of midperpendiculars.

midpoint(ae, A, E); coordinates(ae);

S1 := solve({x = 1/4, ((xC+1)*(1/3))*(x-(xC+1)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

BTW, Maple can't create the midperpendiculars in this case.

point(O1, op(map(rhs, S1)));

Simple details are omitted in the above. The coordinates of the centers of the two next described circles are found similarly.
coordinates(midpoint(mce, MC, E));

S2 := solve({x = 3/4, ((-1/2+xC)*(1/3))*(x-5/12-(1/6)*xC)+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O2, op(map(rhs, S2)));

coordinates(midpoint(bma, B, MA)); coordinates(midpoint(be, B, E));


S3 := solve({(xC-1)*(x-(xC+3)*(1/4))+yC*(y-(1/4)*yC) = 0, ((-2+xC)*(1/3))*(x-(4+xC)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O3, op(map(rhs, S3)));


Now we find the equation of the circumference which passes through O1, O2, and O3.

eq := a*x+b*y+x^2+y^2+c = 0:
sol := solve({eval(eq, S1), eval(eq, S2), eval(eq, S3)}, {a, b, c});

A long output can be seen in the attached .mw file.

eq1 := eval(eq, sol);

  Now we find (in suspense)  the coordinates of the next center and verify whether it belongs to the sircumference O1O2O3.

coordinates(midpoint(mac, C, MA)); coordinates(midpoint(ec, E, C)); S4 := solve({(xC-1)*(x-(3*xC+1)*(1/4))+yC*(y-3*yC*(1/4)) = 0, ((2*xC-1)*(1/3))*(x-(4*xC+1)*(1/6))+(2*yC*(1/3))*(y-4*yC*(1/6)) = 0}, {x, y});

 point(O4, op(map(rhs, S4)));

simplify(eval(eq1, S4));

                             0 = 0

Hope the reader will have a real pleasure to find the two residuary centers and to verify these on his/her own.





It is well known that the medians of a triangle divide it into 6 triangles.
It is less known that the centers of their circumscribed circles belong to one circumference as drawn below

This remarkable theorem  was proved in the 21st century! Unfortunately, I lost its source.
I can't prove this difficult  theorem by hand. However, I can prove it with Maple.
The aim of this post is to expose these proofs. Everybody knows that it is scarcely possible
to construct a general triangle with help of the geometry package of Maple.
Without loss of generality one may assume that the vertex A is placed at the origin,
the vertex B is placed at (1,0), and the vertex C(xC,yC). We firstly consider the theorem
in the case of concrete values of xC and yC.

restart; with(geometry):with(plots):
point(A, 0, 0);
point(B, 1, 0);
xC := 15*(1/10); yC := sqrt(3); point(C, xC, yC);
triangle(T, [A, B, C]);
median(mA, A, T, MA);
median(mB, B, T, MB);
median(mC, C, T, MC);
line(m1, [A, MA]);
line(m2, [B, MB]);
intersection(E, m1, m2);
triangle(AEMB, [A, E, MB]);
circumcircle(c1, AEMB, 'centername' = C1);
circumcircle(c2, triangle(CEMB, [C, E, MB]), 'centername' = C2);
circumcircle(c3, triangle(CEMA, [C, E, MA]), 'centername' = C3);
circumcircle(c4, triangle(BEMA, [B, E, MA]), 'centername' = C4);
circumcircle(c5, triangle(BEMC, [B, E, MC]), 'centername' = C5);
circumcircle(c6, triangle(AEMC, [A, E, MC]), 'centername' = C6);
circle(CC, [C1, C2, C3]);
IsOnCircle(C4, CC);

IsOnCircle(C5, CC);
IsOnCircle(C6, CC);
display([draw([T(color = black), mA(color = black), mB(color = black), mC(color = black), C1(color = blue), C2(color = blue), C3(color = blue), C4(color = blue), C5(color = blue), C6(color = blue), CC(color = red)], symbol = solidcircle, symbolsize = 15, thickness = 2, scaling = constrained), textplot({[-0.5e-1, 0.5e-1, "A"], [.95, 0.5e-1, "B"], [xC-0.5e-1, yC+0.5e-1, "C"]})], axes = frame, view = [-.1 .. max(1, xC)+.1, 0 .. yC+.1]);

This can be done as a procedure in such a way.

restart; SixPoints := proc (xC, yC) geometry:-point(A, 0, 0); geometry:-point(B, 1, 0); geometry:-point(C, xC, yC); geometry:-triangle(T, [A, B, C]); geometry:-median(mA, A, T, MA); geometry:-median(mB, B, T, MB); geometry:-median(mC, C, T, MC); geometry:-line(m1, [A, MA]); geometry:-line(m2, [B, MB]); geometry:-intersection(E, m1, m2); geometry:-triangle(AEMB, [A, E, MB]); geometry:-circumcircle(c1, AEMB, 'centername' = C1); geometry:-circumcircle(c2, geometry:-triangle(CEMB, [C, E, MB]), 'centername' = C2); geometry:-circumcircle(c3, geometry:-triangle(CEMA, [C, E, MA]), 'centername' = C3); geometry:-circumcircle(c4, geometry:-triangle(BEMA, [B, E, MA]), 'centername' = C4); geometry:-circumcircle(c5, geometry:-triangle(BEMC, [B, E, MC]), 'centername' = C5); geometry:-circumcircle(c6, geometry:-triangle(AEMC, [A, E, MC]), 'centername' = C6); geometry:-circle(CC, [C1, C2, C3]); return geometry:-IsOnCircle(C4, CC), geometry:-IsOnCircle(C5, CC), geometry:-IsOnCircle(C6, CC), geometry:-draw([CC(color = blue), C1(color = red), C2(color = red), C3(color = red), C4(color = red), C5(color = red), C6(color = red), T(color = black), mA(color = black), mB(color = black), mC(color = black), c1(color = green), c4(color = green), c2(color = green), c3(color = green), c5(color = green), c6(color = green)], symbol = solidcircle, symbolsize = 15, thickness = 2) end proc;
SixPoints(1.5, 1.2);

true, true, true, PLOT(...)
 SixPoints(1.5, 1.2)[4];


To be continued (The general case will be considered in  part 2 .).




I'd like to pay attention to an application "Periodicity of Sunspots " by Samir Khan, where a real data is analysed. That application can be used in teaching statistics.

PS. The code by Samir Khan works well for me.

The Stone-Weierstass theorem  in its simplest form asserts that every continuous function defined on a closed interval [a,b] can be uniformly approximated as closely as desired by a polynomial function. Let us consider a concrete function (say, arcsin(sqrt(x))) on a concrete interval (for example,[0,1]) and a concrete rate (for instance, 0.01). The question arises: what can be  the degree of an approximating polynomial?
Looking in the constructive proof of the Weierstrass theorem (for example, see
W. Rudin, Principles of mathematical analysis. Third Ed. McGraw-Hill Inc. New York-...-Toronto. 1976, pp. 159-160 SWT.docx), we find the inequality for degree n in terms of the modulus of the  continuity delta and the maximum of the modulus M of a function f on [0,1]: 4*M*sqrt(n)*(1-delta^2)^n < epsilon/2.
Next, we find the modulus of the continuity of arcsin(sqrt(x)) with help of Maple (namely, the DirectSearch package):
>CM := proc (delta) DirectSearch:-Search(abs(arcsin((x+delta)^(1/2))-arcsin(x^(1/2))),
 {0 <= x, 0 <= x+delta, x <= 1, x+delta <= 1}, maximize)
end proc
. Now delta is fitting to satisfy CM(delta) < 0.01:
>Digits := 15: CM(0.9999640e-4);

[0.999995686126010e-2, [x = .999900003599999], 18].
At last, we find the required degree, taking into account M=Pi/2 for arcsin(sqrt(x)) on [0,1]:
>DirectSearch:-SolveEquations((4*Pi*(1/2))*sqrt(n)*(1-0.9999640e-4^2)^n = (1/2)*10^(-2), {n >= 10^9}, tolerances = 10^(-8));

[3.68635028417869*10^(-35), Vector(1, {(1) = -0.607153216591882e-17}),[n = 1.77870508105403*10^9], 74]
The obtained result is unexpected and impressive. However, this is only an estimate of the degree for the chosen construction. There are different ways to construct an approximating polynomial. For example, let us take the interpolating polynomial.
>with(CurveFitting): Digits := 200: P := PolynomialInterpolation([seq([(1/200)*j,
evalf(arcsin(sqrt((1/200)*j)), 180)], j = 0 .. 200)], x);

The whole long output of sort(P) can be seen in the attached file.
>DirectSearch:-Search(abs(arcsin(sqrt(x))-P), {x >= 0, x <= 1}, maximize, tolerances = 10^(-10));


033259753063018233397798614e-2, [x = .999760629733897552108099038488344\



678796478147136266075441732651036025656505033942652374763794644368578081487], 22]

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