Advanced Search

Ronan

1376 Reputation

16 Badges

13 years, 270 days
East Grinstead, United Kingdom
Contact Ronan

MaplePrimes Activity


These are answers submitted by Ronan

Partial solution...

Ronan 1376
November 21 2025
1 0

I can find integer values for the vertex coordinates and the angle bisector.  Could not find side lengths with integer values.

restart

with(LinearAlgebra); with(plots); with(plottools)

v1 := `<,>`(a, b)

Vector[column](%id = 36893489400601263868)

 (1)

v2 := `<,>`(c, d)

Vector[column](%id = 36893489400601251100)

 (2)

v3 := v2-v1

Vector[column](%id = 36893489400601245804)

 (3)

v4 := `<,>`(e, f)

Vector[column](%id = 36893489400601240868)

 (4)

Non*Isosceles

Eq1 := v1[1]^2+v1[2]^2 <> v2[1]^2+v2[2]^2

a^2+b^2 <> c^2+d^2

 (5)

Eq2 := v1[1]^2+v1[2]^2 <> v3[1]^2+v3[2]^2

a^2+b^2 <> (c-a)^2+(d-b)^2

 (6)

Eq3 := v2[1]^2+v2[2]^2 <> v3[1]^2+v3[2]^2

c^2+d^2 <> (c-a)^2+(d-b)^2

 (7)

"Dot product of 2 vectors <>0 if they are not perpendicular "

Eq4 := `assuming`([v1.v2 <> 0], [real])

a*c+b*d <> 0

 (8)

Eq5 := `assuming`([simplify(v1.v3 <> 0)], [real])

-a^2+a*c-b^2+b*d <> 0

 (9)

Eq6 := `assuming`([simplify(v2.v3 <> 0)], [real])

-a*c-b*d+c^2+d^2 <> 0

 (10)

Angle*bisector

Eq7 := v1+alpha*(v2-v1) = v4

Vector[column](%id = 36893489400642032932) = Vector[column](%id = 36893489400601240868)

 (11)

Eq8 := a+alpha*(a-c) = e

a+alpha*(a-c) = e

 (12)

Eq9 := b+alpha*(b-d) = f

b+alpha*(b-d) = f

 (13)

"dot product of two vectors = the magnitude times the cos of the angle between them. angle between v1 and v4 = angle between v2 and v4 Here I have squared the dot product to remove square roots "

Eq10 := `assuming`([(v1.v4)^2 = ((v1[1]^2+v1[2]^2)*(v1[1]^2+v1[2]^2))*cos^2*theta], [real])

(a*e+b*f)^2 = (a^2+b^2)^2*cos^2*theta

 (14)

Eq11 := `assuming`([(v2.v4)^2 = (v2[1]^2+v2[2]^2)*(v1[1]^2+v1[2]^2)*cos^2*theta], [real])

(c*e+d*f)^2 = (c^2+d^2)*(a^2+b^2)*cos^2*theta

 (15)

"Eliminate cos^(2)theta from Eq10 and Eq11"

NULL

Eq12 := `assuming`([(v1.v4)^2/((v1[1]^2+v1[2]^2)*(v4[1]^2+v4[2]^2)) = (v2.v4)^2/((v2[1]^2+v2[2]^2)*(v4[1]^2+v4[2]^2))], [real])

(a*e+b*f)^2/((a^2+b^2)*(e^2+f^2)) = (c*e+d*f)^2/((c^2+d^2)*(e^2+f^2))

 (16)

````

sol := isolve({Eq1, Eq12, Eq2, Eq3, Eq4, Eq5, Eq6})

{a = _Z1*((1/2)*(_Z1*_Z2^2-_Z1*_Z3^2+1)*_Z2/_Z3-(1/2)*(_Z1*_Z2^2-_Z1*_Z3^2+1)*_Z3/_Z2+2*_Z1*_Z2*_Z3), b = _Z1, c = (1/2)*(_Z1*_Z2^2-_Z1*_Z3^2+1)/(_Z2*_Z3), d = _Z1, e = _Z2, f = _Z3}, {a = _Z2*_Z1, b = _Z1, c = _Z2, d = 1, e = _Z4, f = _Z5}

 (17)

NULL

"Pick some values for Z1, 2, 3 "

sol1 := eval(sol, [_Z1 = 1, _Z2 = 1, _Z3 = 1])

{a = 2, b = 1, c = 1/2, d = 1, e = 1, f = 1}, {a = 1, b = 1, c = 1, d = 1, e = _Z4, f = _Z5}

 (18)

assign(sol1[1])

evala(v1)

Vector[column](%id = 36893489400647704572)

 (19)

evala(v2)

Vector[column](%id = 36893489400647706492)

 (20)

evala(v3)

Vector[column](%id = 36893489400647708412)

 (21)

evala(v4)

Vector[column](%id = 36893489400647710348)

 (22)

evala(Eq12)

9/10 = 9/10

 (23)

"Scale the points by a factor of 2 to get integers"

display(line(`<,>`(0, 0), 2*evala(v1)), line(`<,>`(0, 0), 2*evala(v2)), line(2*evala(v1), 2*evala(v2)), line(`<,>`(0, 0), 2*evala(v4)), scaling = constrained)

  

Eq8

2+(3/2)*alpha = 1

 (24)

"(->)"

alpha = -2/3

 (25)

eval(evala(Eq7), alpha = -2/3)

Vector[column](%id = 36893489400647799740) = Vector[column](%id = 36893489400647799620)

 (26)

``

Download 2025-11-22_A_Special_Triangle.mw

Styles and Manage Styles Sets...

Ronan 1376
October 19 2025
1 1

  This might help. In Maple 2024 under Format Styles  you can alter the font and other settings. Different Styles sets can be exported and imported using Manage Styles Sets.

Something like this...

Ronan 1376
October 12 2025
1 1

I am not exactly sure frm you wording of "k consecutive random characters". Is this what you want?

restart

NULL

randomize

randomize

 (1)

[seq(StringTools:-Char(i), i = 30 .. 128)]

["", "", " ", "!", """, "#", "$", "%", "&", "'", "(", ")", "*", "+", ",", "-", ".", "/", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", ":", ";", "<", "=", ">", "?", "@", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "[", "\", "]", "^", "_", "`", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "{", "|", "}", "~", "", "�"]

 (2)

num := rand(30 .. 128)

proc () (proc () option builtin = RandNumberInterface; end proc)(6, 99, 7)+30 end proc

 (3)

k := 25

25

 (4)

[seq(StringTools:-Char(num()), i = 1 .. k)]

["O", "L", "%", "K", "I", "&", "j", "X", "-", "", "c", "j", "D", "y", "d", "`", "k", "P", "R", ")", "0", "\", "E", "w", " "]

 (5)

NULL

Download 2025-10-12_A_Random_Characters.mw

use evalc...

Ronan 1376
October 03 2025
2 2

You can do this .

evalc(D11)

To get real part only.

evalc(Re(D11))

Or imaginary part only.

evalc(Im(D11))

LinearAlgebra:-Equal...

Ronan 1376
September 28 2025
2 2

It i seasiest to use Equal fr0m the LlinearAlgebra package for this.

V1 := <2, 3, -7>;
                             [    2     ]
                             [          ]
                       V1 := [    3     ]
                             [          ]
                             [&uminus0;7]

V2 := <2, 3, -7>;
                             [    2     ]
                             [          ]
                       V2 := [    3     ]
                             [          ]
                             [&uminus0;7]

Equal(V1, V2);
                              true

delay evaluation quotes help...

Ronan 1376
June 11 2025
0 1

The is something to do with how maple evaluates Vectors and Matrices.  delay evaluation quotes  '......'  help here. There may be nicer ways to do this though.

{'Matrix'([1])} union {'Matrix'([1])};
                         {Matrix([1])}

eval({'Matrix'([1])} union {'Matrix'([1])});
                             {[1]}

 

Sums of pairs...

Ronan 1376
May 05 2025
1 0

One of your simpler puzzles. 

restart

NULL

NULL

A := Matrix(4, 7, {(1, 1) = 343, (1, 2) = 0, (1, 3) = 15625, (1, 4) = 0, (1, 5) = 32, (1, 6) = 0, (1, 7) = 36, (2, 1) = 0, (2, 2) = x, (2, 3) = 0, (2, 4) = 15657, (2, 5) = 0, (2, 6) = 68, (2, 7) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 31625, (3, 4) = 0, (3, 5) = y, (3, 6) = 0, (3, 7) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = z, (4, 5) = 0, (4, 6) = 0, (4, 7) = 0})

Matrix(%id = 36893489885763238716)

 (1)

The rows appear to be sums of of pairs above.

x := A[1, 1]+A[1, 3]

15968

 (2)

x+A[2, 4]

31625

 (3)

y := A[2, 4]+A[2, 6]

15725

 (4)

z := A[3, 3]+y

47350

 (5)

evalm(A)

array( 1 .. 4, 1 .. 7, [( 4, 5 ) = (0), ( 1, 2 ) = (0), ( 2, 4 ) = (15657), ( 1, 5 ) = (32), ( 1, 6 ) = (0), ( 3, 6 ) = (0), ( 1, 3 ) = (15625), ( 3, 4 ) = (0), ( 3, 2 ) = (0), ( 2, 3 ) = (0), ( 4, 2 ) = (0), ( 4, 1 ) = (0), ( 1, 1 ) = (343), ( 3, 7 ) = (0), ( 2, 2 ) = (15968), ( 4, 6 ) = (0), ( 2, 1 ) = (0), ( 2, 6 ) = (68), ( 2, 5 ) = (0), ( 2, 7 ) = (0), ( 1, 7 ) = (36), ( 4, 3 ) = (0), ( 1, 4 ) = (0), ( 3, 1 ) = (0), ( 3, 5 ) = (15725), ( 4, 4 ) = (47350), ( 4, 7 ) = (0), ( 3, 3 ) = (31625) ] )

 (6)

NULL

Download 5-05-25_A_sums.mw

plot(rhs(sol_1).......

Ronan 1376
April 19 2025
1 4

This works. Nothing fancy.

plot(rhs(sol_1),x=0.15..5)

Yes in most cases...

Ronan 1376
March 27 2025
1 1

If the file doesn't use new features it willl open with a warning that it was created in a newer version of Maple. It should run ok. I opened one from a 2025 answer earlier.

I get 23...

Ronan 1376
March 21 2025
2 0

9/2=4.5   Gives a similar triangle to the yellow one

sqrt(4.5/2)=sqrt( 2.25)=1.5   square root of ratio of areas gives sides ratio

(1.5+1)^2*2=12.5 ( total ratio of side length to mid point)^2 times yellow area gives size of triangle to midpoint of side AD.

12.5+12.5-2 =23 (green area)  

It is quite straight forward...

Ronan 1376
January 07 2025
0 0

For some reason I can't upload the worksheet.
It is just the 3,4,5 triangle.

restart

sph:=(x-3)^2+(y-7)^2+(z-4)^2-25

plxy:=z=0
plyz:=x=0
ctr:=[3,7,4]
R:=5
ctrplxy:=[3,7,0]
4^2+Rxy^2=5^2
fsolve( ?? );
Rxy:=3
ctrplyz:=[0,7,4]
3^2+Ryz^2=5^2

fsolve( ?? );
Ryz:=4
#Center  and Radius on xy plane
ctrplxy;
Rxy
#Center  and Radius on yz plane
ctrplyz;
Ryz

If it is meant to be inscribed in a circ...

Ronan 1376
January 07 2025
0 1

This is just a quick test. I drew it in a CAD system to see. And  another possible assumption is the side slopes are equal.

 

My backup location...

Ronan 1376
January 01 2025
0 5

On my pc Maple docs are here. Or are you asking about windows File History?

Home User...

Ronan 1376
December 31 2024
0 0

If memory serves me correctly, I think I specified Home User. As I have a Personal Edition license.

Possibly...

Ronan 1376
December 27 2024
1 0

Hope this is correct
 

> 

restart
> 

> 

L:=1

1

 (1)
> 

E:=[0,0]

[0, 0]

 (2)
> 

A:=[-1/2*L,L*sqrt(3)/2]
> 

 

[-1/2, (1/2)*3^(1/2)]

 (3)
> 

D1:=[L,0]

[1, 0]

 (4)
> 

C:=[L,L]

[1, 1]

 (5)
> 

AC:=C-A

[3/2, -(1/2)*3^(1/2)+1]

 (6)
> 

AB:=L

1

 (7)
> 

BC:=L

1

 (8)
> 

expn1:=2*L*cos(theta)=sqrt(AC[1]^2+AC[2]^2)

2*cos(theta) = (1/2)*(9+4*(-(1/2)*3^(1/2)+1)^2)^(1/2)

 (9)
> 

theta:=solve(expn1,theta)

arccos((1/4)*(9+4*(-(1/2)*3^(1/2)+1)^2)^(1/2))

 (10)
> 

b:=2*Pi-(Pi-2*theta)

Pi+2*arccos((1/4)*(9+4*(-(1/2)*3^(1/2)+1)^2)^(1/2))

 (11)
> 

simplify( (11) );

2*Pi-2*arcsin((1/2)*(4-3^(1/2))^(1/2))

 (12)
> 

expn2:=L-L*cos(c)=L*sin(2*Pi/3)-L*sin(a+Pi/3)

1-cos(c) = (1/2)*3^(1/2)-sin(a+(1/3)*Pi)

 (13)
> 

expn3:=L-L*sin(c)=L*cos(2*Pi/3)+L(a+Pi/3)

1-sin(c) = 1/2

 (14)
> 

c:=solve(expn3,c)

(1/6)*Pi

 (15)
> 

a:=solve(expn2,a)

-(1/3)*Pi+arcsin(3^(1/2)-1)

 (16)
> 

simplify( (16) );

-(1/3)*Pi+arcsin(3^(1/2)-1)

 (17)
> 

B[1]:=L-L*sin(c)

1/2

 (18)
> 

B[2]:=L-L*cos(c)

-(1/2)*3^(1/2)+1

 (19)
> 

B:=[B[1],B[2]]

[1/2, -(1/2)*3^(1/2)+1]

 (20)
> 


plots:-display(plottools:-polygon([E,D1,C,B,A]),colour=yellow)

1

  

 
> 

 


 

Download Fun-Angles.mw

1 2 3 4 5 6 7 Page 1 of 8