## Problem by plotting triangle in the pyramid...

Hello guys,

I´ve a little bit problem by plotting an pyramid with an triangle in it. When I want to display both

parts, in the drawing the left and the right part and the top of the pyramid is missing. What is the problem ???
Abituraufgabe_2016_B2_Pyramide.mw

Thanks

## To paint a triangle on a sphere...

My profile picture was formerly animation and looked like this:

It would be interesting to paint a triangle on a sphere. How can I do that?

## Simple operation on geometry (area, equation)...

Hello :)

When points of triangle, A, B, C (R^2) are given, how to find area of triangle; equation and radius of circle which passes traingle(A,B,C) and height(AH).

or where can i find information about how to do this?

thank you so much:)

## Possible to plot/draw triangles knowing only the s...

Hi, I was wondering, as stated in the title, if it is possible to plot/draw a triangle knowing only the sides and angles, and not the coordinates for each point making up its corners. If not, what would be the easiest way, to calculate the coordinates using the sides and angles (assuming I know the value for each side and corner) and then plot/draw it?

I'm rather green when it comes to using Maple, so if you could explain it in a simple way that would be appreciated.

## Geometry of triangle

by: Maple

The geometry of the triangle
Romanova Elena,  8 class,  school 57, Kazan, Russia

Construction of triangle and calculation its angles

Construction of  bisectors

Construction of medians

Construction of altitudes

> restart:with(geometry):

The setting of the height of the triandle and let's call it "Т"
> triangle(T,[point(A,4,6),point(B,-3,-5),point(C,-4,8)]);

T

Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);

Construction of the triangle АВС

> draw({T(color=gold,thickness=3)},printtext=true,axes=NONE);
Calculation of the distance between heights А and В - the length of a side АВ

> d1:=distance(A,B);

d1 := sqrt(170)

Calculation of the distance between heights В and С - the length of a side ВС
> d2:=distance(B,C);

d2 := sqrt(170)

The setting of line which passes through two points А and В
> line(l1,[A,B]);

l1

Display the equation of line l1
> Equation(l1);
> x;
> y;

-2 + 11 x - 7 y = 0

The setting of line which passes through two points А and С
> line(l2,[A,C]);

l2

Display the equation of line l2
> Equation(l2);
> x;
> y;

56 - 2 x - 8 y = 0

The setting of line which passes through two points В and С
> line(l3,[B,C]);

l3

Display the equation of line l3
> Equation(l3);
> x;
> y;

-44 - 13 x - y = 0

Check the point А lies on line l1
> IsOnLine(A,l1);

true

Check the point А lies on line l1
> IsOnLine(B,l1);

true

Calculation of the andle between lines l1 and l2
> FindAngle(l1,l2);

arctan(3)

The conversion of result to degrees
> b1:=convert(arctan(97/14),degrees);

97
arctan(--) degrees
14
b1 := 180 ------------------
Pi

Calculation of decimal value of this angle
> b2:=evalf(b1);

b2 := 81.78721981 degrees

Calculation of the andle between lines l1 and l3
> FindAngle(l1,l3);

arctan(3/4)

The conversion of result to degrees
> b3:=convert(arctan(97/99),degrees);

97
arctan(--) degrees
99
b3 := 180 ------------------
Pi

Calculation of decimal value of this angle
> b4:=evalf(b3);

b4 := 44.41536947 degrees

Calculation of the angle between lines l2 and l3
> FindAngle(l2,l3);

arctan(3)

The conversion of  result to degrees
> b5:=convert(arctan(97/71),degrees);

97
arctan(--) degrees
71
b5 := 180 ------------------
Pi

Calculation of decimal value of  this angle
> b6:=evalf(b5);

b6 := 53.79741070 degrees

Check the sum of all the angles of the triangle
> b2+b4+b6;

180.0000000 degrees

Analytical information about the point А
> detail(A);
name of the object: A
form of the object: point2d
coordinates of the point: [4, 6]
Analytical information about the point В
> detail(B);
name of the object: B
form of the object: point2d
coordinates of the point: [-3, -5]
Analytical information about the point С
> detail(C);
name of the object: C
form of the object: point2d
coordinates of the point: [-4, 8]

The setting of heights of the triangle points A,B,C and let's call it "Т"

with(geometry):
> triangle(ABC, [point(A,7,8), point(B,6,-7), point(C,-6,7)]):
The setting of the bisector of angle А in triandle АВС
> bisector(bA, A, ABC);

bA

Analytical information about the bisector of angle А in the triandle
> detail(bA);
name of the object: bA
form of the object: line2d
assume that the name of the horizonal and vertical                    axis are _x and _y
equation of the line: (15*170^(1/2)+226^(1/2))*_x+(-13*226^(1/2)-170^(1/2))*_y+97*226^(1/2)-97*170^(1/2) = 0

Construction of the triangle
> draw(ABC,axes=normal,view=[-8..8,-8..8]);

Construction of the triangle ABC

> draw({ABC(color=gold,thickness=3)},printtext=true,axes=NONE);

Construction of the bisector of angle А

> draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3)},printtext=true,axes=NONE);

The setting of the bisector of angle В in the triangle АВС

> bisector(bB, B, ABC);

bB

Analytical information about the bisector of angle B in the triandle
> detail(bB);
name of the object: bB
form of the object: line2d
assume that the name of the horizonal and vertical                    axis are _x and _y
equation of the line: (-15*340^(1/2)-14*226^(1/2))*_x+(-12*226^(1/2)+340^(1/2))*_y+97*340^(1/2) = 0

Construction of the bisector of angle В
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3)},printtext=true,axes=NONE);

The setting of the bisector of angle С in the triangle АВС

> bisector(bC, C, ABC);

bC

Analytical information about the bisector of angle С in the triangle
> detail(bC);
name of the object: bC
form of the object: line2d
assume that the name of the horizonal and vertical                    axis are _x and _y
equation of the line: (14*170^(1/2)-340^(1/2))*_x+(13*340^(1/2)+12*170^(1/2))*_y-97*340^(1/2) = 0

Construction of the bisector of angle С
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3)},printtext=true,axes=NONE);

Calculation of the point of intersection of the bisectors and let's call it "О"

> intersection(O,bA,bB,bC);coordinates(O);

O

7 sqrt(85) - 3 sqrt(2) sqrt(113) + 3 sqrt(85) sqrt(2)
[2 -----------------------------------------------------,
sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

-16 sqrt(85) - 7 sqrt(2) sqrt(113) + 7 sqrt(85) sqrt(2)
- -------------------------------------------------------]
sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

Construction of the bisectors and  marking of the point of intersection  "О" in the triandle
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3),O},printtext=true,axes=NONE);
> restart:
> with(geometry):
The setting of the heights of the triangle points A,B,C and let's call it "Т"
> point(A,7,8),point(B,6,-7),point(C,-6,7);

A, B, C

Let's call "Т1"
> triangle(T1,[A,B,C]);

T1

Construction of "Т1"
> draw(T1(color=gold,thickness=3),axes=NONE,printtext=true);
The setting of the median from the point В in the trianglemedian(mB,B,T1,B1);
> median(mb,B,T1);

mB

mb

Construction of the median from the point В
> draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mb},printtext=true,axes=NONE);

The setting of the median from the point А in the trianglemedian(mA,A,T1,A1);
> median(ma,A,T1);

mA

ma

Construction of the median from the point А
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),ma},printtext=true,axes=NONE);
The setting of the median from the point С in the trianglemedian(mC,C,T1,C1);
> median(mc,C,T1);

mC

mc

Costruction of the median from the point С
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=maroon,thickness=3)},printtext=true,axes=NONE);

Calculation of the point of  intersection of the median and let's call it "О"

>intersection(O,ma,mb,mC);coordinates(O);

O

[7/3, 8/3]

Construction of medians and marking of the point of  intersection "О" in the triangle
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=violet,thickness=3),O},printtext=true,axes=NONE);
> restart:with(geometry):
> _EnvHorizontalName:=x:_EnvVerticalName=y:       The setting of the heights of the triangle points A, B, C  and let's call it "Т"
> triangle(T,[point(A,7,8),point(B,6,-7),point(C,-6,7)]);

T

Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);

The setting of the altitude in the triangle from the point Сaltitude(hC1,C,T,C1);
> altitude(hC,C,T);

hC1

hC

Analytical information about the altitude hC from the point С in the triangle
> detail(hC);
name of the object: hC
form of the object: line2d
assume that the name of the horizonal and vertical                    axis are _x and _y
equation of the line: -99+_x+15*_y = 0

Construction of the altitude from the point С
> draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hC},printtext=true,axes=NONE);

The setting of the altitude in the triangle from the point Аaltitude(hA1,A,T,A1);
> altitude(hA,A,T);

hA1

hA

Analytical information about the altitude hA from the point А in the triangle
> detail(hA);
name of the object: hA
form of the object: line2d
assume that the name of the horizonal and vertical                    axis are _x and _y
equation of the line: -28-12*_x+14*_y = 0

Construction of the altitude from the point А
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hA1},printtext=true,axes=NONE);       The setting of the altitude from the point В

> altitude(hB1,B,T,B1);
> altitude(hB,B,T);

hB1

hB

Analytical information about the altitude hB from the point В in the triangle
> detail(hB);
name of the object: hB
form of the object: line2d
assume that the name of the horizonal and vertical                    axis are _x and _y
equation of the line: -71+13*_x+_y = 0

Consruction of the altitude from the point В
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1},printtext=true,axes=NONE);
Calculation of the point of intersection of altitudes and let's call it "О"

>intersection(O,hB,hA,hC);coordinates(O);

O

483  608
[---, ---]
97   97

Construction of altitudes and marking of the point of intersection "О" in the triangle
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1,O},printtext=true,axes=NONE);

## How to fill a region between three lines?...

Hi all,

I have three points in 3d space say A1=[a11, a12, a13]; A2=[a21, a22, a23] and A3=[a31, a32, a33]. I want to fill the triangle formed by these points. How can I do that?

## How to draw these geometric objects?...

Hi, there

How can I draw the excircles, incircles,circumcircle and their centers of a triangle simultaneously with maple13 in a geometric plot? please specify the commands.

M.R.Yegan

## Procedure for triangle...

Hello everyone!

i want to make a triangle which height's is a. Sorry, my English is not very good!

And I writed it by C language, but It is not right in maple. and I don't know. Can you help me? thank you very much! triangle.mw

## Questions re an irregular solid angle...

The downloaded worksheet below displays 3 points on the unit sphere which define a solid angle with a triangular face. The sides of the solid angle's are red arcs on the surface of the sphere and red radii which outline the planar sides within the sphere.

Three questions:

1. Is there a way to make the surfaces of the solid triangle more apparent by filling them with color?

2. Is there a way to calculate the area of the face on the surface of the sphere?

3. Is there a way to calculate the volume of the solid triangle?

## Fix a syntax error in my simple function (please h...

firstly apologies in advance for stuff in this question such as "triangle symbol",  my computer is pretty old.

ok so i was confused a bit here, what i'm trying to do is write a maple procedure that computes Af for a given f contained in V . except we only need to correct the bug in the script below. This script demonstrates such a procedure in the case that omega is a square. The domain is given here as the negative set of a function F contained in V .  I have left in notes where/what i think we need to do but i dunno how to...

N:=10 ; # Global Var
F:=(x,y)->sgn(abs(x-N/2)+abs(y-N/2)-N/4);
Average := proc(F, f0) local f, i, j;
f := f0; # !!!!!!!!!!!!!! something is bad here...
for i to N do for j to N do
if F(i, j) < 0 then
f[i, j] := (f0[i - 1, j] + f0[i + 1, j] + f0[i, j + 1] + f0[i, j - 1])/4 ;
end if;
end do;end do;
return f;
end proc;
f0:=Matrix(N,F); # just to have something to test the procedure
Average(F,f0); # does not return the expected average, modifies f0

the necessary information we were given to produce this so far was..

Let N be a positive integer and [N] = {i contained in N | 1<= i <=N }  Let "Omega" C {(i,j) contained in [N] x [N] | 2<=i,j<=N-1} be a subset. Let V = R^([N]x[N]) be the vector space of real valued functions [N]x[N] -> R
and A, "triangle symbol":V->V (average) and "triangle symbole" (Laplacian) be the linear maps such that
[Af](i; j) = f(i; j)      if (i; j) not contained in "Omega"   OR

[f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 if (i,j) is contained in "Omega"

["traingle symbol"f](i,j) =  0 if (i,j) isnt contained in "Omega"   OR

( f(i,j) - [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 )    if (i,j) is contained in "Omega"

## Another proof of six points theorem

by:

In this post we present another compact proof of this remarkable theorem without using  geometry package.
The proof uses a procedure called  Cc , which for three points returns a list of the coordinates of the center and the radius of the circumscribed circle.

restart;

Cc:=proc(A,B,C)

local x1, y1, x2, y2, x3, y3, x, y;

x1,y1:=op(A);  x2,y2:=op(B);  x3,y3:=op(C);

solve({(x2-x1)*(x-(x1+x2)/2)+(y2-y1)*(y-(y1+y2)/2)=0, (x2-x3)*(x-(x2+x3)/2)+(y2-y3)*(y-(y2+y3)/2)=0},{x,y});

assign(%);

[simplify([x,y]), simplify(sqrt((x-x1)^2+(y-y1)^2))];

end proc:

Proof for arbitrary triangle:

A, B, C:=[x1,y1], [x2,y2], [x3,y3]:

A1, B1, C1, M:=(B+C)/2, (A+C)/2, (A+B)/2, (A+B+C)/3:

P1:=Cc(A,M,B1)[1]: P2:=Cc(B1,M,C)[1]: P3:=Cc(C,M,A1)[1]:

P4:=Cc(A1,M,B)[1]: P5:=Cc(B,M,C1)[1]: P6:=Cc(C1,M,A)[1]:

Cc1:=Cc(P1,P2,P3):  Cc2:=Cc(P4,P5,P6):

is(Cc1=Cc2);

true

## plotting a triangular solid...

Hello!

I'm trying to plot the 3d graph of a solid in Maple 16. The solid is generated by equilateral triangles. One vertex is  A(x,0,0), and the other one, B, moves on the circunference described by: y^2+x^2=121, z=0. The AB segment is always parallel to the Y axis and the triangles are in the yz plane. In other words, one side (of each triangle) is sqrt(121-x^2).

I tried using the spacecurve command to start plotting lines but then I didn't know how...

## How do I allow sqrt in my procedure?...

`I've tried all sorts of different assignments for a,b and c but all still give me an error `
`for root 2. `
`have tried posint, int, nonnegative, positve but none work `
`any help would be appreciated. `
` `
`isTriangle:=proc(a::posint,b::posint,c::posint)if (a+b) > c then print (a,b,c,`are lengths of a triangle`)#this is according to the triangle inequality theoremelif (a+b) < c then print ...`

## How to write a procedure to determine if values ar...

`Q3) Write a procedure, isTriangle, to determine if any three given`
` numbers represent the lengths of the sides of a triangle.       The procedure should be able to deal with any input.I have done the following and get an error`
` `
`isTriangle:=proc(a,b,c::nonnegative and numeric)if (a^2+b^2)=c^2 then print (a,b,c "represent the sides of a triangle")end if;end proc;Error, unexpected string`