Euler has some useful functions for geometry. Most functions are implemented in numerical and also in symbolic form. You have to load the geometry file before using the functions.

>load geometry

Numerical and symbolic geometry.

For a demonstration, we compute and plot the Euler line in a triangle. The following construction has been done with C.a.R. and shows the result we wish to achieve.

First, we define the corners of the triangle in Euler. We use a definition, which is visible in symbolic expressions.

>A::=[-1,-1]; B::=[2,0]; C::=[1,2];

To plot geometric objects, we setup a plot area, and add the points to it. All plots of geometric objects are added to the current plot.

>setPlotRange(3); plotPoint(A,"A"); plotPoint(B,"B"); plotPoint(C,"C");

We can also add the sides of the triangle.

>plotSegment(A,B,""); plotSegment(B,C,""); plotSegment(C,A,""):

Here is the area of the triangle, using the determinant formula. Of course, we have to take the absolute value of this result.

>&areaTriangle(A,B,C)

7 - - 2

We can compute the coefficients of the side c.

>c &= lineThrough(A,B)

[- 1, 3, - 2]

And also get a formula for this line.

>&getLineEquation(c,x,y)

3 y - x = - 2

For the Hesse form, we need to specify a point, such that the point is on the positive side of the Hesseform. Inserting the point yields the positive distance to the line.

>&getHesseForm(c,x,y,C), &at(%,[x=C[1],y=C[2]])

3 y - x + 2 ----------- sqrt(10) 7 -------- sqrt(10)

Now we compute the circumcircle of ABC.

>Cu &= circleThrough(A,B,C); &getCircleEquation(Cu,x,y)

5 2 3 2 325 (y - --) + (x - --) = --- 14 14 98

>U &= getCircleCenter(Cu)

3 5 [--, --] 14 14

Plot the circle and its center. Cu and U are symbolic. We evaluate these expressions for Euler.

>plotCircle(Cu()); plotPoint(U(),"U"):

We can compute the intersection of the heights in ABC (the orthocenter) numerically with the following command.

>H &= lineIntersection(perpendicular(A,lineThrough(C,B)),... perpendicular(B,lineThrough(A,C)))

11 2 [--, -] 7 7

Now we can compute the Euler line of the triangle.

>el &= lineThrough(H,getCircleCenter(Cu)); &getLineEquation(el,x,y)

19 y x 1 - ---- - -- = - - 14 14 2

Add it to our plot.

>plotPoint(H(),"H"); plotLine(el()):

The gravitational center should be on this line.

>S &= (A+B+C)/3; &getLineEquation(el,x,y) with [x=S[1],y=S[2]]

1 1 - - = - - 2 2

>plotPoint(S(),"S"):

The theory tells us SH=2*SU. We need to simplify with radcan to achieve this.

>&distance(S,H)/distance(S,U)|radcan

2

The functions include functions for angles too.

>&computeAngle(A,C,B), deg(%())

4 acos(----------------) sqrt(5) sqrt(13) 60.2551187031

The equation for the center of the incircle is not very nice.

>Is &= lineIntersection(angleBisector(A,C,B),angleBisector(C,B,A))|radcan

3/2 (2 + 1) sqrt(5) sqrt(13) - 15 sqrt(2) + 3 [--------------------------------------------, 14 3/2 (sqrt(2) - 3) sqrt(5) sqrt(13) + 5 2 + 5 -------------------------------------------] 14

In Latex, this prints as

Let us compute also the expression for the radius of the inscribed circle.

>ri &= distance(Is,projectToLine(Is,lineThrough(A,B)))|ratsimp

sqrt((- 41 sqrt(2) - 31) sqrt(5) sqrt(13) + 115 sqrt(2) + 614) -------------------------------------------------------------- 7 sqrt(2)

>Ci &= circleWithCenter(Is,ri);

Let us add this to the plot.

>color(5); plotCircle(Ci()):

Next, we find an equation for the points with equal distance from C to AB.

>eq &= getHesseForm(lineThrough(A,B),x,y,C)-distance([x,y],C)

3 y - x + 2 2 2 ----------- - sqrt((2 - y) + (1 - x) ) sqrt(10)

Using Euler's level plot, we can add this to plot.

>plot2d(eq,level=0,add=1,contourcolor=6):

This should be some function, but the default solver of Maxima can find the solution only, if we square the equation. Consequently, we get a fake solution.

>seq &= solve(getHesseForm(lineThrough(A,B),x,y,C)^2-distance([x,y],C)^2,y)

[y = - 3 x - sqrt(70) sqrt(9 - 2 x) + 26, y = - 3 x + sqrt(70) sqrt(9 - 2 x) + 26]

The first solution is

Adding the first solution to the plot show, that it is indeed the path we are looking for. The theory tells us that it is a rotated parabola.

>plot2d(&rhs(seq[1]),add=1):