Mathematics with Maple.
Answers to Lesson 6.

S.Duzhin

Preface.
I tried very hard to find solutions to Exercises 2 and 5, using "geometry", but Maple was not powerful enough to prove these theorems using my plan. (Although Maple could prove a more difficult Pascal's theorem -- exercise 6).

> with(geometry):


Exercise 1. Prove the theorem that the three medians of a triangle are concurrent.
("sankakukei-no mittsu-no chuusen wa 1-ten-de majiwaru"). If you forget what is a median, here is a definition: it is a line that passes through a vertex and the midpoint of the opposite side ("hitotsu no chouten to mukaiau hen no chuuten wo musubu chokusen").

Solution: just replace "altitude" by "median" in Example 1:

> point(A,xa,ya);

> point(B,xb,yb);

> point(C,xc,yc);

> triangle(ABC,[A,B,C]);

> median(ABC, A, mA);

> median(ABC, B, mB);

> median(ABC, C, mC);

> are_concurrent (mA,mB,mC);


Exercise 2 (to be done at home).
Try to complete Example 2, using other procedures from the geometry package
(note: I do not know the answer to this exercise yet).


Exercise 3. Let ABCD be an arbitrary quadrangle ("shikakukei") and
M,N,P,Q the midpoints of its sides AB,BC,CD,DA. Prove that the midpoints of the segments MP and NQ are the same.

> point(A,x1,y1);

> point(B,x2,y2);

> point(C,x3,y3);

> point(D,x4,y4);

> midpoint(A,B,M);

> midpoint(B,C,N);

> midpoint(C,D,P);

> midpoint(D,A,Q);

> midpoint(M,P,X);

> midpoint(N,Q,Y);

> coordinates(X);

> coordinates(Y);

> evalb(coordinates(X)=coordinates(Y));


Exercise 4.
If AD is the bisector of the triangle ABC, then DB/DC = AB/AC.

> point(A,x1,y1);

> point(B,x2,y2);

> point(C,x3,y3);

> triangle(ABC,[A,B,C]);

> bisector(ABC, A, bA);

> D:=inter(bA,line(BC,[B,C]));

> d1:=distance(D,B):

> d2:=distance(D,C):

> a1:=distance(A,B):

> a2:=distance(A,C):

> simplify(d1/d2 - a1/a2);


Exercise 5. How to draw the perpendicular from a point P outside a circle c onto its diameter AB using only a ruler (i.e. if you are allowed to draw only straight lines).
Construct the line PA which meets the circle in point P1 and the line PB which meets the circle in point P2. Then draw the lines [A P2] and [B P1].
Let Q be the intersection point of these two lines. Then PQ is perpendicular to AB.

> point(A,[x1,x2]);

> point(B,[x2,y2]);

> point(P,[x3,y3]);

> circle(c, [A, B, diameter]);

> line(AP,[A,P]);

> AA:=inter(c,AP);

> A1:=AA[1];

> coordinates(A1);

This means that A1=A.

> P1:=AA[2];

> line(BP,[B,P]);

> BB:=inter(c,BP);

> B1:=BB[1];

> coordinates(B1);

OK, this is different from B, so:

> P2:=BB[1];

> line(AP2,[A,P2]);

> line(BP1,[B,P1]);

> Q:=inter(AP2,BP1);

> line(PQ,[P,Q]);

> line(AB,[A,B]);

The last thing we have to check is that the two lines PQ and AB are perpendicular.
There is a special command for that in the "geometry" package, but in my case,
it could not finish the computations and Maple crashed!
So be careful before putting your cursor into the following line and hitting <Return>

> are_perpendicular(PQ,AB);


Exercise 6.

> for i from 0 to 5 do point(A.i,cos (t.i),sin(t.i)) od;

> for i from 0 to 2 do line(L.i,[A.i,A.(i+1)]);line(M.i,[A.(i+3) ,A.(i+4 mod 6)]);

> point(P.i,coordinates(inter(L.i,M.i)));od;

> triangle(T,[P0,P1,P2]);

> area(T);

> simplify(");


Meaning of this theorem:
If A0,A1,A2,A3,A4,A5 are any points on a circle and P0, P1, P2 are the intersection points of the lines A0,A1 and A3,A4; A1,A2 and A4,A5; A2,A3 and A5,A0, then these three points P0, P1, P2 belong to one line.