history of mathematics - National STEM Centre

Figure 6.2
A Xj B~
Figure 6.3
1
6 The approach of Descartes
A proof that your solution is correct is only possible if you were able to draw the set
of points. But how did you get the idea that you must have this particular set of
points: how did you arrive at your solution?
If you have not been able to answer question 2, do not worry. You will return to it
later. But first work on question 3. It appears to resemble question 1, but it has a
quite different solution. Again it is based on the two points A and B in Figure 6.1.
3 Construct the set of points such that the distance of each point X from A equals
twice its distance from B. If you cannot find all the points that satisfy the condition,
trying finding some individual points that do.
4 If you found or guessed a solution to question 3, prove that your set of points
satisfies the condition.
It is quite likely that you cannot find the solution set for question 3. Although
questions 1 and 3 are very similar, the solutions and the strategies to find them are
quite different, unless you introduce a coordinate system and use algebra. There is
no single straight-forward geometric method for solving both problems.
A construction problem
Here is a typical solution to a geometrical construction problem to illustrate the
point made in the last paragraph. Return to question 3 of Activity 6.2.
The problem
A and B are two given points. Construct the set of points X such that the distance of
X from A is twice the distance of X from B, that is, AX = 2BX.
Experimenting
First, try to find, somehow, some individual points X that satisfy the condition.
The point X,, shown in Figure 6.3, is on AB, two-thirds of the way from A, so it
satisfies the condition. To construct Xj take a half-line m, which passes through A
at an angle to AB. On m, starting from A, mark three equal line-segments AC, CD
and DE. Draw EB. Draw a line through D parallel to EB; this line intersects AB in
the required point X,. Notice that AXj is two-thirds of AB because triangles ABE
and AX,D are similar.
I———————I———I———————————I
Figure 6.4
Figure 6.4 shows that on AB produced there is a second point X2 which satisfies the
condition AX = 2BX, since, if BX 2 equals AB, then AX 2 = 2BX 2 .
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