By Julian Lowell Coolidge

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Furthermore, it looks like we have found F. For, by (12), we have so we might very well guess 10F = 1A + 9C (see Fig. 43). But we insist on proceeding algebraically and we do not wish to pull F from out of the blue, so we state the hypotheses on F and proceed. F is on segment AC, and therefore (m and n are unknowns), Next, P is on FB. Therefore, To avoid too many variables, we note that we can always divide (14) by (m + n) and (15) by (p + q). This has the effect of giving F and P mass 1. Therefore, we simply assume m + n = 1, p + q = 1, and we have where we have taken the liberty of using the same letters m and p for the new coefficients.

Draw a few diagrams giving different possibilities for the location of the points of intersection of the transversal with the sides (extended). 3. In Fig. 58, show how barycentric coordinates (some possibly negative, but whose sum is 1) may be assigned to all points of the plane. In particular, distinguish the parts I to VII in the figure according to the signs of the barycentric coordinates. 59 4. In Fig. † 5. Generalize Exercise 3 to space. In particular, characterize the 15 parts of space determined by four points according to barycentric coordinates.

V. Doubling all amounts to be mixed does not affect the final shade, but does double the final amount (the homogeneity law). VI. This axiom is false. We simply ask how it is possible to obtain 5 gallons of orange paint by adding something to 3 gallons of purple in order to realize that the equation 5 Orange = 3 Purple + 2X has no solution X. However, we may substitute the cancellation law for axiom VI, and it turns out to be true. For later convenience we put this in the form of an axiom: VI′ (Cancellation Law).

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Categories: Geometry Topology