P. 17, no. 6: Prove that if S is on segment \overline{PR} and T is on segment \overline{QR}, the segments \overline{PT} and \overline{QS} intersect.
Response. The exercise is mainly meaningful when S and T are not endpoints of the segments on which they lie and when P, Q, R are not collinear. If that is the case, then there are numbers s, t with 0 < s, t < 1 such that S = (1 - s) R + s P and T = (1 - t) R + t Q. Moreover, a point on the line PT has the form (1 - x) P + x T for some x, while a point on the line QS has the form (1 - y) Q + y S for some y. The lines PT and QS meet if and only if there are numbers x, y for which the two previous expressions are equal. The question of whether such values of x, y exist (and, hence, the lines intersect) is addressed algebraically.
If those expressions are expanded using the formulas for S and T, then their equality becomes the relation
| (1 - x) P + x t Q + x (1 - t) R = y s P + (1 - y) Q + y (1 - s) R . |
With the assumption that P, Q, R are not collinear, hence, barycentrically independent, the corresponding coefficients of P, Q, R in this relation must be equal. Hence,
| 1 - x = s y , t x = 1 - y , (1 - t) x = (1 - s) y . |
Solving these equations simultaneously for x, y one finds
| x = {1 - s}/{1 - s t} , y = {1 - t}/{1 - s t} . |
The fact that these solutions exist means that the lines PT and QS intersect. Moreover, from the fact that 0 < s, t < 1 it is clear that 0 < x, y < 1, and, therefore, that the point where the lines intersect is the intersection of the segments \overline{PT} and \overline{QS}.
P. 31, no. 4: P is a point inside a given triangle ABC, and F is the point on the side AB where the line CP meets AB. D is the point of intersection with AC of the line through P parallel to BC, and E is the point of intersection with BC of the line through P parallel to AC. Prove that |AF| . |CD| . |BC| = |BF| . |CE| . |AC| .
Response. If the vertices A, B, C are arranged clockwise, then each of the sides of the triangle is divided into two segments by the points F, E, D. Each corresponding length a, b, c is then decomposed: a = a' + a'', b = b' + b'', and c = c' + c'', where a' = |CE| , b' = |AD| , and c' = |BF| . With this notation the task is to show that a b'' c'' = b c' a'.
Let P = u A + v B + w C. Since the point F has unique barycentric coordinates with respect to A, B, C and is both a barycentric combination of the two points C, P and also a barycentric combination of the two points A, B, one sees that
| F = {u}/{u + v} A + {v}/{u + v} B . |
Let D = (1 - s) C + s A and E = (1 - t) C + t B. By the parallelogram law of addition
| P = D + E - C , |
which leads to a second barycentric expression for P relative to the three vertices:
| P = s A + t B + (1 - s - t) C . |
Hence, s = u, t = v, and, therefore,
| a' = v a, a'' = (1 - v) a, b' = (1 - u)b, b'' = u b , |
while
| c' = {u}/{u + v} c, c'' = {v}/{u + v} c . |
Thus,
| a b'' c'' = {uv}/{u + v} a b c = b c' a' . |