Selected Homework Exercise Solutions Math 331, Transformation Geometry

February 6, 2002

P. 17, no. 6: Prove that if S is on segment \overline{P R} and T is on segment \overline{QR}, the segments \overline{P T} 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 + sP and T = (1 − t)R + tQ. Moreover, a point on the line P T has the form (1 − x)P + xT for some x, while a point on the line QS has the form (1 − y)Q + yS for some y. The lines P T 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 + xtQ + x(1 − t)R = ysP + (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 = sy , tx = 1 − y , (1 − t)x = (1 − s)y .

Solving these equations simultaneously for x, y one finds

x = \frac{1 − s}{1 − st} , y = \frac{1 − t}{1 − st} .

The fact that these solutions exist means that the lines P T 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{P T} 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 ab′′c′′ = bc′a′.

Let P = uA + vB + wC. 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 = \frac{u}{u + v} A + \frac{v}{u + v} B .

Let D = (1 − s)C + sA and E = (1 − t)C + tB. 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 = sA + tB + (1 − s − t)C .

Hence, s = u, t = v, and, therefore,

a′ = va, a′′ = (1 − v)a, b′ = (1 − u)b, b′′ = ub ,

while

c′ = \frac{u}{u + v} c, c′′ = \frac{v}{u + v} c .

Thus,

ab′′c′′ = \frac{uv}{u + v} abc = bc′a′ .