Math 445 Quiz 1 ANSWERS

Problem 1

(a)    In the triangle below, AD = b, AE = c, BD = 1 and CE = 1. Find the barycentric coordinates of F.

Let (x,y,z) be the barycentric coordinates with respect to ABC. Then considering the ratios on the sides AB and AC, we get

• On AB: y/x = b/1 implies y = bx.
• On AC: z/x = c/1 implies z= cx.

Then taking the sum =1, we get x + bx + cx = x(1+b+c) = 1, so

• x = 1/(1+b+c)
• y = bx = b/(1+b+c)
• z = cx = c/(1+b+c)

We can also write [x, y, z] = [1/(1+b+c), b/(1+b+c), c/(1+b+c)] or (1/(1+b+c))[1, b, c].

(b)   If the area of triangle ABC = 10, what is the area of triangle FBC?

The area = x10 = 10/(1+b+c).

Problem 2

Let A = (1, 2, 3) and B = (81, 42, -29). 

(a)    Find the point C on line AB for which AC/CB = 5/3.

C = sA + tB, with s+t=1. By the law of levers, this C can be interpreted as the center of mass for mass s at A and mass t at B. We can also write C = (1-t)A + tB and C = A + t(B-A). All these ways are equivalent.

According to Archimedes, we have t/s = 5/3. We can solve this several ways:

1. Take preliminary t' = 5and s' = 3; then sum s'+t' = 5+3 = 8, so t = t'/8 = 5/8 and s = s'/8 = 3/8. Thus C = (1/8){(3, 6, 9) + (405, 210, -145)} = (1/8)(408, 216, -136) = (51, 27, -17).
2. Get t the same way, but rewrite C = A + t(B-A) = (1, 2, 3) + (5/8)(80, 40, -32) = (1, 2, 3) + (50, 25, -20) = (51, 27, -17). There are fewer calculations with fractions.
3. We can take preliminary s' =1 and then t' = 5/3. then sum s'+t' = 1+(5/3)= 8/3, so t = t'(3/8) = 5/8 and s = s'(3/8) = 3/8. Then continue as before.

(b)   Find the point D on line AB for which AD/DB = -5/3.

We use the same methods.

1. Take preliminary t' = -5and s' = 3; then sum s'+t' = -5+3 = -2, so t = -t'/2 = 5/2 and s = -s'/2 = -3/2. Thus C = (1/2){(-3, -6, -9) + (405, 210, -145)} = (1/2)(402, 204, -154) = (201, 102, -77).
2. Get t the same way, but rewrite C = A + t(B-A) = (1, 2, 3) + (5/2)(80, 40, -32) = (1, 2, 3) + (200, 100, -80) = (201, 102, -77). There are fewer calculations with fractions.
3. We can take preliminary s' =1 and then t' = -5/3. then sum s'+t' = 1-(5/3)= -2/3, so t = t'(-3/2) = 5/2 and s = s'(-3/2) = -3/2. Then continue as before.

(c)    Find the point E where line AB intersects the (x, y, 0) plane (i.e., the coordinate plane containing the x and y axes).

E = A + t(B-A) = (1, 2, 3) + t(80, 40, -32). The z coordinate of E = 3 - 32t which = 0 when t = 3/32.

Thus E = (1, 2, 3) + (3/32)(80, 40, -32) -- this is already a correct answer --

E = (1, 2, 3) + (3/4)(10, 5, -4) = (17/2, 23/4, 0) = (8.5, 5.75, 0)

Problem 3

Suppose that point D has barycentric coordinates (2, 1, -2) with respect to triangle ABC. 

• If T is an affine transformation with T(A) = (1,1), T(B) = (-1,3), T(C) = (5,4), what point is T(D)? (The answer should be numerical, with no letters.)

The image of D will have the same barycentric coordinates with respect to T(A) T(B) T(C) as D has with ABC, so T(D) = 2T(A) + 1 T(B) + (-2) T(C) = (-9, -3).