Week 1 Standings

Here are the point totals after the first week. If your score doesn't look right to you, send me an email and let me know.

MK: 14
SS: 13
AG: 12
JP: 6
TM: 4
PG: 4
BW: 3
ML: 3
SD: 3
MD: 1

Solution #5

The maximal distance will occur when we draw the line segment connecting the two centers, and then extend it so that it intersects the two spheres on their opposite sides.

The distance will then be the sum of (i) the distance between the two centers, (ii) the radius of the larger sphere, and (iii) the radius of the smaller sphere. The two radii are 19 and 87, and their sum is 106.

To find the distance between the two centers, we use the three-dimensional distance formula. The distance is

	_______
√	(12 - -2)2 + (8 - -10)2 + (-16 - 5)2

= √ 196+324+441  = √ 961  =31.

Thus the maximal distance is 106+31 = 137.

Solution #4

We calculate the slopes between pairs of points:

(-3,-7) to (15,4): slope = 18/11
(-3,-7) to (3,-3): slope = 6/4 = 3/2
(-3,-7) to (12,3): slope = 15/10 = 3/2

So (-3,-7), (3,-3), and (12,3) are all on a line of slope 3/2. Thus (15,4) is not on a line with the other three.

Problem #5

[This problem is worth 3 points.]

What is the greatest possible distance between two points, one on a sphere of radius 19 centered at (-2,-10,5) and the other on a sphere of radius 87 centered at (12,8,-16)?

Solution #3

Because AEB and DFC are both 5-12-13 triangles, they are both right. Now add corresponding triangles on sides AD and BC:

The resulting quadrilateral is a rectangle, because all of its angles are right. And it is a square, because each of its sides is composed of two right triangle legs, one of length 5 and one of length 12.

Thus it is a square of side length 17. EF is a diagonal of that square, and hence has a length of 17 √ 2 

Solution #2

We want to find x³+y³. Factoring, we have:

x³+y³=(x+y)(x²-xy+y²)

Since x + y = 1, this is equal to x²-xy+y².

We then have x²-xy+y²=(x²+2xy+y²)-3xy

=(x+y)²-3xy

=1²-3*1

=-2

Alternatively, (x+y)³ = 1³=1. Expanding, (x+y)³ = x³ + 3x² + 3xy² + y³ = (x³+y³) + 3xy(x+y) = (x³+y³) + 3.
So (x³+y³) +3 = 1, and x³+y³=-2.

Problem #4

[This problem is worth 1 point.]

Which of the following four points is not on the same line as the other three: (-3,-7), (15,4), (3,-3), or (12,3)?