2

1. The region in which the two squares overlap with the rectangle
is a right triangle, which we shall call T. Because the squares
each have area 1, the area of T is also 1. However, upon closer
inspection we see that the base and height of the rectangle are the same
as the (slanted) base and height of triangle T. Since area(T)=bh/2=1,
we find bh=2, so the area of the rectangle is 2. Alternate
solution: can the reader figure out how to dissect the two squares into
pieces that can be reassembled into the rectangle, thereby showing that
they must have the same area?

.
60

2. Since 30 must divide 72N, there must be factors of 2, 3, and
5 in 72N. The 2 and 3 are already present in 72, so N
must be divisible by 5. In the same manner, since 30N is a multiple
of 72, N must be a multiple of 12. The smallest positive integer
divisible by both 5 and 12 is 60. As a check, we note that 60 divides
evenly into (30)(72), so N=60 satisfies all the requirements.

.
70

3. In the time it takes the older sister to reach the top,
the up escalator has carried her forward 30 steps, because
the total length of the escalator is 40 steps and she took
10 steps herself. Therefore in this same time the down
escalator pushes the younger sister back 30 steps. To make
up this setback and cover the original 40 steps separating
her from the top the younger sister must make a total of
70 steps.

.
1/12

4. If I choose a shot that I will make with probability p (where
p is between 0 and 1/3), then Michael Jordan will make the same
shot with probability 3p. Hence the probability that I make a
shot that Jordan subsequently misses is p(13p). The
graph of this function is a parabola which equals zero when p=0
and p=1/3. By symmetry the vertex (maximum) is midway between
the two, at p=1/6. Hence the best chance I have of winning the
game is 1/12. Alternate solution: try applying the arithmeticgeometric
mean inequality to the numbers 3p and (13p) to find
the maximal value of p(13p).

.
29pi

5. Notice that a right triangle with legs whose squares are 17 and 99
has the same hypotenuse as a right triangle with legs whose squares are
19 and 97, namely the square root of 116 in both cases. Now place these
two right triangles next to one another (not overlapping) so that their
hypotenuses coincide. The circle with the common hypotenuse as diameter
will neatly circumscribe the resulting quadrilateral since any angle inscribed
in a semicircle must be a right angle, and vice versa. To summarize: we
have created a quadrilateral with the appropriate side lengths inscribed
in a circle with diameter equal to the square root of 116, and hence with
area 29pi.

.
12

6. It turns out to be advantageous to analyze this problem from a polar
coordinate perspective. For starters, line l_{1} is angled
60 degrees from the horizontal, while line l_{2} is inclined
30 degrees, because the slopes of those lines create 306090 triangles.
Also, reflecting a point through either line does not change
its distance from the origin. (Why?) Finally, if a point P has
polar angle x, then the reflection of P through line l_{1}
has angle 120x. This formula stems from the fact that the average
of the polar angles before and after reflecting comes to exactly 60 degrees,
the inclination of line l_{1}. Similarly, reflecting
P through line l_{2} results in an angle of
60x.
Therefore we choose to keep track of points in set S according
to their polar angle. Let x be the angle out
to the point (6,5), which is a little less than 45 degrees. We claim that
exactly twelve different angles arise by reflecting the point (6,5) over
the two lines repeatedly. They are x,
60+x, 120+x, ...,
300+x, x, 60x,
..., and 300x. Clearly these angles are all
distinct. Furthermore, reflecting any of these points over either line
yields a polar angle already listed in our set. For example reflecting
the point with angle 300+x in line l_{2}
produces a point with angle 60(300+x) = 240x
= 120x. Finally, all of these points can be
reached through successive reflections, since reflecting in line l_{2}
followed by line l_{1} produces a net change of 120(60x)
= 60+x. In other words, if a point with angle
x is in set S, then so is the point
with angle 60 degrees greater. Both x and 60x
are in set S, so the remaining ten angles must be also, for a
total of 12 points. 
.
1/2

7. One effective technique for predicting the next value of a polynomial
function is the method of finite differences. For ease of presentation
let us suppose that the first three values of f(x) are
f(0)=a, f(1)=b, and f(2)=0.
Writing these values out as the first row in a table gives (a,
b, 0). Subtracting each number from the one immediately following
it yields (b–a, b) in the second row.
Subtracting these differences produces the third row containing just a2b.
One main result in the theory of finite differences states that for a
quadratic function, this third row will contain all the same numbers.
Therefore the next entry in the third row is also a2b,
so we may work backwards to find that the next entries in the second and
first rows are both a3b. In other words, f(3)=a–3b.
In our case a=cos^{3}40 and b=(cos40)(sin^{2}40),
so f(3)=cos^{3}40–3(cos40)(sin^{2}40).
This is in fact the triple angle formula for cosine (check!), so f(3)=cos
120= 1/2.
