Answer
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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 set-back 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(1-3p). 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 arithmetic-geometric
mean inequality to the numbers 3p and (1-3p) to find the
maximal value of p(1-3p).
12
5. It turns out to be advantageous to analyze this problem
from a polar coordinate perspective. For starters, line l is
angled 60 degrees from the horizontal, while line m is
inclined 30 degrees, because the slopes of those lines
create 30-60-90 triangles. (The reader may wish to show that
in general the slope of a line is equal to the tangent of
the angle of inclination.) Also, reflecting a point through
line l or m does not change its distance from the origin.
(Why?) Finally, if a point P has polar angle Y, then the
reflection of P through line l has angle 120-Y. 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. Similarly, reflecting P through line
m results in an angle of 60-Y. (Recall that the polar angle
measures how far around the origin counter-clockwise from
the positive x-axis our point is located, from 0 to 360
degrees.)
Therefore we choose to keep track of points in set S according to their polar angle. Let Z 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) through lines l and m repeatedly. They are Z, 60+Z, 120+Z, ..., 300+Z, -Z, 60-Z, ..., and 300-Z. Clearly these angles are all distinct. Furthermore, reflecting any of these points through line l or m yields a polar angle already listed in our set. For example reflecting the point with angle 300+Z in line m produces a point with angle 60-(300+Z) = -240-Z = 120-Z. Finally, all of these points can be reached through successive reflections, since reflecting in line m followed by line l produces a net change of 120-(60-Y) = 60+Y. In other words, if a point with angle Y is in set S, then so is the point with angle 60 degrees greater. Both Z and 60-Z are in set S, so the remaining ten angles must be also, for a total of 12 points.
-1/2
6. 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 P(x) are P(0)=A, P(1)=B, and P(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 A-2B. 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 A-2B, so we may work backwards to find that the next
entries in the second and first rows are A-3B and A-3B,
respectively. In other words, P(3)=A-3B.
In our case A=cos340 and
B=(cos40)(sin240), so P(3) = cos340 -
3(cos40)(sin240). This is in fact the triple
angle formula for cosine (check!), so P(3)=cos
120=-1/2.
29pi
7. 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.
It is a curious fact that as long as the side lengths remain the same, the order of the sides does not alter whether or not a polygon can be inscribed in a circle, nor does it alter the size of the circumscribed circle. To see this, verify that if a polygon is inscribed in a circle then switching two adjacent sides produces another polygon which can be inscribed in the same circle. Use this fact to prove our claim. Where did we use this fact above?