2009 Math Bowl Selected Solutions
JV1.2 Since the recipe requires 4 eggs for 2 pies, and Cookie needs to make this recipe
50 times, 200 eggs are required. Since eggs are sold by the dozen, we determine 200/12
– this number is between 16 and 17, so rounding up, the answer is 17.
JV1.3 The Multiplication Principle states that we just need to multiply the number of
possibilities for each choice together. We have 3 possible sizes, 2 possible crusts, 10
different toppings. The product is 60.
JV1.4 Did you know Bakersfield only has one officially sanctioned table tennis club –
you can find it online by following links from the national table tennis organization
www.usatt.org In any case, here we just need to realize that the ratio of the old radius to
the new radius is 19/20=95%. (Of course, you could also start with 38/40 since radius is
proportional to diameter).
JV2.5 Algebraically, this says x 2  x  12 . This is equivalent to ( x  3)( x  4)  0 .
Hence the two solutions 3 and -4.
JV2.10 Of course, the given angle measurements are irrelevant. The key fact is that in
any polygon, the sum of the exterior angles is 360 degrees. A good analogy – if you were
driving around the polygon, you would have rotated 360 degrees by the time you got
back to where you started. In this picture, I just drew in the angles in a slightly
asymmetric manner.
JV3.2. When this division is performed, one obtains a quotient polynomial q(x) and a
remainder polynomial r(x). Actually, since the remainder must have degree less than that
of the polynomial by which we are dividing, namely x-1, the remainder must be a
constant. In equation form, this is x 4  2 x3  3x 2  4 x  5  ( x  1)q( x)  r
It is completely unnecessary to calculate q(x), since at this point we can substitute x=1, to
obtain the answer 1+2+3+4+5=r.
JV3.5 Many people do this problem quickly by saying that the two hands will point in the
same direction exactly once an hour, hence the answer must be twelve. The easiest way
to see that this reasoning is incorrect is to note that the time in between these occurrences
is more than an hour; therefore there is no way that they can occur 12 times in a twelve
hour period. Answer: 11
JV3.7 Here, of course, one can rationalize the denominator, but the numerator and
denominator look so similar, it pays to take a couple of seconds and look for a short cut.
Since the numerator is i times the denominator. Answer: i.
JV3.9 A discount of 20% means you have to pay 80% of the value of the dress. Three
such discounts implies that one must pay 80% of 80% of 80% = (.8)^3=.512=51.2% of
the dress’s original price. Thus, the answer is 100%-51.2%=48.8%.
JV3.10 The original pyramid has volume 10x10x6/3=200. After cutting, the top part has
volume 5x5x3/3=25. So the lower, larger part has volume 175. Alternatively – and more
quickly, one could just say the top part is similar to the whole, with a 1-dimensional
proportionality ratio of ½. Therefore the 3-dimensional ratio is 1/8. So, from this one
knows the top part has volume 200/8=25 and therefore the answer is 175.
JV4.1 There are 5! =120 possible orders to play the 5 songs. To count the number of
ways that the two PiHedz songs can be played consecutively, we consider these songs are
stuck together to make one song, so there are now 4! = 24 ways to order the 4 songs.
But do not forget that there are 2 ways to stick the songs together. Therefore there are
actually 48 ways to play the two songs together. The probability that the two PiHedz
songs will be played together is 2x4!/5!=2/5. Thus, the desired answer is 3/5.
JV4.3 Of course this is again the multiplication principle at work. This time we have to
choose one of 3 sizes, one of two crusts, and one of 10C3 (10 choose 3) possible toppings
combinations. Since 10C3 =10x9x8/(3x2), the final answer is 3x2x10x9x8/(3x2)=720
(note the cancellation).
JV4.6 The method of completing the square gives y  (3x  11)2  23 , which should make
it clear that the smallest possible value for y is 23.
JV4.7 This is one of the those questions that is worded in a way to make it seem much
harder than it actually is. The key facts are that there are 11 stripes on a field. A
rectangle that is also on the field can intersect any number of these stripes including zero
of them. Thus it is possible to put down 12 rectangles that all intersect a different number
of the stripes.
JV4.8 A ring will not intersect a stripe if its center is more than one yard away from
every stripe. There are 10 yards between stripes. If the ring is centered anywhere in the
8 yard middle region between two stripes it won’t intersect. Answer: 8/10=80% or 4/5.
JV4.9 We need to use the Pythagorean Theorem twice. First, slant height of this pyramid
is
62  72  85 . So the desired distance is
62  ( 85)2  36  85  11 .
V1.2 Of course, this problem is just asking can you add 12  22  32  42  52  62 , you
should also know there is a shortcut. Especially if you have already studied integrals in
calculus class then you should know there is a formula for the sum of the first n squares.
This formula is n(n+1)(2n+1)/6. If you put in n=6, then the denominator cancels and
you are left with 7x13=91.
V1.4 Hopefully, it is clear that the given angle 1/e is irrelevant here. Especially, when
you remember that the “co” in cosine stands for “complement, you should realize for that
this question is asking for the sum of an angle and its complement = pi/2.
V1.12. What’s going on here is just an simple application of the ei  cos  i sin 
formula. In this formula, if you let    , you get the familiar ei  1 result. Similarly,
if you let    / 2 , you obtain ei / 2  i . Thus the given value can be simplified as
2
follows: ii  (ei / 2 )i  ei  / 2  e / 2 . The phrase “principle value” is necessary in this
problem, because there are other possible answers once you realize that we could have
substituted    / 2  2k for any integer value of k. Thus for all possible values, we
have ii  (ei (4k 1) / 2 )i  e(4 k 1) / 2 . The principle value means simply that we let k=0.
V2.9 This just comes down to figuring out how many divisors 360 has. The
multiplication principle gives a simple formula for this problem for any number – simple
e
e
e
if you can determine its prime factorization. If n  p1 1 p2 2 pk k , then the number of
divisor of n is (e1  1)(e2  2) (ek  1) . So 360  23  32  5 has 4x3x2=24 factors. But
two of these correspond to only 1 or 2 people eating pie. Thus the final answer is 22.
V2,10 This is a typical “unit conversion” problem. One should take the given
information and multiply the given information as follows:
32 clicks 1 rotation 12 inches 11 feet



. At least for me, it helps considerably to
1 rotation   24 inches 1 foot second
keep the units in place, at least for the first step. Now to get the desired numerical
answer, put in pi=22/7 and simplify:
32  7 12 11
 56 . Again, note the convenient cancellation.
22  24
V3.2 Whenever you have to minimize a quantity that is squared, you should first check
to see if it is possible that the quantity can be zero – is so you are done. Here, it is the
case that 3pi/4 does the trick.
n2
2  2  3  3  4  4 9  9 10 10
V3.3 The given product can be written: 
.

1 3  2  4  3  5 8 10  9 11
n2 (n  1)(n  1)
All the factors in the middle cancel. The ones that do not cancel are a 2 and a 10 on top
and a 1 and 11 on the bottom. Hence, the product is 20/11.
10
V3.4 : Let s be the number of squares. It is given that at each vertex of the polyhedron
three faces come together. Thus 3x24 = 72 is the total number of vertices of all the faces.
Thus one merely needs to solve 8  6  s  4  72 . Answer: s=6.
V3.5 I thought this problem was simpler than the previous one. Of course it is only
simple if you know the basic Euler’s Characteristic Formula: V-E+F=2, valid for any
polyhedron . In this case you are given that the number of vertices (V) is 24 and the
number of faces (F) is 14. Thus the number of edges (E) must be 36.
V3.7 Here the preferred method of solution is to count out the number of points either on
the positive x-axis or in the first quadrant. Then take this number and multiply by 4 and
finally add 1 for the point at the origin. This gives 81.
V3.9 The number of subsets of an n element set is 2^n. Thus there are 2^10=1024
possible sets of toppings. From this number we subtract 1 since we don’t want to count
the empty set (the pizzas with no toppings). Since there are 6 possible size and crust
variations, all we need to do is find 1023x6=6138.
V3.12 Here we just need to sum an infinite geometric series with first term 63. The trick
is that the common ratio is not .8 – rather it is .8^2=.64 because the given .8 is the 1dimensional proportionality constant, but the question asks about area – a 2-dimensional
1
100 7  9 100
 63 

 7  25  175 .
quantity. So, answer is 63 
1  .64
36
94
V4.3 The vectors from (3,4) to the other two points are <3,5> and <8,-2>. The area of
this triangle is ½ the absolute value of the determinant of the matrix composed of these
1 3 5 | 6  40 |

 23 .
two vectors. I.e. Area =
2 8 2
2
V4.5 It doesn’t matter what the first card is, the probability that the second card matches
it is 3/51=1/17.
V4.9 It takes 35 “chicken-days” (you might have heard the term “man-hour” – this is the
same idea) to produce 2 dozen eggs. Since we want 30 dozen eggs, we will need 35x15
chicken-days. Thus if d is the number of days, we need to solve 21d=35x15. We get
d=25. Note, because of the cancellation is not necessary to multiply out 31x15.
V4.10 The key here is that 3 is the greatest common divisor 45 and 12. So, before the
ball gets back to Randy, every third person will get the ball once. Thus, there are 30
people left who do not touch the ball.