50th ANNUAL
MASSACHUSETTS MATHEMATICS OLYMPIAD
2013 – 2014
A High School Competition Conducted by
THE MASSACHUSETTS ASSOCIATION
OF MATHEMATICS LEAGUES (MAML)
And Sponsored by
THE ACTUARIES’ CLUB OF BOSTON
FIRST-LEVEL EXAMINATION
*SOLUTIONS*
Thursday, October 24, 2013
SOLUTIONS TO THE MASSACHUSETTS OLYMPIAD LEVEL I 2013 – 2014
4t  3
. If t  2 then u  1 and if t  7
5
then u  5 . This gives the numbers 21 and 75 for a sum of 96 .
1.
From 10t  u  6(t  u )  3 we have 4t  3  5u  u 
2.
18  y
x
8
18  x 1
4
.
  x  8 and
 4  y  6 . Therefore,


12  x 2
3
12  y
y
6
3.
There are 3.78 liters per gallon. Let TL stand for Turkish liras.
 7 dollars 
3.78 liters 7 dollars
3.5 TL / liter
7
Then
 dollars/liter . Finally,

 1.26  
 =
1.5 TL / dollar 3
gallon
3 liters
 gallon 
$8.82 / gallon .
4.
A number is divisible by 24 if it is divisible by 8 and 3. A number is divisible by 8 if the
number formed by the last three digits is divisible by 8. Since 680/8 = 85 and 688/8 = 86,
68B is divisible by 8 if B = 0 or 8. A number is divisible by 3 if the sum of its digits is
divisible by 3. Let B = 0. Then we have A034680 and the sum of the digits is 21 + A. So
A must be divisible by 3, making A = 3, 6, or 9. If B = 8 we have A834688 and the sum
of the digits is 37 + A. So A must be 2 more than a multiple of 3, making A  2, 5, or 8 .
For distinct A and B we have the following ordered pairs (A, B) = (3, 0), (6, 0), (9, 0),
(2,8), and (5,8) . Thus, there are 5 ordered pairs.
5.
1

 1  ab 
 a 1  b   a b   a  b   a 2  b2   a
 ab
=
 1      
 1  
 1  2 2 =


1
 b a


 a  1   ab   1  ab  a  b
 ab

b 

 b

b
ab
a b
ab
b 
.



  1 
a b
a (a  b)(a  b)
 a  (a  b)(a  b)
6.
Let the numbers be a  40, a  20, a, a  20, and a  40 . As long as the sum of the two
smallest exceeds the largest, then any three lengths will form a triangle. From
(a  40)  (a  20)  (a  40) we obtain a  100 . Thus, the least value of a is 101 .
7.
The problem is easier if we solve the general case. Let AB  n , BC  n  2 , and
AC  n  1. If AD  x and BD  h , then we have (n  1)  x   h2  (n  2)2 and
2
x 2  h 2  n 2 . Subtracting the second from the first yields
(n  1)2  2 x(n  1)  4(n  1) giving n  1  2x  4 , so x 
AD 
2011  3
 1004 .
2
n3
. In this case
2
8.
Since the least common multiple of three and four is twelve, write out the first 12 digits of
each fraction in the difference .2007  .101 . Thus, .20072007 2007 – .101101101 101
= . 099618960906 . This is the repeating block of the difference. Multiply by 1036 so that
the entire block is shifted 3 blocks to the left. Multiplying next by 10 4 puts the decimal
place in front of the 1. Ans: 1 .
9.
Let mPCB   , then the rest of the angles are
as marked because BC || AD and triangles ABC
and DCB are isosceles. Since the diagonals of an
isosceles trapezoid are congruent,
BAD  CDA by SSS, making   60   so
  30 . This means that triangles PTC, CBD,
and RPD are 30-60-90 right triangles. Let
PT  1, then TC  3 , PC  2 , PD  4 , and
BC 1
RD  2 3 . So
 .
AD 2
B
T
60 - q
60
P
q
A
C
q
60 - q
60 - q
q
D
R
10.
Given A(3, –5), B(–9, t), C(w, 10) with AB : BC = 3:2, the ratio of AB to BC will be the
same as the ratio of the difference in x-coordinates as well as the ratio of the difference in
AB 3
3  9 3
12
3
 
 
  24  3w  27 
y-coordinates. Thus, we have
BC 2
9  w 2
w9 2
5  t 3
  t  4 . Thus, t  w  13 .
w  17 . Also
t  10 2
11.
The side of the small hexagon is 1 so the radius of the inscribed circle is
3
, making the
2
2
 3
9
sum of the areas of the three small circles 3    


 2 
4


5 3
equilateral triangle is 5 so the radius of the large circle is
2
2
. The side of the large

2 5 3

. Its area is
3
3
25
9 25
27
 5 

  3 . The ratio of the sum to the large is 4  3  100 .
 3
12.
Since (6, z ) lies on both lines we have 6a  3  7 and 6  3az  4 . Multiply the first
equation by a giving the system 6a 2  3az  7a and 6  3az  4 . Adding gives the
quadratic equation 6a 2  6  7a  4  6a 2  7a  2  0 . Factoring gives
2
1
7
.
(3a  2)(2a  1)  0  a  or . The sum is
3
2
6
13.
From 2 log( xy )  (log x)(log y ) we obtain 2 log x  2 log y  (log x)(log y) . To see the
2a
algebra let a  log x and b  log y . Then 2a  2b  ab  b 
. To more easily
a2
4
find integral solutions let t  a  2 giving b  2  . If t  1, 2, or 4 we have
t
(a, b)  (3, 6), (4, 4), (6,3) respectively. Thus,  log x, log y   (3,6),(4, 4),(6,3) giving




( x, y )  103 , 106 , 104 , 104 , 106 , 103 . There are 3 ordered pairs of integral solutions.
14.
The answer is 9! . Note that there is a one-to-one correspondence between each ordering
of 9 knights and one of Arthur's choices: the first is the head knight of the first mission, the
second is the knight-errant of the first mission, the third is the knight bachelor of the first
mission, the fourth is the head knight of the second mission, and so on. Notice that we
 9  6  3 
3
could have alternatively calculated      3! , which also equals 9! .
 6  3  3 
15.
Let x = probability that Louie wins and 1  x the probability that he loses. Then
6 4
5
2  6 5
   x 1  x      x 1  x   15(1  x)  6 x  x  . Then
7
 4
 5
 6   5 2  2 4
    
 2  7   7 
6  5 
  
 3  7 
16.
3
2
 
7
3

3
15 2 / 7
=
.

10
20 5 / 7
Pairing terms in the numerator and denominator we have
So the expression collapses to
2  2 tan A
1  tan 2 A
 2 tan 2 A
2 tan 2 2 tan 4  2 tan 20
 210  1024 .
tan 2 tan 4  tan 20
17.
Using points (–2, 0, 3) and (3, 10, –7), the vector equation of the line is
r  2,0,3  s 5,10, 10 . This line can be written as r  2,0,3  t 1, 2, 2 . From
1,6, k  2,0,3  t 1, 2, 2 we obtain 1  2  t  t  3 and 6  0  2t  t  3 . Thus,
k  3  (2)3 = 3 .
18.
 2  6i 2   1  i  a  bi   32  24i   a  b    a  b  i .
Solving the system a  b  32
and a  b  24 gives (a, b)  (28, 4) . Using the arithmetic sequence 4  6i, 7  2i , x  yi ,
then 2  7  2i    4  6i    x  yi   10  2i  x  yi . Thus ( x, y )  (10, 2) . We have
a  b 28  4

 4 .
x  y 10  2
19.
If m is the length and n is the width at any point in this procedure, then the area of the
m n

shaded region taken from the m by n rectangle would be
, the dimensions of the
10 10
9
9
new rectangle would be
m by
n , and the area of the shaded region in the m by n
10
10
rectangle would be one-hundredth of that rectangle's area. Thus the sum of the areas of the
10 20 
1 
1   81   162 
shaded rectangles is
This is a geometric

  9   18    


10 10  10  
10   100   100 
2
200
2
9

series. The first term is 2 and the common ratio is   . The sum is
.
19
1  .81
 10 
20.
Since the smaller radius is 2 2 , the edge of the cube is of
length 4. Note that the larger radius R of the frustum is
the sum of r and an additional length x. By using the
2
Pythagorean Theorem, we obtain x 
h
2
5  4  3.
r
Thus, R  2 2  3 . We now set up similar triangles,
letting the height of the full cone be h. Clearly,


2 2 3
h4
r
2 2
. Thus h  4 
.


3
h
R 2 2 3

 4 2 2 3
The volume is
2 2 3 

3
3




2
  


5
4
4
x
R

4 2 2  3
9

3
.
21.
Notice that
x 2 y 2  2 x 2 y  x 2  2 xy 2  4 xy  2 x  y 2  2 y  1 



 x  12  y  12
 36.
 

x2 y 2  2 y  1  2 x y 2  2 y  1  1 y 2  2 y  1 
If  x  1
2
 y  12
 36 , then  x  1  1, 4, 9, or 36 and  y  1  36, 9, 4, or 1,
2
respectively. As a result  x  1, y  1 
2
  1,  6 ,   2,  3 ,   3,  2 , or   6,  1 .
Hence, the answer is 16 .
22.
Since the angle of incidence equals the angle of
F
A
B
reflection, AFP  BFC , but since AB || CD ,
AFP  FPC and BFC  FCP , so
FPC  FCP , making PFC isosceles. The
D
P
altitude from F bisects PC at G. Placing P at D
and sliding P toward C we see that G can always
G
A
C
FB
bisect PC until P passes the midpoint of DC .
Thus, P can lie anywhere on the first half of DC .
D
C
P
G
Ans: 1/2 .
23.
If he puts 8 numbers into 7 non-empty sets, then the Pigeonhole principle applies, meaning
that exactly one set must have two elements. So we count the number of different ways to
select two objects from 8: 8 C2 
87
 28 . If he puts 8 numbers into two non-empty
2
sets then there are two choices for each number, either it goes in set A or in set B. That
gives 28 possibilities. But the sets are non-empty so he can't put all the numbers into A or
all the numbers into B, giving 28  2 possibilities. Finally, this number of possibilities
double counts since it distinguishes between sets A and B, so the result is
28  2
 27  1  127 . Total: 127 + 28 = 155 .
2
24.
Let mCAD   . We have AB  AC  7 ,
E
mBAD  mCAE  60   , and AD  AE  13 , so
BAD  CAE by SAS, making CE  BD . Use the
13
Law of Cosines to compute BD:
A
AD2  AB2  ( BC  CD)2  2( AB)( BC  CD) cos 60
7 60 q
. Let CD  x , getting 169  49  ( x  7)  7(7  x) .
2
This simplifies to x 2  7 x  120  0 which factors
60
13
7
B
7
C
x
D
as ( x  15)( x  8)  0  x  8 . Since CD  8 , BD  BC  CD  7  8  15 . Since
CE  BD , then CE = 15 .
25.
Given that c  a, b , then from 3(a  b  c)  abc we obtain the following inequality:
3(c  c  c)  abc , giving 9c  abc . Since c  0 this means that ab  9  ab  8 . We
obtain a lower bound for ab by noting that since a  b  c  c then 3(a  b  c)  3c .
Substituting abc for 3(a + b + c) gives abc  3c  ab  3  ab  4 . From
4  ab  8 , we obtain the following ordered pairs for (a, b): (1, 4), (1, 5), (1, 6), (1, 7),
(1, 8), (2, 3), and (2, 4). From (1, 4) we obtain (1, 4, 15), (1, 5) gives (1, 5, 9), (1, 6) gives
(1, 6, 7), and (2, 3) gives (2, 3, 5). The others fail for different reasons: (1, 7) gives
(1, 7, 6) making b  c while (1, 8) and (2, 4) give non-integer values for c. So there are 4
ordered triples.
ANSWERS:
1.
2.
3
4.
5.
6.
7.
8.
E
E
D
C
A
D
D
B
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
'22.
23.
24.
25.
A
B
D
D
C
E
B
C
B
E
E
C
E
A
A
C
C