New York State Mathematics
Association of Two-Year Colleges
Math League Contest ~ Spring 2016 ~ Solutions
1.
Since,
x
y
z
x y z
and
, then
 1 , for x  0 , and the same for
  could be  3 , 1 , 1 , or 3.
x
y
z
x y z
Also,
x yz
x yz
 1, provided x yz  0. However,
 1 only if one or all three of the variables
x yz
x yz
are negative, making
x y z
equal 1 or  3. Thus, making f ( x, y, z )  0 or  4. A similar
 
x y z
line of reasoning shows that when
x yz
 1 , f ( x, y, z )  0 or 4.
x yz
Answer: B
2.
We can simplify the problem by dividing all values in the table by 2, to obtain the counting numbers,
then determine the location of 1008 (i.e. 2016  2). 1008  11 = 91 with a remainder of 7, which
tells us that we have to count a full 91 rows, then go 7 positions on the 92nd row. Thus, m  92 and
Answer: 99
n  7 , making m  n  99.
3.
The graph of
f ( x)  x 2  20 x  2016 is a parabola with a minimum at
10, 2116)

x   2201  10 and y  f (10)  100  200  2016   2116. Therefore, the graph
of g ( x)  x 2  20 x  2016 appears as shown. Hence, x 2  20 x  2016  k will
have exactly three real solutions when the horizontal line y  k passes through ( 10, 2116), i.e.
4.
when k  2116.
Answer: 2116
We can draw radii from the center of the circle to each point of contact with the two
3
squares, and then form an equilateral triangle (as shown) since the radii and the side of
the square all have length 1. The height of the triangle is 23 . Thus, the center of the
circle
1
5.
3
2

is
1
3
2
 radius  1 

units
3
2
1 
from
the
3
.
2
 
"floor,"

and
the
distance
we
seek
2
is
Answer: 23

log 120 2  log(120)  log 21 2  log 10  2 2  3  12 log(2)  log(10)  2 log(2)  log(3)  12 log(2)
 
 1  52 log(2)  log(3) . Now use log(8)  x and log(9)  y to obtain: log 2 3  x or log(2)  13 x
 
 
and log 3 2  y or log(3)  12 y. Thus, we get: 1  52 13 x  12 y.
6.




Answer: 1  56 x  12 y
2
n 4  n 2  1  n 4  2 n 2  n 2  1  n 4  2 n 2  1  n 2  n 2  1  n 2 , which is the difference of



two squares and factors as: n 2  1  n n 2  1  n . When n  1 , we get 3 (a prime number).
For any other integer, we get a non-prime. Note: n  0 yields 1, which is not prime.
Answer: B
7.
7
 2x 
x
2
7  x 1
14 x  8



 x ( x  8)  14  72
72
x
72
x
 x 2  8x  14  72  0  x 2  8x  28  36  0
 ( x  28)( x  36)  0  x  28 or x   36
8.
 statue
7
Letting x = the side length of the playground. The description of the situation
yields the diagram at right. Using similar triangles, we get:
1
72
Answer: 28
Letting x  the number of minutes past 8 PM one person shows up, and
y  the number of minutes past 8 PM the other person shows, they will
meet only if x  y  10 or 10  x  y  10. Graphing this region between
0  x  60 and 0  y  60 , i.e. the one-hour interval they each have to
arrive, gives the region illustrated. The area of the entire square region is
60 2  3600 , while the area of the non-shaded region consists of two
isosceles triangles each with an area of 12  50  50  1250 for a total area of
2500. Thus, the area of the shaded region is 3600  2500  1100, making the probability that the
two friends meet 1100
Answer: C
 11
.
3600
36
9.
The identity 1  tan 2 ( x)  sec2 ( x) gives: sec2 ( x)  tan 2 ( x)  1 , which factors as
[sec( x)  tan( x)][sec( x)  tan( x)]  1 , now use sec( x)  tan( x)  2016 to obtain
[sec( x)  tan( x)]  2016  1  sec( x)  tan( x) 
Answer: C
1
2016
10. A cross-section through the center of each sphere yields the following
diagram, giving two small right triangles where r is the radius of the
circle of we seek. The Pythagorean Theorem applied to the two right
triangles gives:
x 2  r 2  30 2 and (50  x) 2  r 2  40 2. Subtracting
the first from the second gives: (50  x)  x  40  30 , which
simplifies to: 2500  100 x  700  x  18. Substituting x  18 into
2
2
2
30
r
40
50  x
x
2
x 2  r 2  30 2 yields: 18 2  r 2  30 2  r 2  30 2  18 2  (30  18)(30  18)
 r 2  12  48  24  24  24 2  r  24.
Answer: 24
Alternate solution: Using similar triangles (since the large 30-40-50 triangle is also a right triangle),
r  30  r  3040  120  24.
we obtain: 40
50
50
5
11. Using the change of base formula, we can rewrite the equation as:

log(10) log(100)

 10
log(b) log b 1/ 2
 
1
2
1
4
5
1
1
 10 

 10 
 10  log(b) 
log(b) 2 log(b)
log(b) log(b)
log(b)
2
Hence, b  101/ 2  10.
12. The solved 4⨯4 problem is:
Answer:
10
Answer: 5
13.
 
2 74,207,281  1  2107,420,728  210
7,421,000
 
 1024 7,421,000  10 3
7,421,000
 10 22,263,000 ,
which is a 1 followed by approximately 22.3 million zeros.
Alternate solution:
Answer: B
210  1024  10log(2)  log(1024)  10log(2)  log(1000)  10log(2)  3

 log(2)  0.3  2  10 0.3  2 74,207,281  1  10 0.3

74,210,000
 10 22,263,000
106
14.
k  86 
 n  2016  86  87  88 
 105  106  2016.
The only other solution
n  86
would have to be when k  0 , so that we have enough negative integers to "cancel" the sum
Thus,
the
other
solution
must
be
since
k   85 ,
1  2  3   84  85.
Answer:  85
(85)  (84)  (83)   83  84  85  0.
15. The 51-second (24+3+24 seconds) time-line shows 0
24 27
51
34
44
17
the light-cycle, where changes occur at the t  24, 27
...
7
7
7
and 51 second marks (assuming we start the instant
the light turned green at t  0 ). We can see the range of 7-second intervals for which we would see
a light change, i.e. any time t  (17,34) or t  (44,51]. That is a 17-second interval plus a 7-second
interval on the 51-second cycle. Thus, the probability of observing a light change is
177
51

24
51
38
8.
 3 17  17
Answer: D
16. Since the graph of x 2  y 2  x  y is symmetric about both the x-axis and y-axis, it
is sufficient to determine the area bound in the first-quadrant then multiply by four
to obtain the entire area. In the first-quadrant, we can dispense with the absolute
values to get: x 2  y 2  x  y. Rearranging and completing the square for x and y :
 x  12    y  12   12 . Thus, in the first quadrant, the graph is a part of the circle
centered at  12 , 12  with a radius of 22 , as shown (the full graph is also shown for
2
2
reference). This makes the area, in the first quadrant, consist of the semi-circle with
radius
1
2
 
2
2
2
2
2
and a right triangle with a base and height of length 1, i.e.
 12 11  14   12 . Therefore, the entire enclosed area is 4  14   12     2.
Answer: D
17. A cross-section through the cone, perpendicular to the base and through its apex,
gives the diagram shown, where the height is represented by the dotted line and
2
the radius of the base is designated by r. Thus, r  1 by noticing we have a
3
30-60-90 right triangle or using the Pythagorean Theorem. Therefore, the
r
circumference of the base of the cone is 2 1  2 . This tells us that the arc length
subtended by θ from the original circle is 2 . Since the original circle has a circumference of
2  2  4 ,  
2
4
 360 o  180 o.
Answer: 180

 
18. Letting A  arctan 12
along the coordinate axes: the hypotenuse of the right triangle with angle A
terminates at (2,1) and the hypotenuse of the right triangle with angle B
terminates at (1,3). The slope of the line segment joining them is 1231  2, while
the slope of the hypotenuse of the right triangle with angle A is 12 . Thus, the
triangle formed between the other two is also a right triangle with legs of length:
12  22  5 and
(1,3)
and B  arctan 13 , construct two right triangles each
B
(2,1)
θ
A
(2  1)2  (1  3)2  5 , showing that the right triangle is isosceles. Hence,
  45 o. Therefore, A  B  90 o  45 o  45 o or
 radians.
4
Answer: B
Alternate solution:
Using the sine formula for the sum of angles, i.e. sin( A  B)  cos( A)sin( B)  cos( B)sin( A). From
the triangles formed above, we obtain: cos( A)  2 , cos( B)  3 , sin( A)  1 , and sin( B)  1 .
5
Thus, sin( A  B) 
2
5

1
10

3
10

1
5

5
50

1 .
2
10
5

Hence, A  B  4 .
obtained, of course, using cos( A  B)  cos( A)cos( B)  sin( A)sin( B).
19. Drawing a chord (dotted line) spanning the two smaller chords
and a radius to the point where the two smaller chords meet,
forms two small congruent right triangles with hypotenuses
of length 2 and a right triangle with a hypotenuse of length 3. The
Pythagorean Theorem gives:

4  a2
  (3  a)
2
2
a b  2
2
 32  a 
2
2
 b  4a
2
 b  4  94 
3
2
3
2
and
10
The same result can be
b
a
2
x
3 a
b
3
2. Use the Pythagorean Theorem again
for the right triangle whose hypotenuse is the diameter of the semi-circle with legs 2b and

x (the length in question): 2  23 2

2
 x 2  6 2  x 2  36  16
 196
 x  14
.
9
9
3
Answer: C
20. Letting x  the number of first place votes the player received, y  the number of second place
votes received, and z  the number of third place votes received.
Thus,
5 x  3 y  z  37 , with x  y  z  10. The table lists the only solutions. In each case,
exactly 9 of the 10 writers voted for the player.
Answer: C