2011 AMTNJ Contest
Solutions
1) 20. Only integers which end in 1 or 9 will have a 1 in the units place
after squaring. Therefore, there will be 20.
2) 80 Draw the picture. CDB will be isosceles with m B  m DCB  40 .
Therefore m BDC  100 and m ADC  80 .
3)
34650
4)
2010
2011
11!
because of the repeating letters.
4!4!2!
Given Ax2  Bx  C  0 , the sum of the roots is
B
A
5) 2, or 6 Solving both 6  2 x  x2  3x and (6  2 x  x 2 )  3x gives x = -3,
-1, 2 or 6. Checking shows that only 2 and 6 check.
6) 24 Setting AP = 1x, PB = 5x, PQ = 2y, and QB = 3y, and solving gives
x = y =4. Therefore AB = 24.
7)
19 yards
Let L = Layla rate, J = Julian rate, and P = Pressley rate.
d 100 90
100 90
100 90
100 90


 , solving the systems


and t2 
and
r
L
J
J
P
L
J
J
P
100 81
 , therefore Layla should give Pressley a 19 yard headstart.
gives
L
P
t1 
8) 13
Using the fact that sides AB and CD are not adjacent and the
Triangle Inequality Theorem on all 4 triangle faces of the tetrahedron, leads to
only one possible solution. If AB = 41, then CD = 13.
9)
Using m 
4 points
281  17 88
and starting at (3, 17) leads to (18,

48  3 15
105), (33, 193), and (48, 281) as the only 4 possible points.
10) 89 Solving the systems; A+B+C=180, A+B+D=197, B+C+D=208, and
A+C+D=222, gives a maximum vale of 89.
11) 5.77
Draw the picture and label the point of intersection D. By the
Law of Cosines AC  32  42  283848cos(120)  6.0828 . By the Law of Sines
CAB  sin 1 (4*sin(120) / 6.0828)  34.715 , DAC  90  CAB  55.285.
D  360  (90  90  120)  60. And finally, by the Law of Sines
CD  6.0828*sin(55.285) / sin(60)  5.77 .
12)
1
1
(1  50 )
2
30
Let p be the probability that the total number of heads is
even and q be the probability that the total number of heads if odd.
0
50
2
48
50
0
 50   2   1 
 50   2   1 
 50   2   1 
p               ...      and
 0  3   3 
 2  3   3 
 50   3   3 
1
49
3
47
49
1
 50   2   1 
 50   2   1 
 50   2   1 
q               ...      . Note that
 1  3   3 
 3  3   3 
 49   3   3 
50
1
1
1 
 2 1
p  q      50 . Since p+q = 1, we get p  1  50 
3
2 3 
 3 3
13) 100
They both start a new cycle every 20 days. Bob’s schedule is
WWWRWWWRWWWRWWWRWWWR. Bill’s schedule is
WWWWWWWRRRWWWWWWWRRR. They both rest on the same day
twice every 20 days.
1000
 50 , therefore 50*2  100 .
20
14)
3
2
By the double angle formula A  120 . Therefore
sin( A)  sin(120) 
3
.
2
15) 10log(2)
 A3 B 4 
3log x A  4log x B  7log x C  log x  7  
 C 
10
 103204 
 10324104 
 24 
log x 

log

log

log
2
 10log x 2
x
x  14 
x


7
7
7
40
4
10
2






 