Download 2010DiscoveryRound1Solutions

Rocket City Math League
2009-2010
Round 1
Discovery Test Solutions
1. The number of ways he can choose 2 of his 10 friends =
=
= 45
45
2.
, so
, and so on. The pattern
repeats
every four terms. 2009/4 has a remainder of one, so the answer will be the first term in the
pattern. Therefore,
OR
i 2009  i 2008 i1  (1)(i)  i

3.
190
4. Factor by grouping
i
 
to base 10.
=
=
to get
. So, the roots of
the equation are
and -3. The rational root is -3.
5. The probability that the bridge collapses and Zinke gets the treasure = (0.7)(0.1) = 0.07 =
The probability that the bridge does not collapse and Zinke gets the treasure = (1-.7)(.95) =
0.285 =
190
-3
.
So, the probability that the bridge collapsed, given that he got the treasure, =
=
6. Use the law of cosines to find the angle.
.
. So,
. Sketch the triangle with a hypotenuse of 16 and
8
an adjacent leg of 5 to find the sine of the angle. To find the length of the other leg, use the
formula
, so b =
7. To find the number of integers less than 2009 divisible by 3, divide 2008 by 3. 2008/3 =
669.33, so there are 669 multiples of 3. 2008/4 = 502, so there are 502 multiples of 4. You must
subtract the multiples of 12 so that no integer is counted twice. 2008/12 = 167.33, so there are
167 multiples of 12. The total number of integers less than 2009 divisible by 3 or 4 = 669 + 502
- 167 = 1004.
8. R: The sum of the first 13 cubes equals the sum of the first 13 numbers squared. For


2
example: 13  2 3  33  36 , and 1  2  3 =36. This isn’t necessarily a shortcut, but it’s
faster than calculating all the cubes and then adding them because you can use the
 n(n  1) 
 13(14) 
2
summation formula. 
 
  91  8281
2
2




2
C: The form of a harmonic mean is
2
n
where n is the number of terms, and a and b are
1 1

a b
n
2
3

the terms themselves. So
=
1 1 1 1 2


a b 1 3
M: The geometric mean is the nth root of product of n numbers, so since there are 2
numbers in this case, it is the square root of them =
=
= 24.
L: The arithmetic mean of 3 and 5 = (3+5)/2 = 4.
R+C+M+L = 8281+3/2+ 24+4=
1004
9. Absolute value is defined as a point’s distance from zero, or in this case the origin.
Therefore, the inequality |z|≤ 1 is the set of all complex numbers whose distance from the
origin is less than or equal to one, which is a circle of radius one whose center is the origin.
Similarly, |z-|≤1 is another circle of radius one that is shifted to the right by 2. The graph
looks like this:
B
We are looking for the area of the middle
1
region. So create a triangle by connecting the
1/2
center of one circle to the two points of intersection
A
D
E
(ABC). You know that AB is 1 because it is the
radius of circle A. You also know that AE is ½
because AE is congruent to DE and AD is 1 (a radius
of
either circle). Angle AEB is a right angle, so use the
C
3
Pythagorean Theorem and you find the BE=
. Using inverse trig or knowing your 30-602
90 ratios, you see that angle BAE is 60 degrees and angle ABE is 30 degrees. Angle CAE is
congruent to angle BAE, so angle CAB=60+60=120 degrees. The area of sector CAB is
n deg rees
120

  r 2 
   12 
360
360
3
1
1 1
3
3

Now find that the area of triangle BEA=  bh   
. The area of triangle CAB
2
2 2 2
8
3
is twice that, or
.
4
To find the entire middle region, subtract the area of the triangle from the area of the sector,
then you will be left with the area of half of the region. Multiply this by two to get the whole
area.
4  3 3
6
or
2
3

3
2

3 
2
3 4  3 3
2   




4 
3
2
6
3
10. The area of the circle =
. The area of the ellipse =
. a = the radius
of the circle = 12. The minor axis is one half the length of the major axis, so b = a/2 = 6.
Therefore, A =
=
. The area outside the ellipse but inside the circle =
11. The number of ways to get from P to S can be modeled by looking at the number of ways to
get to each intersection on the grid, and building from the
S
starting point toward the goal. Any point along the bottom row
will only have a 1 because Zinke can only get there by moving
right. Simlarly, any point along the left side will also be a 1,
because Zinke can only get there by moving up. To determine
the number of ways to get to the interior points, you sum the
A
number of ways to get to the two points before it. For example,
there are 2 ways to get to point A from point P because Zinke
can go right then up, or up then right, and this is represented by P
the sum of 1+1 from the 1 at each of those points. Following this logic, you will have
1 1 1 1 1 1 1 1 1 along the first row of intersections
1 2 3 4 5 6 7 8 9 along the second
1 3 6 10 15 21 28 36 45 along the third
3 3 3 18 39 75 120 along the fourth
And 1 4 7 25 64 139 259 along the top row.
Since 259 corresponds to point S, there are 259 ways for Zinke to get from point P to point S.
259
12. The given series can be made into an infinite series of infinite series.
1
1
1
1
1  

 ...
3
9
27
81
1 1 1
1
  ...
+  
3 9 27 81
1 1
1
  ...
+ 
9 27 81
1
1
  ...
+
27 81
1
+  ...
81
2 3 4
5
  ... = 1  2  1  4  ... which is the given series.
=1   
3 9 27 81
3 3 27
Now to find the sum of each of these infinite series. The sum of any infinite geometric
a
series with the magnitude of the common ratio less than 1, is equal to
where a is the
1 r
first term, and r is the common ratio. A proof of this fact is shown below (if you already
know this, then you can skip to the *):
Infinite geometric series with first term a and common ratio r has a sum of S means:
a  ar  ar 2  ar 3  ...  S Factor out an r from the left
a

r   a  ar  ar 2  ar 3  ...   S Divide by r.
r

S
a
2
3
and a  ar  ar 2  ar 3  ...  S from the first equation, so
   a  ar  ar  ar 
r
r
 
9
4
S
a
Multiply by r.
  x 
r
r
a  Sr  S Now rearrange.
a  S  Sr Solve for x.
a
S
1 r
*Now using this summation formula, you can say that
1 1 1
1
1
1
3
1  
  ... 


3 9 27 81
1  1/ 3 2 / 3 2
1 1 1
1
1/ 3
1/ 3 1
 
  ... 


3 9 27 81
1  1/ 3 2 / 3 2
1 1
1
1/ 9
1/ 9 1

  ... 


9 27 81
1  1/ 3 2 / 3 6
Following the same procedure as above, the next summations will be 1/18 and 1/54. As you
can see, the summations of these form another infinite series with a first term of 3/2 and a
common ratio of 1/3. Using our formula for the summation of an infinite geometric series
3/ 2
3/ 2 9


we get that the final answer is
1  1/ 3 2 / 3 4
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