Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event A
Question #1 is intended to be a quickie and is worth 1 point. Each of the next three questions is worth 2 points.
Place your answer to each question on the line provided. You have 12 minutes for this event.
NO CALCULATORS are allowed on this event.
For the purposes of this event, a “simpli@ied” answer (or portion of an answer) will be de@ined as:
a quotient of two polynomials having no common factor.
x
2
+
x + 3 x − 2
1.
Simplify: 2.
( x − 8)( x − x − 2)
Simplify: ( x − 2)( x − 4)( x + 2x + 4x )
3.
⎛ 1⎞
⎛ 1⎞
Determine exactly all solutions to the equation 6
⎜⎝ x ⎟⎠ − 29⎜⎝ x ⎟⎠ + 35 = 0 .
4.
Express 3
2
2
3
2
2
x = (x
2x −1
2
)(
+ x +1 x +1
)
as the sum of two simpli*ied rational expressions, one of which has a quadratic denominator while the other has a linear denominator.
Name: ___________________________________ Team: ___________________________________
Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event A
SOLUTIONS
NO CALCULATORS are allowed on this event.
For the purposes of this event, a “simpli@ied” answer (or portion of an answer) will be de@ined as:
a quotient of two polynomials having no common factor.
x2 + 6
2
x + x − 6
x
2
Simplify: +
x + 3 x − 2
1.
Graders: A factored denominator is okay.
x +1
2
x + 2x
Graders: award 1 point per correct solution.
3x + 2
−3
+
x + x +1 x +1
2
2.
3x + 2
3
−
x + x +1 x +1
(
)
2 x +3
+
=
x 2 −2x +2x + 6
( x + 3)( x − 2) ( x + 3)( x − 2) ( x + 3)( x − 2)
=
x2 + 6
x2 + x − 6
2
2
3
2
( x − 8)( x − x − 2) = ( x − 2) ( x + 2x + 4) ( x − 2) ( x +1) = x +1 .
( x − 2)( x − 4)( x + 2x + 4x ) ( x − 2) ( x − 2) ( x + 2)( x ) ( x + 2x + 4) x + 2x
3
2
2
3
2
2
2
2
⎛ 1⎞
⎛ 1⎞
Determine exactly all solutions to the equation 6
⎜⎝ x ⎟⎠ − 29⎜⎝ x ⎟⎠ + 35 = 0 .
3.
Let y =
4.
Also accept:
2
)
( x − 8)( x − x − 2)
Simplify: ( x − 2)( x − 4)( x + 2x + 4x )
2
⎧2 3 ⎫
x ∈ ⎨ , ⎬
5 7⎭
⎩
=
3
Graders: A factored denominator is okay.
(
x x −2
1
5
7
2
3
. Then 6 y 2 − 29 y + 35 = 0 ⇒ 2 y −5 3 y − 7 = 0 ⇒ y = or
⇒ x = or .
x
2
3
5
7
(
Express (x
2x −1
2
)(
+ x +1 x +1
)
)(
)
as the sum of two simpli*ied rational expressions, one of which has a quadratic denominator while the other has a linear denominator.
Using partial fraction decomposition, (
(
2x −1
)(
)
=
Ax + B
C
+
. Therefore, 2
x + x +1 x +1
x 2 + x +1 x +1
Ax + B x +1 + C x 2 + x +1 = 2x −1 . Substituting x = 0 yields B + C = –1, while
)(
) (
)
substituting x = –1 yields C = –3. This means B = 2. Since there is no x 2 term in the @inal (
)(
) (
) (
)
(
) (
)
numerator, we have Ax + B x +1 + C x 2 + x +1 = A + C x 2 + A + B + C x + B + C , so A + C = 0, and A = 3. The resulting decomposition is (x
2x −1
2
)(
+ x +1 x +1
)
=
3x + 2
−3
+
.
x + x +1 x +1
2
.
Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event B
Question #1 is intended to be a quickie and is worth 1 point. Each of the next three questions is worth 2 points.
Place your answer to each question on the line provided. You have 12 minutes for this event.
1.
An equilateral triangle is both inscribed and circumscribed by circles (Figure 1). Determine exactly the ratio of the length of the outer circle’s radius to the length of the inner circle’s radius.
2.
Figure 2 shows four circles of radius 1 that are externally tangent to each other and internally tangent to a larger circle. Determine exactly the length of the larger circle’s radius.
Figure 1
Figure 2
3.
Four tennis balls Kit perfectly inside a cylindrical container (Fig. 3). What fraction of the container’s volume is not occupied by balls?
Figure 3
QD = 4.
A circle inscribed into trapezoid ABCD (with AD  BC ) intersects sides AB and CD at points P and Q, respectively. If it is known that AP = 6, PB = 8, and QC = 12, determine exactly the length QD.
Name: ___________________________________ Team: ___________________________________
Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event B
SOLUTIONS
2 :1
1.
2
or
or 2
1
An equilateral triangle is both inscribed and circumscribed by circles (Figure 1). Determine exactly the ratio of the length of the outer circle’s radius to the length of the inner circle’s radius.
2r
r 3
The two radii are the hypotenuse and short leg of a 30-­‐60-­‐90 triangle, and therefore, one is twice the length of the other.
1+ 2
2.
Figure 1
Figure 2 shows four circles of radius 1 that are externally tangent to each other and internally tangent to a larger circle. Determine exactly the length of the larger circle’s radius.
Connect the centers of the four smaller circles to create a square of side length 2. The diagonal of this square has 1
1
length 2 2 , so the diameter of the larger circle is 2+ 2 2 , and its radius has half that length, or 1+ 2 .
1
3
3.
r
Figure 2
1
Four tennis balls Kit perfectly inside a cylindrical container (Fig. 3). What fraction of the container’s volume is not occupied by balls?
Figure 3
also accept:
33.333%
0.333
0.3
QD = 4
No matter how many tennis balls we stack, the ratio of air : cylinder will remain the same. Consider, then, a single tennis ball @it into a cylinder with radius r and height 2r: Volume of sphere = 43 π r 3 ; Volume of cylinder = π r 2h = π r 2 2r = 2π r 3 . Volume of air = 1
V(cylinder) – V(sphere) = 23 π r 3 , which is the volume of the cylinder.
3
( )
4.
A circle inscribed into trapezoid ABCD (with AD  BC ) intersects sides AB and CD at points P and Q, respectively. If it is known that AP = 6, PB = 8, and QC = 12, determine exactly the length QD.
x
6
D
A
Draw and label the trapezoid as shown in Figure 4, using liberally the fact that two tangents to a circle from the same point are equal in length. Drop altitudes from both A and D. Using the left altitude: h + 2 = 14 ⇒ h = 192 .
2
(
2
) (
2
2
2
)
Q
P
h
h
12
8
2
Using the right altitude: h2 + 12− x = 12+ x
2
⇒
192
+144
−
24x
+x
= 144 + 24x +x 2 ⇒ x = DQ = 4 .
x
6
B
2
12 - x
6
8
12
Figure 4
C
Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event C
Question #1 is intended to be a quickie and is worth 1 point. Each of the next three questions is worth 2 points.
Place your answer to each question on the line provided. You have 12 minutes for this event.
x =
a =
2
1. In an arithmetic sequence, several terms are scratched out, leaving:
. . . , 3, ____ , x, ____ , ____ , 58, . . .
Determine exactly the value of the term x.
2.
When the Kirst term is removed from an inKinite geometric series, the sum of the series changes from 5 to 2. Determine exactly the second term of the original series.
3.
Find all ordered pairs of digits (A, B) for which A 1 = B ! + 1 .
4.
Find all ordered triples of integers (x, m, n), with x, m, n ≥ 0, for which
m ⎛
n ⎛
⎞
⎞
n−k
∑ ⎜ m ⎟ x k 3 m−k = 1024 and ∑ ⎜ n ⎟ x k −2
= 729 .
k =0 ⎝ k ⎠
k =0 ⎝ k ⎠
( A,B ) =
( )
2
( x , m, n) =
( )
Name: ___________________________________ Team: ___________________________________
Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event C
SOLUTIONS
x = 25
1. In an arithmetic sequence, several terms are scratched out, leaving:
. . . , 3, ____ , x, ____ , ____ , 58, . . .
Determine exactly the value of the term x.
58 − 3 55
=
= 11 , so x = 3+ 2 11 = 3+ 22 = 25 .
The common difference is 5
5
( )
6
a =
2 5
1
or 1.2
5
or 1
2.
When the Kirst term is removed from an inKinite geometric series, the sum of the series changes from 5 to 2. Determine exactly the second term of the original series.
a
⎛ 2⎞ 6
3
2
⇒ 5−5r = 3 ⇒ r = , and a2 = 3⎜ ⎟ = .
The @irst term must be 3, so S∞ = 1 ⇔ 5 =
1− r
1− r
5
⎝ 5⎠ 5
( A,B ) =
(1, 5) and (7, 7)
Graders: 1 point per correct pair; give a maximum of 1 point on this problem if any incorrect pairs are listed.
3.
( )
2
Find all ordered pairs of digits (A, B) for which A 1 = B ! + 1 .
Since all integers that end in 1 will have also have a last digit of 1 when squared, we are looking for factorials that end in 0. The candidates are 5! = 120 , 6! = 720 , 7! = 5040 , 8!= 40320 , and 2
2
9! = 362880 . Only 5! + 1 = 121 = 11 and 7! + 1 = 5041 = 71 satisfy the given conditions, so the ordered pairs are (1, 5) and (7, 7).
( x , m, n) =
(29, 2, 2)
Graders: award only 1 point if exactly one extra triple is listed; zero points for two or more additional triples
4.
Find all ordered triples of integers (x, m, n), with x, m, n ≥ 0, for which
m ⎛
n ⎛
⎞
⎞
n−k
∑ ⎜ m ⎟ x k 3 m−k = 1024 and ∑ ⎜ n ⎟ x k −2
= 729 .
k =0 ⎝ k ⎠
k =0 ⎝ k ⎠
( )
(
)
m
The given sigma notations are representative of the binomial expansions of x + 3 and m
n
10
2
5
x − 2 respectively, so we have x + 3 = 1024 = ±2 = 4 = ±32 = 10241 and n
6
2
x − 2 = 729 = ±3 = 93 = ±27 = 7291 . The @irst set of equalities gives us x = –5, –1, 1, 29, –35, and 1021. The second set of equalities yields x = –1, 5, 11, –25, 29, and 731. (
(
)
)
( )
(
(
)
)
( )
(
)
Since x ≥ 0, x = 29, and the only ordered triple that works is (x, m, n) = (29, 2, 2).
Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event D
Question #1 is intended to be a quickie and is worth 1 point. Each of the next three questions is worth 2 points.
Place your answer to each question on the line provided. You have 12 minutes for this event.
1. Often thought of as a rational function, the equation xy = k , where k is non-­‐zero, actually describes the graph of a hyperbola. What are the equations of its asymptotes?
y − int =
2. Write an equation for the parabola whose directrix is the line x = –3 and whose focus is located at (3, 0).
3.
To facilitate water run-­‐off, roads are often designed with parabolic-­‐shaped cross-­‐sections.
One such road is 32 feet wide, and 0.4 foot higher in the center than at the edges. If we place the origin at the high point (center) of the road, write an equation for this parabola.
4.
2
2
In the coordinate plane, the graphs of x + y − 4x + 6 y −12 = 0 and the parabola
( )
y = A x + bx + c, A ≠ 0 have exactly three points in common. If two of these points are (–3, –3) and (7, –3), determine exactly the y-­‐intercept of the parabola.
2
2
Name: ___________________________________ Team: ___________________________________
Minnesota State High School Mathematics League
2012-13 Meet 4, Individual Event D
SOLUTIONS
x = 0, y = 0
1.
Often thought of as a rational function, the equation xy = k , where k is non-­‐zero, also accept:
actually describes the graph of a hyperbola. the x-­‐ and y-­‐axes
1 2
x=
y
12
What are the equations of its asymptotes?
2. Write an equation for the parabola whose directrix is the line x = –3 and whose focus is located at (3, 0).
or equivalent
The parabola opens to the right (away from the directrix and toward the focus), so its equation 1 2
must be of the form x =
y , where p is the 4p
1 2
focal length. p = 3, so we have x =
y .
12
y=
−1 2
x
640
3.
or equivalent
To facilitate water run-­‐off, roads are often designed with parabolic-­‐shaped cross-­‐sections.
One such road is 32 feet wide, and 0.4 foot higher in the center than at the edges. If we place the origin at the high point (center) of the road, write an equation for this parabola.
Using the given conditions, the edges of the road can be marked by the points (16, –0.4) and
(–16, –0.4). Substitute the @irst of those points into the general equation y = ax to obtain
2
( )
−0.4 = a 16
36
y − int = − 5
4.
2
⇒ − 0.4 = 256a ⇒ a =
−0.4
−4
−1
−1 2
x .
=
=
. The equation is y =
256 2560 640
640
2
2
In the coordinate plane, the graphs of x + y − 4x + 6 y −12 = 0 and the parabola
( )
y = A2 x 2 + bx + c, A ≠ 0 have exactly three points in common. If two of these points are (–3, –3) and (7, –3), determine exactly the y-­‐intercept of the parabola.
1
or −7
or −7.2
5
(
)
2
(
)
2
(
) (
2
)
2
x 2 + y 2 − 4x + 6 y −12 = 0 ⇔ x − 2 − 4 + y + 3 − 9−12 = 0 ⇔ x − 2 + y + 3 = 25 , which is a circle centered at (2, –3) with radius 5. The two given points on the parabola are symmetric −3+ 7
= 2 . This means the third point about the axis of symmetry, which must be located at x =
2
must be a point of tangency at either the top or bottom of the circle! Since the parabola has a positive x 2 coef@icient, it faces upward, and its vertex must be located at (2, –8). Starting from (
)
2
y = a x − 2 − 8 , we use (–3, –3) to @ind that a = 0.2; substitute x = 0 to @ind the y-­‐int = –7.2.
Minnesota State High School Mathematics League
2012-13 Meet 4, Team Event
Each question is worth 4 points. Team members may cooperate in any way, but at the end of 20 minutes,
submit only one set of answers. Place your answer to each question on the line provided.
( A, B, C , D) =
1.
(
)
2
x 3 + 8x + 32
2
The rational expression , where P(x)= x + 4 and Q(x)= x +2 , can be P(x)⋅Q(x)
Ax + B Cx + D
expressed in the form +
, where A, B, C, and D are integers. Do so.
Q(x)
P(x)
inches 2.
In Event D, a parabolic road surface was described as 32 feet wide and 0.4 foot higher in the center than at the edges. A car drives in the “middle” of the right-­‐hand lane, so that the midlines of its left tires are 5 feet (measured horizontally) from the center of the road, while the midlines of its right tires are 5 feet (measured horizontally) from the right edge of the road. Calculate how many inches higher the car’s left tires are than its right tires.
(
)(
)(
)(
)
(
)(
)
k =
3.
Let X k = 2 2 − 2 2 2 + 2 + 3 3 − 3 3 3 + 3 +…+ k k − k k k + k . What is the smallest integer k for which X k is a multiple of 2012?
BD =
4.
Point D is located on side BC of ABC such that the inscribed circles of ABD and ACD are tangent to each other. If AB = 13, BC = 15, and AC = 18, determine BD exactly.
c =
5.
6 ⎛
⎞
Determine exactly all real numbers c for which ∑ ⎜ 6 ⎟ 3n c 6−n = 1000 .
n=0 ⎝ n ⎠
T =
6.
Alec wrote a computer program to compute the sum 13 + 23 + 33 +…+ n3 for any given 


positive integer n > 100. Unfortunately, an error in the program code causes the program to omit the Kirst T terms. The resulting output of the Klawed program is now equivalent to (
)
a polynomial of the form 14 n2 + n + 42 ⋅p(n) , where p(n) is a quadratic polynomial with integer coefKicients. What is the value of T?
Team: ___________________________________
Minnesota State High School Mathematics League
2012-13 Meet 4, Team Event
SOLUTIONS (page 1)
( A, B, C , D) =
( −2, 1, 3, 7)
1.
Graders: Award 1 point for each correct value
1.8 inches
(
)
2
x 3 + 8x + 32
2
The rational expression , where P(x)= x + 4 and Q(x)= x +2 , can be P(x)⋅Q(x)
Ax + B Cx + D
expressed in the form +
, where A, B, C, and D are integers. Do so.
Q(x)
P(x)
2.
In Event D, a parabolic road surface was described as 32 feet wide and 0.4 foot higher in the center than at the edges. A car drives in the “middle” of the right-­‐hand lane, so that the midlines of its left tires are 5 feet (measured horizontally) from the center of the road, while the midlines of its right tires are 5 feet (measured horizontally) from the right edge of the road. Calculate how many inches higher the car’s left tires are than its right tires.
Figure 2
k = 670
3.
(
)(
)(
)(
)
(
)(
)
Let X k = 2 2 − 2 2 2 + 2 + 3 3 − 3 3 3 + 3 +…+ k k − k k k + k . What is the smallest integer k for which X k is a multiple of 2012?
A
BD = 5
4.

Point D is located on side BC of ABC such 
that the inscribed circles of ABD and ACD are tangent to each other. If AB = 13, BC = 15, and AC = 18, determine BD exactly.

13-x
x
B x
c = −3± 10
18-(13-x)
y
yD y
15
5+x
C
5.
6 ⎛
⎞
Determine exactly all real numbers c for which ∑ ⎜ 6 ⎟ 3n c 6−n = 1000 .
n=0 ⎝ n ⎠
6.
Alec wrote a computer program to compute the sum 13 + 23 + 33 +…+ n3 for any given Graders: Award 2 points for each correct value
T = 6
Figure 4
13-x
positive integer n > 100. Unfortunately, an error in the program code causes the program to omit the Kirst T terms. The resulting output of the Klawed program is now equivalent to (
)
a polynomial of the form 14 n2 + n + 42 ⋅p(n) , where p(n) is a quadratic polynomial with integer coefKicients. What is the value of T?
Minnesota State High School Mathematics League
2012-13 Meet 4, Team Event
SOLUTIONS (page 2)
1.
) (
)(
2
Ax + B Cx + D
x 3 + 8x + 32
2
3
+
=
, so Ax + B x + 2 + x + 4 Cx + D = x + 8x + 32 . Expanding the left side, we 2
2
2
x +4
x +2
x2 + 4 x + 2
(
) (
)(
(
)
)(
)
(
)
(
)
(
) (
have: Ax 3 + Bx 2 + 4Ax 2 + 4Bx + 4Ax + 4B + Cx 3 + Dx 2 + 4Cx + 4D = A + C x 3 + 4A + B + D x 2 + 4A + 4B + 4C x + 4B + 4D
)
⎧
A+C =1
⎪
⎪
3
= x + 8x + 32 ⇒ ⎨ 4A + B + D = 0 ⇒ 4A + 8 = 0 ⇒ A = −2 ⇒ C = 3 ⇒ B = 1 ⇒ D = 7 .
⎪ A+ B +C = 2
⎪⎩
B+D=8
()
2.
Event D participants should be able to provide the equation for the parabola: y =
for the x-­‐values of 5 and 11: y =
96
=
3
640 20
3. −1
640
(5 )
2
−25
=
640
and y =
−1
(11)
2
640
, but remember that this value is in feet! Multiply by (
=
12 in.
1 ft.
) (
(
)
−121
640
−1
640
x 2 . Now calculate the y-­‐coordinates . The difference in these two y-­‐coordinates is to convert to inches: 3
20
⋅12 =
9
5
= 1.8 inches .
)
X k = 8 − 4 + 27 − 9 +…+ k 3 − k 2 . If we append a trivial term of (1 – 1) to the front of this series, we can rewrite it as (
⎛ k k +1
∑ n3 − ∑ n2 = ⎜⎜ 2
n=1
n=1
⎝
k
k
) ⎞⎟
2
⎟⎠
−
(
)(
k k +1 2k +1
6
) = k (k +1) ⎛⎜ k (k +1) − 2k +1 ⎞⎟ = k (k +1) ⎛ 3k
2
⎜⎝
3 ⎟⎠
2
2
⎜
⎝
2
(
)(
)(
)
− k − 2 ⎞ k k +1 3k + 2 k −1
⋅
. ⎟=
6
2
6
⎠
2
2012 = 2 ⋅503 , and these factors must all appear among the numerators’ factors. The minimum k occurs when 3k + 2 = 503 ⇒ k = 167 , which makes k + 1 and k – 1 both even to generate the two necessary factors of 2.
4.

As shown in Figure 4, label the tangent from B to the incircle of ABD as length x, and the tangent from D to that incircle as length y. Repeatedly using the fact that two tangents to a circle from the same point are congruent. chase lengths around ABC as shown. Note that the three tangents drawn from D are all congruent, so that BC = 15 = x + y + y + (5 + x).

Solve for (x + y) to @ind that BD = 5 .
5.
⎛ 6 ⎞ n 6−n ⎛ 6 ⎞ 0 6 ⎛ 6 ⎞ 1 5
⎛ 6 ⎞ 5 1 ⎛ 6 ⎞ 6 0
6
⎟⎠ 3 c = ⎜⎝
⎟⎠ 3 c + ⎜⎝
⎟⎠ 3 c +…+ ⎜⎝
⎟⎠ 3 c + ⎜⎝
⎟⎠ 3 c , which is the binomial expansion of 3+ c . n
0
1
5
6
n=0
(
6
∑ ⎜⎝
(
So 3+ c
6. )
6
(
= 1000 = 103 ⇒ 3+ c
)
2
= 10 ⇒ 3+ c = ± 10 ⇒ c = −3± 10 .
(
⎛ n n +1
1 + 2 + 3 +…+ n − 1 + 2 + 3 +…+T = ⎜
⎜⎝ 2
3
3
3
3
(
3
3
3
3
)
) ⎞⎟
2
(
⎛ T T +1
−⎜
⎟⎠ ⎜⎝
2
) ⎞⎟
(
2
)
2
(
)
2
⎡n n +1 ⎤ − ⎡T T +1 ⎤
⎦ ⎣
⎦ . Applying the difference of two =⎣
⎟⎠
4
1
n + n − (T + T ))( n + n + (T + T )) = ⋅ p(n)⋅ ( n + n + 42) . (
4
4
1
2
2
2
2
squares to the numerator, we have 2
2
So T + T = 42 ⇒ T + T − 42 = 0 ⇒ T + 7 T − 6 = 0 ⇒ T = −7 or 6 .
)
(
)(
)
2