minnesota state high school mathematics league ·...

2013-14 Meet 4, Individual Event AQuestion #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. Determine exactly the value of n for which 4−n =8 .

Name: ___________________________________ Team: ___________________________________

Minnesota State High School Mathematics League

2. Simplify the expression x + y( )−1 x −1 + y−1( ) so that it no longer involves any addition or

negative exponents.

4. For all real numbers a and b, the function g satisDies the equation g ab( ) = g a( )⋅ g b( ) .

If g 0( )≠0 , determine exactly the value of g 2013( )+ g 2014( ) .

n =

NO CALCULATORS are allowed on this event.

x =

g 2013( )+ g 2014( ) =

3. Given that f x( ) = x

x −1, x >0 , determine exactly all values of x for which

f x2( ) = 5

24.

SOLUTIONS

Minnesota State High School Mathematics League2013-14 Meet 4, Individual Event A

1xy

f x2( ) = x2

x2 −1= xx2 −1

= 524

. Noting that x ≠1 , cross-‐multiply to get 5x2 −5=24x . Moving all

terms to the left side yields 5x2 −24x −5=0 , which can be factored into 5x +1( ) x −5( ) =0 . This

suggests solutions of x = −1

5or x =5 , but since x must be positive,

x = −1

5 is an extraneous

solution. The only valid solution is x = 5.

5

Let a = 0. Then g 0⋅b( ) = g 0( )⋅ g b( ) ⇒ g 0( ) = g 0( )⋅ g b( ) . We’re told that g 0( )≠0 , so we can

legally use the Division Property of Equality to say that 1= g b( ) . In other words, g is a constant

function, with a value of 1 everywhere! Thus g 2013( )+ g 2014( ) =1+1=2 .

NO CALCULATORS are allowed on this event.

4−n =8 ⇒ 4−n=64 ⇒ n= 4−64 = −60 .

x + y( )−1 x −1 + y−1( ) = 1

x + y1x+ 1y

⎛⎝⎜

⎞⎠⎟= 1x + y

x + yxy

⎛

⎝⎜⎜

⎞

⎠⎟⎟= 1xy.

−60 1. Determine exactly the value of n for which 4−n =8 .

2. Simplify the expression x + y( )−1 x −1 + y−1( ) so that it no longer involves any addition or

negative exponents.

3. Given that f x( ) = x

x −1, x >0 , determine exactly all values of x for which

f x2( ) = 5

24.

4. For all real numbers a and b, the function g satisDies the equation g ab( ) = g a( )⋅ g b( ) .

If g 0( )≠0 , determine exactly the value of g 2013( )+ g 2014( ) .

n =

x =

Graders:When checking for equivalents, please note the restrictions

given in the problem.

2 g 2013( )+ g 2014( ) =

1. Using a circle of radius 3 as a base, a right circular cone of height 6 is constructed. Determine exactly the volume of this cone.

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.

Name: ___________________________________ Team: ___________________________________

Minnesota State High School Mathematics League2013-14 Meet 4, Individual Event B

2. Figure 2 shows a circle with radius OB = 3. This radius is currently parallel to the base line that is tangent to the circle at A. If the circle rolls to the right (without slipping) until point B touches the base line, determine exactly the distance point O will have moved.

4. Square PQRS, with side length 1, is placed with P at the origin and PQ on the x-‐axis (Figure 4). The square then “rolls” to the right, Dirst by rotating about vertex Q to the position indicated by the dotted lines so that QR is on the x-‐axis. It continues to roll in this manner, rotating about vertex R, and then S, and Dinally P, so that each vertex has served as the center of rotation exactly once. Determine exactly the total length of the path traced by point P.

Volume =

OB

A

Figure 2

3. VABC is inscribed in a semicircle with diameter AC . The triangle has side lengths AB = 7 and BC = 3. Determine exactly the length of the altitude dropped from B to AC .

P

S R

Q x

y

Figure 4

Minnesota State High School Mathematics League2013-14 Meet 4, Individual Event B

SOLUTIONS

9π2

O will travel exactly the horizontal distance that the exterior of the circle covers on the baseline as it rolls.

This is exactly the length of arc AB, or 342π ⋅3( ) = 9π2 .

Any triangle inscribed in a semicircle is a right triangle, so we can use the Pythagorean Theorem to determine that AC = 4. Then, treating AB as the base,

Area VABC⎡⎣ ⎤⎦ =

12

7( ) 3( ) , and using AC as the base, Area VABC⎡⎣ ⎤⎦ =

12AC ⋅h

Setting these equal, 7( ) 3( ) = AC ⋅h ⇒ h= 3 7

AC= 3 7

4.

18π

3 74

Volume =

π 1+ 2

2

⎛

⎝⎜

⎞

⎠⎟

or equivalent

1. Using a circle of radius 3 as a base, a right circular cone of height 6 is constructed. Determine exactly the volume of this cone.

2. Figure 2 shows a circle with radius OB = 3. This radius is currently parallel to the base line that is tangent to the circle at A. If the circle rolls to the right (without slipping) until point B touches the base line, determine exactly the distance point O will have moved.

4. Square PQRS, with side length 1, is placed with P at the origin and PQ on the x-‐axis (Figure 4). The square then “rolls” to the right, Dirst by rotating about vertex Q to the position indicated by the dotted lines so that QR is on the x-‐axis. It continues to roll in this manner, rotating about vertex R, and then S, and Dinally P, so that each vertex has served as the center of rotation exactly once. Determine exactly the total length of the path traced by point P.

Figure 2

3. VABC is inscribed in a semicircle with diameter AC . The triangle has side lengths AB = 7 and BC = 3. Determine exactly the length of the altitude dropped from B to AC .

OB

A B'

O'A'

h 37

A C

B

Figure 3

Volume of cone = 13πr

2h= 13π 3( )2 6( ) =18π .

P

P' P''

P'''=P''''x

y

12

1

Figure 4

In the initial rotation, P moves along a quarter-‐circle of radius 1. In the next rotation, it moves along a quarter-‐circle of radius 2 . In the third rotation, it again moves along a quarter-‐circle of radius 1, and in the last rotation, P is itself the center, and thus doesn’t move. The length of

P’s path is 14 2π ⋅1( )+ 1

4 2π ⋅ 2( )+ 14 2π ⋅1( ) =π 1+ 2

2( ) .

1. Determine exactly the sum of the inDinite series 5+

53+59+…


Name: ___________________________________ Team: ___________________________________

Minnesota State High School Mathematics League2013-14 Meet 4, Individual Event C

2. Four distinct integers, chosen from among 1 through 10 inclusive, form an arithmetic sequence. Three of those integers also form a geometric sequence. List both possible sets of four integers that satisfy these conditions. (1 point per correct set)

3. Find the sum of all integers between 1 and 100 inclusive that do not use 2 as a digit.

4. A sequence of integers an , with n ≥ 1, is deDined recursively by an+2 =5an+1 −6an .

If a1 = −11 and a5 =19 , Dind the value of a3 . a3 =

Minnesota State High School Mathematics League2013-14 Meet 4, Individual Event C

SOLUTIONS

This is an inXinite geometric series with a1 =5 and r =

13. Its sum is

a11− r

=51− 1

3= 5

23( ) =

152.

−29

152

There are only three possible 3-‐integer geometric sequences contained within 1 through 10:{1, 2, 4}, {2, 4, 8}, and {1, 3, 9}. In the Xirst sequence, we can insert the integer “3” to make it arithmetic; in the second, we can insert “6”; in the third, there is no way to create a sequence with a common difference. The two possible sets of integers are {1, 2, 3, 4} and {2, 4, 6, 8}.

4357

We can use the Principle of Inclusion/Exclusion. Begin with the sum 1 + 2 + . . . + 100 = 5050. Sum numbers that end in 2: 2 + 12 + . . . + 92 = 2(10) + (10 + 20 + . . . + 90) = 20 + 450 = 470. Sum those that begin with 2: 20 + 21 + . . . + 29 = 20(10) + (1 + 2 + . . . + 9) = 200 + 45 = 245. Subtract the latter two sums from the Xirst sum, adding back in 22, which was subtracted twice:5050 – 470 – 245 + 22 = 5050 – 715 + 22 = 4335 + 22 = 4357.

1, 2, 3, 4

Let x = a3 . Then x =5a2 −6a1 =5a2 −6 −11( ) =5a2 +66 ; a4 =5x −6a2 =5x −6

x −665

⎛⎝⎜

⎞⎠⎟;

and a5 =5a4 −6x ⇒ 19=5 5x −6

x −665

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ −6x ⇒ 19= 25x −6x +396−6x .

Solving for x, 13x +396 =19 ⇒ 13x = −377 ⇒ x = −29 .

1. Determine exactly the sum of the inDinite series 5+

53+59+…

3. Find the sum of all integers between 1 and 100 inclusive that do not use 2 as a digit.

4. A sequence of integers an , with n ≥ 1, is deDined recursively by an+2 =5an+1 −6an .

If a1 = −11 and a5 =19 , Dind the value of a3 . a3 =

or 7.5

2. Four distinct integers, chosen from among 1 through 10 inclusive, form an arithmetic sequence. Three of those integers also form a geometric sequence. List both possible sets of four integers that satisfy these conditions. (1 point per correct set)

and

2, 4, 6, 8


Name: ___________________________________ Team: ___________________________________

Minnesota State High School Mathematics League2013-14 Meet 4, Individual Event D

1. Calculate the length of the radius of the circle x2 +20x + y2 +14 y =20 .

3. A parabola has its focus at (10, 7) and its directrix is the line y = –3. Determine exactly the parabola’s y-‐intercept.

2. Determine exactly the coordinates of both foci of the ellipse x2 +4 y2 =16 .

4. An ellipse centered at the origin has one focus at (3, 1) and intersects the positive y-‐axis at (0, 5). Determine exactly the length of the ellipse’s major axis.

y − int =

r =

x , y( ) =

x , y( ) =

5 10 15 20 25-5

5

-5

-10

-15

-20

5 10 15 20 25-5

5

-5

-10

-15

-20

(3,1)

(-3,-1)

(0,5)

Minnesota State High School Mathematics League

Completing the square on both x and y, we have x2 +20x +100+ y2 +14 y +49= 20+100+49

⇔ x +10( )2 + y +7( )2 =169=132 , so the radius has length 13.

SOLUTIONS2013-14 Meet 4, Individual Event D

Dividing through by 16 yields x2

16+y2

4=1 . The Pythagorean relationship in an ellipse indicates

that c2 = a2 −b2 =16−4 =12 , so c = 12 = 2 3 , and the foci are located at ±2 3, 0( ) .

13

See Figure 3. The directrix is a horizontal line, and the focus lies above it, so this is an upward facing parabola with

equation of the form y =

14p

x −h( )2 + k . The parabola’s vertex is located halfway between the focus and the directrix,

at (10, 2). Thus p = 5, and y =

120

x −10( )2 +2 . Substituting x = 0 reveals the

y − int =

120

−10( )2 +2= 120

100( )+2= 7 .

See Figure 4. The two foci are equidistant and in opposite directions from the center, so the second focus is located at (–3, –1). Using the fact that each ellipse is the locus of points, the sum of whose distances from two foci is a constant, it must be true that the sum of distances from (0, 5) to (3, 1) and (–3, –1) is equal to the length of the major axis. In other words,

0−3( )2 + 5−1( )2 + 0− −3( )2 + 5− −1( )2 = 25 + 45 =5+3 5 .

2 3, 0( )

5+3 5

0, 7( )

Graders: 1 point per correct ordered pair.

1. Calculate the length of the radius of the circle x2 +20x + y2 +14 y =20 .

3. A parabola has its focus at (10, 7) and its directrix is the line y = –3. Determine exactly the parabola’s y-‐intercept.

2. Determine exactly the coordinates of both foci of the ellipse x2 +4 y2 =16 .

4. An ellipse centered at the origin has one focus at (3, 1) and intersects the positive y-‐axis at (0, 5). Determine exactly the length of the ellipse’s major axis.

y − int =

r =

x , y( ) =

x , y( ) = −2 3, 0( )

7

or

Figure 3

Figure 4

5 10 15 20 25 30 35 40 45

5

-5

-10

-15

-20

-25

-30

5 10 15 20 25 30 35 40 45

5

-5

-10

-15

-20

-25

-30

F (10, 7)

V (10, 2)

y = -3

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.

Team: ___________________________________

Minnesota State High School Mathematics League2013-14 Meet 4, Team Event

6. Figure 6 shows two congruent circles of radius r < 1 tangent to each other at C, and internally tangent at A and B to a circle of

radius 2 centered at K. If AC)

+ CB)

= AB)

, (where AC)

represents

arc length AC), express r as a function of central angle θ =∠AKB .

2. Calculate the value of 299 100

1( )−298 1002( )+297 100

3( )+…−22 10098( )+21 100

99( ) .You may express your answer as a power or binomial coefDicient if necessary.

3. Two congruent circles, each of area K, Dit inside the region enclosed by the curve

25x2 + y2 =100 . They are internally tangent to the curve, and externally tangent to

each other. Determine K exactly.

4. Calculate the number of consecutive zeroes that occur at the end of the integer (2014!).

1. If 4a−1( ) =7, 7 b−1( ) =11, 11 c−1( ) =13, and 13 d−1( ) =16,determine exactly the value abcd.

5. Let S be the conic section deDined by the set of points (x, y) that are twice as far from the point (0, 2) as they are from the line y = –1. List the coordinates of any and all foci of S.

abcd =

K =

r =

θ

B

K

A

C

Figure 6


501

SOLUTIONS (page 1)

0, −6( )

9625

π

2θπ +θ

1,267,650,600,228,229,401,496,703,205,376

12

1. If 4a−1( ) =7, 7 b−1( ) =11, 11 c−1( ) =13, and 13 d−1( ) =16,

determine exactly the value abcd.

abcd =

K = 3. Two congruent circles, each of area K, Dit inside the region enclosed by the curve 25x

2 + y2 =100 . They are internally tangent to the curve, and externally tangent to each other. Determine K exactly.

5 10 15 20 25 30

5

10

-5

-10

-15

5 10 15 20 25 30

5

10

-5

-10

-15

x, y0, r-x, y

Figure 3

4. Calculate the number of consecutive zeroes that occur at the end of the integer (2014!).

5. Let S be the conic section deDined by the set of points (x, y) that are twice as far from the point (0, 2) as they are from the line y = –1. List the coordinates of any and all foci of S.

and 0, 2( )

Graders: 1 point per correct ordered pair;

–1 point for each extraneous focus.

r = 6. Figure 6 shows two congruent circles of radius r < 1 tangent to each other at C, and internally tangent at A and B to a circle of

radius 2 centered at K. If AC)

+ CB)

= AB)

, (where AC)

represents

arc length AC), express r as a function of central angle θ =∠AKB .

β βrr

length))=)2θ

θ

B

K

AC

Figure 6

2. Calculate the value of 299 100

1( )−298 1002( )+297 100

3( )+…

−22 100

98( )+21 10099( ) . You may express your answer as a

power or binomial coefDicient if necessary.

2100or

6. Let θ be measured in radians. Then using the arc length formula s = rθ, AB)

= 2θ . An isosceles triangle is formed using

two radii of circle K and the radii of the smaller circles that meet at C. Labeling this triangle’s base angles as β, we have

θ +2β = π ⇒ β =

π −θ

2. This means the central angles of the smaller circles that correspond to AC

)and CB)each

measure π −

π −θ

2=π +θ

2, and

AC)

= CB)

= r ⋅ π +θ

2.

AC)

+ CB)

= AB)

⇒ 2r ⋅π +θ

2= 2θ ⇒ r =

2θ

π +θ .


SOLUTIONS (page 2)

2. This looks binomial. 2+ −1( )( )100 = 100

0( ) 2100( ) −1( )0 + 1001( ) 299( ) −1( )1 +…+ 100

99( ) 21( ) −1( )99 + 100100( ) 20( ) −1( )100 ,

so let x = the desired quantity: 1( )100 = 2100 − 100

1( ) 299( )−…+ 10099( ) 21( )⎡

⎣⎤⎦+1 ⇒ 1= 2100 − x⎡⎣ ⎤⎦+1 ⇒ x = 2100 .

1. log

47 =

1

a, log

711 =

1

b, log

1113=

1

c, log

1316 =

1

d, so using change of base yields log4

7( ) log7 11( ) log1113( ) log1316( ) =

log7

log4⋅log11

log7⋅log13

log11⋅log16

log13=log16

log4= log

416 = 2, which equals

1

a

⎛⎝⎜

⎞⎠⎟

1

b

⎛⎝⎜

⎞⎠⎟

1

c

⎛⎝⎜

⎞⎠⎟

1

d

⎛⎝⎜

⎞⎠⎟=

1

abcd. Therefore, abcd =

12 .

3. 25x2 + y2 = 100 ⇒ x2

4 + y2100 = 1 , so the region is an ellipse with major axis extending from (0, –10) to (0, 10) and

minor axis from (–2, 0) to (2, 0). The region is symmetric about the origin (see Figure 3), so the two circles are tangent to

each other at the origin. Focus on the upper circle. Its center is at (0, r), so its equation is x2 + y − r( )2 = r2 . Expanding

and substituting for x2 in the original equation of the ellipse gives us 25 2ry − y 2( )+ y 2 = 100 ⇒ 12y 2 −25ry +50 = 0 .

The two intersection points deXined by this equation have equal y-‐coordinates, so we want the discriminant of this equation

to equal zero; i.e., −25r( )2 − 4 12( ) 50( ) = 625r2 −2400 = 0 ⇒ r2 = 2400625 = 96

25 . Thus K = πr2 =

9625π .

5. The distance from (x, y) to (0, 2) is x 2 + y −2( )2 , while the distance from (x, y) to y = –1 is y − −1( ) = y +1 . The Xirst

distance is twice the second; i.e. x 2 + y −2( )2 = 2 y +1 ⇒ x2 + y −2( )2 = 4 y +1( )2 ⇒ x2 + y2 − 4 y + 4 = 4 y2 +8 y + 4

⇒ 3y2 +12y − x2 = 0 ⇒ 3 y +2( )2 − x2 = 12 . Dividing through by 12 reveals a vertically oriented hyperbola centered at (0, –2):

y+2( )24 − x2

12 = 1 ⇒ c2 = a2 + b2 = 4 +12= 16 ⇒ c = 4, so the two foci are located at 0, −6( ) and 0, 2( ) .

4. Every trailing zero indicates a factor of 10, which must be created by the product of a factor of 2 and a factor of 5. Factors of 2 occur much more frequently, so it is sufXicient to count the factors of 5. Each multiple of 5 of course creates a factor of 5; there are 2014 ÷5⎢⎣ ⎥⎦ = 402of these. Each multiple of 25 adds an additional factor of 5 (we have already counted the Xirst); there are 2014 ÷25⎢⎣ ⎥⎦ = 80 of these. Similarly, there are 2014 ÷125⎢⎣ ⎥⎦ = 16multiples of 5

3 and 2014 ÷625⎢⎣ ⎥⎦ = 3 multiples of 54 . The total number of factors of 5 (= factors of 10 = trailing zeroes) is 402 + 80 + 16 + 3 = 501 .

minnesota state high school mathematics league ·...

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