1991 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If for any three distinct numbers

a,

b,

and

c

we define

\boxed{a,b,c}=\frac{c+a}{c-b},

then

\boxed{1,-2,-3}=

-2

-\frac25

-\frac14

\frac25

2

Concepts:defined operation substitution

Difficulty rating: 840

Small hint:

Substitute

a=1,

b=-2,

and

c=-3

into the definition

Big hint:

Be careful with the two negative signs in the denominator

Solution:

Direct substitution gives

\begin{aligned} \boxed{1,-2,-3} &=\frac{-3+1}{-3-(-2)}\\ &=\frac{-2}{-1}=2. \end{aligned}

Thus the correct answer is E.

2.

|3-\pi|=

\frac17

0.14

3-\pi

3+\pi

\pi-3

Concepts:absolute value comparing constants

Difficulty rating: 770

Small hint:

Recall that

\pi\gt3

Big hint:

The absolute value of a negative number is its opposite

Solution:

Since

\pi\gt3,

the quantity

3-\pi

is negative. Therefore

|3-\pi|=-(3-\pi)=\pi-3.

Thus the correct answer is E.

3.

\left(4^{-1}-3^{-1}\right)^{-1}=

-12

-1

\frac1{12}

1

12

Concepts:negative exponent reciprocal fraction subtraction

Difficulty rating: 840

Small hint:

Rewrite the negative first powers as reciprocals

Big hint:

First simplify

\frac14-\frac13

, then take its reciprocal

Solution:

We have

4^{-1}-3^{-1}=\frac14-\frac13=-\frac1{12}.

Raising this result to the power

-1

takes its reciprocal, giving

-12.

Thus the correct answer is A.

4.

Which of the following triangles cannot exist?

An acute isosceles triangle

An isosceles right triangle

An obtuse right triangle

A scalene right triangle

A scalene obtuse triangle

Concepts:triangle angle classification

Difficulty rating: 800

Small hint:

A right triangle already contains an angle of

90^\circ

Big hint:

An obtuse triangle must contain an angle greater than

90^\circ

Solution:

A right triangle has one

90^\circ

angle, leaving only

90^\circ

for its other two positive angles together. Neither remaining angle can therefore be obtuse. Hence a triangle cannot be both right and obtuse.

Thus the correct answer is C.

5.

In the arrow-shaped polygon shown, the angles at vertices

A,

C,

D,

E,

and

F

are right angles,

BC=FG=5,

CD=FE=20,

DE=10,

and

AB=AG.

The area of the polygon is closest to

288

291

294

297

300

Concepts:polygon area decomposition isosceles right triangle

Difficulty rating: 1360

Small hint:

Separate the arrowhead from the rectangular shaft

Big hint:

Use

AB=AG

and the two right angles at

C

and

F

to determine the arrowhead

Solution:

The shaft

CDEF

is a

20

-by-

10

rectangle, so its area is

200.

Also

BG=BC+CF+FG

=5+10+5=20.

Because

\angle BAG=90^\circ

and

AB=AG,

triangle

ABG

is an isosceles right triangle with hypotenuse

20.

Its legs are

10\sqrt2,

so its area is

\frac12(10\sqrt2)^2=100.

The total area is

200+100=300.

Thus the correct answer is E.

6.

If

x\ge0,

then

\sqrt{x\sqrt{x\sqrt{x}}}=

x\sqrt{x}

x\sqrt[4]{x}

\sqrt[8]{x}

\sqrt[8]{x^3}

\sqrt[8]{x^7}

Concepts:nested radical fractional exponent

Difficulty rating: 1260

Small hint:

Work from the innermost square root outward

Big hint:

Rewrite each square root as a power of

\frac{1}{2}

Solution:

Starting inside and using fractional exponents,

\begin{aligned} \sqrt{x\sqrt{x\sqrt{x}}} &=\sqrt{x\cdot x^{\frac{3}{4}}}\\ &=x^{\frac{7}{8}} =\sqrt[8]{x^7}. \end{aligned}

Thus the correct answer is E.

7.

If

x=\frac{a}{b},

a\ne b,

and

b\ne0,

then

\dfrac{a+b}{a-b}=

\frac{x}{x+1}

\frac{x+1}{x-1}

1

x-\frac1x

x+\frac1x

Concepts:algebraic substitution rational expression

Difficulty rating: 1030

Small hint:

Divide the numerator and denominator by

b

Big hint:

Replace each occurrence of

\frac{a}{b}

by

x

Solution:

Dividing both numerator and denominator by the nonzero number

b

gives

\frac{a+b}{a-b}=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{x+1}{x-1}.

Thus the correct answer is B.

8.

Liquid

X

does not mix with water. Unless obstructed, it spreads out on the surface of water to form a circular film

0.1

cm thick. A rectangular box measuring

6

cm by

3

cm by

12

cm is filled with liquid

X.

Its contents are poured onto a large body of water. What will be the radius, in centimeters, of the resulting circular film?

\frac{\sqrt{216}}{\pi}

\sqrt{\frac{216}{\pi}}

\sqrt{\frac{2160}{\pi}}

\frac{216}{\pi}

\frac{2160}{\pi}

Concepts:volume conservation cylinder circle area

Difficulty rating: 1260

Small hint:

Equate the volume in the box to the volume of the thin circular film

Big hint:

The film volume is

\pi r^2(0.1)

Solution:

The liquid volume is

6\cdot3\cdot12=216

cubic centimeters. Thus

0.1\pi r^2=216,

so

r^2=\frac{2160}{\pi}

and

r=\sqrt{\frac{2160}{\pi}}.

Thus the correct answer is C.

9.

From time

t=0

to time

t=1

a population increased by

i\%,

and from time

t=1

to time

t=2

the population increased by

j\%.

Therefore, from time

t=0

to time

t=2

the population increased by

(i+j)\%

ij\%

(i+ij)\%

\left(i+j+\frac{ij}{100}\right)\%

\left(i+j+\frac{i+j}{100}\right)\%

Concepts:percent change compound growth

Difficulty rating: 1310

Small hint:

Represent the two increases by multiplication factors

Big hint:

Expand

\left(1+\frac{i}{100}\right)\left(1+\frac{j}{100}\right)

Solution:

The combined growth factor is

\begin{aligned} &\left(1+\frac{i}{100}\right) \left(1+\frac{j}{100}\right)\\ &\qquad=1+\frac{i+j}{100} +\frac{ij}{10000}. \end{aligned}

Hence the percent increase is

i+j+\frac{ij}{100}.

Thus the correct answer is D.

10.

Point

P

is

9

units from the center of a circle of radius

15.

How many different chords of the circle contain

P

and have integer lengths?

11

12

13

14

29

Concepts:circle chord distance counting

Difficulty rating: 1610

Small hint:

Find the shortest and longest chord through

P

Big hint:

Except at the extreme lengths, each attainable chord length occurs in two directions

Solution:

The longest chord through

P

is a diameter of length

30.

The shortest is perpendicular to the radius through

P,

and has length

2\sqrt{15^2-9^2}=24.

As the chord rotates between these positions, its length varies continuously from

24

to

30.

The endpoint lengths each occur once, while each of the five interior integer lengths

25,26,27,28,29

occurs twice. Thus the count is

1+2(5)+1=12.

Thus the correct answer is B.

11.

Jack and Jill run

10

kilometers. They start at the same point, run

5

kilometers up a hill, and return to the starting point by the same route. Jack has a

10

-minute head start and runs at the rate of

15

km/hr uphill and

20

km/hr downhill. Jill runs

16

km/hr uphill and

22

km/hr downhill. How far from the top of the hill are they when they pass going in opposite directions?

\frac54

km

\frac{35}{27}

km

\frac{27}{20}

km

\frac73

km

\frac{28}{9}

km

Concepts:relative motion rates distance

Difficulty rating: 1680

Small hint:

Determine Jill’s position when Jack reaches the top

Big hint:

After Jack turns around, use their combined closing speed

Solution:

Jack reaches the top in

\frac{5}{15}=\frac{1}{3}

hour. Jill has then run for

\frac{1}{3}-\frac{1}{6}=\frac{1}{6}

hour and is

\frac{16}{6}=\frac{8}{3}

km uphill, so they are

5-\frac{8}{3}=\frac{7}{3}

km apart. Their closing speed is

20+16=36

km/hr, so they meet after

\frac{\frac{7}{3}}{36}=\frac{7}{108}

hour. Jack descends

20\left(\frac7{108}\right)=\frac{35}{27}

kilometers.

Thus the correct answer is B.

12.

The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of positive integers. Let

m^\circ

be the measure of the largest interior angle of the hexagon. The largest possible value of

m^\circ

is

165^\circ

167^\circ

170^\circ

175^\circ

179^\circ

Concepts:polygon angle sum arithmetic sequence divisibility

Difficulty rating: 1630

Small hint:

Write the angles as

a,a+d,\ldots,a+5d

and use their sum

Big hint:

The equation forces

d

to be even, and convexity bounds the largest angle below

180^\circ

Solution:

The angle sum is

720^\circ,

so

\begin{aligned} 6a+15d&=720,\\ 2a+5d&=240. \end{aligned}

Hence

d

is even. The largest angle is

a+5d=120+\frac52d,

which must be less than

180.

Thus

d\lt24,

and the largest allowable even value is

d=22.

It gives

a=65

and

m=65+5(22)=175.

Thus the correct answer is D.

13.

Horses

X,

Y,

and

Z

are entered in a three-horse race in which ties are not possible. If the odds against

X

winning are

3

-to-

1

and the odds against

Y

winning are

2

-to-

3,

what are the odds against

Z

winning? (By “odds against

H

winning are

p

-to-

q

” we mean that the probability of

H

winning the race is

\frac{q}{p+q}.

)

3

-to-

20

5

-to-

6

8

-to-

5

17

-to-

3

20

-to-

3

Concepts:probability odds complementary events

Difficulty rating: 1520

Small hint:

Convert each set of odds to a probability

Big hint:

The three winning probabilities sum to

1

Solution:

The given probabilities are

P(X)=\frac14

and

P(Y)=\frac35.

Therefore

P(Z)=1-\frac14-\frac35=\frac3{20}.

The probability that

Z

loses is

\frac{17}{20},

so the odds against

Z

are

17

-to-

3.

Thus the correct answer is D.

14.

If

x

is the cube of a positive integer and

d

is the number of positive integers that are divisors of

x,

then

d

could be

200

201

202

203

204

Concepts:divisor count prime factorization perfect cube

Difficulty rating: 1750

Small hint:

In the prime factorization of a cube, every exponent is divisible by

3

Big hint:

Try a cube having only one prime factor

Solution:

For any prime

p,

take

x=p^{201}=(p^{67})^3.

This is a cube, and its positive divisors are

1,p,p^2,\ldots,p^{201},

a total of

202.

Thus

202

can occur.

Thus the correct answer is C.

15.

A circular table has exactly

60

chairs around it. There are

N

people seated at this table in such a way that the next person to be seated must sit next to someone. The smallest possible value of

N

is

15

20

30

40

58

Concepts:circular arrangement covering extremal counting

Difficulty rating: 1490

Small hint:

No empty chair can have both neighboring chairs empty

Big hint:

Between consecutive occupied chairs there can be at most two empty chairs

Solution:

The condition says that every empty chair has an occupied neighbor, so no three consecutive chairs may all be empty. Therefore each occupied chair can account for at most itself and the two empty chairs following it, giving

60\le3N

and

N\ge20.

Repeating the pattern occupied-empty-empty around the table attains

N=20.

Thus the correct answer is B.

16.

One hundred students at Century High School participated in the AHSME last year, and their mean score was

100.

The number of non-seniors taking the AHSME was

50\%

more than the number of seniors, and the mean score of the seniors was

50\%

higher than that of the non-seniors. What was the mean score of the seniors?

100

112.5

120

125

150

Concepts:weighted average percent linear equation

Difficulty rating: 1410

Small hint:

First determine the numbers of seniors and non-seniors

Big hint:

Let the non-senior mean be

x

and form a weighted-average equation

Solution:

If there are

s

seniors, then there are

1.5s

non-seniors, so

2.5s=100

and

s=40.

Let the non-senior mean be

x;

the senior mean is

1.5x.

The total score equation is

60x+40(1.5x)=100(100),

so

120x=10000,

and the senior mean is

1.5x=125.

Thus the correct answer is D.

17.

A positive integer

N

is a palindrome if the integer obtained by reversing the sequence of digits of

N

is equal to

N.

The year

1991

is the only year in the current century with the following two properties:

(a) It is a palindrome.

(b) It factors as a product of a

2

-digit prime palindrome and a

3

-digit prime palindrome.

How many years in the millennium between

1000

and

2000

(including the year

1991

) have properties (a) and (b)?

1

2

3

4

5

Concepts:palindrome prime factorization divisibility

Difficulty rating: 1880

Small hint:

Every four-digit palindrome from

1000

to

2000

has the form

1dd1

Big hint:

Factor

1dd1

as

11(91+10d)

and test when the three-digit factor is prime

Solution:

The palindromes in the interval are

1dd1,

where

d

is a digit. Algebraically,

\begin{aligned} 1dd1&=1001+110d\\ &=11(91+10d). \end{aligned}

The two-digit prime palindrome must therefore be

11,

and

91+10d

must be a three-digit prime palindrome. For

d=0,1,\ldots,9,

the prime values occur for

d=1,4,6,9,

giving

101,131,151,181.

Thus there are

4

years.

Thus the correct answer is D.

18.

If

S

is the set of points

z

in the complex plane such that

(3+4i)z

is a real number, then

S

is a

right triangle

circle

hyperbola

line

parabola

Concepts:complex plane real and imaginary parts locus

Difficulty rating: 1780

Small hint:

Write

z=x+yi

and expand the product

Big hint:

Set the imaginary part of

(3+4i)(x+yi)

equal to zero

Solution:

Writing

z=x+yi,

\begin{aligned} (3+4i)z &=(3x-4y)\\ &\quad +(4x+3y)i. \end{aligned}

This is real exactly when

4x+3y=0,

which is the equation of a line through the origin.

Thus the correct answer is D.

19.

Triangle

ABC

has a right angle at

C,

AC=3,

and

BC=4.

Triangle

ABD

has a right angle at

A

and

AD=12.

Points

C

and

D

are on opposite sides of

AB.

The line through

D

parallel to

AC

meets

CB

extended at

E.

If

\frac{DE}{DB}=\frac mn,

where

m

and

n

are relatively prime positive integers, then

m+n=

25

128

153

243

256

Concepts:coordinate geometry perpendicular vectors distance

Difficulty rating: 2110

Small hint:

Place

C=(0,0),

A=(0,3),

and

B=(4,0)

Big hint:

Find

D

by using a length-

12

vector perpendicular to

\overrightarrow{AB}

Solution:

Place

C=(0,0),

A=(0,3),

and

B=(4,0).

Since

\overrightarrow{AB}=(4,-3),

the length-

12

vector perpendicular to

\overrightarrow{AB}

and directed away from

C

is

(\frac{36}{5},\frac{48}{5}).

Hence

D=(\frac{36}{5},\frac{63}{5}).

Since

DE\parallel AC,

we have

E=(\frac{36}{5},0),

so

DE=\frac{63}{5}

and

DB=13.

Therefore

\frac{DE}{DB}=\frac{\frac{63}{5}}{13}=\frac{63}{65}.

Thus

m+n=63+65=128.

Thus the correct answer is B.

20.

The sum of all real

x

such that

\begin{aligned} &(2^x-4)^3+(4^x-2)^3\\ &\qquad=(4^x+2^x-6)^3 \end{aligned}

is

\frac32

2

\frac52

3

\frac72

Concepts:sum of cubes exponential equation factoring

Difficulty rating: 1950

Small hint:

Set

a=2^x-4

and

b=4^x-2

Big hint:

Factor

a^3+b^3-(a+b)^3

as

-3ab(a+b)

Solution:

Let

a=2^x-4

and

b=4^x-2.

The right side is

(a+b)^3,

so

a^3+b^3=(a+b)^3

is equivalent to

ab(a+b)=0.

The cases give

2^x=4,

4^x=2,

or

4^x+2^x=6,

whose real solutions are

x=2,\frac12,

and

1,

respectively. Their sum is

2+\frac12+1=\frac72.

Thus the correct answer is E.

21.

If

f\left(\frac{x}{x-1}\right)=\frac1x

for all

x\notin\{0,1\},

and

0\lt\theta\lt\frac{\pi}{2},

then

f(\sec^2\theta)=

\sin^2\theta

\cos^2\theta

\tan^2\theta

\cot^2\theta

\csc^2\theta

Concepts:function substitution trigonometric identity

Difficulty rating: 1800

Small hint:

Choose

x

so that

\frac{1}{x}=\sec^2\theta

Big hint:

Use

1-\cos^2\theta=\sin^2\theta

Solution:

Since

\sec^2\theta=\frac1{\cos^2\theta} =\frac{\csc^2\theta}{\csc^2\theta-1},

choose

x=\csc^2\theta

in the definition. Then

f(\sec^2\theta)=\frac1{\csc^2\theta}=\sin^2\theta.

Thus the correct answer is A.

22.

Two circles are externally tangent. Lines

PAB

and

PA'B'

are common tangents with

A

and

A'

on the smaller circle and

B

and

B'

on the larger circle. If

PA=AB=4,

then the area of the smaller circle is

1.44\pi

2\pi

2.56\pi

\sqrt8\pi

4\pi

Concepts:tangent circles homothety power of a point

Difficulty rating: 2160

Small hint:

The two circles are related by a dilation centered at

P

Big hint:

Use

PA=4,

PB=8

to find the dilation ratio, then apply the tangent-length relation to the smaller circle

Solution:

Since

PA=4

and

AB=4,

we have

PB=8.

The homothety centered at

P

taking the smaller circle to the larger therefore has ratio

2.

If the smaller radius is

r,

its center is

3r

from

P:

the centers are on the same ray, their distances from

P

have ratio

2,

and their difference is

r+2r=3r.

Applying the right triangle formed by

P,

the small center, and

A

,

PA^2=(3r)^2-r^2=8r^2.

Thus

16=8r^2,

so

r^2=2

and the area is

2\pi.

Thus the correct answer is B.

23.

If

ABCD

is a

2\times2

square,

E

is the midpoint of

AB,

F

is the midpoint of

BC,

AF

and

DE

intersect at

I,

and

BD

and

AF

intersect at

H,

then the area of quadrilateral

BEIH

is

\frac13

\frac25

\frac7{15}

\frac8{15}

\frac35

Concepts:coordinate geometry line intersection polygon area

Difficulty rating: 2250

Small hint:

Assign coordinates to the square and write equations for

AF,

DE,

and

BD

Big hint:

Find

I

and

H,

then use the shoelace formula on

B,E,I,H

Solution:

Set

B=(0,0),

C=(2,0),

D=(2,2),

and

A=(0,2).

Then

E=(0,1),

F=(1,0),

and the line intersections are

\begin{aligned} I&=AF\cap DE =\left(\frac25,\frac65\right),\\ H&=AF\cap BD =\left(\frac23,\frac23\right). \end{aligned}

The shoelace formula gives

\begin{aligned} [BEIH] &=\frac12\left| \frac4{15}-\frac25-\frac45 \right|\\ &=\frac7{15}. \end{aligned}

Thus the correct answer is C.

24.

The graph,

G,

of

y=\log_{10}x

is rotated

90^\circ

counter-clockwise about the origin to obtain a new graph

G'.

Which of the following is an equation for

G'?

y=\log_{10}\left(\frac{x+90}{9}\right)

y=\log_x10

y=\frac1{x+1}

y=10^{-x}

y=10^x

Concepts:graph transformation rotation logarithm exponential

Difficulty rating: 1710

Small hint:

A

90^\circ

counter-clockwise rotation sends

(x,y)

to

(-y,x)

Big hint:

Rename the rotated coordinates and solve the logarithmic relation for the new

y

Solution:

A point

(u,\log_{10}u)

rotates to

(x,y)=(-\log_{10}u,u).

Hence

x=-\log_{10}y,

so

\log_{10}y=-x

and

y=10^{-x}.

Thus the correct answer is D.

25.

If

T_n=1+2+3+\cdots+n

and

\begin{aligned} P_n&=\frac{T_2}{T_2-1} \cdot\frac{T_3}{T_3-1}\\ &\quad\cdot\frac{T_4}{T_4-1} \cdots\frac{T_n}{T_n-1} \end{aligned}

for

n=2,

3,

4,

\ldots,

then

P_{1991}

is closest to which of the following numbers?

2.0

2.3

2.6

2.9

3.2

Concepts:telescoping product triangular number approximation

Difficulty rating: 1950

Small hint:

Factor

T_k-1

and simplify each factor

\frac{T_k}{T_k-1}

Big hint:

Write each factor as

\frac{k}{k-1}\cdot\frac{k+1}{k+2}

and telescope

Solution:

Since

T_k=\frac{k(k+1)}{2}

and

T_k-1=\frac{(k-1)(k+2)}{2},

\frac{T_k}{T_k-1} =\frac{k}{k-1}\cdot\frac{k+1}{k+2}.

The product telescopes:

\begin{aligned} P_n &=\left(\prod_{k=2}^n\frac{k}{k-1}\right) \left(\prod_{k=2}^n\frac{k+1}{k+2}\right)\\ &=n\cdot\frac3{n+2}\\ &=\frac{3n}{n+2}. \end{aligned}

For

n=1991,

this is just under

3

and is closest to

2.9.

Thus the correct answer is D.

26.

An

n

-digit positive integer is cute if its

n

digits are an arrangement of the set

\{1,2,\ldots,n\}

and its first

k

digits form an integer that is divisible by

k,

for

k=1,

2,

\ldots,

n.

For example,

321

is a cute

3

-digit integer because

1

divides

3,

2

divides

32,

and

3

divides

321.

How many cute

6

-digit integers are there?

0

1

2

3

4

Concepts:divisibility permutations case analysis

Difficulty rating: 2270

Small hint:

The second digit must be even, the third-prefix digit sum must be divisible by

3,

and the fourth prefix must be divisible by

4

Big hint:

The fifth digit must be

5,

and the full number must be even and divisible by

3

Solution:

Divisibility by

5

forces the fifth digit to be

5.

Divisibility by

2

and

6

forces the second and sixth digits to be even. Now apply the divisibility tests successively: the first three digits must have sum divisible by

3,

and the two-digit number formed by the third and fourth digits must be divisible by

4.

Checking the remaining choices from

\{1,2,3,4,6\}

leaves

123654

and

321654.

Each number directly satisfies all six prefix divisibility conditions, so there are

2.

Thus the correct answer is C.

27.

If

\begin{aligned} x+\sqrt{x^2-1} &+\frac1{x-\sqrt{x^2-1}}\\ &=20, \end{aligned}

then

\begin{aligned} x^2+\sqrt{x^4-1} &+\frac1{x^2+\sqrt{x^4-1}}\\ &= \end{aligned}

5.05

20

51.005

61.25

400

Concepts:conjugate radicals reciprocal algebraic identity

Difficulty rating: 2110

Small hint:

Rationalize

\frac1{x-\sqrt{x^2-1}}

Big hint:

The last two terms of the requested expression are conjugates

Solution:

Because

\frac1{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1},

the given equation implies

x+\sqrt{x^2-1}=10.

Its reciprocal is

x-\sqrt{x^2-1}=\frac{1}{10},

so adding yields

2x=10.1

and

x=5.05.

Also

\frac1{x^2+\sqrt{x^4-1}}=x^2-\sqrt{x^4-1}.

Therefore the requested expression is

2x^2=2(5.05)^2=51.005.

Thus the correct answer is C.

28.

Initially an urn contains

100

black marbles and

100

white marbles. Repeatedly, three marbles are removed from the urn and replaced from a pile outside the urn as follows:

Marbles removed	Replaced with
$3$ black	$1$ black
$2$ black, $1$ white	$1$ black, $1$ white
$1$ black, $2$ white	$2$ white
$3$ white	$1$ black, $1$ white

Which of the following sets of marbles could be the contents of the urn after repeated applications of this procedure?

2

black marbles

2

white marbles

1

black marble

1

black and

1

white marble

1

white marble

Concepts:invariant constructive process parity

Difficulty rating: 2330

Small hint:

The number of white marbles always changes by

0

or

2

Big hint:

When an operation first leaves at most two marbles, inspect the possible outputs of that final operation

Solution:

The parity of the number of white marbles never changes, so choices with one white marble are impossible. Also a final operation starting with three marbles can leave

1

black,

1

black and

1

white, or

2

white, but never

2

black. The state with

2

white is attainable: repeatedly replace

3

black by

1

black until

2

black remain; twice use the

1

-black-

2

-white rule, leaving

100

white; then alternate the

3

-white rule with the

1

-black-

2

-white rule, reducing the number of white marbles by

2

per pair until

2

remain.

Thus the correct answer is B.

29.

Equilateral triangle

ABC

has been creased and folded so that vertex

A

now rests at

A'

on

BC

as shown. If

BA'=1

and

A'C=2,

then the length of crease

PQ

is

\frac85

\frac7{20}\sqrt{21}

\frac{1+\sqrt5}{2}

\frac{13}{8}

\sqrt3

Concepts:folding perpendicular bisector coordinate geometry distance

Difficulty rating: 2360

Small hint:

A fold crease is the perpendicular bisector of the segment joining a point to its image

Big hint:

Place

B=(0,0),

C=(3,0),

A=(\frac{3}{2},\frac{3\sqrt3}{2}),

and

A'=(1,0)

Solution:

With the coordinates in the hint, the perpendicular-bisector condition for

X=(x,y)

is

|X-A|=|X-A'|,

which simplifies to

x+3\sqrt3\,y=8.

Intersecting this line with

AB,\ y=\sqrt3x,

gives

P=(\frac{4}{5},\frac{4\sqrt3}{5}).

Intersecting it with

AC,\ y=\sqrt3(3-x),

gives

Q=(\frac{19}{8},\frac{5\sqrt3}{8}).

Hence

\begin{aligned} PQ &=\sqrt{\left(\frac{63}{40}\right)^2 +\left(\frac{7\sqrt3}{40}\right)^2}\\ &=\frac7{20}\sqrt{21}. \end{aligned}

Thus the correct answer is B.

30.

For any set

S,

let

|S|

denote the number of elements in

S,

and let

n(S)

be the number of subsets of

S,

including the empty set and the set

S

itself. If

A,

B,

and

C

are sets for which

\begin{gathered} n(A)+n(B)+n(C)\\ {}=n(A\cup B\cup C),\\ |A|=|B|=100, \end{gathered}

then what is the minimum possible value of

|A\cap B\cap C|?

96

97

98

99

100

Concepts:set cardinality subsets powers of two extremal counting

Difficulty rating: 2430

Small hint:

Use

n(S)=2^{|S|}

and determine

|C|

and

|A\cup B\cup C|

Big hint:

Within the union, count how many elements each of

A,

B,

and

C

can omit

Solution:

Let

c=|C|

and

u=|A\cup B\cup C|.

Since

n(S)=2^{|S|},

the equation becomes

2^{101}+2^c=2^u.

The two summands must be equal, so

c=101

and

u=102.

Within the union,

A,

B,

and

C

omit

2,

2,

and

1

elements. Thus at most five distinct elements are absent from the triple intersection, giving

|A\cap B\cap C|\ge97.

Equality is attained by making those omissions distinct.

Thus the correct answer is B.