1994 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

4^4\cdot9^4\cdot4^9\cdot9^9=

13^{13}

13^{36}

36^{13}

36^{36}

1296^{26}

Concepts:laws of exponents factoring

Difficulty rating: 960

Small hint:

Group the powers of

4

and the powers of

9

Big hint:

Both bases have total exponent

4+9

Solution:

Combining like bases and then pairing them,

\begin{aligned} 4^4\cdot4^9\cdot9^4\cdot9^9 &=4^{13}9^{13}\\ &=(4\cdot9)^{13}\\ &=36^{13}. \end{aligned}

Thus the correct answer is C.

2.

A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle?

$6$	$14$
?	$35$

10

15

20

21

25

Concepts:rectangle areas proportionality

Difficulty rating: 960

Small hint:

Represent the two column widths and two row heights by variables

Big hint:

Products of diagonally opposite areas are equal

Solution:

Let the column widths be

u,v

and the row heights be

s,t.

The displayed areas give

us=6,

vs=14,

and

vt=35.

Therefore the missing area is

ut=\frac{(us)(vt)}{vs}=\frac{6\cdot35}{14}=15.

Thus the correct answer is B.

3.

How many of the following are equal to

x^x+x^x

for all

x\gt0?

\begin{aligned} \mathrm{I:}\quad&2x^x\\ \mathrm{II:}\quad&x^{2x}\\ \mathrm{III:}\quad&(2x)^x\\ \mathrm{IV:}\quad&(2x)^{2x} \end{aligned}

0

1

2

3

4

Concepts:exponential expressions identities

Difficulty rating: 1260

Small hint:

First combine the two identical terms

Big hint:

Test whether changing the base or exponent preserves

2x^x

for every

x

Solution:

The given sum is

2x^x,

so expression I is always equal to it. Expression II fails at

x=1,

where its value is

1

instead of

2.

Expression III fails at

x=2,

where its value is

16

instead of

8.

Expression IV fails at

x=1,

where its value is

4

instead of

2.

Hence only one expression works. Thus the correct answer is B.

4.

In the

xy

-plane, the segment with endpoints

(-5,0)

and

(25,0)

is the diameter of a circle. If the point

(x,15)

is on the circle, then

x=

10

12.5

15

17.5

20

Concepts:circle equation diameter

Difficulty rating: 1070

Small hint:

Find the center and radius from the diameter endpoints

Big hint:

The point’s vertical distance from the center already equals the radius

Solution:

The center is

(10,0)

and the radius is

15.

Thus

(x-10)^2+15^2=15^2,

so

x=10.

Thus the correct answer is A.

5.

Pat intended to multiply a number by

6

but instead divided by

6.

Pat then meant to add

14

but instead subtracted

14.

After these mistakes, the result was

16.

If the correct operations had been used, the value produced would have been

less than

400

between

400

and

600

between

600

and

800

between

800

and

1000

greater than

1000

Concepts:reverse operations linear equations

Difficulty rating: 890

Small hint:

Undo the mistaken subtraction and division to recover the starting number

Big hint:

Apply

6x+14

to that starting number

Solution:

If the starting number is

x,

the mistaken calculation gives

\frac{x}{6}-14=16,

so

x=180.

The intended calculation would produce

6(180)+14=1094,

which is greater than

1000.

Thus the correct answer is E.

6.

In the sequence

\ldots,a,b,c,d,0,1,1,2,3,5,8,\ldots

each term is the sum of the two terms to its left. Find

a.

-3

-1

0

1

3

Concepts:Fibonacci recurrence working backward

Difficulty rating: 1150

Small hint:

Work backward by subtracting the preceding known term

Big hint:

Determine

d,c,b

in that order before finding

a

Solution:

Working backward,

d+0=1

gives

d=1,

then

c+d=0

gives

c=-1,

and

b+c=d

gives

b=2.

Finally

a+b=c,

so

a=-1-2=-3.

Thus the correct answer is A.

7.

Squares

ABCD

and

EFGH

are congruent,

AB=10,

and

G

is the center of square

ABCD.

The area of the region in the plane covered by these squares is

75

100

125

150

175

Concepts:overlapping squares area diagonals

Difficulty rating: 1460

Small hint:

The overlap is triangle

GAB

Big hint:

The distance from the center

G

to side

\overline{AB}

is

5

Solution:

Each square has area

100.

Their overlap is

\triangle GAB,

whose base is

AB=10

and whose altitude from the center

G

to

\overline{AB}

is

5.

Its area is

\frac12(10)(5)=25.

Hence the union has area

100+100-25=175.

Thus the correct answer is E.

8.

In the polygon shown, each side is perpendicular to its adjacent sides, and all

28

of the sides are congruent. The perimeter of the polygon is

56.

The area of the region bounded by the polygon is

84

96

100

112

196

Concepts:orthogonal polygon area decomposition

Difficulty rating: 1180

Small hint:

Each side has length

\frac{56}{28}

Big hint:

In units of one side length, split the shape into horizontal strips of widths

1,3,5,7,5,3,1

Solution:

Each side has length

2.

Measured in side-length units, the seven horizontal strips have widths

1,3,5,7,5,3,1,

totaling

25

unit squares. Each such square has area

2^2=4,

so the polygon’s area is

25\cdot4=100.

Thus the correct answer is C.

9.

If

\angle A

is four times

\angle B,

and the complement of

\angle B

is four times the complement of

\angle A,

then

\angle B=

10^\circ

12^\circ

15^\circ

18^\circ

22.5^\circ

Concepts:complementary angles linear equations

Difficulty rating: 1260

Small hint:

Let

\angle B=x

and write

\angle A=4x

Big hint:

Translate the second condition as

90-x=4(90-4x)

Solution:

Let

\angle B=x,

so

\angle A=4x.

The complement condition gives

90-x=4(90-4x).

Hence

15x=270

and

x=18^\circ.

Thus the correct answer is D.

10.

For distinct real numbers

x

and

y,

let

M(x,y)

be the larger of

x

and

y

and let

m(x,y)

be the smaller of

x

and

y.

If

a\lt b\lt c\lt d\lt e,

then

\begin{aligned} &M\bigl(M(a,m(b,c)),\\ &\qquad m(d,m(a,e))\bigr)= \end{aligned}

a

b

c

d

e

Concepts:nested minimum and maximum inequalities

Difficulty rating: 1590

Small hint:

Evaluate the innermost

m

expressions first

Big hint:

The two arguments of the outermost

M

reduce to

b

and

a

Solution:

From the ordering,

m(b,c)=b,

so

M(a,m(b,c))=M(a,b)=b.

Also

m(a,e)=a,

hence

m(d,m(a,e))=m(d,a)=a.

The outer expression is therefore

M(b,a)=b.

Thus the correct answer is B.

11.

Three cubes of volume

1,

8

and

27

are glued together at their faces. The smallest possible surface area of the resulting configuration is

36

56

70

72

74

Concepts:surface area cubes maximizing contact area

Difficulty rating: 1900

Small hint:

The cube side lengths are

1,2,

and

3

Big hint:

Minimize exposed area by maximizing the total areas of glued face portions

Solution:

Before gluing, the total surface area is

6(1^2+2^2+3^2)=84.

The side-

2

cube can share area

4

with the side-

3

cube, while the side-

1

cube is placed at their common edge so that it shares area

1

with each larger cube. Thus the total contact area is

6.

Each glued area removes two exposed copies, giving

84-2(6)=72.

No pair can share more than the smaller face, so this is maximal contact and minimal surface area. Thus the correct answer is D.

12.

If

i^2=-1,

then

\left(i-i^{-1}\right)^{-1}=

0

-2i

2i

-\frac{i}{2}

\frac{i}{2}

Concepts:complex numbers reciprocals

Difficulty rating: 1570

Small hint:

Use

i^{-1}=-i

Big hint:

After simplifying the parentheses, rationalize

\frac{1}{2i}

Solution:

Because

i^{-1}=-i,

\begin{aligned} \left(i-i^{-1}\right)^{-1} &=(i+i)^{-1}\\ &=\frac1{2i}\\ &=-\frac i2. \end{aligned}

Thus the correct answer is D.

13.

In triangle

ABC,

AB=AC.

If there is a point

P

strictly between

A

and

B

such that

AP=PC=CB,

then

\angle A=

30^\circ

36^\circ

48^\circ

60^\circ

72^\circ

Concepts:isosceles triangles angle chasing

Difficulty rating: 1960

Small hint:

Let

\angle A=\theta

and use

AP=PC

in

\triangle APC

Big hint:

Use

PC=CB

to relate

\angle PCB

to a base angle of

\triangle ABC

Solution:

Let

\angle A=\theta.

Since

AP=PC,

triangle

APC

has

\angle PAC=\angle PCA=\theta.

Since

AB=AC,

each base angle of

\triangle ABC

is

\frac{180^\circ-\theta}{2}.

Also

PC=CB,

so the base angles of

\triangle PCB

at

P

and

B

are equal; the one at

B

is

\frac{180^\circ-\theta}{2}.

Hence

\angle PCB=\theta.

At

C,

2\theta=\frac{180^\circ-\theta}{2},

giving

5\theta=180^\circ

and

\theta=36^\circ.

Thus the correct answer is B.

14.

Find the sum of the arithmetic series

20+20\frac15+20\frac25+\cdots+40.

3000

3030

3150

4100

6000

Concepts:arithmetic series mixed numbers

Difficulty rating: 1420

Small hint:

The common difference is

\frac{1}{5}

Big hint:

Find the number of terms, then multiply by the average of the endpoints

Solution:

The number of terms is

\frac{40-20}{\frac{1}{5}}+1=101.

Their average is

\frac{20+40}{2}=30,

so the sum is

101\cdot30=3030.

Thus the correct answer is B.

15.

For how many

n

in

\{1,2,3,\ldots,100\}

is the tens digit of

n^2

odd?

10

20

30

40

50

Concepts:squares modulo 100 decimal digits

Difficulty rating: 1750

Small hint:

The parity of the tens digit of

n^2

depends only on the units digit of

n

Big hint:

Check the squares of the ten possible units digits

Solution:

Write

n=10a+b.

Modulo

100,

n^2\equiv20ab+b^2.

The term

20ab

changes the tens digit by an even amount, so only the tens digit of

b^2

matters. Among

b=0,1,\ldots,9,

it is odd only for

b=4

and

b=6.

Each units digit occurs

10

times from

1

through

100,

giving

2\cdot10=20.

Thus the correct answer is B.

16.

Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?

8

22

36

57

71

Concepts:systems of equations proportions

Difficulty rating: 1510

Small hint:

Let

R

be the red count and

T

the total count

Big hint:

Translate the two experiments into

\frac{R-1}{T-1}=\frac{1}{7}

and

\frac{R}{T-2}=\frac{1}{5}

Solution:

The two conditions give

\frac{R-1}{T-1}=\frac17

and

\frac{R}{T-2}=\frac15.

Thus

T=7R-6

and

T=5R+2.

Hence

R=4

and

T=22.

Thus the correct answer is B.

17.

An

8

by

2\sqrt2

rectangle has the same center as a circle of radius

2.

The area of the region common to both the rectangle and the circle is

2\pi

2\pi+2

4\pi-4

2\pi+4

4\pi-2

Concepts:circle segments area symmetry

Difficulty rating: 1960

Small hint:

The rectangle removes two congruent caps from the circle

Big hint:

The cap chord is at distance

\sqrt2

from the center, so its half central angle is

45^\circ

Solution:

The rectangle is wider than the circle, so the common region is the circle with the top and bottom caps beyond distance

\sqrt2

from the center removed. One cap has area

\begin{aligned} 2^2\left(\frac{\pi}{4}\right) &-\sqrt2\sqrt{2^2-(\sqrt2)^2}\\ &=\pi-2. \end{aligned}

Therefore the common area is

4\pi-2(\pi-2)=2\pi+4.

Thus the correct answer is D.

18.

Triangle

ABC

is inscribed in a circle, and

\angle B=\angle C=4\angle A.

If

B

and

C

are adjacent vertices of a regular polygon of

n

sides inscribed in this circle, then

n=

5

7

9

15

18

Concepts:inscribed angles regular polygons

Difficulty rating: 1800

Small hint:

Use the triangle angle sum to find

\angle A

Big hint:

The central angle subtending

\overline{BC}

is twice

\angle A

Solution:

The angle sum gives

9\angle A=180^\circ,

so

\angle A=20^\circ.

The central angle subtending chord

\overline{BC}

is therefore

40^\circ.

Adjacent vertices of a regular

n

-gon subtend

\frac{360^\circ}{n},

so

\frac{360^\circ}{n}=40^\circ,

and

n=9.

Thus the correct answer is C.

19.

Label one disk “

1

,” two disks “

2

,” three disks “

3

,”

\ldots,

fifty disks “

50

.” Put these

1+2+3+\cdots+50=1275

labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is

10

51

415

451

501

Concepts:pigeonhole principle extremal counting

Difficulty rating: 1750

Small hint:

Count the most disks that can be drawn while taking at most nine of each label

Big hint:

All disks labeled

1

through

9

may be drawn, but only nine of each label

10

through

50

Solution:

To avoid ten equal labels, one may draw all

1+2+\cdots+9=45

disks with labels below

10,

and at most

9

from each of the

41

labels

10

through

50.

Thus

45+9(41)=414

disks can be drawn without forcing ten alike, and the next draw guarantees them. The minimum is

415.

Thus the correct answer is C.

20.

Suppose

x,

y,

z

is a geometric sequence with common ratio

r

and

x\ne y.

If

x,

2y,

3z

is an arithmetic sequence, then

r

is

\frac14

\frac13

\frac12

2

4

Concepts:geometric sequences arithmetic sequences quadratic equations

Difficulty rating: 1570

Small hint:

Write

y=xr

and

z=xr^2

Big hint:

The arithmetic-sequence condition is

4y=x+3z

Solution:

Because

y=xr

and

z=xr^2,

the arithmetic-sequence condition gives

4xr=x+3xr^2.

Here

x\ne0,

so

3r^2-4r+1=0,

yielding

r=1

or

r=\frac{1}{3}.

The condition

x\ne y

excludes

r=1,

leaving

r=\frac{1}{3}.

Thus the correct answer is B.

21.

Find the number of counterexamples to the statement:

“If

N

is an odd positive integer the sum of whose digits is

4

and none of whose digits is

0,

then

N

is prime.”

0

1

2

3

4

Concepts:digit sums primality enumeration

Difficulty rating: 1900

Small hint:

A number satisfying the digit conditions has at most four digits

Big hint:

List the odd possibilities by composing

4

into positive digits

Solution:

The odd possibilities are

13,31,121,211,

and

1111.

The numbers

13,31,

and

211

are prime, while

121=11^2

and

1111=11\cdot101.

Thus there are

2

counterexamples. Thus the correct answer is C.

22.

Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?

12

36

60

84

630

Concepts:seating arrangements nonconsecutive selections

Difficulty rating: 1900

Small hint:

Professor chairs cannot be endpoints or adjacent to one another

Big hint:

First choose three nonconsecutive positions from chairs

2

through

8

, then assign the professors

Solution:

The professors must occupy three nonconsecutive positions among chairs

2,3,\ldots,8.

The number of such

3

-subsets of

7

consecutive positions is

\binom{7-3+1}{3}=\binom53=10.

The three distinct professors can be assigned to the selected chairs in

3!=6

ways, for

10\cdot6=60.

Thus the correct answer is C.

23.

In the

xy

-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at

(0,0),

(0,3),

(3,3),

(3,1),

(5,1)

and

(5,0).

The slope of the line through the origin that divides the area of this region exactly in half is

\frac27

\frac13

\frac23

\frac34

\frac79

Concepts:coordinate geometry area bisectors slope

Difficulty rating: 1960

Small hint:

The L-shaped region has total area

11

Big hint:

For the desired slope

m,

the area below the line is the

2

-unit extension plus a triangle of base

3

Solution:

The region has area

3\cdot3+2\cdot1=11,

so each half has area

\frac{11}{2}.

If the line is

y=mx,

the part below it consists of the entire

2

-unit rectangle from

x=3

to

x=5,

together with a triangle of base

3

and height

3m.

Therefore

2+\frac12(3)(3m)=\frac{11}{2}.

This gives

m=\frac{7}{9},

which indeed lies in the assumed range. Thus the correct answer is E.

24.

A sample consisting of five observations has an arithmetic mean of

10

and a median of

12.

The smallest value that the range (largest observation minus smallest) can assume for such a sample is

2

3

5

7

10

Concepts:mean median range extremal construction

Difficulty rating: 1960

Small hint:

Order the observations as

a\le b\le12\le d\le e

Big hint:

Use the total sum

50

, then test whether a range below

5

is possible

Solution:

The sample

7,7,12,12,12

has sum

50,

median

12,

and range

5,

so range

5

is possible. Suppose the range were at most

4.

If the largest value were at most

12,

then the top three values would all be

12,

leaving the bottom two with sum

14;

yet each would be at least

8,

a contradiction. If the largest value exceeded

12,

then the smallest would exceed

8,

making the total exceed

8+8+12+12+12=52.

Thus no smaller range works. The correct answer is C.

25.

If

x

and

y

are non-zero real numbers such that

|x|+y=3

and

|x|y+x^3=0,

then the integer nearest to

x-y

is

-3

-1

2

3

5

Concepts:absolute value systems of equations

Difficulty rating: 2080

Small hint:

Consider the signs of

x

separately

Big hint:

For

x\lt0,

put

t=-x

and combine

y=t^2

with

t+y=3

Solution:

If

x\gt0,

the second equation gives

y=-x^2,

while the first gives

y=3-x;

these would require

x^2-x+3=0,

which has no real root. Hence

x\lt0.

Put

t=-x\gt0.

Then the equations become

t+y=3

and

ty-t^3=0,

so

y=t^2.

Therefore

x-y=-t-t^2=-3,

already an integer. Thus the correct answer is A.

26.

A regular polygon of

m

sides is exactly enclosed (no overlaps, no gaps) by

m

regular polygons of

n

sides each. (Shown here for

m=4,

n=8.

) If

m=10,

what is the value of

n?

5

6

14

20

26

Concepts:regular polygons interior angles angle sums

Difficulty rating: 1900

Small hint:

At each vertex of the inner polygon, one

m

-gon angle and two

n

-gon angles fill

360^\circ

Big hint:

Use the regular

k

-gon interior angle

180^\circ(1-\frac{2}{k})

Solution:

At a vertex of the inner polygon, its interior angle and two angles from the surrounding polygons total

360^\circ.

Thus

\begin{gathered} 180^\circ\left(1-\frac2m\right) \\ {}+2\cdot180^\circ\left(1-\frac2n\right)\\ {}=360^\circ. \end{gathered}

This simplifies to

\frac{1}{m}+\frac{2}{n}=\frac{1}{2}.

With

m=10,

we get

\frac{2}{n}=\frac{2}{5},

so

n=5.

Thus the correct answer is A.

27.

A bag of popping corn contains

\frac{2}{3}

white kernels and

\frac{1}{3}

yellow kernels. Only

\frac{1}{2}

of the white kernels will pop, whereas

\frac{2}{3}

of the yellow ones will pop. A kernel is selected at random from the bag, and pops when placed in the popper. What is the probability that the kernel selected was white?

\frac12

\frac59

\frac47

\frac35

\frac23

Concepts:conditional probability Bayes’ theorem

Difficulty rating: 1750

Small hint:

Compute the probabilities of selecting-and-popping for each color

Big hint:

Condition the white-and-popped probability on the total probability of popping

Solution:

The probabilities of selecting a kernel that pops are

\begin{gathered} P(W\text{ and pop})=\frac23\cdot\frac12=\frac13, \\ P(Y\text{ and pop})=\frac13\cdot\frac23=\frac29. \end{gathered}

Hence

P(\text{pop})=\frac{5}{9},

and

P(W\mid\text{pop})=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac35.

Thus the correct answer is D.

28.

In the

xy

-plane, how many lines whose

x

-intercept is a positive prime number and whose

y

-intercept is a positive integer pass through the point

(4,3)?

0

1

2

3

4

Concepts:intercept form primes divisibility

Difficulty rating: 1960

Small hint:

Let the intercepts be

p

and

q

, and use

\frac{4}{p}+\frac{3}{q}=1

Big hint:

Rearrange to

q=3+\frac{12}{p-4}

Solution:

Let the

x

-intercept be the prime

p

and the

y

-intercept be the positive integer

q.

Intercept form gives

\begin{gathered} \frac4p+\frac3q=1, \\ q=\frac{3p}{p-4}=3+\frac{12}{p-4}. \end{gathered}

Thus

p-4

is a positive divisor of

12.

Testing

1,2,3,4,6,12

gives prime

p

only for

p=5

and

p=7.

Therefore there are

2

lines. Thus the correct answer is C.

29.

Points

A,

B

and

C

on a circle of radius

r

are situated so that

AB=AC,

AB\gt r,

and the length of minor arc

\overset{\frown}{BC}

is

r.

If angles are measured in radians, then

\frac{AB}{BC}=

\frac12\csc\frac14

2\cos\frac12

4\sin\frac12

\csc\frac12

2\sec\frac12

Concepts:arc length chord length trigonometric identities

Difficulty rating: 2280

Small hint:

The minor arc

\overset{\frown}{BC}

subtends a central angle of

1

radian

Big hint:

Use the chord formula

2r\sin(\frac{\theta}{2})

for both

BC

and

AB

Solution:

The central angle subtending minor arc

\overset{\frown}{BC}

is

\frac{r}{r}=1,

so

BC=2r\sin\frac12.

Since

AB=AC

and

AB\gt r,

point

A

is the midpoint of the major arc

BC.

The minor central angle from

A

to

B

is

\pi-\frac12,

so

AB=2r

\sin\left(\frac{\pi-\frac{1}{2}}{2}\right)=2r\cos\frac14.

Therefore

\frac{AB}{BC}=\frac{\cos(\frac{1}{4})}{\sin(\frac{1}{2})}

=\frac1{2\sin(\frac{1}{4})}=\frac12\csc\frac14.

Thus the correct answer is A.

30.

When

n

standard

6

-sided dice are rolled, the probability of obtaining a sum of

1994

is greater than zero and is the same as the probability of obtaining a sum of

S.

The smallest possible value of

S

is

333

335

337

339

341

Concepts:dice sums symmetry extremal values

Difficulty rating: 2150

Small hint:

A sum of

1994

is possible only when

6n\ge1994

Big hint:

Replacing every die value

d

by

7-d

pairs sums

T

and

7n-T

Solution:

The smallest feasible number of dice is

n=\left\lceil\frac{1994}{6}\right\rceil=333.

Replacing every die result

d

by

7-d

is a bijection between outcomes of sum

T

and outcomes of sum

7n-T.

Thus for

n=333,

the sum paired with

1994

is

S=7(333)-1994=337.

Dice-sum counts increase strictly from the minimum sum up to the mean, so for any feasible

n

no sum below

337

can match the count at

1994.

Thus the smallest value is

337,

and the correct answer is C.