1984 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

\frac{1000^2}{252^2-248^2}

equals

62{,}500

1000

500

250

\frac12

Concepts:difference of squares algebraic manipulation

Difficulty rating: 1210

Small hint:

Factor the denominator as a difference of squares

Big hint:

Use

252+248=500

and

252-248=4

Solution:

Factoring the denominator gives

\begin{aligned} 252^2-248^2 &=(252+248)\\ &\quad{}\cdot(252-248)\\ &=500\cdot4=2000. \end{aligned}

Thus the given quotient is

\frac{1000^2}{2000}=500.

Therefore, the correct answer is C.

2.

If

x,

y

and

y-\frac1x

are not

0,

then

\frac{x-\frac1y}{y-\frac1x}

equals

1

\frac{x}{y}

\frac{y}{x}

\frac{x}{y}-\frac{y}{x}

xy-\frac1{xy}

Concepts:fraction algebraic manipulation

Difficulty rating: 1330

Small hint:

Write the numerator and denominator as single fractions

Big hint:

Both parts contain the common factor

xy-1

Solution:

We have

x-\frac1y=\frac{xy-1}{y}

and

y-\frac1x=\frac{xy-1}{x}.

The hypotheses make the needed quantities nonzero, so their quotient is

\frac{\frac{xy-1}{y}}{\frac{xy-1}{x}}=\frac{x}{y}.

Therefore, the correct answer is B.

3.

Let

n

be the smallest nonprime integer greater than

1

with no prime factor less than

10.

Then

100\lt n\le110

110\lt n\le120

120\lt n\le130

130\lt n\le140

140\lt n\le150

Concepts:prime factorization extremal argument

Difficulty rating: 1770

Small hint:

The smallest permitted prime factor is

11

Big hint:

To make the smallest composite, use the two smallest permitted prime factors

Solution:

Every prime factor of

n

is at least

11.

The smallest composite with that property is

11\cdot11=121,

which lies in

120\lt n\le130.

Therefore, the correct answer is C.

4.

A rectangle intersects a circle as shown:

AB=4,

BC=5

and

DE=3.

Then

EF

equals

6

7

\frac{20}{3}

8

9

Concepts:chord perpendicular bisector rectangle

Difficulty rating: 1720

Small hint:

The perpendicular bisectors of the two parallel chords pass through the same center

Big hint:

Therefore the midpoints of

BC

and

EF

have the same horizontal position

Solution:

The line through the circle’s center perpendicular to the parallel chords

BC

and

EF

bisects both. Measured from the rectangle’s left side, the midpoint of

BC

is

4+\frac52,

while the midpoint of

EF

is

3+\frac{EF}{2}.

Thus

4+\frac52=3+\frac{EF}{2},

giving

EF=7.

Therefore, the correct answer is B.

5.

The largest integer

n

for which

n^{200}\lt5^{300}

is

8

9

10

11

12

Concepts:exponent perfect square inequality

Difficulty rating: 1630

Small hint:

Take the positive one-hundredth root of both sides

Big hint:

Compare

n^2

with

5^3

Solution:

Taking positive one-hundredth roots gives

n^2\lt5^3=125.

Since

11^2=121\lt125

but

12^2=144\gt125,

the largest possible integer is

11.

Therefore, the correct answer is D.

6.

In a certain school, there are three times as many boys as girls and nine times as many girls as teachers. Using the letters

b,

g,

t

to represent the number of boys, girls and teachers, respectively, then the total number of boys, girls and teachers can be represented by the expression

31b

\frac{37}{27}b

13g

\frac{37}{27}g

\frac{37}{27}t

Concepts:ratio and proportion algebraic manipulation

Difficulty rating: 1290

Small hint:

Express both

g

and

t

in terms of

b

Big hint:

The given ratios imply

g=\frac{b}{3}

and

t=\frac{b}{27}

Solution:

Because

b=3g

and

g=9t,

we have

g=\frac{b}{3}

and

t=\frac{b}{27}.

Hence

b+g+t=b+\frac b3+\frac b{27}=\frac{37}{27}b.

Therefore, the correct answer is B.

7.

When Dave walks to school, he averages

90

steps per minute, each of his steps

75

cm long. It takes him

16

minutes to get to school. His brother, Jack, going to the same school by the same route, averages

100

steps per minute, but his steps are only

60

cm long. How long does it take Jack to get to school?

14\frac29

min.

15

min.

18

min.

20

min.

22\frac29

min.

Concepts:distance rate and time unit conversion

Difficulty rating: 1290

Small hint:

Find the route’s length from Dave’s step rate, step length, and time

Big hint:

Jack covers

100\cdot60

centimeters each minute

Solution:

The route is

16\cdot90\cdot75

centimeters long. Jack covers

100\cdot60

centimeters per minute, so his time is

\frac{16\cdot90\cdot75}{100\cdot60}=18

minutes. Therefore, the correct answer is C.

8.

Figure

ABCD

is a trapezoid with

AB\parallel DC,

AB=5,

BC=3\sqrt2,

\angle BCD=45^\circ

and

\angle CDA=60^\circ.

The length of

DC

is

7+\frac23\sqrt3

8

9\frac12

8+\sqrt3

8+3\sqrt3

Concepts:trapezoid special right triangle area decomposition

Difficulty rating: 1800

Small hint:

Drop perpendiculars from

A

and

B

to

DC

Big hint:

The right-hand triangle is

45^\circ

-

45^\circ

-

90^\circ,

and the left-hand triangle is

30^\circ

-

60^\circ

-

90^\circ

Solution:

Drop perpendiculars from

A

and

B

to

DC.

Since

BC=3\sqrt2

and the angle at

C

is

45^\circ,

both the height and the right horizontal offset are

3.

The left

30^\circ

-

60^\circ

-

90^\circ

triangle has height

3,

so its horizontal offset is

\sqrt3.

Therefore

DC=\sqrt3+AB+3=8+\sqrt3.

The correct answer is D.

9.

The number of digits in

4^{16}5^{25}

(when written in the usual base

10

form) is

31

30

29

28

27

Concepts:exponent digits power of 2

Difficulty rating: 1750

Small hint:

Rewrite the power of

4

as a power of

2

Big hint:

Pair

25

factors of

2

with the

25

factors of

5

Solution:

We have

\begin{aligned} 4^{16}5^{25} &=2^{32}5^{25}\\ &=2^7\cdot10^{25}\\ &=128\cdot10^{25}. \end{aligned}

This is

128

followed by

25

zeros, so it has

3+25=28

digits. Therefore, the correct answer is D.

10.

Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are

1+2i,

-2+i

and

-1-2i.

The fourth number is

2+i

2-i

1-2i

-1+2i

-2-i

Concepts:complex number square (geometry)symmetry

Difficulty rating: 1680

Small hint:

Two of the listed vertices are opposites

Big hint:

Their midpoint is the square’s center, so reflect the remaining given vertex through that point

Solution:

The points

1+2i

and

-1-2i

are opposite vertices, so their midpoint

0

is the center of the square. The fourth vertex is therefore the reflection of

-2+i

through the origin, namely

2-i.

Therefore, the correct answer is B.

11.

A calculator has a key which replaces the displayed entry with its square, and another key which replaces the displayed entry with its reciprocal. Let

y

be the final result if one starts with an entry

x\ne0

and alternately squares and reciprocates

n

times each. Assuming the calculator is completely accurate (e.g., no roundoff or overflow), then

y

equals

x^{(-2)^n}

x^{2n}

x^{-2n}

x^{-2^n}

x^{(-1)^n2n}

Concepts:exponent process simulation pattern recognition

Difficulty rating: 1750

Small hint:

Track only the exponent of

x

Big hint:

One squaring-reciprocating pair multiplies the exponent by

-2

Solution:

Write the displayed value as

x^e.

Squaring changes

e

to

2e,

and then reciprocating changes it to

-2e.

Starting from

e=1,

after

n

such pairs the exponent is

(-2)^n.

Thus

y=x^{(-2)^n}.

Therefore, the correct answer is A.

12.

If the sequence

\{a_n\}

is defined by

\begin{aligned} a_1&=2,\\ a_{n+1}&=a_n+2n\quad(n\ge1), \end{aligned}

then

a_{100}

equals

9900

9902

9904

10100

10102

Concepts:recursion arithmetic sequence summation

Difficulty rating: 1800

Small hint:

Add all increments from

a_1

through

a_{100}

Big hint:

Use

a_{100}=2+2(1+2+\cdots+99)

Solution:

Telescoping the recurrence gives

\begin{aligned} a_{100} &=a_1+2\sum_{n=1}^{99}n\\ &=2+2\cdot\frac{99\cdot100}{2}\\ &=9902. \end{aligned}

Therefore, the correct answer is B.

13.

\frac{2\sqrt6}{\sqrt2+\sqrt3+\sqrt5}

equals

\sqrt2+\sqrt3-\sqrt5

4-\sqrt2-\sqrt3

\sqrt2+\sqrt3+\sqrt6-5

\frac12(\sqrt2+\sqrt5-\sqrt3)

\frac13(\sqrt3+\sqrt5-\sqrt2)

Concepts:radical rationalizing denominator algebraic manipulation

Difficulty rating: 2150

Small hint:

Group the first two radicals in the denominator

Big hint:

Multiply

\sqrt2+\sqrt3+\sqrt5

by

\sqrt2+\sqrt3-\sqrt5

Solution:

Observe that

\begin{aligned} &(\sqrt2+\sqrt3+\sqrt5)\\ &\quad{}\cdot(\sqrt2+\sqrt3-\sqrt5)\\ &=(\sqrt2+\sqrt3)^2-5\\ &=2\sqrt6. \end{aligned}

Hence the quotient is

\sqrt2+\sqrt3-\sqrt5.

Therefore, the correct answer is A.

14.

The product of all real roots of the equation

x^{\log_{10}x}=10

is

1

-1

10

10^{-1}

none of these

Concepts:logarithm quadratic substitution

Difficulty rating: 1960

Small hint:

Take the base-

10

logarithm of both sides

Big hint:

Let

t=\log_{10}x

and solve the resulting equation in

t

Solution:

The logarithm requires

x\gt0.

Taking base-

10

logarithms gives

(\log_{10}x)^2=1.

Thus

\log_{10}x=\pm1,

so the two roots are

10

and

10^{-1},

whose product is

1.

Therefore, the correct answer is A.

15.

If

\sin2x\sin3x=\cos2x\cos3x,

then one value for

x

is

18^\circ

30^\circ

36^\circ

45^\circ

60^\circ

Concepts:trigonometric identity angle sum

Difficulty rating: 1800

Small hint:

Move all terms to one side and recognize a cosine addition identity

Big hint:

Use

\cos2x\cos3x-\sin2x\sin3x

=\cos5x

Solution:

The equation is equivalent to

\begin{aligned} &\cos2x\cos3x\\ &\quad-\sin2x\sin3x\\ &=\cos5x=0. \end{aligned}

Therefore

5x=90^\circ+180^\circ k.

Taking

k=0

gives

x=18^\circ,

the listed value.

Therefore, the correct answer is A.

16.

The function

f(x)

satisfies

f(2+x)=f(2-x)

for all real numbers

x.

If the equation

f(x)=0

has exactly four distinct real roots, then the sum of these roots is

0

2

4

6

8

Concepts:functional equation symmetry (algebra)pairing and grouping

Difficulty rating: 1800

Small hint:

The equation makes the graph symmetric about

x=2

Big hint:

Pair each root

2+r

with its reflected root

2-r

Solution:

The relation shows that every root

2+r

is paired with

2-r.

Four distinct roots therefore form two such pairs, and each pair has sum

4.

The sum of all four roots is

2\cdot4=8.

Therefore, the correct answer is E.

17.

A right triangle

ABC

with hypotenuse

AB

has side

AC=15.

Altitude

CH

divides

AB

into segments

AH

and

HB,

with

HB=16.

The area of

\triangle ABC

is

120

144

150

216

144\sqrt5

Concepts:right triangle altitude similarity triangle area

Difficulty rating: 1920

Small hint:

Use the right-triangle projection relation

AC^2=AH\cdot AB

Big hint:

After finding

AH,

use

CH^2=AH\cdot HB

Solution:

Let

AH=h.

Similarity in the right triangle gives

\begin{aligned} AC^2&=AH\cdot AB,\\ 225&=h(h+16), \end{aligned}

so

h=9.

Hence

AB=25.

Also

CH=\sqrt{AH\cdot HB},

so

CH=\sqrt{9\cdot16}=12.

The area is

\frac12\cdot25\cdot12=150.

Therefore, the correct answer is C.

18.

A point

(x,y)

is to be chosen in the coordinate plane so that it is equally distant from the

x

-axis, the

y

-axis, and the line

x+y=2.

Then

x

is

\sqrt2-1

\frac12

2-\sqrt2

1

not uniquely determined

Concepts:distance formula angle bisector counterexample

Difficulty rating: 2130

Small hint:

Equal distance from the two axes forces

|x|=|y|

Big hint:

Check both lines

y=x

and

y=-x

Solution:

Equal distance from the axes gives

y=x

or

y=-x.

On

y=-x,

the distance to the line

x+y=2

is

\sqrt2,

so both

(\sqrt2,-\sqrt2)

and

(-\sqrt2,\sqrt2)

satisfy all three distance conditions. Their

x

-coordinates differ, so

x

is not uniquely determined.

Therefore, the correct answer is E.

19.

A box contains

11

balls, numbered

1,

2,

3,

\ldots,

11.

If

6

balls are drawn simultaneously at random, what is the probability that the sum of the numbers on the balls drawn is odd?

\frac{100}{231}

\frac{115}{231}

\frac12

\frac{118}{231}

\frac6{11}

Concepts:combinations parity sampling without replacement

Difficulty rating: 2070

Small hint:

There are

6

odd-numbered balls and

5

even-numbered balls

Big hint:

Count selections containing

1,

3,

or

5

odd balls

Solution:

An odd sum requires an odd number of the six selected balls to be odd. There are

6

odd and

5

even balls, so the favorable count is

\begin{aligned} &\binom61\binom55+\binom63\binom53\\ &\quad+\binom65\binom51\\ &=6+200+30=236. \end{aligned}

Of the

\binom{11}{6}=462

selections, the desired probability is

\frac{236}{462}=\frac{118}{231}.

Therefore, the correct answer is D.

20.

The number of distinct solutions of the equation

\left|x-|2x+1|\right|=3

is

0

1

2

3

4

Concepts:absolute value casework linear equation

Difficulty rating: 1720

Small hint:

Split the outer absolute value into two equations

Big hint:

For each equation, isolate

|2x+1|

before checking its two cases

Solution:

If

x-|2x+1|=3,

then

|2x+1|=x-3,

which requires

x\ge3

and yields no solution. If

x-|2x+1|=-3,

then

|2x+1|=x+3.

Its two linear cases give

x=2

and

x=-\frac{4}{3},

both valid. Thus there are

2

distinct solutions.

Therefore, the correct answer is C.

21.

The number of triples

(a,b,c)

of positive integers which satisfy the simultaneous equations

\begin{aligned} ab+bc&=44,\\ ac+bc&=23. \end{aligned}

is

0

1

2

3

4

Concepts:Diophantine Equation factor substitution

Difficulty rating: 1960

Small hint:

Factor the second equation as

c(a+b)=23

Big hint:

Use the primality of

23

and positivity to determine

c

and

a+b

Solution:

Since

c(a+b)=23

and all variables are positive integers,

c=1

and

a+b=23.

Put

b=23-a

in

ab+b=44.

Then

(23-a)(a+1)=44,

or

a^2-22a+21=0,

so

a=1

or

21.

Each gives a positive

b,

producing exactly two triples.

Therefore, the correct answer is C.

22.

Let

a

and

c

be fixed positive numbers. For each real number

t

let

(x_t,y_t)

be the vertex of the parabola

y=ax^2+tx+c.

If the set of vertices

(x_t,y_t)

for all real values of

t

is graphed in the plane, the graph is

a straight line

a parabola

part, but not all, of a parabola

one branch of a hyperbola

none of these

Concepts:parabola completing the square coordinate geometry

Difficulty rating: 1960

Small hint:

Write the vertex coordinates in terms of

t

Big hint:

Use

x_t=-\frac{t}{2a}

to eliminate

t

from

y_t

Solution:

The vertex coordinates are

x_t=-\frac{t}{2a},\qquad y_t=c-\frac{t^2}{4a}.

Since

t=-2ax_t,

eliminating

t

gives

y_t=c-ax_t^2.

As

t

ranges over all reals, so does

x_t,

so the entire parabola is traced.

Therefore, the correct answer is B.

23.

\frac{\sin10^\circ+\sin20^\circ}{\cos10^\circ+\cos20^\circ}

equals

\tan10^\circ+\tan20^\circ

\tan30^\circ

\frac12(\tan10^\circ+\tan20^\circ)

\tan15^\circ

\frac14\tan60^\circ

Concepts:trigonometric identity algebraic manipulation

Difficulty rating: 2130

Small hint:

Apply the sum-to-product identities to both numerator and denominator

Big hint:

Both transformed sums contain the same

\cos5^\circ

factor

Solution:

Let

R

denote the given ratio. By the sum-to-product identities,

\begin{aligned} R&=\frac{\sin10^\circ+\sin20^\circ} {\cos10^\circ+\cos20^\circ},\\ R&=\frac{2\sin15^\circ\cos5^\circ} {2\cos15^\circ\cos5^\circ}\\ &=\tan15^\circ. \end{aligned}

Therefore, the correct answer is D.

24.

If

a

and

b

are positive real numbers and each of the equations

\begin{aligned} x^2+ax+2b&=0,\\ x^2+2bx+a&=0 \end{aligned}

has real roots, then the smallest possible value of

a+b

is

2

3

4

5

6

Concepts:quadratic inequality

Difficulty rating: 2240

Small hint:

Require both quadratic discriminants to be nonnegative

Big hint:

Combine

a^2\ge8b

and

b^2\ge a

to bound

a

and

b

Solution:

The two discriminants give

a^2\ge8b

and

b^2\ge a.

Therefore

a^4\ge64b^2\ge64a.

Since

a\gt0,

this yields

a\ge4,

and then

b^2\ge a\ge4

gives

b\ge2.

Thus

a+b\ge6.

Equality occurs at

a=4,b=2,

for which both discriminants are zero.

Therefore, the correct answer is E.

25.

The total area of all the faces of a rectangular solid is

22\text{ cm}^2,

and the total length of all its edges is

24\text{ cm}.

Then the length in cm of any one of its internal diagonals is

\sqrt{11}

\sqrt{12}

\sqrt{13}

\sqrt{14}

not uniquely determined

Concepts:rectangular prism surface area diagonal

Difficulty rating: 1770

Small hint:

Let the side lengths be

x,y,z

and translate both totals into equations

Big hint:

Expand

(x+y+z)^2

to find

x^2+y^2+z^2

Solution:

The conditions give

2(xy+xz+yz)=22

and

4(x+y+z)=24,

so

xy+xz+yz=11

and

x+y+z=6.

Hence the square of an internal diagonal is

x^2+y^2+z^2=6^2-2(11)=14.

Its length is

\sqrt{14}.

Therefore, the correct answer is D.

26.

In the obtuse triangle

ABC,

AM=MB,

MD\perp BC,

EC\perp BC.

If the area of

\triangle ABC

is

24,

then the area of

\triangle BED

is

9

12

15

18

not uniquely determined

Concepts:triangle area midpoint area decomposition

Difficulty rating: 1960

Small hint:

Draw segment

MC

and compare

\triangle DMC

with

\triangle DME

Big hint:

Those triangles have the same base

MD

and equal altitudes; also

M

is the midpoint of

AB

Solution:

Draw

MC.

Since

EC\parallel MD,

points

E

and

C

have equal perpendicular distances from

MD,

so

[DME]=[DMC].

Therefore

\begin{aligned} [BED]&=[BMD]+[DME]\\ &=[BMD]+[DMC]\\ &=[BMC]. \end{aligned}

Because

M

is the midpoint of

AB,

triangle

BMC

has half the area of

ABC.

Thus

[BED]=12.

Therefore, the correct answer is B.

27.

In

\triangle ABC,

D

is on

AC

and

F

is on

BC.

Also,

AB\perp AC,

AF\perp BC,

and

BD=DC=FC=1.

Find

AC.

\sqrt2

\sqrt3

\sqrt[3]{2}

\sqrt[3]{3}

\sqrt[4]{3}

Concepts:right triangle similarity law of cosines

Difficulty rating: 2350

Small hint:

Let

AC=s

and use the altitude-to-hypotenuse similarity relation

Big hint:

Then compare

\cos\angle ACB

in right triangle

ABC

and isosceles triangle

BDC

Solution:

Let

AC=s.

The right-triangle projection relation gives

AC^2=FC\cdot BC,

so

BC=s^2.

Since

BD=DC=1,

the numerator and denominator in the Law of Cosines satisfy

\begin{aligned} BC^2+DC^2-BD^2&=s^4,\\ 2(BC)(DC)&=2s^2. \end{aligned}

Hence

\cos\angle BCD=\frac{s^2}{2}.

But

D

lies on

AC,

and in right triangle

ABC,

\cos\angle BCA=\frac{AC}{BC}=\frac{1}{s}.

Thus

\frac{s^2}{2}=\frac{1}{s},

so

s^3=2

and

AC=\sqrt[3]{2}.

Therefore, the correct answer is C.

28.

The number of distinct pairs of integers

(x,y)

such that

\begin{aligned} 0&\lt x\lt y,\\ \sqrt{1984}&=\sqrt{x}+\sqrt{y} \end{aligned}

is

0

1

2

3

7

Concepts:radical perfect square counting pairs

Difficulty rating: 2380

Small hint:

Use

1984=31\cdot8^2

and write both radicands with a common squarefree part

Big hint:

Set

x=31u^2,\ y=31v^2

and count positive integers

u\lt v

with

u+v=8

Solution:

Because

\sqrt{x}+\sqrt{y}=8\sqrt{31},

the two radicals must have common squarefree part

31.

Write

x=31u^2

and

y=31v^2

for positive integers

u,v.

Then

u+v=8,\qquad u\lt v.

The possibilities are

(u,v)=(1,7),

(2,6),

and

(3,5),

producing three distinct pairs

(x,y).

Therefore, the correct answer is D.

29.

Find the largest value of

\frac{y}{x}

for pairs of real numbers

(x,y)

which satisfy

(x-3)^2+(y-3)^2=6.

3+2\sqrt2

2+\sqrt3

3\sqrt3

6

6+2\sqrt3

Concepts:circle slope tangent line optimization

Difficulty rating: 2240

Small hint:

Interpret

\frac{y}{x}

as the slope of a line through the origin

Big hint:

At an extreme slope, the line

y=mx

is tangent to the circle

Solution:

The circle lies entirely where

x\gt0,

so

\frac{y}{x}

is the slope

m

of the line

y=mx

through a point on it. At an extreme, this line is tangent. The distance from the center

(3,3)

to

mx-y=0

must equal

\sqrt6,

so

\frac{|3m-3|}{\sqrt{m^2+1}}=\sqrt6.

Squaring and simplifying gives

m^2-6m+1=0,

hence

m=3\pm2\sqrt2.

The larger value is

3+2\sqrt2.

Therefore, the correct answer is A.

30.

For any complex number

w=a+bi,

|w|

is defined to be the real number

\sqrt{a^2+b^2}.

If

w=\cos40^\circ+i\sin40^\circ,

then

|w+2w^2+3w^3+\cdots+9w^9|^{-1}

equals

\frac19\sin40^\circ

\frac29\sin20^\circ

\frac19\cos40^\circ

\frac1{18}\cos20^\circ

none of these

Concepts:complex number roots of unity arithmetico-geometric series

Difficulty rating: 2460

Small hint:

Let

S=w+2w^2+\cdots+9w^9

and subtract

wS

from

S

Big hint:

Use

w^9=1

and

|1-e^{i\theta}|=2\sin(\frac{\theta}{2})

Solution:

Let

S=\sum_{k=1}^9kw^k.

Since

w^9=1

and

w^{10}=w,

\begin{aligned} (1-w)S &=(w+w^2+\cdots+w^9)\\ &\quad-9w^{10}\\ &=-9w. \end{aligned}

Therefore

|S|=\frac{9}{|1-w|},

because

|w|=1.

Thus

|S|^{-1}=\frac{|1-w|}{9} =\frac{2}{9}\sin20^\circ.

Therefore, the correct answer is B.