1963 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Which one of the following points is not on the graph of

y=\dfrac{x}{x+1}?

(0,0)

\left(-\dfrac12,-1\right)

\left(\dfrac12,\dfrac13\right)

(-1,1)

(-2,2)

Concepts:function substitution

Difficulty rating: 1320

Small hint:

Check the domain before substituting coordinates

Big hint:

The denominator vanishes at one of the listed

x

-coordinates

Solution:

The expression

\frac{x}{x+1}

is undefined when

x=-1.

Therefore no point whose first coordinate is

-1

can lie on its graph.

Thus, the correct answer is D.

2.

Let

n=x-y^{x-y}.

Find

n

when

x=2

and

y=-2.

-14

0

1

18

256

Concepts:substitution exponent order of operations

Difficulty rating: 1180

Small hint:

First compute the exponent

x-y

Big hint:

Evaluate the power before subtracting it from

x

Solution:

Here

x-y=2-(-2)=4,

so

n=2-(-2)^4=2-16=-14.

Therefore, the correct answer is A.

3.

If the reciprocal of

x+1

is

x-1,

then

x

equals:

0

1

-1

\pm1

none of these

Concepts:rational equation quadratic

Difficulty rating: 1320

Small hint:

Translate “the reciprocal of

x+1

” into an equation

Big hint:

Multiplying by

x+1

leads to a difference of squares

Solution:

The equation is

\frac1{x+1}=x-1.

Hence

1=x^2-1,

so

x=\pm\sqrt2.

Neither value is listed.

Thus, the correct answer is E.

4.

For what value(s) of

k

does the pair of equations

y=x^2

and

y=3x+k

have two identical solutions?

\dfrac49

-\dfrac49

\dfrac94

-\dfrac94

\pm\dfrac94

Concepts:parabola tangent line quadratic

Difficulty rating: 1280

Small hint:

Set the two expressions for

y

equal

Big hint:

A repeated intersection makes the resulting quadratic’s discriminant zero

Solution:

Intersections satisfy

x^2-3x-k=0.

Two identical solutions require

(-3)^2-4(1)(-k)=9+4k=0,

giving

k=-\frac{9}{4}.

Therefore, the correct answer is D.

5.

If

x

and

\log_{10}x

are real numbers and

\log_{10}x\lt0,

then:

x\lt0

-1\lt x\lt1

0\lt x\leq1

-1\lt x\lt0

0\lt x\lt1

Concepts:logarithm inequality

Difficulty rating: 940

Small hint:

A real logarithm requires a positive argument

Big hint:

Compare

x

with

10^0

Solution:

The logarithm is real only for

x\gt0.

Since base

10

is greater than

1,

\log_{10}x\lt0=\log_{10}1

implies

x\lt1.

Thus, the correct answer is E.

6.

Triangle

BAD

is right-angled at

B.

On

AD

there is a point

C

for which

AC=CD

and

AB=BC.

The magnitude of angle

DAB,

in degrees, is:

67\dfrac12

60

45

30

22\dfrac12

Concepts:right triangle midpoint equilateral triangle

Difficulty rating: 1500

Small hint:

A midpoint of a right triangle’s hypotenuse is equidistant from all three vertices

Big hint:

Use

AB=BC=AC

Solution:

Since

C

is the midpoint of hypotenuse

AD,

AC=BC=CD.

Given

AB=BC,

triangle

ABC

is equilateral. Thus

\angle DAB=\angle CAB=60^\circ.

Therefore, the correct answer is B.

7.

Given the four equations:

\begin{aligned} (1)\quad&3y-2x=12,\\ (2)\quad&-2x-3y=10,\\ (3)\quad&3y+2x=12,\\ (4)\quad&2y+3x=10. \end{aligned}

The pair representing perpendicular lines is:

(1)

and

(4)

(1)

and

(3)

(1)

and

(2)

(2)

and

(4)

(2)

and

(3)

Concepts:linear equation slope

Difficulty rating: 1080

Small hint:

Rewrite each equation in slope-intercept form

Big hint:

Perpendicular nonvertical lines have slopes whose product is

-1

Solution:

Line

(1)

has slope

\frac{2}{3},

and line

(4)

has slope

-\frac{3}{2}.

Their product is

-1,

so these two lines are perpendicular.

Thus, the correct answer is A.

8.

The smallest positive integer

x

for which

1260x=N^3,

where

N

is an integer, is:

1050

1260

1260^2

7350

44{,}100

Concepts:prime factorization perfect power

Difficulty rating: 1680

Small hint:

Factor

1260

into primes

Big hint:

Raise every prime exponent to the next multiple of

3

Solution:

Since

1260=2^2\cdot3^2\cdot5\cdot7,

the least multiplier making every exponent divisible by

3

is

2\cdot3\cdot5^2\cdot7^2=7350.

Thus, the correct answer is D.

9.

In the expansion of

\left(a-\dfrac1{\sqrt a}\right)^7

the coefficient of

a^{-\frac{1}{2}}

is:

-7

7

-21

21

35

Concepts:binomial theorem exponent

Difficulty rating: 1850

Small hint:

The term choosing the second summand

r

times has exponent

7-\frac{3r}{2}

Big hint:

Set that exponent equal to

-\frac{1}{2}

before finding the binomial coefficient and sign

Solution:

The term using

-a^{-\frac{1}{2}}

exactly

r

times has exponent

7-r-\frac{r}{2}=7-\frac{3r}{2}.

Setting this to

-\frac{1}{2}

gives

r=5.

Its coefficient is

\binom75(-1)^5=-21.

Therefore, the correct answer is C.

10.

Point

P

is taken interior to a square with side-length

a

and such that it is equally distant from two consecutive vertices and from the side opposite these vertices. If

d

represents the common distance, then

d

equals:

\dfrac{3a}{5}

\dfrac{5a}{8}

\dfrac{3a}{8}

\dfrac{a\sqrt2}{2}

\dfrac a2

Concepts:square (geometry)perpendicular bisector distance formula

Difficulty rating: 1730

Small hint:

The equal distances to two consecutive vertices put

P

on their perpendicular bisector

Big hint:

If the distance to the opposite side is

d,

use a right triangle with legs

\frac{a}{2}

and

a-d

Solution:

Place the opposite side at height

0

and the two vertices at height

a.

Then

P=(\frac{a}{2},d).

Equality of the vertex distance and

d

gives

\left(\frac a2\right)^2+(a-d)^2=d^2,

so

\frac{5a^2}{4}=2ad

and

d=\frac{5a}{8}.

Thus, the correct answer is B.

11.

The arithmetic mean of a set of

50

numbers is

38.

If two numbers of the set, namely

45

and

55,

are discarded, the arithmetic mean of the remaining set of numbers is:

38.5

37.5

37

36.5

36

Concepts:mean whole number operations

Difficulty rating: 1030

Small hint:

Recover the original sum from the mean

Big hint:

Subtract

45+55

and divide by

48

Solution:

The original sum is

50\cdot38=1900.

After removing

45+55=100,

the remaining sum is

1800,

so the new mean is

\frac{1800}{48}=37.5.

Thus, the correct answer is B.

12.

Three vertices of parallelogram

PQRS

are

P(-3,-2),

Q(1,-5),

R(9,1)

with

P

and

R

diagonally opposite. The sum of the coordinates of vertex

S

is:

13

12

11

10

9

Concepts:parallelogram coordinate geometry midpoint

Difficulty rating: 1290

Small hint:

The diagonals of a parallelogram bisect each other

Big hint:

Use the vector relation

P+R=Q+S

Solution:

Since

P+R=Q+S,

\begin{aligned} S&=P+R-Q\\ &=(-3,-2)+(9,1)-(1,-5)\\ &=(5,4). \end{aligned}

The coordinate sum is

5+4=9.

Therefore, the correct answer is E.

13.

If

2^a+2^b=3^c+3^d,

the number of integers

a,

b,

c,

d

which can possibly be negative is, at most:

4

3

2

1

0

Concepts:exponent divisibility parity

Difficulty rating: 2160

Small hint:

A reduced denominator on the left can contain only

2,

while one on the right can contain only

3

Big hint:

Conclude both sides must be integers, then examine when a sum of two negative powers can be integral

Solution:

In lowest terms, the left side can have only a power of

2

in its denominator, while the right side can have only a power of

3.

Equality therefore forces both sides to be integers. If either

c

or

d

were negative,

3^c+3^d

would retain a factor

3

in its denominator. Thus

c,d\geq0.

Likewise, negative

a

or

b

can give an integer only in the exceptional case

a=b=-1,

whose sum is

1.

But

3^c+3^d\geq2.

Hence

a,b\geq0

as well, so none can be negative.

Thus, the correct answer is E.

14.

Given the equations

x^2+kx+6=0

and

x^2-kx+6=0.

If, when the roots of the equations are suitably listed, each root of the second equation is

5

more than the corresponding root of the first equation, then

k

equals:

5

-5

7

-7

none of these

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1500

Small hint:

Compare the sums of the two pairs of roots

Big hint:

Adding

5

to each root increases their sum by

10

Solution:

The first pair of roots has sum

-k,

while the second has sum

k.

Since both roots increase by

5,

k=-k+10,

so

k=5.

Therefore, the correct answer is A.

15.

A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is:

\sqrt3:1

\sqrt3:\sqrt2

3\sqrt3:2

3:\sqrt2

3:2\sqrt2

Concepts:equilateral triangle incircle, incenter, and inradius square (geometry)area ratio

Difficulty rating: 1570

Small hint:

Express both areas using the circle’s radius

r

Big hint:

The triangle’s side is

2\sqrt3r

and the square’s diagonal is

2r

Solution:

If the circle has radius

r,

the equilateral triangle has side

2\sqrt3r

and area

3\sqrt3r^2.

The square has side

r\sqrt2

and area

2r^2.

The ratio is

3\sqrt3:2.

Thus, the correct answer is C.

16.

Three numbers

a,

b,

c,

none zero, form an arithmetic progression. Increasing

a

by

1

or increasing

c

by

2

results in a geometric progression. Then

b

equals:

16

14

12

10

8

Concepts:arithmetic sequence geometric sequence system of equations

Difficulty rating: 1920

Small hint:

Translate the three progression conditions into equations involving

b^2

and

a+c

Big hint:

Compare

(a+1)c=b^2=a(c+2)

first

Solution:

The conditions give

\begin{gathered} 2b=a+c,\\ b^2=(a+1)c,\qquad b^2=a(c+2). \end{gathered}

Comparing the last two yields

c=2a.

Thus

b=\frac{3a}{2},

and

\frac{9a^2}{4}=2a(a+1).

Since

a\ne0,

a=8,

so

b=12.

Therefore, the correct answer is C.

17.

The expression

\frac{\dfrac{a}{a+y}+\dfrac{y}{a-y}} {\dfrac{y}{a+y}-\dfrac{a}{a-y}},

a

real,

a\ne0,

has the value

-1

for:

all but two real values of

y

only two real values of

y

all real values of

y

only one real value of

y

no real values of

y

Concepts:fraction algebraic manipulation

Difficulty rating: 1570

Small hint:

Combine the two fractions in the numerator and denominator separately

Big hint:

Keep track of the excluded values

y=\pm a

Solution:

For

y\ne\pm a,

the numerator simplifies to

\frac{a^2+y^2}{a^2-y^2}

and the denominator to its negative. Their quotient is therefore

-1.

The two values

y=a

and

y=-a

are undefined.

Thus, the correct answer is A.

18.

Chord

EF

is the perpendicular bisector of chord

BC,

intersecting it in

M.

Between

B

and

M

point

U

is taken, and

EU

extended meets the circle in

A.

Then, for any selection of

U,

as described, triangle

EUM

is similar to triangle:

EFA

EFC

ABM

ABU

FMC

Concepts:chord circle similarity

Difficulty rating: 1990

Small hint:

Because the perpendicular bisector of a chord passes through the center,

EF

is a diameter

Big hint:

Compare the right angles and the shared angle at

E

Solution:

Since

EF

is a diameter,

\angle EAF=90^\circ.

Also

EF\perp BC,

so

\angle EMU=90^\circ.

Because

E,U,A

are collinear, the two triangles share the same acute angle at

E.

Hence

\triangle EUM\sim\triangle EFA.

Therefore, the correct answer is A.

19.

In counting

n

colored balls, some red and some black, it was found that

49

of the first

50

counted were red. Thereafter,

7

out of every

8

counted were red. If, in all,

90\%

or more of the balls counted were red, the maximum value of

n

is:

225

210

200

180

175

Concepts:percentage inequality rate

Difficulty rating: 1640

Small hint:

Write the number of red balls as

49+\frac78(n-50)

Big hint:

Require that quantity to be at least

0.9n

Solution:

The condition is

49+\frac78(n-50)\geq\frac9{10}n.

Simplifying gives

\frac{21}{4}\geq \frac{n}{40},

so

n\leq210.

The maximum is

210.

Thus, the correct answer is B.

20.

Two men at points

R

and

S,

76

miles apart, set out at the same time to walk towards each other. The man at

R

walks uniformly at

4\dfrac12

miles per hour; the man at

S

walks at

3\dfrac14

miles per hour for the first hour, at

3\dfrac34

miles per hour for the second hour, and so on, in arithmetic progression. If the men meet

x

miles nearer

R

than

S

in an integral number of hours, then

x

is:

10

8

6

4

2

Concepts:arithmetic sequence summation distance rate and time

Difficulty rating: 1920

Small hint:

Let the integral meeting time be

h

hours and sum the second man’s hourly distances

Big hint:

Their combined distance equation simplifies to

h^2+30h-304=0

Solution:

In

h

hours the first man walks

\frac{9h}{2}.

The second walks an arithmetic-series total

\begin{aligned} &\frac h2\left(2\cdot\frac{13}{4} +(h-1)\frac12\right)\\ &\qquad=3h+\frac{h^2}{4}. \end{aligned}

Their distances sum to

76,

giving

h^2+30h-304=0,

whose positive root is

h=8.

They walk

36

and

40

miles, so the meeting point is

4

miles nearer

R

than

S.

Thus, the correct answer is D.

21.

The expression

x^2-y^2-z^2+2yz+x+y-z

has:

no linear factor with integer coefficients and integer exponents

the factor

-x+y+z

the factor

x-y-z+1

the factor

x+y-z+1

the factor

x-y+z+1

Concepts:difference of squares factoring substitution

Difficulty rating: 1520

Small hint:

Group

y-z

as a single expression

Big hint:

Rewrite the polynomial as

x^2-(y-z)^2+x+(y-z)

Solution:

Let

u=y-z.

The two groups factor as

\begin{gathered} x^2-u^2=(x+u)(x-u),\\ x+u=(x+u)\cdot1. \end{gathered}

Thus the whole expression is

(x+u)(x-u+1),

and one factor is

x-y+z+1.

Therefore, the correct answer is E.

22.

Acute-angled triangle

ABC

is inscribed in a circle with center at

O;

\overset{\frown}{AB}=120^\circ

and

\overset{\frown}{BC}=72^\circ.

A point

E

is taken in minor arc

AC

such that

OE

is perpendicular to

AC.

Then the ratio of the magnitudes of angles

OBE

and

BAC

is:

\dfrac5{18}

\dfrac29

\dfrac14

\dfrac13

\dfrac49

Concepts:arc inscribed angle isosceles triangle

Difficulty rating: 2030

Small hint:

Find minor arc

AC,

then use

OE\perp AC

to locate

E

at its midpoint

Big hint:

Compute the central angle

BOE

and the inscribed angle

BAC

Solution:

Minor arc

AC

has measure

360^\circ-120^\circ-72^\circ=168^\circ.

Since

OE\perp AC,

E

bisects that arc, so

\angle BOE=72^\circ+84^\circ=156^\circ.

Isosceles triangle

OBE

gives

\angle OBE=12^\circ,

while

\angle BAC=\frac{72^\circ}{2}=36^\circ.

Their ratio is

\frac{1}{3}.

Thus, the correct answer is D.

23.

A

gives

B

as many cents as

B

has and

C

as many cents as

C

has. Similarly,

B

then gives

A

and

C

as many cents as each then has.

C

, similarly, then gives

A

and

B

as many cents as each then has. If each finally has

16

cents, with how many cents does

A

start?

24

26

28

30

32

Concepts:money work backwards

Difficulty rating: 1780

Small hint:

Work backward from the final holdings

Big hint:

Immediately before someone gives, each recipient must have half of the amount held just after that gift

Solution:

Reverse the transactions. Before

C

gives, the holdings are

(8,8,32).

Before

B

gives, they are

(4,28,16).

Before

A

gives,

B

and

C

must have had

14

and

8,

while

A

had

4+14+8=26.

Therefore, the correct answer is B.

24.

Consider equations of the form

x^2+bx+c=0.

How many such equations have real roots and have coefficients

b

and

c

selected from the set of integers

\{1,2,3,4,5,6\}?

20

19

18

17

16

Concepts:quadratic basic counting inequality

Difficulty rating: 1570

Small hint:

Real roots require

b^2\geq4c

Big hint:

For each

c=1,\ldots,6,

count the allowed values of

b

Solution:

For

c

equal to

1,

2,

3,

4,

5,

and

6,

the condition

b^2\geq4c

permits respectively

5,

4,

3,

3,

2,

and

2

values of

b

in the given set. Their sum is

5+4+3+3+2+2=19.

Thus, the correct answer is B.

25.

Point

F

is taken in side

AD

of square

ABCD.

At

C

a perpendicular is drawn to

CF,

meeting

AB

extended at

E.

The area of

ABCD

is

256

square inches and the area of triangle

CEF

is

200

square inches. Then the number of inches in

BE

is:

12

14

15

16

20

Concepts:square (geometry)congruence (geometry)triangle area Pythagorean Theorem

Difficulty rating: 1900

Small hint:

Show that right triangles

CDF

and

CBE

are congruent

Big hint:

Then

CF=CE,

so use the area of right triangle

CEF

Solution:

The complementary acute angles and the equal square sides give

\triangle CDF\cong\triangle CBE,

hence

CF=CE.

Since

CF\perp CE,

\begin{aligned} 200=[CEF] &=\frac12 CE\cdot CF\\ &=\frac12CE^2, \end{aligned}

so

CE=20.

The square side is

16,

and right triangle

CBE

gives

BE=\sqrt{20^2-16^2}=12.

Therefore, the correct answer is A.

26.

Form I. Consider the statements

\begin{aligned} (1)\quad&p\land\neg q\land r,\\ (2)\quad&\neg p\land\neg q\land r,\\ (3)\quad&p\land\neg q\land\neg r,\\ (4)\quad&\neg p\land q\land r, \end{aligned}

where

p,

q,

r

are propositions. How many of these imply the truth of

(p\to q)\to r?

Form II. Consider the statements

(1)

p

and

r

are true and

q

is false,

(2)

r

is true and

p

and

q

are false,

(3)

p

is true and

q

and

r

are false,

(4)

q

and

r

are true and

p

is false. How many of these imply the truth of the statement “

r

is implied by the statement that

p

implies

q

”?

0

1

2

3

4

Concepts:logical deduction

Difficulty rating: 1880

Small hint:

An implication is false only when its antecedent is true and its consequent is false

Big hint:

Evaluate

p\to q

first in each of the four assignments

Solution:

In cases

(1)

and

(3),

p\to q

is false, so the outer implication is true. In cases

(2)

and

(4),

p\to q

is true, but

r

is also true, so the outer implication is again true. All four statements imply it.

Thus, the correct answer is E.

27.

Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:

16

20

22

24

26

Concepts:counting regions counting intersections summation

Difficulty rating: 1520

Small hint:

The

k

th line is cut into

k

pieces by the previous lines

Big hint:

Start with one region and add

1+2+\cdots+6

Solution:

Successive lines create

1,2,\ldots,6

new regions. Thus the total is

1+\sum_{k=1}^{6}k=1+21=22.

Therefore, the correct answer is C.

28.

Given the equation

3x^2-4x+k=0

with real roots. The value of

k

for which the product of the roots of the equation is a maximum is:

\dfrac{16}{9}

\dfrac{16}{3}

\dfrac49

\dfrac43

-\dfrac43

Concepts:quadratic Vieta’s Formulas optimization

Difficulty rating: 1210

Small hint:

The product of the roots is

\frac{k}{3}

Big hint:

Use the discriminant condition to find the largest allowed

k

Solution:

Real roots require

16-12k\geq0,

so

k\leq\frac{4}{3}.

The root product is

\frac{k}{3},

which increases with

k.

It is therefore largest at

k=\frac{4}{3}.

Thus, the correct answer is D.

29.

A particle projected vertically upward reaches, at the end of

t

seconds, an elevation of

s

feet where

s=160t-16t^2.

The highest elevation is:

800

640

400

320

160

Concepts:quadratic optimization

Difficulty rating: 1320

Small hint:

The height is a downward-opening quadratic

Big hint:

Find its vertex time using

-\frac{b}{2a}

Solution:

The vertex occurs at

t=-\frac{160}{2\cdot-16}=5.

Then

s(5)=160(5)-16(25)=400.

Thus, the correct answer is C.

30.

Let

F=\log\frac{1+x}{1-x}.

Form a new function

G

by replacing each

x

in

F

by

\frac{3x+x^3}{1+3x^2},

and simplify. The simplified expression

G

is equal to:

-F

F

3F

F^3

F^3-F

Concepts:logarithm factoring algebraic manipulation

Difficulty rating: 2020

Small hint:

Call the substituted fraction

u

and simplify

\frac{1+u}{1-u}

Big hint:

Its numerator and denominator factor as cubes

Solution:

For

u=\frac{3x+x^3}{1+3x^2},

\begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x+3x^2+x^3} {1-3x+3x^2-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3. \end{aligned}

Therefore

G=\log(\frac{1+x}{1-x})^3,

so

G=3F.

Thus, the correct answer is C.

31.

The number of solutions in positive integers of

2x+3y=763

is:

255

254

128

127

0

Concepts:Diophantine Equation parity counting integers in a range

Difficulty rating: 1710

Small hint:

Solve for

x=\frac{763-3y}{2}

Big hint:

Positive integral

x

requires

y

to be an odd positive integer below

\frac{763}{3}

Solution:

We need

y

odd so that

763-3y

is even, and positivity gives

1\leq y\leq253.

The odd values

1,

3,

\ldots,

and

253

number

\frac{253+1}{2}=127.

Therefore, the correct answer is D.

32.

The dimensions of a rectangle

R

are

a

and

b,

a\lt b.

It is required to obtain a rectangle with dimensions

x

and

y,

x\lt a,

y\lt a,

so that its perimeter is one-third that of

R,

and its area is one-third that of

R.

The number of such (different) rectangles is:

0

1

2

4

infinitely many

Concepts:rectangle inequality algebraic manipulation

Difficulty rating: 1990

Small hint:

Translate the conditions into

3(x+y)=a+b

and

3xy=ab

Big hint:

Divide the sum equation by the product equation and compare reciprocals using

x\lt a,

y\lt a,

and

a\lt b

Solution:

The conditions give

3(x+y)=a+b

and

3xy=ab.

Dividing yields

\frac1x+\frac1y=\frac1a+\frac1b.

But

x\lt a

and

y\lt a

make the left side greater than

\frac{2}{a},

while

a\lt b

makes the right side less than

\frac{2}{a}.

This is impossible.

Thus, the correct answer is A.

33.

Given the line

y=\dfrac34x+6

and a line

L

parallel to the given line and

4

units from it. A possible equation for

L

is:

y=\dfrac34x+1

y=\dfrac34x

y=\dfrac34x-\dfrac23

y=\dfrac34x-1

y=\dfrac34x+2

Concepts:parallel lines distance formula slope

Difficulty rating: 1750

Small hint:

Write parallel lines as

3x-4y+c=0

Big hint:

The distance equals the absolute difference of constants divided by

5

Solution:

The given line is

3x-4y+24=0.

A parallel line

y=\frac{3x}{4}+b

is

3x-4y+4b=0,

so the distance is

\frac{|24-4b|}{5}.

Setting this equal to

4

gives

|6-b|=5,

hence

b=1

or

11.

Choice A is possible.

Thus, the correct answer is A.

34.

In triangle

ABC,

side

a=\sqrt3,

side

b=\sqrt3,

and side

c\gt3.

Let

x

be the largest number such that the magnitude, in degrees, of the angle opposite side

c

exceeds

x.

Then

x

equals:

150

120

105

90

60

Concepts:law of cosines inequality bounding to limit cases

Difficulty rating: 1920

Small hint:

Use the law of cosines for the angle

C

opposite

c

Big hint:

Compare

c\gt3

with the limiting case

c=3

Solution:

The law of cosines gives

\cos C=\frac{3+3-c^2}{2\sqrt3\sqrt3}=\frac{6-c^2}{6}.

Since

c\gt3,

\cos C\lt-\frac{1}{2},

so

C\gt120^\circ.

Values can approach

120^\circ

as

c

approaches

3

from above, so the largest guaranteed bound is

120.

Therefore, the correct answer is B.

35.

The lengths of the sides of a triangle are integers, and its area is also an integer. One side is

21

and the perimeter is

48.

The shortest side is:

8

10

12

14

16

Concepts:Heron’s Formula casework triangle inequality

Difficulty rating: 2100

Small hint:

Write the other sides as

x

and

27-x

, with semiperimeter

24

Big hint:

Heron’s formula makes the squared area

72(24-x)(x-3)

Solution:

Let the other sides be

x

and

27-x,

with

x\leq27-x.

The semiperimeter is

24,

so Heron’s formula gives

\begin{aligned} K^2 &=24\cdot3(24-x)(x-3)\\ &=72(24-x)(x-3). \end{aligned}

The triangle inequalities give

4\leq x\leq13.

Checking these integers, the expression is not a square for

x=4,\ldots,9,

while at

x=10

it is

72\cdot14\cdot7=7056=84^2.

Thus the shortest side is

10.

Therefore, the correct answer is B.

36.

A person starting with

64

cents and making

6

bets, wins three times and loses three times, the wins and losses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:

a loss of

27

¢

a gain of

27

¢

a loss of

37

¢

neither a gain nor a loss

a gain or a loss depending upon the order in which the wins and losses occur

Concepts:exponent process simulation

Difficulty rating: 1470

Small hint:

A win multiplies the current amount by

\frac{3}{2}

, while a loss multiplies it by

\frac{1}{2}

Big hint:

Multiplication makes the order irrelevant

Solution:

After three wins and three losses, the amount, in cents, is

\begin{aligned} 64\left(\frac32\right)^3 \left(\frac12\right)^3 &=64\cdot\frac{27}{64}\\ &=27. \end{aligned}

The loss is

64-27=37

cents, regardless of order.

Thus, the correct answer is C.

37.

Given points

P_1,

P_2,

\ldots,

P_7

on a straight line, in the order stated (not necessarily evenly spaced). Let

P

be an arbitrarily selected point on the line and let

s

be the sum of the undirected lengths

PP_1,\ PP_2,\ \ldots,\ PP_7.

Then

s

is smallest if and only if the point

P

is:

midway between

P_1

and

P_7

midway between

P_2

and

P_6

midway between

P_3

and

P_5

at

P_4

at

P_1

Concepts:optimization pairing and grouping

Difficulty rating: 1640

Small hint:

Pair the distances to

P_1,P_7

, then

P_2,P_6

, then

P_3,P_5

Big hint:

Each paired sum is minimized throughout its intervening segment; the unpaired middle point selects one location

Solution:

For any pair

P_i,P_{8-i},

the sum

PP_i+PP_{8-i}

is minimized when

P

lies between them. The three such intervals all contain

P_4.

The remaining term

PP_4

is uniquely minimized at

P=P_4.

Thus, the correct answer is D.

38.

Point

F

is taken on the extension of side

AD

of parallelogram

ABCD.

BF

intersects diagonal

AC

at

E

and side

DC

at

G.

If

EF=32

and

GF=24,

then

BE

equals:

4

8

10

12

16

Concepts:parallelogram vector ratio and proportion

Difficulty rating: 2150

Small hint:

Parameterize points on

FB

by their fraction of the distance from

F

to

B

Big hint:

If

F=tD

in affine coordinates based at

A,

the parameters of

G

and

E

are

\frac{t-1}{t}

and

\frac{t}{t+1}

Solution:

Use affine coordinates

A=0,

B=u,

D=v,

C=u+v,

and

F=tv.

A point on

FB

is

su+t(1-s)v.

Intersecting

DC

gives

s_G=\frac{t-1}{t},

while intersecting

AC

gives

s_E=\frac{t}{t+1}.

Therefore

\begin{aligned} \frac{GF}{EF} &=\frac{s_G}{s_E} =\frac{t^2-1}{t^2}\\ &=\frac{24}{32}=\frac34, \end{aligned}

so

t=2

and

s_E=\frac{2}{3}.

Thus

EF=(\frac{2}{3})FB=32,

making

FB=48

and

BE=16.

Therefore, the correct answer is E.

39.

In triangle

ABC

lines

CE

and

AD

are drawn so that

\dfrac{CD}{DB}=\dfrac31

and

\dfrac{AE}{EB}=\dfrac32.

Let

r=\dfrac{CP}{PE},

where

P

is the intersection point of

CE

and

AD.

Then

r

equals:

3

\dfrac32

4

5

\dfrac52

Concepts:mass points ratio and proportion

Difficulty rating: 2030

Small hint:

Assign endpoint masses inversely proportional to the given side ratios

Big hint:

Choose masses

m_A=2,

m_B=3,

and

m_C=1

, then find the mass at

E

Solution:

The ratio

AE:EB=3:2

is represented by masses

m_A=2

and

m_B=3.

The ratio

CD:DB=3:1

then gives

m_C=1.

Point

E

has mass

m_A+m_B=5,

so along cevian

CE,

\frac{CP}{PE}=\frac{m_E}{m_C}=\frac51=5.

Thus, the correct answer is D.

40.

If

x

is a number satisfying the equation

\sqrt[3]{x+9}-\sqrt[3]{x-9}=3,

then

x^2

is between:

55

and

65

65

and

75

75

and

85

85

and

95

95

and

105

Concepts:radical sum and difference of cubes symmetry (algebra)

Difficulty rating: 2290

Small hint:

Set

u=\sqrt[3]{x+9}

and

v=\sqrt[3]{x-9}

Big hint:

Use both

u-v=3

and

u^3-v^3=18

to find

uv

Solution:

Let

u=\sqrt[3]{x+9}

and

v=\sqrt[3]{x-9}.

Then

u-v=3

and

\begin{aligned} 18&=u^3-v^3\\ &=(u-v)(u^2+uv+v^2), \end{aligned}

so

u^2+uv+v^2=6.

Comparing with

(u-v)^2=9

gives

uv=-1.

Hence

(u+v)^2=5.

Cubing

2u=3\pm\sqrt5

gives

x=\pm4\sqrt5,

and either way

x^2=80,

between

75

and

85.

Therefore, the correct answer is C.