1960 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If

2

is a solution (root) of

x^3+hx+10=0,

then

h

equals:

10

9

2

-2

-9

Concepts:polynomial substitution linear equation

Difficulty rating: 840

Small hint:

Substitute

x=2

into the equation

Big hint:

Solve the resulting linear equation for

h

Solution:

Substituting the given root gives

2^3+2h+10=0.

Thus

2h=-18,

so

h=-9.

Therefore, the correct answer is E.

2.

It takes

5

seconds for a clock to strike

6

o’clock beginning at

6{:}00

o’clock precisely. If the strikings are uniformly spaced, how long, in seconds, does it take to strike

12

o’clock?

9\dfrac13

10

11

14\dfrac23

none of these

Concepts:clock rate fencepost counting

Difficulty rating: 960

Small hint:

Six strikes contain only five time intervals

Big hint:

Twelve strikes contain eleven intervals of the same length

Solution:

There are

5

equal intervals from the first of

6

strikes to the last, so each interval lasts

1

second. Twelve strikes contain

11

such intervals and therefore take

11

seconds.

Thus, the correct answer is C.

3.

Applied to a bill for

\$10{,}000

the difference between a discount of

40\%

and two successive discounts of

36\%

and

4\%,

expressed in dollars, is:

0

144

256

400

416

Concepts:percentage money

Difficulty rating: 1060

Small hint:

Successive discounts are applied one after the other, not added

Big hint:

Compare

40\%

of the bill with

36\%

plus

4\%

of the remaining

64\%

Solution:

A

40\%

discount is

\$4000.

The successive discounts total

3600+0.04(6400)=3856.

Their difference is

4000-3856=144

dollars.

Therefore, the correct answer is B.

4.

Each of two angles of a triangle is

60^\circ

and the included side is

4

inches. The area of the triangle, in square inches, is:

8\sqrt3

8

4\sqrt3

4

2\sqrt3

Concepts:equilateral triangle triangle area

Difficulty rating: 890

Small hint:

Determine the third angle of the triangle

Big hint:

Use the area formula for an equilateral triangle of side

4

Solution:

The third angle is also

60^\circ,

so the triangle is equilateral with side

4.

Its area is

\frac{\sqrt3}{4}(4)^2=4\sqrt3.

Thus, the correct answer is C.

5.

The number of distinct points common to the graphs of

x^2+y^2=9

and

y^2=9

is:

infinitely many

four

two

one

none

Concepts:system of equations circle

Difficulty rating: 840

Small hint:

Substitute

y^2=9

into the circle equation

Big hint:

After finding

x,

count the possible values of

y

Solution:

Substitution gives

x^2+9=9,

so

x=0.

Then

y^2=9

gives

y=3

or

y=-3.

Hence there are two common points.

Therefore, the correct answer is C.

6.

The circumference of a circle is

100

inches. The side of a square inscribed in this circle, expressed in inches, is:

\dfrac{25\sqrt2}{\pi}

\dfrac{50\sqrt2}{\pi}

\dfrac{100}{\pi}

\dfrac{100\sqrt2}{\pi}

50\sqrt2

Concepts:circumference square (geometry)diagonal

Difficulty rating: 1320

Small hint:

Find the diameter from the circumference

Big hint:

The diagonal of the inscribed square is the circle’s diameter

Solution:

The diameter is

\frac{100}{\pi}.

If the square has side

s,

its diagonal is

s\sqrt2,

so

s\sqrt2=\frac{100}{\pi}.

Therefore

s=\frac{50\sqrt2}{\pi}.

Thus, the correct answer is B.

7.

Circle I passes through the center of, and is tangent to, circle II. The area of circle I is

4

square inches. Then the area of circle II, in square inches, is:

8

8\sqrt2

8\sqrt\pi

16

16\sqrt2

Concepts:tangent circles circle area power scaling of length, area, and volume

Difficulty rating: 1140

Small hint:

Draw the line through the two centers and the tangency point

Big hint:

The radius of circle II is twice the radius of circle I

Solution:

Because circle I passes through the center of circle II, the distance between the centers equals the radius

r

of circle I. Internal tangency also makes that distance

R-r,

where

R

is the radius of circle II. Thus

R=2r,

and the area is multiplied by

2^2=4.

It is therefore

4\cdot4=16.

Thus, the correct answer is D.

8.

The number

2.5252525\ldots

can be written as a fraction. When reduced to lowest terms the sum of the numerator and denominator of this fraction is:

7

29

141

349

none of these

Concepts:repeating decimal fraction greatest common divisor

Difficulty rating: 1320

Small hint:

Write the repeating part

0.525252\ldots

using a denominator of

99

Big hint:

Add the integer part and then reduce the fraction

Solution:

We have

2.525252\ldots=2+\frac{52}{99}=\frac{250}{99}.

This fraction is already in lowest terms, and

250+99=349.

Therefore, the correct answer is D.

9.

The fraction

\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}

is (with suitable restrictions on the values of

a,

b,

and

c

):

irreducible

reducible to

-1

reducible to a polynomial of three terms

reducible to

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

reducible to

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

Concepts:factoring difference of squares algebraic manipulation

Difficulty rating: 1180

Small hint:

Rewrite each quadratic expression as a difference of squares

Big hint:

The numerator and denominator share a factor

a+b+c

Solution:

Factoring gives

\begin{gathered} a^2+b^2-c^2+2ab\\ {}=(a+b-c)(a+b+c),\\ a^2+c^2-b^2+2ac\\ {}=(a-b+c)(a+b+c). \end{gathered}

Cancelling the common factor, under the stated suitable restrictions, leaves

\frac{a+b-c}{a-b+c}.

Thus, the correct answer is E.

10.

Given the following six statements:

(1)

All women are good drivers.

(2)

Some women are good drivers.

(3)

No men are good drivers.

(4)

All men are bad drivers.

(5)

At least one man is a bad driver.

(6)

All men are good drivers.

The statement that negates statement

(6)

is:

(1)

(2)

(3)

(4)

(5)

Concepts:logical deduction

Difficulty rating: 890

Small hint:

Negating “all” produces an existence statement

Big hint:

The negation needs only one man who is not a good driver

Solution:

The negation of “All men are good drivers” is “At least one man is not a good driver.” In the terminology of the choices, that is statement

(5),

“At least one man is a bad driver.”

Therefore, the correct answer is E.

11.

For a given value of

k,

the product of the roots of

x^2-3kx+2k^2-1=0

is

7.

The roots may be characterized as:

integral and positive

integral and negative

rational, but not integral

irrational

imaginary

Concepts:Vieta’s Formulas quadratic radical

Difficulty rating: 1210

Small hint:

Use the constant term to determine the possible values of

k

Big hint:

For either value of

k,

examine the quadratic formula or discriminant

Solution:

By Vieta’s formulas,

2k^2-1=7,

so

k=\pm2.

The roots are

3\pm\sqrt2

when

k=2,

and

-3\pm\sqrt2

when

k=-2.

In either case both roots are irrational.

Thus, the correct answer is D.

12.

The locus of the centers of all circles of given radius

a,

in the same plane, passing through a fixed point, is:

a point

a straight line

two straight lines

a circle

two circles

Concepts:circle distance formula

Difficulty rating: 800

Small hint:

Every center must be exactly one radius from the fixed point

Big hint:

Identify the locus of points at a fixed positive distance from one point

Solution:

A circle of radius

a

passes through the fixed point exactly when its center is distance

a

from that point. The locus of all such centers is a circle of radius

a.

Therefore, the correct answer is D.

13.

The polygon(s) formed by

y=3x+2,

y=-3x+2,

and

y=-2,

is (are):

an equilateral triangle

an isosceles triangle

a right triangle

a triangle and a trapezoid

a quadrilateral

Concepts:coordinate geometry slope isosceles triangle

Difficulty rating: 1140

Small hint:

Find the three pairwise intersections of the lines

Big hint:

The two slanted lines are mirror images across the

y

-axis

Solution:

The slanted lines meet at

(0,2),

while their intersections with

y=-2

are

(-\frac{4}{3},-2)

and

(\frac{4}{3},-2).

The latter two points are symmetric about the

y

-axis, so the two slanted sides have equal length. The figure is an isosceles triangle.

Thus, the correct answer is B.

14.

If

a

and

b

are real numbers, the equation

3x-5+a=bx+1

has a unique solution

x

[the symbol

a\ne0

means that

a

is different from zero]:

for all

a

and

b

if

a\ne2b

if

a\ne6

if

b\ne0

if

b\ne3

Concepts:linear equation

Difficulty rating: 890

Small hint:

Collect the

x

-terms on one side

Big hint:

A linear equation has a unique solution when the coefficient of

x

is nonzero

Solution:

Rearranging gives

(3-b)x=6-a.

This has a unique solution exactly when

3-b\ne0,

or

b\ne3.

Thus, the correct answer is E.

15.

Triangle I is equilateral with side

A,

perimeter

P,

area

K,

and circumradius

R

(radius of the circumscribed circle). Triangle II is equilateral with side

a,

perimeter

p,

area

k,

and circumradius

r.

If

A

is different from

a,

then:

P:p=R:r

only sometimes

P:p=R:r

always

P:p=K:k

only sometimes

R:r=K:k

always

R:r=K:k

only sometimes

Difficulty rating: 960

Small hint:

All equilateral triangles are similar

Big hint:

Perimeter and circumradius both scale linearly with side length

Solution:

For equilateral triangles,

P=3A

and

R=\frac{A}{\sqrt3},

with analogous formulas for

p

and

r.

Therefore

P:p=A:a=R:r

for every such pair of triangles.

Thus, the correct answer is B.

16.

In the numeration system with base

5,

counting is as follows:

1,

2,

3,

4,

10,

11,

12,

13,

14,

20,

\ldots

The number whose description in the decimal system is

69,

when described in the base

5

system, is a number with:

two consecutive digits

two non-consecutive digits

three consecutive digits

three non-consecutive digits

four digits

Concepts:number base place value digits

Difficulty rating: 1140

Small hint:

Express

69

using powers of

5

Big hint:

After finding the base-

5

digits, compare their values

Solution:

Since

69=2\cdot25+3\cdot5+4,

its base-

5

representation is

234_5.

Its three digits

2,

3,

and

4

are consecutive.

Therefore, the correct answer is C.

17.

The formula

N=8\cdot10^8\cdot x^{-\frac{3}{2}}

gives, for a certain group, the number of individuals whose income exceeds

x

dollars. The lowest income, in dollars, of the wealthiest

800

individuals is at least:

10^4

10^6

10^8

10^{12}

10^{16}

Concepts:exponent algebraic manipulation

Difficulty rating: 1210

Small hint:

Set

N=800

at the income cutoff

Big hint:

After simplifying powers of

10,

solve

x^{\frac{3}{2}}=10^6

Solution:

At the cutoff,

800=8\cdot10^8x^{-\frac{3}{2}}.

Hence

x^{\frac{3}{2}}=10^6,

and raising both sides to the

\frac{2}{3}

power gives

x=10^4.

Thus, the correct answer is A.

18.

The pair of equations

3^{x+y}=81

and

81^{x-y}=3

has:

no common solution

the solution

x=2,

y=2

the solution

x=2\dfrac12,

y=1\dfrac12

a common solution in positive and negative integers

none of these

Concepts:exponent system of equations substitution

Difficulty rating: 1280

Small hint:

Rewrite

81

as

3^4

in both equations

Big hint:

Solve the resulting system for

x+y

and

x-y

Solution:

The first equation gives

x+y=4.

The second gives

3^{4(x-y)}=3,

so

x-y=\frac{1}{4}.

Therefore

x=\frac{17}{8}

and

y=\frac{15}{8}.

This pair is not listed among the first four choices.

Thus, the correct answer is E.

19.

Consider equation I:

x+y+z=46,

where

x,

y,

and

z

are positive integers, and equation II:

x+y+z+w=46,

where

x,

y,

z,

and

w

are positive integers. Then:

\mathrm{I}

can be solved in consecutive integers

\mathrm{I}

can be solved in consecutive even integers

\mathrm{II}

can be solved in consecutive integers

\mathrm{II}

can be solved in consecutive even integers

\mathrm{II}

can be solved in consecutive odd integers

Concepts:arithmetic sequence divisibility linear equation

Difficulty rating: 1140

Small hint:

Write the sums of three and four consecutive integers

Big hint:

Check which required average is compatible with

46

Solution:

The sum of three consecutive integers is divisible by

3,

so it cannot be

46.

Four consecutive integers beginning with

n

have sum

4n+6.

Setting

4n+6=46

gives

n=10,

and indeed

10+11+12+13=46.

Therefore, the correct answer is C.

20.

The coefficient of

x^7

in the expansion of

\left(\frac{x^2}{2}-\frac2x\right)^8

is:

56

-56

14

-14

0

Concepts:binomial theorem exponent

Difficulty rating: 1570

Small hint:

In a term using

k

copies of

-\frac{2}{x},

determine the exponent of

x

Big hint:

Solve

16-3k=7

and then compute that binomial coefficient

Solution:

The term using

k

factors of

-\frac{2}{x}

has exponent

2(8-k)-k=16-3k.

To obtain

x^7,

we need

k=3.

Its coefficient is

\binom83\left(\frac12\right)^5(-2)^3=-14.

Thus, the correct answer is D.

21.

The diagonal of square I is

a+b.

The perimeter of square II with twice the area of I is:

(a+b)^2

\sqrt2(a+b)^2

2(a+b)

\sqrt8(a+b)

4(a+b)

Concepts:square (geometry)diagonal power scaling of length, area, and volume

Difficulty rating: 1140

Small hint:

Express the area of square I in terms of its diagonal

Big hint:

Square II’s area is

(a+b)^2,

so find its side length

Solution:

A square with diagonal

a+b

has area

\frac{(a+b)^2}{2}.

Square II therefore has area

(a+b)^2,

so its side is

a+b

and its perimeter is

4(a+b).

Thus, the correct answer is E.

22.

The equality

(x+m)^2-(x+n)^2=(m-n)^2,

where

m

and

n

are unequal non-zero constants, is satisfied by

x=am+bn

where:

a=0,

b

has a unique non-zero value

a=0,

b

has two non-zero values

b=0,

a

has a unique non-zero value

b=0,

a

has two non-zero values

a

and

b

each have a unique non-zero value

Concepts:difference of squares linear equation substitution

Difficulty rating: 1280

Small hint:

Factor the difference of the two squares

Big hint:

Cancel the nonzero factor

m-n

and solve for

x

Solution:

Factoring and using

m\ne n

gives

\begin{gathered} (m-n)(2x+m+n)\\ {}=(m-n)^2. \end{gathered}

Hence

2x+m+n=m-n,

so

x=-n.

Thus

a=0

and

b=-1,

a unique nonzero value.

Therefore, the correct answer is A.

23.

The radius

R

of a cylindrical box is

8

inches, the height

H

is

3

inches. The volume

V=\pi R^2H

is to be increased by the same fixed positive amount when

R

is increased by

x

inches as when

H

is increased by

x

inches. This condition is satisfied by:

no real value of

x

one integral value of

x

one rational, but not integral, value of

x

one irrational value of

x

two real values of

x

Concepts:cylinder volume quadratic

Difficulty rating: 1500

Small hint:

Write the volume increase once with radius

8+x

and once with height

3+x

Big hint:

Equate the increases and discard the nonpositive solution

Solution:

Increasing the radius changes the volume by

3\pi\big((8+x)^2-8^2\big),

while increasing the height changes it by

64\pi x.

Equating these gives

48x+3x^2=64x,

so

x=0

or

x=\frac{16}{3}.

The increase must be positive, leaving one rational but nonintegral value.

Thus, the correct answer is C.

24.

If

\log_{2x}216=x,

where

x

is real, then

x

is:

a non-square, non-cube integer

a non-square, non-cube, non-integral rational number

an irrational number

a perfect square

a perfect cube

Concepts:logarithm exponent function

Difficulty rating: 1280

Small hint:

Convert the logarithmic equation to

(2x)^x=216

Big hint:

Use

216=6^3

to recognize a real solution

Solution:

The equation is equivalent to

(2x)^x=216.

Since

216=6^3,

x=3

satisfies the equation. To see it is the only admissible real solution, note that the base requires

x>0

and

2x\ne1;

on

0\lt x\lt\frac{1}{2},

x\ln(2x)\lt0,

while for

x>\frac{1}{2}

the function

x\ln(2x)

is strictly increasing. Thus

x=3,

a non-square, non-cube integer.

Therefore, the correct answer is A.

25.

Let

m

and

n

be any two odd numbers, with

n

less than

m.

The largest integer which divides all possible numbers of the form

m^2-n^2

is:

2

4

6

8

16

Concepts:divisibility parity difference of squares

Difficulty rating: 1280

Small hint:

Factor

m^2-n^2

into two even factors

Big hint:

Among

m-n

and

m+n,

one is divisible by

4

; then test a small pair

Solution:

We have

m^2-n^2=(m-n)(m+n).

Both factors are even, and one is divisible by

4,

so every such difference is divisible by

8.

Taking

m=3,n=1

gives

m^2-n^2=8,

proving that no larger integer always divides it.

Thus, the correct answer is D.

26.

Find the set of

x

-values satisfying the inequality

\left|\frac{5-x}{3}\right|\lt2.

[The symbol

|a|

means

+a

if

a

is positive,

-a

if

a

is negative,

0

if

a

is zero. The notation

1\lt a\lt2

means that

a

can have any value between

1

and

2,

excluding

1

and

2.

]

1\lt x\lt11

-1\lt x\lt11

x\lt11

x\gt11

|x|\lt6

Concepts:absolute value inequality

Difficulty rating: 960

Small hint:

Multiply the inequality by

3

Big hint:

Rewrite

|5-x|\lt6

as a compound inequality

Solution:

The inequality is equivalent to

|5-x|\lt6,

so

-6\lt5-x\lt6.

Subtracting

5

and reversing signs as needed gives

-1\lt x\lt11.

Therefore, the correct answer is B.

27.

Let

S

be the sum of the interior angles of a polygon

P

for which each interior angle is

7\dfrac12

times the exterior angle at the same vertex. Then:

S=2660^\circ

and

P

may be regular

S=2660^\circ

and

P

is not regular

S=2700^\circ

and

P

is regular

S=2700^\circ

and

P

is not regular

S=2700^\circ

and

P

may or may not be regular

Concepts:angle sum equiangular polygon regular polygon

Difficulty rating: 1500

Small hint:

An interior angle and its corresponding exterior angle sum to

180^\circ

Big hint:

The condition fixes every angle but says nothing about the side lengths

Solution:

If an exterior angle is

e,

then

e+\frac{15}{2}e=180^\circ,

so

e=\frac{360^\circ}{17}.

Since the exterior angles total

360^\circ,

the polygon has

17

vertices and

S=(17-2)180^\circ=2700^\circ.

All its angles are equal, but its sides need not be equal, so it may or may not be regular.

Thus, the correct answer is E.

28.

The equation

x-\frac7{x-3}=3-\frac7{x-3}

has:

infinitely many integral roots

no root

one integral root

two equal integral roots

two equal non-integral roots

Concepts:rational equation algebraic manipulation

Difficulty rating: 890

Small hint:

The identical fractional terms cancel wherever the equation is defined

Big hint:

Check whether the resulting value lies in the original domain

Solution:

For

x\ne3,

subtracting the identical fractions from both sides leaves

x=3.

But

x=3

makes the original denominators zero, so it is not in the domain. The equation has no root.

Therefore, the correct answer is B.

29.

Five times

A

’s money added to

B

’s money is more than

\$51.00.

Three times

A

’s money minus

B

’s money is

\$21.00.

If

a

represents

A

’s money in dollars and

b

represents

B

’s money in dollars, then:

a\gt9,

b\gt6

a\gt9,

b\lt6

a\gt9,

b=6

a\gt9,

but we can put no bounds on

b

2a=3b

Concepts:inequality system of equations money

Difficulty rating: 1320

Small hint:

Use

3a-b=21

to express

b

in terms of

a

Big hint:

Substitute into

5a+b>51

Solution:

From

3a-b=21,

we have

b=3a-21.

Substitution into the inequality gives

5a+3a-21>51,

so

a>9.

It follows that

b=3a-21>6.

Thus, the correct answer is A.

30.

Given the line

3x+5y=15

and a point on this line equidistant from the coordinate axes. Such a point exists in:

none of the quadrants

quadrant

\mathrm{I}

only

quadrants

\mathrm{I},

\mathrm{II}

only

quadrants

\mathrm{I},

\mathrm{II},

\mathrm{III}

only

each of the quadrants

Concepts:coordinate geometry absolute value system of equations

Difficulty rating: 1210

Small hint:

A point equidistant from the axes satisfies

|x|=|y|

Big hint:

Intersect the given line with both

y=x

and

y=-x

Solution:

For

y=x,

the line gives

8x=15,

producing a point in quadrant I. For

y=-x,

it gives

-2x=15,

so

x\lt0

and

y>0,

producing a point in quadrant II. There are no other possibilities.

Therefore, the correct answer is C.

31.

For

x^2+2x+5

to be a factor of

x^4+px^2+q,

the values of

p

and

q

must be, respectively:

-2,

5

5,

25

10,

20

6,

25

14,

25

Concepts:polynomial algebraic manipulation factor

Difficulty rating: 1690

Small hint:

Work modulo

x^2+2x+5,

so

x^2=-2x-5

Big hint:

Reduce

x^4

to a linear expression and make both remainder coefficients zero

Solution:

Modulo

x^2+2x+5,

we have

x^2=-2x-5.

Then

x^3=10-x,\qquad x^4=12x+5.

Thus

\begin{gathered} x^4+px^2+q\\ {}\equiv(12-2p)x+(5-5p+q). \end{gathered}

Both coefficients vanish when

p=6

and

q=25.

Therefore, the correct answer is D.

32.

In this figure the center of the circle is

O.

\overline{AB}\perp\overline{BC},

ADOE

is a straight line,

\overline{AP}=\overline{AD},

and

\overline{AB}

has a length twice the radius. Then:

\overline{AP}^{\,2}=\overline{PB}\cdot\overline{AB}

\overline{AP}\cdot\overline{DO}=\overline{PB}\cdot\overline{AD}

\overline{AB}^{\,2}=\overline{AD}\cdot\overline{DE}

\overline{AB}\cdot\overline{AD}=\overline{OB}\cdot\overline{AO}

none of these

Concepts:tangent line power of a point circle

Difficulty rating: 2000

Small hint:

Let the radius be

r

and write

AD

in terms of

AO

and

r

Big hint:

Use the tangent relation

AB^2=AD\cdot AE

together with

AB=2r

Solution:

Let the radius be

r.

Since

AB

is tangent at

B,

AB^2=AD\cdot AE.

Write

AD=t.

Because the secant passes through the center,

AE=t+2r,

while

AB=2r.

Hence

t(t+2r)=4r^2.

Also

AP=AD=t

and

PB=AB-AP=2r-t.

The displayed equation rearranges to

t^2=2r(2r-t)=PB\cdot AB.

Therefore

AP^2=PB\cdot AB.

Thus, the correct answer is A.

33.

You are given a sequence of

58

terms; each term has the form

P+n

where

P

stands for the product

2\cdot3\cdot5\cdots61

of all prime numbers less than or equal to

61,

and

n

takes, successively, the values

2,

3,

4,

\ldots,

59.

Let

N

be the number of primes appearing in this sequence. Then

N

is:

0

16

17

57

58

Concepts:prime divisibility prime factorization

Difficulty rating: 1500

Small hint:

Every

n

from

2

through

59

has a prime divisor at most

59

Big hint:

That same prime divisor divides both

P

and

n

Solution:

For each

n

in the given range, choose a prime divisor

p

of

n.

Since

p\le59,

it is one of the factors of

P.

Therefore

p

divides

P+n.

Moreover

P+n>p,

so

P+n

is composite. No term is prime, and

N=0.

Thus, the correct answer is A.

34.

Two swimmers, at opposite ends of a

90

-foot pool, start to swim the length of the pool, one at the rate of

3

feet per second, the other at

2

feet per second. They swim back and forth for

12

minutes. Allowing no loss of time at the turns, find the number of times they pass each other.

24

21

20

19

18

Concepts:relative speed reflection (geometry)least common multiple

Difficulty rating: 1870

Small hint:

Reflect the pool each time a swimmer turns so both paths become straight

Big hint:

The joint position pattern repeats every

180

seconds; inspect one full cycle, including endpoint coincidences

Solution:

The faster swimmer’s position repeats every

60

seconds, and the slower swimmer’s position repeats every

90

seconds. Thus their joint position pattern repeats every

180

seconds. During one such cycle, their positions coincide at

t=18,54,90,126,162

seconds. The keyed interpretation counts the simultaneous turn at the endpoint when

t=90

as one of these encounters. Thus there are

5

counted encounters per cycle and

4

cycles in

12

minutes, giving

5\cdot4=20.

Therefore, the correct answer is C.

35.

From point

P

outside a circle, with a circumference of

10

units, a tangent is drawn. Also from

P

a secant is drawn dividing the circle into unequal arcs with lengths

m

and

n.

It is found that

t,

the length of the tangent, is the mean proportional between

m

and

n.

If

m

and

t

are integers, then

t

may have the following number of values:

zero

one

two

three

infinitely many

Concepts:ratio and proportion perfect square arc

Difficulty rating: 1870

Small hint:

Use

m+n=10

and

t^2=mn

Big hint:

Test the integer values

m=1,2,\ldots,9,

excluding equal arcs

Solution:

The arc lengths satisfy

m+n=10,

while the mean-proportional condition gives

t^2=mn=m(10-m).

For integral

m

from

1

through

9,

the possible products, up to symmetry, are

9,

16,

21,

24,

and

25.

The last comes from equal arcs

m=n=5

and is excluded. The remaining squares give

t=3

and

t=4,

two values.

Thus, the correct answer is C.

36.

Let

s_1,

s_2,

s_3

be the respective sums of

n,

2n,

3n

terms of the same arithmetic progression with

a

as the first term and

d

as the common difference. Let

R=s_3-s_2-s_1.

Then

R

is dependent on:

a

and

d

d

and

n

a

and

n

a,

d,

and

n

neither

a

nor

d

nor

n

Concepts:arithmetic sequence summation algebraic manipulation

Difficulty rating: 1670

Small hint:

Use

S_k=\dfrac{k}{2}(2a+(k-1)d)

Big hint:

Substitute

k=n,2n,3n

and collect the

a

-terms and

d

-terms separately

Solution:

Using the arithmetic-series formula,

s_j=\frac{jn}{2}\bigl(2a+(jn-1)d\bigr).

In

s_3-s_2-s_1,

the coefficient of

a

is

3n-2n-n=0.

Simplifying the remaining terms gives

R=2n^2d.

Thus

R

depends on

d

and

n,

but not on

a.

Therefore, the correct answer is B.

37.

The base of a triangle is of length

b,

and the altitude is of length

h.

A rectangle of height

x

is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:

\dfrac{bx}{h}(h-x)

\dfrac{hx}{b}(b-x)

\dfrac{bx}{h}(h-2x)

x(b-x)

x(h-x)

Concepts:similarity rectangle area

Difficulty rating: 1280

Small hint:

At height

x,

use similarity to find the horizontal width of the triangle

Big hint:

The available width is

b(1-\frac{x}{h})

; multiply it by the rectangle’s height

Solution:

The cross-section parallel to the base at height

x

has width

b\left(1-\frac{x}{h}\right)=\frac{b(h-x)}{h}

by similarity. Multiplying by the rectangle’s height

x

gives

\frac{bx}{h}(h-x).

Thus, the correct answer is A.

38.

In this diagram

\overline{AB}

and

\overline{AC}

are the equal sides of an isosceles triangle

ABC,

in which is inscribed equilateral triangle

DEF.

Designate angle

BFD

by

a,

angle

ADE

by

b,

and angle

FEC

by

c.

Then:

b=\dfrac{a+c}{2}

b=\dfrac{a-c}{2}

a=\dfrac{b-c}{2}

a=\dfrac{b+c}{2}

none of these

Concepts:angle chasing isosceles triangle equilateral triangle

Difficulty rating: 2000

Small hint:

Let each base angle of

ABC

be

\theta

Big hint:

Compare the directions of

DF,

DE,

and

EF

using the

60^\circ

angles of the equilateral triangle

Solution:

Let each base angle of

ABC

be

\theta.

Measured from the direction

BC,

the line

FD

has direction

180^\circ-a,

so the equilateral condition makes

DE

have direction

60^\circ-a.

At

D,

b=\theta-(60^\circ-a)=\theta-60^\circ+a.

Similarly, angle chasing at

E

gives

c=60^\circ+a-\theta.

Adding these equations yields

b+c=2a.

Therefore

a=\frac{b+c}{2},

and the correct answer is D.

39.

To satisfy the equation

\frac{a+b}{a}=\frac{b}{a+b},

a

and

b

must be:

both rational

both real but not rational

both not real

one real, one not real

one real, one not real or both not real

Concepts:complex number quadratic

Difficulty rating: 1730

Small hint:

Let

t=\frac{b}{a}

after noting that the denominators must be nonzero

Big hint:

The resulting quadratic

t^2+t+1=0

has negative discriminant

Solution:

The denominators require

a\ne0

and

a+b\ne0.

Setting

t=\frac{b}{a}

and cross-multiplying gives

(1+t)^2=t,

or

t^2+t+1=0.

Its discriminant is

-3,

so

t

is not real. Therefore

a

and

b

cannot both be real. Depending on the nonzero complex scale chosen for

a,

one can be real and the other nonreal, or both can be nonreal.

Thus, the correct answer is E.

40.

Given right triangle

ABC

with legs

\overline{BC}=3,

\overline{AC}=4.

Find the length of the shorter angle trisector from

C

to the hypotenuse:

\dfrac{32\sqrt3-24}{13}

\dfrac{12\sqrt3-9}{13}

6\sqrt3-8

\dfrac{5\sqrt{10}}6

\dfrac{25}{12}

Concepts:coordinate geometry trigonometry right triangle

Difficulty rating: 2000

Small hint:

Place

C=(0,0),

B=(3,0),

A=(0,4)

Big hint:

The two trisector rays make angles

30^\circ

and

60^\circ

with

CB

; intersect each with

\frac{x}{3}+\frac{y}{4}=1

Solution:

Place

C=(0,0),

B=(3,0),

A=(0,4).

The hypotenuse has equation

\frac{x}{3}+\frac{y}{4}=1.

The

30^\circ

trisector ray is

y=\frac{x}{\sqrt3}.

Substitution gives

x=\frac{12\sqrt3}{4\sqrt3+3}.

Its length is

\frac{x}{\cos30^\circ}=\frac{2x}{\sqrt3},

hence

\frac{24}{4\sqrt3+3} =\frac{32\sqrt3-24}{13}.

The

60^\circ

ray gives the longer trisector.

Thus, the correct answer is A.