1966 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Given that the ratio of

3x-4

to

y+15

is constant, and

y=3

when

x=2,

then, when

y=12,

x

equals:

\dfrac18

\dfrac87

\dfrac73

\dfrac72

8

Concepts:linear equation ratio and proportion

Difficulty rating: 960

Small hint:

Use

x=2

and

y=3

to find the constant ratio

Big hint:

Keep

\frac{3x-4}{y+15}

equal to that ratio when

y=12

Solution:

The given pair makes the ratio

\frac{3\cdot2-4}{3+15}=\frac{1}{9}.

Therefore, when

y=12,

\frac{3x-4}{27}=\frac19,

so

3x-4=3

and

x=\frac{7}{3}.

Therefore, the correct answer is C.

2.

When the base of a triangle is increased

10\%

and the altitude to this base is decreased

10\%,

the change in area is:

1\%

increase

\dfrac12\%

increase

0\%

\dfrac12\%

decrease

1\%

decrease

Concepts:percentage triangle area

Difficulty rating: 890

Small hint:

The area is proportional to the product of base and altitude

Big hint:

Multiply the scale factors

1.1

and

0.9

Solution:

The new area is

1.1\cdot0.9=0.99

times the old area. It is therefore

1\%

smaller.

Thus, the correct answer is E.

3.

If the arithmetic mean of two numbers is

6

and their geometric mean is

10,

then an equation with the given two numbers as roots is:

x^2+12x+100=0

x^2+6x+100=0

x^2-12x-10=0

x^2-12x+100=0

x^2-6x+100=0

Concepts:mean Vieta’s Formulas

Difficulty rating: 1180

Small hint:

The two numbers have sum

12

and product

100

Big hint:

Roots with sum

S

and product

P

satisfy

x^2-Sx+P=0

Solution:

If the numbers are

r,

and

s,

then

\frac{r+s}{2}=6

gives

r+s=12,

while

\sqrt{rs}=10

gives

rs=100.

The monic equation with those roots is

x^2-(r+s)x+rs=0,

or

x^2-12x+100=0.

Therefore, the correct answer is D.

4.

Circle

\mathrm{I}

is circumscribed about a given square and circle

\mathrm{II}

is inscribed in the given square. If

r

is the ratio of the area of circle

\mathrm{I}

to that of circle

\mathrm{II},

then

r

equals:

\sqrt2

2

\sqrt3

2\sqrt2

2\sqrt3

Concepts:area ratio circle area square (geometry)

Difficulty rating: 1030

Small hint:

For square side

s,

compare radii

\frac{s}{\sqrt2}

and

\frac{s}{2}

Big hint:

Circle areas are proportional to the squares of their radii

Solution:

For square side

s,

the outer circle has radius

\frac{s}{\sqrt2},

while the inner circle has radius

\frac{s}{2}.

Hence

r=\frac{\pi(\frac{s}{\sqrt2})^2}{\pi(\frac{s}{2})^2}=2.

Therefore, the correct answer is B.

5.

The number of values of

x

satisfying the equation

\frac{2x^2-10x}{x^2-5x}=x-3

is:

zero

one

two

three

an integer greater than

3

Concepts:factoring rational equation

Difficulty rating: 1320

Small hint:

Factor numerator and denominator before solving

Big hint:

Do not admit values that make the original denominator zero

Solution:

The denominator requires

x\ne0

and

x\ne5.

On that domain,

\frac{2x(x-5)}{x(x-5)}=2.

The remaining equation

2=x-3

gives

x=5,

which is excluded. Thus there are no solutions.

Therefore, the correct answer is A.

6.

AB

is a diameter of a circle centered at

O.

C

is a point on the circle such that angle

BOC

is

60^\circ.

If the diameter of the circle is

5

inches, the length of chord

AC,

expressed in inches, is:

3

\dfrac{5\sqrt2}{2}

\dfrac{5\sqrt3}{2}

3\sqrt3

none of these

Concepts:chord special right triangle

Difficulty rating: 1110

Small hint:

Because

AB

is a diameter,

\angle AOC=120^\circ

Big hint:

Split isosceles triangle

AOC

into two

30

-

60

-

90

triangles

Solution:

The radius is

\frac{5}{2},

and

\angle AOC=180^\circ-60^\circ=120^\circ.

A chord subtending

120^\circ

has length

2\left(\frac52\right)\sin60^\circ=\frac{5\sqrt3}{2}.

Therefore, the correct answer is C.

7.

Let

\frac{35x-29}{x^2-3x+2} =\frac{N_1}{x-1}+\frac{N_2}{x-2}

be an identity in

x.

The numerical value of

N_1N_2

is:

-246

-210

-29

210

246

Concepts:linear equation partial fractions

Difficulty rating: 1210

Small hint:

Combine the two fractions over

(x-1)(x-2)

Big hint:

Compare the coefficient of

x

and the constant term

Solution:

Combining the right side gives numerator

\begin{aligned} &N_1(x-2)+N_2(x-1)\\ &\quad=(N_1+N_2)x\\ &\qquad-(2N_1+N_2). \end{aligned}

Thus

N_1+N_2=35

and

2N_1+N_2=29.

Hence

N_1=-6,

N_2=41,

and

N_1N_2=-246.

Therefore, the correct answer is A.

8.

The length of the common chord of two intersecting circles is

16

feet. If the radii are

10

feet and

17

feet, a possible value for the distance between the centers of the circles, expressed in feet, is:

27

21

\sqrt{389}

15

undetermined

Concepts:chord circle Pythagorean Theorem

Difficulty rating: 1450

Small hint:

The line of centers perpendicularly bisects the common chord

Big hint:

Find each center’s distance from the chord using half-chord

8

Solution:

The half-chord has length

8.

The two perpendicular distances from the centers to its line are

\begin{gathered} \sqrt{10^2-8^2}=6,\\ \sqrt{17^2-8^2}=15. \end{gathered}

If the centers lie on opposite sides of the chord, their distance is

6+15=21,

which is a possible value.

Therefore, the correct answer is B.

9.

If

x=(\log_8 2)^{(\log_2 8)},

then

\log_3 x

equals:

-3

-\dfrac13

\dfrac13

3

9

Concepts:exponent logarithm

Difficulty rating: 1180

Small hint:

Evaluate

\log_8 2

and

\log_2 8

separately

Big hint:

Express the resulting power as a power of

3

Solution:

We have

\log_8 2=\frac{1}{3}

and

\log_2 8=3.

Therefore

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

so

\log_3x=-3.

Thus, the correct answer is A.

10.

If the sum of two numbers is

1

and their product is

1,

then the sum of their cubes is, where

i=\sqrt{-1}:

2

-2-\dfrac{3\sqrt3i}{4}

0

-\dfrac{3\sqrt3i}{4}

-2

Concepts:complex number sum and difference of cubes symmetry (algebra)

Difficulty rating: 1110

Small hint:

Use

u^3+v^3=(u+v)^3-3uv(u+v)

Big hint:

Substitute the given sum and product directly

Solution:

If the numbers are

u,

and

v,

then

\begin{aligned} u^3+v^3 &=(u+v)^3-3uv(u+v)\\ &=1^3-3=-2. \end{aligned}

Therefore, the correct answer is E.

11.

The sides of triangle

BAC

are in the ratio

2:3:4.

BD

is the angle-bisector drawn to the shortest side

AC,

dividing it into segments

AD

and

CD.

If the length of

AC

is

10,

then the length of the longer segment of

AC

is:

3\dfrac12

5

5\dfrac57

6

7\dfrac12

Concepts:angle bisector theorem ratio and proportion

Difficulty rating: 1180

Small hint:

Since

AC

is the shortest side, the adjacent sides are in ratio

3:4

Big hint:

Apply the angle-bisector theorem to

AD:DC

Solution:

The sides adjacent to angle

B

are in ratio

3:4.

By the angle-bisector theorem,

AD:DC=3:4.

The longer of the two parts is therefore

\frac4{3+4}\cdot10=\frac{40}{7}=5\frac57.

Therefore, the correct answer is C.

12.

The number of real values of

x

that satisfy the equation

(2^{6x+3})(4^{3x+6})=8^{4x+5}

is:

0

1

2

3

greater than

3

Concepts:algebraic manipulation exponent

Difficulty rating: 1110

Small hint:

Rewrite every term with base

2

Big hint:

Compare the total exponent on each side before trying to solve for

x

Solution:

In base

2,

the left exponent is

\begin{aligned} &(6x+3)+2(3x+6)\\ &\qquad=12x+15, \end{aligned}

and the right exponent is

3(4x+5)=12x+15.

The equation is an identity for every real

x,

so it has more than three real solutions.

Therefore, the correct answer is E.

13.

The number of points with positive rational coordinates selected from the set of points in the

xy

-plane such that

x+y\leq5,

is:

9

10

14

15

infinite

Concepts:coordinate geometry inequality

Difficulty rating: 1030

Small hint:

Try fixing one positive rational coordinate

Big hint:

There are infinitely many positive rational numbers in an interval

Solution:

For example, set

y=1.

Every positive rational

x\leq4

then gives an allowed point

(x,1),

and there are infinitely many such rationals.

Therefore, the correct answer is E.

14.

The length of rectangle

ABCD

is

5

inches and its width is

3

inches. Diagonal

AC

is divided into three equal segments by points

E

and

F.

The area of triangle

BEF,

expressed in square inches, is:

\dfrac32

\dfrac53

\dfrac52

\dfrac13\sqrt{34}

\dfrac13\sqrt{68}

Concepts:coordinate geometry rectangle triangle area

Difficulty rating: 1450

Small hint:

Place the rectangle at

(0,0),

(5,0),

(5,3),

and

(0,3)

Big hint:

The trisection points of diagonal

AC

have easy coordinates

Solution:

Take

A=(0,0),

B=(5,0),

and

C=(5,3).

Then

E=(\frac{5}{3},1)

and

F=(\frac{10}{3},2).

The determinant formula gives

\begin{aligned} [BEF] &=\frac12\left| \left(-\frac{10}{3}\right)2 -\left(-\frac53\right)1 \right|\\ &=\frac52. \end{aligned}

Therefore, the correct answer is C.

15.

If

x-y\gt x

and

x+y\lt y,

then:

y\lt x

x\lt y

x\lt y\lt0

x\lt0,

and

y\lt0

x\lt0,

and

y\gt0

Concepts:algebraic manipulation inequality

Difficulty rating: 960

Small hint:

Subtract

x

from the first inequality

Big hint:

Subtract

y

from the second inequality

Solution:

From

x-y\gt x

we get

-y\gt0,

so

y\lt0.

From

x+y\lt y

we get

x\lt0.

Therefore, the correct answer is D.

16.

If

\begin{gathered} \dfrac{4^x}{2^{x+y}}=8,\\ \dfrac{9^{x+y}}{3^{5y}}=243, \end{gathered}

x

and

y

are real numbers, then

xy

equals:

1\dfrac34

4

6

12

-4

Concepts:exponent system of equations

Difficulty rating: 1430

Small hint:

Rewrite both equations with bases

2

and

3

Big hint:

Equate exponents to obtain two linear equations in

x

and

y

Solution:

The first equation becomes

2^{x-y}=2^3,

so

x-y=3.

The second becomes

3^{2x-3y}=3^5,

so

2x-3y=5.

Solving gives

y=1,

x=4,

and hence

xy=4.

Therefore, the correct answer is B.

17.

The number of distinct points common to the curves

x^2+4y^2=1

and

4x^2+y^2=4

is:

0

1

2

3

4

Concepts:ellipse substitution system of equations

Difficulty rating: 1430

Small hint:

Use the first equation to substitute for

x^2

in the second

Big hint:

After finding

y,

count both possible signs of

x

Solution:

From the first equation,

x^2=1-4y^2.

Substitution into the second gives

4(1-4y^2)+y^2=4,

so

y=0.

Then

x^2=1,

giving the two points

(1,0)

and

(-1,0).

Therefore, the correct answer is C.

18.

In a given arithmetic sequence the first term is

2,

the last term is

29,

and the sum of all the terms is

155.

The common difference is:

3

2

\dfrac{27}{19}

\dfrac{13}{9}

\dfrac{23}{38}

Concepts:arithmetic sequence summation

Difficulty rating: 1180

Small hint:

Use

155=\frac{n(2+29)}{2}

to find the number of terms

Big hint:

Then use

29=2+(n-1)d

Solution:

The sum formula gives

155=\frac{31n}{2},

so

n=10.

Therefore

29=2+9d,

and

d=3.

Thus, the correct answer is A.

19.

Let

s_1

be the sum of the first

n

terms of the arithmetic sequence

8,

12,

\ldots

and let

s_2

be the sum of the first

n

terms of the arithmetic sequence

17,

19,

\ldots.

Then

s_1=s_2

for:

no value of

n

one value of

n

two values of

n

four values of

n

more than four values of

n

Concepts:arithmetic sequence quadratic summation

Difficulty rating: 1210

Small hint:

Write a sum formula for each progression

Big hint:

Cancel the positive factor

n

after equating the sums

Solution:

The two sums are

\begin{gathered} s_1=2n^2+6n,\\ s_2=n^2+16n. \end{gathered}

Equality gives

n(n-10)=0.

Since a number of terms is positive, only

n=10

works, so there is one value.

Therefore, the correct answer is B.

20.

If the proposition “

a=0

” is true, the negation of the proposition “For real values of

a

and

b,

if

a=0,

then

ab=0

” is:

If

a\ne0,

then

ab\ne0

If

a\ne0,

then

ab=0

If

a=0,

then

ab\ne0

If

ab\ne0,

then

a\ne0

If

ab=0,

then

a\ne0

Concepts:logical deduction

Difficulty rating: 1180

Small hint:

The negation of a true conclusion is its opposite statement

Big hint:

Under the stated assumption

a=0,

negate the conclusion

ab=0

Solution:

The stated premise

a=0

is assumed true. Negating the implication therefore negates its conclusion:

ab=0

becomes

ab\ne0.

Thus the required statement is “If

a=0,

then

ab\ne0.

”

Therefore, the correct answer is C.

21.

An “

n

-pointed star” is formed as follows: the sides of a convex polygon are numbered consecutively

1,

2,

\ldots,

k,

\ldots,

n,

n\geq5;

for all

n

values of

k,

sides

k

and

k+2

are non-parallel, sides

n+1

and

n+2

being respectively identical with sides

1

and

2;

prolong the

n

pairs of sides numbered

k

and

k+2

until they meet. (A figure is shown for the case

n=5.

)

Let

S

be the degree-sum of the interior angles at the

n

points of the star; then

S

equals:

180

360

180(n+2)

180(n-2)

180(n-4)

Concepts:angle chasing angle sum

Difficulty rating: 1880

Small hint:

Relate each star-tip angle to two adjacent interior angles of the original polygon

Big hint:

The original polygon’s interior-angle sum is

180(n-2)

degrees

Solution:

Let the original polygon’s interior angles be

a_1,\ldots,a_n.

The star angle made from sides

k

and

k+2

equals

a_k+a_{k+1}-180^\circ.

Summing cyclically counts each

a_k

twice. Hence

\begin{aligned} S&=2\sum_{k=1}^n a_k-180n\\ &=2\cdot180(n-2)-180n\\ &=180(n-4). \end{aligned}

Therefore, the correct answer is E.

22.

Consider the statements:

\text{(I) }\sqrt{a^2+b^2}=0

\text{(II) }\sqrt{a^2+b^2}=ab

\text{(III) }\sqrt{a^2+b^2}=a+b

\text{(IV) }\sqrt{a^2+b^2}=a-b

where we allow

a

and

b

to be real or complex numbers. Those statements for which there exist solutions other than

a=0

and

b=0,

are:

(\mathrm{I}),

(\mathrm{II}),

(\mathrm{III}),

(\mathrm{IV})

(\mathrm{II}),

(\mathrm{III}),

(\mathrm{IV})

only

(\mathrm{I}),

(\mathrm{III}),

(\mathrm{IV})

only

(\mathrm{III}),

(\mathrm{IV})

only

(\mathrm{I})

only

Concepts:complex number counterexample radical

Difficulty rating: 1880

Small hint:

One nonzero example is enough for each statement

Big hint:

Try

b=i

for

(\mathrm{I}),

equal positive values for

(\mathrm{II}),

and

b=0

for

(\mathrm{III})

and

(\mathrm{IV})

Solution:

Every statement has a nonzero solution. For

(\mathrm{I}),

use

(a,b)=(1,i).

For

(\mathrm{II}),

use

a=b=\sqrt2,

giving

\sqrt4=2=ab.

For both

(\mathrm{III})

and

(\mathrm{IV}),

use

(a,b)=(1,0).

Thus all four statements qualify.

Therefore, the correct answer is A.

23.

If

x

is real and

4y^2+4xy+x+6=0,

then the complete set of values of

x

for which

y

is real, is:

x\leq-2

or

x\geq3

x\leq2

or

x\geq3

x\leq-3

or

x\geq2

-3\leq x\leq2

-2\leq x\leq3

Concepts:inequality quadratic

Difficulty rating: 1180

Small hint:

Treat the equation as a quadratic in

y

Big hint:

Require its discriminant

16(x-3)(x+2)

to be nonnegative

Solution:

As a quadratic in

y,

the equation has discriminant

\begin{aligned} &(4x)^2-16(x+6)\\ &\qquad=16(x-3)(x+2). \end{aligned}

This is nonnegative exactly when

x\leq-2

or

x\geq3.

Therefore, the correct answer is A.

24.

If

\log_MN=\log_NM,

M\ne N,

MN\gt0,

and

M\ne1,

N\ne1,

then

MN

equals:

\dfrac12

1

2

10

a number greater than

2

and less than

10

Concepts:algebraic manipulation logarithm

Difficulty rating: 1430

Small hint:

Write both logarithms using natural logs

Big hint:

The equality gives

(\ln M)^2=(\ln N)^2

; use

M\ne N

Solution:

Change of base gives

\frac{\ln N}{\ln M}=\frac{\ln M}{\ln N},

so

(\ln M)^2=(\ln N)^2.

Equality of the logs themselves would give

M=N,

which is excluded. Hence

\ln M=-\ln N,

and

\ln(MN)=0.

Thus

MN=1.

Therefore, the correct answer is B.

25.

If

F(n+1)=\dfrac{2F(n)+1}{2}

for

n=1,

2,

\ldots,

and

F(1)=2,

then

F(101)

equals:

49

50

51

52

53

Concepts:arithmetic sequence recursion

Difficulty rating: 960

Small hint:

Simplify the recurrence to

F(n+1)=F(n)+\frac{1}{2}

Big hint:

Count the increments from

n=1

to

n=101

Solution:

Each step adds

\frac{1}{2}.

There are

100

steps from

F(1)

to

F(101),

so

F(101)=2+\frac{100}{2}=52.

Therefore, the correct answer is D.

26.

Let

m

be a positive integer and let the lines

13x+11y=700

and

y=mx-1

intersect in a point whose coordinates are integers. Then

m

can be:

4

only

5

only

6

only

7

only

one of the integers

4,

5,

6,

7

and one other positive integer

Concepts:Diophantine Equation divisibility linear equation

Difficulty rating: 1650

Small hint:

Substitute

y=mx-1

into the first line

Big hint:

For integer

x,

the number

13+11m

must divide

711

Solution:

Substitution gives

(13+11m)x=711.

Thus

13+11m

must be a positive divisor of

711=3^2\cdot79.

Among divisors greater than

13,

only

79

is congruent to

13\pmod{11}.

Hence

13+11m=79,

so

m=6.

Therefore, the correct answer is C.

27.

At his usual rate a man rows

15

miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream’s current is:

2

\dfrac52

3

\dfrac72

4

Concepts:rational equation relative speed system of equations

Difficulty rating: 1850

Small hint:

Let

v

be the rower’s still-water speed and

c

the current speed

Big hint:

Translate the two time differences using speeds

v\pm c

and

2v\pm c

Solution:

The two time differences give

\begin{aligned} \frac{15}{v-c}-\frac{15}{v+c}&=5,\\ \frac{15}{2v-c}-\frac{15}{2v+c}&=1. \end{aligned}

These simplify to

v^2-c^2=6c

and

4v^2-c^2=30c.

Substituting

v^2=c^2+6c

into the second gives

3c^2-6c=0.

Since

c\gt0,

we have

c=2.

Therefore, the correct answer is A.

28.

Five points

O,

A,

B,

C,

D

are taken in order on a straight line with distances

OA=a,

OB=b,

OC=c,

and

OD=d.

P

is a point on the line between

B

and

C

and such that

AP:PD=BP:PC.

Then

OP

equals:

\dfrac{b^2-bc}{a-b+c-d}

\dfrac{ac-bd}{a-b+c-d}

-\dfrac{bd+ac}{a-b+c-d}

\dfrac{bc+ad}{a+b+c+d}

\dfrac{ac-bd}{a+b+c+d}

Concepts:linear equation ratio and proportion

Difficulty rating: 1850

Small hint:

Write

p=OP.

Then

AP=p-a,

PD=d-p,

BP=p-b,

and

PC=c-p

Big hint:

Cross-multiply the two given ratios after making those substitutions

Solution:

Let

p=OP.

The given ratio becomes

\frac{p-a}{d-p}=\frac{p-b}{c-p}.

Cross-multiplication cancels the

p^2

terms and yields

p(a-b+c-d)=ac-bd.

Therefore

p=\frac{ac-bd}{a-b+c-d}.

Thus, the correct answer is B.

29.

The number of positive integers less than

1000

divisible by neither

5

nor

7

is:

688

686

684

658

630

Concepts:counting integers in a range divisibility inclusion-exclusion

Difficulty rating: 1180

Small hint:

There are

999

positive integers below

1000

Big hint:

Subtract multiples of

5

and

7

, then add back multiples of

35

Solution:

By inclusion-exclusion, the count is

\begin{aligned} &999-\left\lfloor\frac{999}{5}\right\rfloor -\left\lfloor\frac{999}{7}\right\rfloor\\ &\quad+\left\lfloor\frac{999}{35}\right\rfloor\\ &=999-199-142+28\\ &=686. \end{aligned}

Therefore, the correct answer is B.

30.

If three of the roots of

x^4+ax^2+bx+c=0

are

1,

2,

and

3,

then the value of

a+c

is:

35

24

-12

-61

-63

Concepts:polynomial Vieta’s Formulas

Difficulty rating: 1210

Small hint:

The missing

x^3

coefficient makes the sum of all four roots zero

Big hint:

Use the fourth root to compute the pairwise sum

a

and product

c

Solution:

The fourth root is

-6

because the root sum is zero. The sum of pairwise products is

\begin{aligned} a&=2+3-6+6\\ &\quad-12-18=-25, \end{aligned}

while the product is

c=(1)(2)(3)(-6)=-36.

Thus

a+c=-61.

Therefore, the correct answer is D.

31.

Triangle

ABC

is inscribed in a circle with center

O'.

A circle with center

O

is inscribed in triangle

ABC.

AO

is drawn, and extended to intersect the larger circle in

D.

Then we must have:

CD=BD=O'D

AO=CO=OD

CD=CO=BD

CD=OD=BD

O'B=O'C=OD

Concepts:angle chasing circumcircle, circumcenter, and circumradius incircle, incenter, and inradius

Difficulty rating: 1880

Small hint:

Line

AD

bisects

\angle A,

so

D

is the midpoint of arc

BC

Big hint:

Compare angles in triangle

COD

to show

CD=OD

Solution:

Because

AO

is an angle bisector, the inscribed angles

\angle BAD

and

\angle CAD

are equal. Hence arcs

BD

and

CD

, and therefore chords

BD

and

CD,

are equal.

Let

\angle BAD=\alpha

and

\angle BCO=\beta.

Then

\angle OCD=\alpha+\beta.

In triangle

AOC,

the exterior angle

\angle COD

also equals

\alpha+\beta.

Thus

CD=OD,

so

CD=OD=BD.

Therefore, the correct answer is D.

32.

Let

M

be the midpoint of side

AB

of triangle

ABC.

Let

P

be a point on

AB

between

A

and

M,

and let

MD

be drawn parallel to

PC

and intersecting

BC

at

D.

If the ratio of the area of triangle

BPD

to that of triangle

ABC

is denoted by

r,

then:

\dfrac12\lt r\lt1

depending upon the position of

P

r=\dfrac12

independent of the position of

P

\dfrac12\leq r\lt1

depending upon the position of

P

\dfrac13\lt r\lt\dfrac23

depending upon the position of

P

r=\dfrac13

independent of the position of

P

Concepts:area ratio median (geometry)parallel lines

Difficulty rating: 1650

Small hint:

Triangles

MDP

and

MDC

have equal bases on parallel lines

Big hint:

Add their areas to

[BMD]

and use that

CM

is a median

Solution:

Since

MD\parallel PC,

triangles

MDP

and

MDC

have the same base

MD

and equal altitudes, so their areas are equal. Hence

\begin{aligned} [BPD] &=[BMD]+[MDP]\\ &=[BMD]+[MDC]\\ &=[BMC]. \end{aligned}

Because

CM

is a median,

[BMC]=\frac{[ABC]}{2}.

Thus

r=\frac{1}{2}

for every allowed

P.

Therefore, the correct answer is B.

33.

If

ab\ne0

and

|a|\ne|b|,

the number of distinct values of

x

satisfying the equation

\begin{aligned} \frac{x-a}{b}+\frac{x-b}{a} &=\frac{b}{x-a}\\ &\quad+\frac{a}{x-b} \end{aligned}

is:

0

1

2

3

4

Concepts:algebraic manipulation factoring rational equation

Difficulty rating: 1710

Small hint:

Both sides contain the numerator

(a+b)x-(a^2+b^2)

after combining terms

Big hint:

Factor the equation into that numerator times a difference of reciprocals

Solution:

Let

N=(a+b)x-a^2-b^2.

Combining each side gives

N\left(\frac1{ab} -\frac1{(x-a)(x-b)}\right)=0.

The first factor gives

x=\frac{a^2+b^2}{a+b}.

The second gives

(x-a)(x-b)=ab,

or

x(x-a-b)=0,

giving

x=0

and

x=a+b.

The hypotheses ensure that all three are defined and distinct.

Therefore, the correct answer is D.

34.

Let

r

be the speed in miles per hour at which a wheel,

11

feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by

\dfrac14

of a second, the speed

r

is increased by

5

miles per hour. Then

r

is:

9

10

10\dfrac12

11

12

Concepts:distance rate and time rational equation unit conversion

Difficulty rating: 1650

Small hint:

At

r

miles per hour, one rotation takes

\frac{7.5}{r}

seconds

Big hint:

Set

\frac{7.5}{r}-\frac{7.5}{r+5}=\frac{1}{4}

Solution:

Since

11

feet is

\frac{11}{5280}

mile, one rotation at

r

mph takes

\frac{3600\cdot11}{5280r}=\frac{7.5}{r}

seconds. Thus

\frac{7.5}{r}-\frac{7.5}{r+5}=\frac14.

This reduces to

r^2+5r-150=0,

or

(r-10)(r+15)=0.

The positive speed is

r=10.

Therefore, the correct answer is B.

35.

Let

O

be an interior point of triangle

ABC,

and let

s_1=OA+OB+OC.

If

s_2=AB+BC+CA,

then:

for every triangle

s_1\gt\dfrac12s_2,

and

s_1\leq s_2

for every triangle

s_1\geq\dfrac12s_2,

and

s_1\lt s_2

for every triangle

s_1\gt\dfrac12s_2,

and

s_1\lt s_2

for every triangle

s_1\geq\dfrac12s_2,

and

s_1\leq s_2

neither

(A)

nor

(B)

nor

(C)

nor

(D)

applies to every triangle

Concepts:bounding to limit cases triangle inequality

Difficulty rating: 1880

Small hint:

Add

AB\lt OA+OB,

BC\lt OB+OC,

and

CA\lt OC+OA

Big hint:

Extend

AO

to side

BC

to prove

OA+OB\lt AC+CB,

and cycle

Solution:

Adding the three strict triangle inequalities

\begin{gathered} AB\lt OA+OB,\\ BC\lt OB+OC,\\ CA\lt OC+OA \end{gathered}

gives

s_2\lt2s_1.

For the upper bound, extend

AO

to

D

on

BC.

Then

AO+OD\lt AC+CD

and

OB\lt OD+DB,

so cancellation gives

OA+OB\lt AC+CB.

Cycling and adding yields

2s_1\lt2s_2.

Therefore

s_1\gt \frac{s_2}{2}

and

s_1\lt s_2.

Thus, the correct answer is C.

36.

Let

\begin{aligned} (1+x+x^2)^n &=a_0+a_1x\\ &\quad+a_2x^2+\cdots\\ &\quad+a_{2n}x^{2n} \end{aligned}

be an identity in

x.

If we let

s=a_0+a_2+a_4+\cdots+a_{2n},

then

s

equals:

2^n

2^n+1

\dfrac{3^n-1}{2}

\dfrac{3^n}{2}

\dfrac{3^n+1}{2}

Concepts:parity polynomial substitution

Difficulty rating: 1430

Small hint:

Evaluate the polynomial at

x=1

and

x=-1

Big hint:

Adding those two values cancels all odd-degree coefficients

Solution:

Let

P(x)=(1+x+x^2)^n.

Then

P(1)=3^n

is the sum of all coefficients, while

P(-1)=1

is the even-coefficient sum minus the odd-coefficient sum. Therefore

s=\frac{P(1)+P(-1)}2=\frac{3^n+1}{2}.

Therefore, the correct answer is E.

37.

Three men, Alpha, Beta, and Gamma, working together, do a job in

6

hours less time than Alpha alone, in

1

hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let

h

be the number of hours needed by Alpha and Beta, working together, to do the job. Then

h

equals:

\dfrac52

\dfrac32

\dfrac43

\dfrac54

\dfrac34

Concepts:rate rational equation system of equations

Difficulty rating: 1850

Small hint:

Let

t

be the time all three take together

Big hint:

Their individual times are

t+6,

t+1,

and

2t

; add reciprocal rates

Solution:

If all three together take

t

hours, then

\frac1{t+6}+\frac1{t+1}+\frac1{2t}=\frac1t.

This simplifies to

3t^2+7t-6=0,

so

t=\frac{2}{3}.

Alpha and Beta alone take

\frac{20}{3}

and

\frac{5}{3}

hours, so their combined rate is

\frac{3}{20}+\frac{3}{5}=\frac{3}{4}.

Hence

h=\frac{4}{3}.

Therefore, the correct answer is C.

38.

In triangle

ABC

the medians

AM

and

CN

to sides

BC

and

AB,

respectively, intersect in point

O.

P

is the midpoint of side

AC,

and

MP

intersects

CN

in

Q.

If the area of triangle

OMQ

is

n,

then the area of triangle

ABC

is:

16n

18n

21n

24n

27n

Concepts:area ratio centroid coordinate geometry median (geometry)

Difficulty rating: 1650

Small hint:

Use an affine model

A=(0,0),

B=(2,0),

and

C=(0,2)

Big hint:

Find

M,O,P,Q

and compare the two areas

Solution:

Area ratios are affine-invariant, so take

A=(0,0),

B=(2,0),

and

C=(0,2).

Then

\begin{gathered} M=(1,1),\\ O=(\frac{2}{3},\frac{2}{3}),\\ P=(0,1). \end{gathered}

Line

MP

is

y=1,

and it meets median

CN

at

Q=(\frac{1}{2},1).

Thus

[OMQ]=\frac{1}{12},

while

[ABC]=2.

Their ratio is

24,

so

[ABC]=24n.

Therefore, the correct answer is D.

39.

In base

R_1

the expanded fraction

F_1

becomes

0.373737\ldots

and the expanded fraction

F_2

becomes

0.737373\ldots.

In base

R_2

fraction

F_1,

when expanded, becomes

0.252525\ldots,

while fraction

F_2

becomes

0.525252\ldots.

The sum of

R_1

and

R_2,

each written in base ten, is:

24

22

21

20

19

Concepts:number base repeating decimal system of equations

Difficulty rating: 1990

Small hint:

A repeating pair

ab

in base

R

equals

\frac{aR+b}{R^2-1}

Big hint:

Add the equations for

F_1

and

F_2,

then subtract them

Solution:

The two descriptions give

\begin{aligned} F_1&=\frac{3R_1+7}{R_1^2-1}\\ &=\frac{2R_2+5}{R_2^2-1},\\ F_2&=\frac{7R_1+3}{R_1^2-1}\\ &=\frac{5R_2+2}{R_2^2-1}. \end{aligned}

Adding yields

\frac{10}{R_1-1}=\frac{7}{R_2-1},

or

10R_2-7R_1=3.

Subtracting yields

\frac{4}{R_1+1}=\frac{3}{R_2+1},

or

4R_2-3R_1=-1.

Solving gives

R_1=11,

R_2=8,

whose sum is

19.

Therefore, the correct answer is E.

40.

In this figure

AB

is a diameter of a circle, centered at

O,

with radius

a.

A chord

AD

is drawn and extended to meet the tangent to the circle at

B,

in point

C.

Point

E

is taken on

AC

so that

AE=DC.

If the coordinates of

E

are

(x,y),

then:

y^2=\dfrac{x^3}{2a-x}

y^2=\dfrac{x^3}{2a+x}

y^4=\dfrac{x^2}{2a-x}

x^2=\dfrac{y^2}{2a-x}

x^2=\dfrac{y^2}{2a+x}

Concepts:circle coordinate geometry similarity tangent line

Difficulty rating: 2060

Small hint:

Drop perpendiculars from

E

and

D

to diameter

AB

Big hint:

Use

AE=DC

between the two parallel tangents, then combine the altitude theorem with similar triangles

Solution:

Drop perpendiculars

EM

and

DN

to

AB.

Since

AE=DC

and the tangents through

A,B

are parallel, their projections give

NB=x.

Hence

AN=2a-x.

In right triangle

ADB,

the altitude theorem gives

DN^2=x(2a-x).

Similar triangles

AME

and

AND

give

\frac{DN}{2a-x}=\frac{y}{x},

so

DN=\frac{y(2a-x)}{x}.

Substitution and cancellation yield

y^2=\frac{x^3}{2a-x}.

Therefore, the correct answer is A.