1979 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If rectangle

ABCD

has area

72

square meters and

E

and

G

are the midpoints of sides

AD

and

CD,

respectively, then the area of rectangle

DEFG

in square meters is

8

9

12

18

24

Concepts:rectangle midpoint area

Difficulty rating: 840

Small hint:

The two midpoint conditions halve the relevant side lengths

Big hint:

Compare the area scale factor of

DEFG

with that of

ABCD

Solution:

The side lengths of

DEFG

are one half the corresponding side lengths of

ABCD.

Its area is therefore

(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}

of

72,

which is

18.

Therefore, the correct answer is D.

2.

For all nonzero real numbers

x

and

y

such that

x-y=xy,

\frac1x-\frac1y

equals

\frac1{xy}

\frac1{x-y}

0

-1

y-x

Concepts:fraction algebraic manipulation

Difficulty rating: 960

Small hint:

Combine the two reciprocals over the denominator

xy

Big hint:

Use the given equation to replace

y-x

Solution:

We have

\begin{aligned} \frac1x-\frac1y&=\frac{y-x}{xy}\\ &=-\frac{x-y}{xy}\\ &=-1, \end{aligned}

because

x-y=xy.

Therefore, the correct answer is D.

3.

In the adjoining figure,

ABCD

is a square,

ABE

is an equilateral triangle and point

E

is outside square

ABCD.

What is the measure of

\angle AED

in degrees?

10

12.5

15

20

25

Concepts:square (geometry)equilateral triangle isosceles triangle angle chasing

Difficulty rating: 1410

Small hint:

Compare

DA,

AB,

and

AE

Big hint:

Find

\angle DAE

, then use the base angles of isosceles

\triangle DAE

Solution:

The square and equilateral triangle give

DA=AB=AE.

Also

\begin{aligned} \angle DAE &=\angle DAB+\angle BAE\\ &=90^\circ+60^\circ\\ &=150^\circ. \end{aligned}

Thus

\triangle DAE

is isosceles, and each base angle is

\frac{180^\circ-150^\circ}{2}=15^\circ.

Therefore, the correct answer is C.

4.

For all real numbers

x,

x[x\{x(2-x)-4\}+10]+1

equals

-x^4+2x^3+4x^2+10x+1

-x^4-2x^3+4x^2+10x+1

-x^4-2x^3-4x^2+10x+1

-x^4-2x^3-4x^2-10x+1

-x^4+2x^3-4x^2+10x+1

Concepts:polynomial distributive property algebraic manipulation

Difficulty rating: 1410

Small hint:

Expand from the innermost parentheses outward

Big hint:

Track the outer two factors of

x

separately

Solution:

Expanding from the inside gives

\begin{aligned} &x[x\{x(2-x)-4\}+10]+1\\ &\quad=x[x(2x-x^2-4)+10]+1\\ &\quad=-x^4+2x^3-4x^2\\ &\qquad+10x+1. \end{aligned}

Therefore, the correct answer is E.

5.

Find the sum of the digits of the largest even three digit number (in base ten representation) which is not changed when its units and hundreds digits are interchanged.

22

23

24

25

26

Concepts:digits place value parity

Difficulty rating: 960

Small hint:

The hundreds and units digits must be equal

Big hint:

Maximize the equal outer digits subject to the number being even, then maximize the tens digit

Solution:

The equal hundreds and units digit must be even. The largest possible nonzero such digit is

8,

and the tens digit can be

9.

Thus the largest number is

898,

whose digit sum is

8+9+8=25.

Therefore, the correct answer is D.

6.

\frac32+\frac54+\frac98+\frac{17}{16}

+\frac{33}{32}+\frac{65}{64}-7=

-\frac1{64}

-\frac1{16}

0

\frac1{16}

\frac1{64}

Concepts:fraction geometric sequence summation

Difficulty rating: 1540

Small hint:

Write each fraction as

1

plus a reciprocal power of

2

Big hint:

Sum

\frac12+\frac14+\cdots+\frac1{64}

as a finite geometric series

Solution:

Each numerator is one more than its denominator, so the expression is

\begin{aligned} 6+\biggl(&\frac12+\frac14+\frac18\\ &+\frac1{16}+\frac1{32}+\frac1{64}\biggr)-7. \end{aligned}

The parenthesized geometric sum is

1-\frac{1}{64}=\frac{63}{64}.

Therefore the value is

6+\frac{63}{64}-7=-\frac{1}{64}.

Therefore, the correct answer is A.

7.

The square of an integer is called a perfect square. If

x

is a perfect square, the next larger perfect square is

x+1

x^2+1

x^2+2x+1

x^2+x

x+2\sqrt x+1

Concepts:perfect square algebraic manipulation pattern recognition

Difficulty rating: 1100

Small hint:

Write

x=n^2

for a nonnegative integer

n

Big hint:

Expand the next square

(n+1)^2

and replace

n

by

\sqrt x

Solution:

Let

x=n^2

with

n\ge0.

The next larger perfect square is

\begin{aligned} (n+1)^2&=n^2+2n+1\\ &=x+2\sqrt x+1. \end{aligned}

Therefore, the correct answer is E.

8.

Find the area of the smallest region bounded by the graphs of

y=|x|

and

x^2+y^2=4.

\frac\pi4

\frac{3\pi}4

\pi

\frac{3\pi}2

2\pi

Concepts:circle absolute value sector circle area

Difficulty rating: 1540

Small hint:

The two rays of

y=|x|

meet at the center of the circle

Big hint:

Find the smaller central angle between the rays

y=x

and

y=-x

Solution:

The two rays meet the circle at angles

45^\circ

and

135^\circ,

so the smallest bounded region is a

90^\circ

sector of the radius-

2

circle. Its area is

\frac{90^\circ}{360^\circ}\pi(2)^2=\pi.

Therefore, the correct answer is C.

9.

The product of

\sqrt[3]{4}

and

\sqrt[4]{8}

equals

\sqrt[7]{12}

2\sqrt[7]{12}

\sqrt[7]{32}

\sqrt[12]{32}

2\sqrt[12]{32}

Concepts:radical exponent algebraic manipulation

Difficulty rating: 1590

Small hint:

Rewrite both radicands as powers of

2

Big hint:

Add the exponents

\frac23

and

\frac34

, then separate the integer part

Solution:

Writing both radicals as powers of

2

gives

\begin{aligned} 4^{\frac{1}{3}}8^{\frac{1}{4}} &=2^{\frac{2}{3}}2^{\frac{3}{4}}\\ &=2^{\frac{17}{12}}\\ &=2\sqrt[12]{2^5}\\ &=2\sqrt[12]{32}. \end{aligned}

Therefore, the correct answer is E.

10.

If

P_1P_2P_3P_4P_5P_6

is a regular hexagon whose apothem (distance from the center to the midpoint of a side) is

2,

and

Q_i

is the midpoint of side

P_iP_{i+1}

for

i=1,

2,

3,

4,

then the area of quadrilateral

Q_1Q_2Q_3Q_4

is

6

2\sqrt6

\frac{8\sqrt3}{3}

3\sqrt3

4\sqrt3

Concepts:regular polygon equilateral triangle area decomposition

Difficulty rating: 1780

Small hint:

The points

Q_i

lie on a circle of radius equal to the apothem

Big hint:

Join the center to the four consecutive side midpoints and decompose the quadrilateral

Solution:

Let

O

be the center. Consecutive apothems

OQ_i

and

OQ_{i+1}

have length

2

and form a

60^\circ

angle, so each

\triangle OQ_iQ_{i+1}

is equilateral of side

2.

The quadrilateral is the union of three such triangles, and hence has area

3\left(\frac{\sqrt3}{4}\cdot2^2\right)=3\sqrt3.

Therefore, the correct answer is D.

11.

Find a positive integral solution to the equation

\frac{\substack{1+3+5+\cdots\\{}+(2n-1)}} {\substack{2+4+6+\cdots\\{}+2n}}=\frac{115}{116}.

110

115

116

231

The equation has no positive integral solutions.

Concepts:sum of first n odd numbers summation linear equation

Difficulty rating: 1540

Small hint:

Use the sums of the first

n

odd and first

n

even positive integers

Big hint:

The fraction simplifies to

\frac{n}{n+1}

Solution:

The numerator is

n^2,

and the denominator is

n(n+1).

Thus

\frac{n}{n+1}=\frac{115}{116},

which gives

116n=115n+115,

so

n=115.

Therefore, the correct answer is B.

12.

In the adjoining figure,

CD

is the diameter of a semicircle with center

O.

Point

A

lies on the extension of

DC

past

C;

point

E

lies on the semicircle, and

B

is the point of intersection (distinct from

E

) of line segment

AE

with the semicircle. If length

AB

equals length

OD,

and the measure of

\angle EOD

is

45^\circ,

then the measure of

\angle BAO

is

10^\circ

15^\circ

20^\circ

25^\circ

30^\circ

Concepts:circle isosceles triangle angle chasing

Difficulty rating: 1980

Small hint:

Let

y=\angle BAO

and use

AB=OB=OE

Big hint:

Express

\angle BOE

in two ways using the isosceles triangles and the

45^\circ

central angle

Solution:

Let

y=\angle BAO.

Since

AB=OB,

\angle AOB=y,

so

\angle EBO=2y.

Also

OB=OE,

hence

\angle BEO=2y

and

\angle BOE=180^\circ-4y.

On the other hand,

\angle BOD=180^\circ-y,

so

\angle BOE=135^\circ-y.

Therefore

180^\circ-4y=135^\circ-y,

giving

y=15^\circ.

Therefore, the correct answer is B.

13.

The inequality

y-x\lt\sqrt{x^2}

is satisfied if and only if

y\lt0

or

y\lt2x

(or both inequalities hold)

y\gt0

or

y\lt2x

(or both inequalities hold)

y^2\lt2xy

y\lt0

x\gt0

and

y\lt2x

Concepts:absolute value inequality casework

Difficulty rating: 1650

Small hint:

Replace

\sqrt{x^2}

by

|x|

Big hint:

Separate the cases

x\ge0

and

x\lt0

, then combine the resulting regions

Solution:

If

x\ge0,

the inequality becomes

y-x\lt x,

or

y\lt2x.

If

x\lt0,

it becomes

y-x\lt-x,

or

y\lt0.

Combining the two cases gives exactly

y\lt0

or

y\lt2x,

with overlap allowed.

Therefore, the correct answer is A.

14.

In a certain sequence of numbers, the first number is

1,

and, for all

n\ge2,

the product of the first

n

numbers in the sequence is

n^2.

The sum of the third and the fifth numbers in the sequence is

\frac{25}{9}

\frac{31}{15}

\frac{61}{16}

\frac{576}{225}

34

Concepts:algebraic manipulation fraction pattern recognition

Difficulty rating: 1590

Small hint:

Divide the product of the first

n

terms by the product of the first

n-1

terms

Big hint:

This gives the

n

th term directly as a ratio of two squares

Solution:

For

n\ge2,

the

n

th term is

a_n=\frac{n^2}{(n-1)^2}.

Therefore

a_3=\frac{9}{4}

and

a_5=\frac{25}{16},

whose sum is

\frac{36}{16}+\frac{25}{16}=\frac{61}{16}.

Therefore, the correct answer is C.

15.

Two identical jars are filled with alcohol solutions, the ratio of the volume of alcohol to the volume of water being

p:1

in one jar and

q:1

in the other jar. If the entire contents of the two jars are mixed together, the ratio of the volume of alcohol to the volume of water in the mixture is

\frac{p+q}{2}

\frac{p^2+q^2}{p+q}

\frac{2pq}{p+q}

\frac{2(p^2+pq+q^2)}{3(p+q)}

\frac{p+q+2pq}{p+q+2}

Concepts:mixture ratio and proportion fraction

Difficulty rating: 1860

Small hint:

Assign the same total volume to each jar and convert each ratio to alcohol and water fractions

Big hint:

Add the two alcohol amounts and the two water amounts before forming their ratio

Solution:

For equal total volumes, the alcohol fractions are

\frac{p}{p+1}

and

\frac{q}{q+1},

while the water fractions are

\frac{1}{p+1}

and

\frac{1}{q+1}.

Their ratio is

\frac{\frac p{p+1}+\frac q{q+1}} {\frac1{p+1}+\frac1{q+1}} =\frac{p+q+2pq}{p+q+2}.

Therefore, the correct answer is E.

16.

A circle with area

A_1

is contained in the interior of a larger circle with area

A_1+A_2.

If the radius of the larger circle is

3,

and if

A_1,

A_2,

A_1+A_2

is an arithmetic progression, then the radius of the smaller circle is

\frac{\sqrt3}{2}

1

\frac2{\sqrt3}

\frac32

\sqrt3

Concepts:circle area arithmetic sequence algebraic manipulation

Difficulty rating: 1590

Small hint:

Use the middle-term condition for the three areas in arithmetic progression

Big hint:

Express the larger circle’s area as a multiple of

A_1

Solution:

The progression condition gives

2A_2=A_1+(A_1+A_2),

so

A_2=2A_1.

Hence the larger area is

3A_1=9\pi,

and

A_1=3\pi.

If the smaller radius is

r,

then

\pi r^2=3\pi,

so

r=\sqrt3.

Therefore, the correct answer is E.

17.

Points

A,

B,

C,

and

D

are distinct and lie, in the given order, on a straight line. Line segments

AB,

AC,

and

AD

have lengths

x,

y,

and

z,

respectively. If line segments

AB

and

CD

may be rotated about points

B

and

C,

respectively, so that points

A

and

D

coincide, to form a triangle with positive area, then which of the following three inequalities must be satisfied?

\begin{aligned} &\text{I. }x\lt\frac z2\\ &\text{II. }y\lt x+\frac z2\\ &\text{III. }y\lt\frac z2 \end{aligned}

\mathrm{I}

only

\mathrm{II}

only

\mathrm{I}

and

\mathrm{II}

only

\mathrm{II}

and

\mathrm{III}

only

\mathrm{I},

\mathrm{II}

and

\mathrm{III}

Concepts:triangle inequality casework logical deduction

Difficulty rating: 2040

Small hint:

After the rotations, identify the three side lengths of the new triangle

Big hint:

Apply all three strict triangle inequalities to

x,

y-x,

and

z-y

Solution:

The new triangle has side lengths

x,

y-x,

and

z-y.

Its triangle inequalities give

\begin{aligned} x&\lt(y-x)+(z-y),\\ &\Longrightarrow x\lt \frac{z}{2};\\ y-x&\lt x+(z-y),\\ &\Longrightarrow y\lt x+\frac{z}{2};\\ z-y&\lt x+(y-x),\\ &\Longrightarrow y\gt \frac{z}{2}. \end{aligned}

Thus I and II must hold, while III must fail.

Therefore, the correct answer is C.

18.

To the nearest thousandth,

\log_{10}2

is

0.301

and

\log_{10}3

is

0.477.

Which of the following is the best approximation of

\log_5 10?

\frac87

\frac97

\frac{10}{7}

\frac{11}{7}

\frac{12}{7}

Concepts:logarithm estimation fraction

Difficulty rating: 1540

Small hint:

Use

\log_{10}5=1-\log_{10}2

Big hint:

Apply change of base, then compare the decimal result with the five fractions

Solution:

We have

\log_{10}5\approx1-0.301=0.699.

Therefore

\begin{aligned} \log_5 10&=\frac1{\log_{10}5}\\ &\approx\frac1{0.699}\\ &\approx1.431. \end{aligned}

Of the choices,

\frac{10}{7}\approx1.429

is closest.

Therefore, the correct answer is C.

19.

Find the sum of the squares of all real numbers satisfying the equation

x^{256}-256^{32}=0.

8

128

512

65{,}536

2(256^{32})

Concepts:exponent perfect power zero product property

Difficulty rating: 1650

Small hint:

Rewrite

256^{32}

as a single power with exponent

256

Big hint:

Determine the two real values whose

256

th powers equal

2^{256}

Solution:

Since

256=2^8,

256^{32}=2^{256}.

Thus

(\frac{x}{2})^{256}=1.

The real solutions are

x=2

and

x=-2,

and the sum of their squares is

4+4=8.

Therefore, the correct answer is A.

20.

If

a=\frac12

and

(a+1)(b+1)=2,

then the radian measure of

\operatorname{Arctan}a+\operatorname{Arctan}b

equals

\frac\pi2

\frac\pi3

\frac\pi4

\frac\pi5

\frac\pi6

Concepts:trigonometric identity algebraic manipulation

Difficulty rating: 1780

Small hint:

First solve the given product equation for

b

Big hint:

Use the tangent addition formula on the two positive inverse-tangent angles

Solution:

Substituting

a=\frac{1}{2}

gives

b=\frac{1}{3}.

If

u=\operatorname{Arctan}a

and

v=\operatorname{Arctan}b,

then

\begin{aligned} \tan(u+v)&=\frac{a+b}{1-ab}\\ &=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\\ &=1. \end{aligned}

Both angles are positive and their sum is less than

\frac{\pi}{2},

so

u+v=\frac{\pi}{4}.

Therefore, the correct answer is C.

21.

The length of the hypotenuse of a right triangle is

h,

and the radius of the inscribed circle is

r.

The ratio of the area of the circle to the area of the triangle is

\frac{\pi r}{h+2r}

\frac{\pi r}{h+r}

\frac{\pi r}{2h+r}

\frac{\pi r^2}{h^2+r^2}

none of these

Concepts:right triangle incircle, incenter, and inradius triangle area

Difficulty rating: 1870

Small hint:

Use the formula

K=rs

for a triangle’s area

Big hint:

For a right triangle, relate the sum of the legs to

h

and

r

Solution:

If the legs are

x

and

y,

then the right-triangle inradius identity is

x+y=h+2r.

Hence the semiperimeter is

s=\frac{x+y+h}{2}=h+r.

The triangle’s area is

rs=r(h+r),

while the circle’s area is

\pi r^2.

Their ratio is

\frac{\pi r}{h+r}.

Therefore, the correct answer is B.

22.

Find the number of pairs

(m,n)

of integers which satisfy the equation

\begin{aligned} m^3+6m^2+5m &=27n^3+9n^2\\ &\quad+9n+1. \end{aligned}

0

1

3

9

infinitely many

Concepts:divisibility modular arithmetic factoring

Difficulty rating: 1860

Small hint:

Factor the left side as

m(m+1)(m+5)

Big hint:

Compare the two sides modulo

3

Solution:

The left side is

\begin{gathered} m(m+1)(m+5)\\ =m(m+1)(m+2)\\ \quad+3m(m+1), \end{gathered}

which is divisible by

3

for every integer

m.

The right side is

3(9n^3+3n^2+3n)+1,

which is congruent to

1\pmod3.

Therefore no integer pair satisfies the equation.

Therefore, the correct answer is A.

23.

The edges of a regular tetrahedron with vertices

A,

B,

C,

and

D

each have length one. Find the least possible distance between a pair of points

P

and

Q,

where

P

is on edge

AB

and

Q

is on edge

CD.

\frac12

\frac34

\frac{\sqrt2}{2}

\frac{\sqrt3}{2}

\frac{\sqrt3}{3}

Concepts:3D geometry distance formula optimization

Difficulty rating: 2100

Small hint:

Use symmetric coordinates for a regular tetrahedron with vertices

(\pm1,\pm1,\pm1)

Big hint:

Parameterize points on the two opposite edges and minimize the squared distance

Solution:

Before scaling, take

\begin{aligned} A&=(1,1,1),\\ B&=(1,-1,-1),\\ C&=(-1,1,-1),\\ D&=(-1,-1,1). \end{aligned}

These edges have length

2\sqrt2,

so scale by

\frac{1}{2\sqrt2}.

Points on

AB

and

CD

have forms

P=(1,t,t)

and

Q=(-1,u,-u).

Before scaling,

\begin{aligned} PQ^2&=4+(t-u)^2+(t+u)^2\\ &=4+2t^2+2u^2, \end{aligned}

minimized at

t=u=0.

After scaling, the minimum squared distance is

\frac{4}{8}=\frac{1}{2},

so the distance is

\frac{\sqrt2}{2}.

Therefore, the correct answer is C.

24.

A polygon is called “simple” if it is not self-intersecting. Sides

AB,

BC,

and

CD

of simple quadrilateral

ABCD

have lengths

4,

5,

and

20,

respectively. If vertex angles

B

and

C

are obtuse and

\sin C=-\cos B=\frac35,

then side

AD

has length

24

24.5

24.6

24.8

25

Concepts:coordinate geometry trigonometry distance formula

Difficulty rating: 2040

Small hint:

Place

B=(0,0)

and

C=(5,0)

, then use the signs forced by the obtuse angles

Big hint:

Determine coordinates for

A

and

D

from the

3

-

4

-

5

ratios

Solution:

Put

B=(0,0)

and

C=(5,0).

Since

B

is obtuse and

-\cos B=\frac{3}{5},

we may take

\begin{aligned} A&=4(-\frac{3}{5},\frac{4}{5})\\ &=(-\frac{12}{5},\frac{16}{5}). \end{aligned}

The simple configuration and the obtuse angle at

C

give

\begin{aligned} D&=C+20(\frac{4}{5},\frac{3}{5})\\ &=(21,12). \end{aligned}

Thus

\begin{aligned} AD&=\sqrt{\begin{gathered} (21+\frac{12}{5})^2\\ {}+(12-\frac{16}{5})^2 \end{gathered}}\\ &=\sqrt{(\frac{117}{5})^2+(\frac{44}{5})^2}\\ &=25. \end{aligned}

Therefore, the correct answer is E.

25.

If

q_1(x)

and

r_1

are the quotient and remainder, respectively, when the polynomial

x^8

is divided by

x+\frac12,

and if

q_2(x)

and

r_2

are the quotient and remainder, respectively, when

q_1(x)

is divided by

x+\frac12,

then

r_2

equals

\frac1{256}

-\frac1{16}

1

-16

256

Concepts:polynomial factoring substitution

Difficulty rating: 2040

Small hint:

Let

a=-\frac12

and express the first quotient using

x^8-a^8

Big hint:

Apply the remainder theorem to

q_1(x)

at

x=a

Solution:

Let

a=-\frac{1}{2}.

Since the first remainder is

a^8,

\begin{aligned} q_1(x)&=\frac{x^8-a^8}{x-a}\\ &=x^7+ax^6+\cdots+a^7. \end{aligned}

Therefore the second remainder is

\begin{aligned} q_1(a)&=8a^7\\ &=8(-\frac{1}{2})^7\\ &=-\frac{1}{16}. \end{aligned}

Therefore, the correct answer is B.

26.

The function

f

satisfies the functional equation

f(x)+f(y)=f(x+y)-xy-1

for every pair

x,

y

of real numbers. If

f(1)=1,

then the number of integers

n\ne1

for which

f(n)=n

is

0

1

2

3

infinite

Concepts:functional equation recursion quadratic

Difficulty rating: 2200

Small hint:

Set one variable equal to

1

to get a recurrence for consecutive integer inputs

Big hint:

Find

f(0)

, solve the recurrence for all integers, and then impose

f(n)=n

Solution:

Setting

x=1

gives

f(y+1)-f(y)=y+2.

Setting

x=y=0

gives

f(0)=-1.

Extending the recurrence in both directions yields, for every integer

n,

f(n)=\frac{n^2+3n-2}{2}.

Thus

f(n)=n

is equivalent to

\begin{aligned} n^2+n-2&=(n+2)(n-1)\\ &=0. \end{aligned}

The integer solutions are

n=-2

and

n=1,

and after excluding

1

exactly one remains.

Therefore, the correct answer is B.

27.

An ordered pair

(b,c)

of integers, each of which has absolute value less than or equal to five, is chosen at random, with each such ordered pair having an equal likelihood of being chosen. What is the probability that the equation

x^2+bx+c=0

will not have distinct positive real roots?

\frac{106}{121}

\frac{108}{121}

\frac{110}{121}

\frac{112}{121}

none of these

Concepts:quadratic complementary probability systematic listing

Difficulty rating: 2200

Small hint:

Distinct positive roots require

b\lt0,

c\gt0,

and

b^2\gt4c

Big hint:

Count the qualifying

c

values for each

b=-1,

b=-2,

\ldots,

b=-5,

then take the complement among

121

pairs

Solution:

There are

11^2=121

ordered pairs. Distinct positive roots require

b\lt0,

c\gt0,

and

b^2\gt4c.

For

b=-1

and

b=-2

there are no choices; for

b=-3,

b=-4,

and

b=-5

there are respectively

2,

3,

and

5

choices of

c.

Thus only

10

pairs produce distinct positive roots, so the requested probability is

1-\frac{10}{121}=\frac{111}{121},

which is not listed.

Therefore, the correct answer is E.

28.

Circles with centers

A,

B,

and

C

each have radius

r,

where

1\lt r\lt2.

The distance between each pair of centers is

2.

If

B'

is the point of intersection of circle

A

and circle

C

which is outside circle

B,

and if

C'

is the point of intersection of circle

A

and circle

B

which is outside circle

C,

then length

B'C'

equals

3r-2

r^2

r+\sqrt{3(r-1)}

1+\sqrt{3(r^2-1)}

none of these

Concepts:circle equilateral triangle similarity symmetry

Difficulty rating: 2200

Small hint:

Introduce the analogous point

A'

outside circle

A

; symmetry makes

A'B'C'

equilateral

Big hint:

Use the common centroid and the altitude

\sqrt{r^2-1}

of an isosceles triangle with sides

r,

r,

and

2

Solution:

Let

A'

be the analogous outer intersection of the circles centered at

B

and

C.

Then

A'B'C'

and

ABC

are concentric equilateral triangles. If

K

is their common centroid and

M

is the midpoint of

BC,

then

\begin{aligned} A'M&=\sqrt{r^2-1},\\ MK&=\frac{\sqrt3}{3},\\ AK&=\frac{2\sqrt3}{3}. \end{aligned}

Similarity gives

\begin{aligned} \frac{B'C'}2&=\frac{A'K}{AK}\\ &=\frac{\sqrt{r^2-1}+\frac{\sqrt3}{3}} {\frac{2\sqrt3}{3}}. \end{aligned}

Hence

B'C'=1+\sqrt{3(r^2-1)}.

Therefore, the correct answer is D.

29.

For each positive number

x,

let

f(x)= \frac{\begin{gathered} \left(x+\frac1x\right)^6\\ {}-\left(x^6+\frac1{x^6}\right)-2 \end{gathered}} {\begin{gathered} \left(x+\frac1x\right)^3\\ {}+\left(x^3+\frac1{x^3}\right) \end{gathered}}.

The minimum value of

f(x)

is

1

2

3

4

6

Concepts:algebraic manipulation difference of squares AM-GM Inequality

Difficulty rating: 2100

Small hint:

Set

t=x+\frac{1}{x}

and write

x^3+x^{-3}

in terms of

t

Big hint:

Factor the numerator as a difference of squares, then use

x+\frac{1}{x}\ge2

Solution:

Let

t=x+\frac{1}{x}

and

u=x^3+x^{-3}=t^3-3t.

Since

x^6+x^{-6}=u^2-2,

the numerator is

\begin{aligned} t^6-u^2&=(t^3-u)(t^3+u)\\ &=3t(t^3+u). \end{aligned}

The denominator is

t^3+u,

so

f(x)=3t=3(x+\frac{1}{x}).

For

x\gt0,

x+\frac{1}{x}\ge2,

with equality at

x=1.

Thus the minimum is

6.

Therefore, the correct answer is E.

30.

In

\triangle ABC,

E

is the midpoint of side

BC

and

D

is on side

AC.

If the length of

AC

is

1

and

\angle BAC=60^\circ,

\angle ABC=100^\circ,

\angle ACB=20^\circ,

and

\angle DEC=80^\circ,

then the area of

\triangle ABC

plus twice the area of

\triangle CDE

equals

\frac14\cos10^\circ

\frac{\sqrt3}{8}

\frac14\cos40^\circ

\frac14\cos50^\circ

\frac18

Concepts:triangle area congruence (geometry)similarity equilateral triangle

Difficulty rating: 2200

Small hint:

Extend

AB

to

F

so that

AF=AC

, creating an equilateral triangle

Big hint:

Choose

G

on

BF

with

\angle BCG=20^\circ

; compare

\triangle BCG

with

\triangle DCE

Solution:

Extend

AB

through

B

to

F

with

AF=AC=1.

Then

\triangle ACF

is equilateral. Choose

G

on

BF

so that

\angle BCG=20^\circ.

The angle conditions give

\triangle FGC\cong\triangle ABC,

while

\triangle BCG\sim\triangle DCE.

Since

E

is the midpoint of

BC,

the similarity scale is

2,

so

[BCG]=4[CDE].

Decomposing the equilateral triangle gives

\frac{\sqrt3}{4} =2[ABC]+4[CDE].

Dividing by

2

yields

[ABC]+2[CDE]=\frac{\sqrt3}{8}.

Therefore, the correct answer is B.