1964 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

What is the value of

\left[\log_{10}\left(5\log_{10}100\right)\right]^2?

\log_{10}50

25

10

2

1

Concepts:logarithm order of operations exponent

Difficulty rating: 1030

Small hint:

Evaluate the inner logarithm first

Big hint:

After multiplying by

5,

take the base-

10

logarithm and then square

Solution:

Since

\log_{10}100=2,

the expression inside the square brackets is

\log_{10}(5\cdot2)=\log_{10}10=1.

Its square is

1.

Therefore, the correct answer is E.

2.

The graph of

x^2-4y^2=0

is:

a parabola

an ellipse

a pair of straight lines

a point

none of these

Concepts:factoring difference of squares coordinate geometry

Difficulty rating: 940

Small hint:

Factor the left side as a difference of squares

Big hint:

Set each linear factor equal to zero

Solution:

Factoring gives

x^2-4y^2=(x-2y)(x+2y).

Thus every point lies on

x=2y

or

x=-2y,

a pair of straight lines.

Therefore, the correct answer is C.

3.

When a positive integer

x

is divided by a positive integer

y,

the quotient is

u

and the remainder is

v,

where

u

and

v

are integers. What is the remainder when

x+2uy

is divided by

y?

0

2u

3u

v

2v

Concepts:modular arithmetic whole number operations substitution

Difficulty rating: 960

Small hint:

Write

x=uy+v

Big hint:

Adding a multiple of

y

does not change the remainder upon division by

y

Solution:

The division algorithm gives

x=uy+v.

Hence

x+2uy=3uy+v.

The first term is divisible by

y,

so the remainder is

v.

Therefore, the correct answer is D.

4.

The expression

\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q},

where

P=x+y

and

Q=x-y,

is equivalent to:

\dfrac{x^2-y^2}{xy}

\dfrac{x^2-y^2}{2xy}

1

\dfrac{x^2+y^2}{xy}

\dfrac{x^2+y^2}{2xy}

Concepts:algebraic manipulation substitution difference of squares

Difficulty rating: 1210

Small hint:

First substitute

P+Q=2x

and

P-Q=2y

Big hint:

Simplify

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

using a common denominator

Solution:

Since

P+Q=2x

and

P-Q=2y,

the expression becomes

\frac{x}{y}-\frac{y}{x}=\frac{x^2-y^2}{xy}.

Therefore, the correct answer is A.

5.

If

y

varies directly as

x,

and if

y=8

when

x=4,

the value of

y

when

x=-8

is:

-16

-4

-2

4k,

k=\pm1,

\pm2,

\ldots

16k,

k=\pm1,

\pm2,

\ldots

Concepts:ratio and proportion rate substitution

Difficulty rating: 890

Small hint:

Write direct variation as

y=kx

Big hint:

Use the first pair to find

k

Solution:

Direct variation gives

y=kx.

From

8=4k,

k=2.

Therefore, when

x=-8,

y=2(-8)=-16.

Thus, the correct answer is A.

6.

If

x,

2x+2,

3x+3,

\ldots

are in geometric progression, the fourth term is:

-27

-13\dfrac12

12

13\dfrac12

27

Concepts:geometric sequence quadratic ratio and proportion

Difficulty rating: 1430

Small hint:

For three consecutive nonzero terms, the middle term squared equals the product of its neighbors

Big hint:

After finding

x,

multiply the third term by the common ratio

Solution:

The geometric-mean relation gives

(2x+2)^2=x(3x+3).

Under the examination’s ratio convention for a geometric progression, the consecutive ratios must be defined, so

x\ne-1.

Dividing by

x+1

yields

4(x+1)=3x

and

x=-4.

The first three terms are

-4,-6,-9,

with ratio

\frac{3}{2}.

The fourth is

-9(\frac{3}{2})=-\frac{27}{2}.

Therefore, the correct answer is B.

7.

Let

n

be the number of real values of

p

for which the roots of

x^2-px+p=0

are equal. Then

n

equals:

0

1

2

a finite number greater than

2

an infinitely large number

Concepts:quadratic zero product property

Difficulty rating: 1140

Small hint:

Equal roots make the discriminant zero

Big hint:

Solve

p^2-4p=0

Solution:

Equal roots require

(-p)^2-4p=p(p-4)=0.

Thus

p=0

or

p=4,

giving two real values.

Therefore, the correct answer is C.

8.

The smaller root of the equation

\begin{aligned} &\left(x-\frac34\right)\left(x-\frac34\right)\\ &\quad+\left(x-\frac34\right) \left(x-\frac12\right)=0 \end{aligned}

is:

-\dfrac34

\dfrac12

\dfrac58

\dfrac34

1

Concepts:factoring linear equation zero product property

Difficulty rating: 1110

Small hint:

Factor out the common expression

x-\frac34

Big hint:

The remaining linear factor is

2x-\frac54

Solution:

Combining the two terms and factoring gives

\left(x-\frac34\right)\left(2x-\frac54\right)=0.

Hence the roots are

\frac{3}{4}

and

\frac{5}{8},

of which

\frac{5}{8}

is smaller.

Therefore, the correct answer is C.

9.

A jobber buys an article at

\$24

less

12\dfrac12\%.

He then wishes to sell the article at a gain of

33\dfrac13\%

of his cost after allowing a

20\%

discount on his marked price. At what price, in dollars, should the article be marked?

25.20

30.00

33.60

40.00

none of these

Concepts:percentage money ratio and proportion

Difficulty rating: 1450

Small hint:

Compute the discounted cost first, then increase it by one third

Big hint:

The selling price is

80\%

of the marked price

Solution:

The cost is

24(1-\frac{1}{8})=21

dollars. A gain of one third makes the desired selling price

21(\frac{4}{3})=28

dollars. If

M

is the marked price, then

0.8M=28,

so

M=35.

This is not listed.

Thus, the correct answer is E.

10.

Given a square with side of length

s.

On a diagonal as base a triangle with three unequal sides is constructed so that its area equals that of the square. The length of the altitude drawn to the base is:

s\sqrt2

\dfrac{s}{\sqrt2}

2s

2\sqrt{s}

\dfrac2{\sqrt{s}}

Concepts:square (geometry)diagonal triangle area radical

Difficulty rating: 1110

Small hint:

The square’s diagonal has length

s\sqrt2

Big hint:

Set

\frac12(s\sqrt2)h=s^2

Solution:

The triangle’s base is the square’s diagonal,

s\sqrt2.

If its altitude is

h,

equality of areas gives

\frac12(s\sqrt2)h=s^2,

so

h=s\sqrt2.

Therefore, the correct answer is A.

11.

Given

2^x=8^{y+1}

and

9^y=3^{x-9},

find the value of

x+y.

18

21

24

27

30

Concepts:exponent system of equations substitution

Difficulty rating: 1430

Small hint:

Rewrite

8

as

2^3

and

9

as

3^2

Big hint:

Equate exponents to obtain two linear equations in

x

and

y

Solution:

Writing both equations with common bases gives

\begin{aligned} x&=3y+3,\\ 2y&=x-9. \end{aligned}

Substitution yields

y=6

and

x=21,

so

x+y=27.

Therefore, the correct answer is D.

12.

Which of the following is the negation of the statement: For all

x

of a certain set,

x^2\gt0?

For all

x,

x^2\lt0

For all

x,

x^2\leq0

For no

x,

x^2\gt0

For some

x,

x^2\gt0

For some

x,

x^2\leq0

Concepts:logical deduction inequality counterexample

Difficulty rating: 1140

Small hint:

The negation of “for all” begins with “for some”

Big hint:

Negating

x^2\gt0

gives

x^2\leq0

Solution:

The negation of a universal statement is an existential statement, and the negation of

x^2\gt0

is

x^2\leq0.

Thus the negation is: for some

x,

x^2\leq0.

Therefore, the correct answer is E.

13.

A circle is inscribed in a triangle with side lengths

8,

13,

and

17.

Let the segments of the side of length

8,

made by a point of tangency, be

r

and

s,

with

r\lt s.

What is the ratio

r:s?

1:3

2:5

1:2

2:3

3:4

Concepts:incircle, incenter, and inradius tangent line ratio and proportion

Difficulty rating: 1210

Small hint:

Compute the semiperimeter of the triangle

Big hint:

A tangency segment adjacent to a vertex equals the semiperimeter minus the opposite side

Solution:

The semiperimeter is

\frac{8+13+17}{2}=19.

At the endpoints of the side of length

8,

the tangent lengths are

19-13=6

and

19-17=2.

Thus

r:s=2:6=1:3.

Therefore, the correct answer is A.

14.

A farmer bought

749

sheep. He sold

700

of them for the price paid for the

749

sheep. The remaining

49

sheep were sold at the same price per head as the other

700.

Based on the cost, the percent gain on the entire transaction is:

6.5

6.75

7

7.5

8

Concepts:percentage rate money

Difficulty rating: 1180

Small hint:

Let the total purchase cost be

C

Big hint:

The selling price per sheep is

\frac{C}{700}

Solution:

Let the total cost be

C.

Since the first

700

sheep sell for

C,

each sheep sells for

\frac{C}{700}.

Total revenue is therefore

\frac{749C}{700}=1.07C,

a

7\%

gain.

Thus, the correct answer is C.

15.

A line through the point

(-a,0)

cuts from the second quadrant a triangular region with area

T.

The equation of the line is:

2Tx+a^2y+2aT=0

2Tx-a^2y+2aT=0

2Tx+a^2y-2aT=0

2Tx-a^2y-2aT=0

none of these

Concepts:slope coordinate geometry triangle area

Difficulty rating: 1500

Small hint:

If the positive

y

-intercept is

b,

then

\frac12ab=T

Big hint:

Use the intercept form

-\frac{x}{a}+\frac{y}{b}=1

Solution:

Let the

y

-intercept be

b.

The second-quadrant triangle has area

\frac{ab}{2}=T,

so

b=\frac{2T}{a}.

The intercept form is

\frac{x}{-a}+\frac{y}{\frac{2T}{a}}=1.

Clearing denominators gives

2Tx-a^2y+2aT=0.

Therefore, the correct answer is B.

16.

Let

f(x)=x^2+3x+2

and let

S

be the set of integers

\{0,1,2,\ldots,25\}.

The number of members

s

of

S

such that

f(s)

has remainder zero when divided by

6

is:

25

22

21

18

17

Concepts:modular arithmetic divisibility counting integers in a range

Difficulty rating: 1450

Small hint:

Factor

f(s)

as

(s+1)(s+2)

Big hint:

The product is always even; determine which residue of

s

modulo

3

fails

Solution:

We have

f(s)=(s+1)(s+2),

a product of consecutive integers, so it is always even. It is divisible by

3

unless

s\equiv0\pmod3.

Among

0,1,\ldots,25,

nine values are multiples of

3.

Thus

26-9=17

values work.

Therefore, the correct answer is E.

17.

Given the distinct points

P(x_1,y_1),

Q(x_2,y_2)

and

R(x_1+x_2,y_1+y_2).

Line segments are drawn connecting these points to each other and to the origin

O.

Of the three possibilities:

(1)

parallelogram,

(2)

straight line,

(3)

trapezoid, figure

OPRQ,

depending upon the location of the points

P,

Q,

and

R,

can be:

(1)

only

(2)

only

(3)

only

(1)

or

(2)

only

all three

Concepts:vector parallelogram logical deduction

Difficulty rating: 1450

Small hint:

Interpret the position vectors as

R=P+Q

Big hint:

Separate the cases in which the vectors

P

and

Q

are independent or dependent

Solution:

The relation

\overrightarrow{OR}=\overrightarrow{OP}+\overrightarrow{OQ}

makes opposite sides of

OPRQ

parallel and equal whenever

P

and

Q

are not collinear, so the figure is a parallelogram. If the two vectors are dependent, all four points lie on a straight line. It cannot be a genuine trapezoid.

Thus, the correct answer is D.

18.

Let

n

be the number of pairs of values of

b

and

c

such that

3x+by+c=0

and

cx-2y+12=0

have the same graph. Then

n

is:

0

1

2

finite but more than

2

greater than any finite number

Concepts:system of equations slope logical deduction

Difficulty rating: 1500

Small hint:

Coincident line equations have proportional coefficient triples

Big hint:

Set

(c,-2,12)=k(3,b,c)

and use the first and third coordinates

Solution:

For the same graph,

(c,-2,12)=k(3,b,c).

Thus

c=3k

and

12=kc=3k^2,

so

k=\pm2.

Each value determines one pair through

b=-\frac{2}{k}

and

c=3k.

Hence there are two pairs.

Therefore, the correct answer is C.

19.

If

2x-3y-z=0

and

x+3y-14z=0,

z\ne0,

the numerical value of

\dfrac{x^2+3xy}{y^2+z^2}

is:

7

2

0

-\dfrac{20}{17}

-2

Concepts:system of equations substitution algebraic manipulation

Difficulty rating: 1210

Small hint:

Solve both linear equations for

x

and

y

in terms of

z

Big hint:

The equations give

y=3z

and

x=5z

Solution:

Solving the equations gives

y=3z

and

x=5z.

Therefore

\frac{x^2+3xy}{y^2+z^2}=\frac{25z^2+45z^2}{9z^2+z^2}=7.

Thus, the correct answer is A.

20.

The sum of the numerical coefficients of all the terms in the expansion of

(x-2y)^{18}

is:

0

1

19

-1

-19

Concepts:polynomial substitution binomial theorem

Difficulty rating: 1030

Small hint:

A polynomial’s coefficient sum is found by setting every variable equal to

1

Big hint:

Evaluate

(1-2)^{18}

Solution:

Set

x=y=1.

The sum of all numerical coefficients is then

(1-2)^{18}=(-1)^{18}=1.

Therefore, the correct answer is B.

21.

If

\log_{b^2}x+\log_{x^2}b=1,

where

b\gt0,

b\ne1,

and

x\ne1,

then

x

equals:

\dfrac1{b^2}

\dfrac1b

b^2

b

\sqrt b

Concepts:logarithm substitution algebraic manipulation

Difficulty rating: 1450

Small hint:

Let

t=\log_bx

Big hint:

The equation becomes

\frac t2+\frac1{2t}=1

Solution:

Let

t=\log_bx.

Then

\log_{b^2}x=\frac{t}{2}

and

\log_{x^2}b=\frac{1}{2t}.

Thus

t+\frac1t=2,

so

(t-1)^2=0

and

t=1.

Hence

x=b.

Therefore, the correct answer is D.

22.

Given parallelogram

ABCD

with

E

the midpoint of diagonal

BD.

Point

E

is connected to a point

F

in

DA

so that

DF=\frac13DA.

What is the ratio of the area of triangle

DFE

to the area of quadrilateral

ABEF?

1:2

1:3

1:5

1:6

1:7

Concepts:parallelogram area ratio midpoint triangle area

Difficulty rating: 1670

Small hint:

Use vectors

A=0,

B=u,

and

D=v

Big hint:

Then

E=\frac{u+v}{2}

and

F=\frac{2v}{3}

; compute both areas as fractions of the parallelogram

Solution:

Let the parallelogram’s area be

S=|u\times v|,

with

A=0,

B=u,

and

D=v.

Then

E=\frac{u+v}{2}

and

F=\frac{2v}{3}.

Determinants give

[DFE]=\frac{S}{12}.

For quadrilateral

ABEF,

the two determinant contributions give

[ABEF]=\frac S4+\frac S6=\frac{5S}{12}.

The desired ratio is

1:5.

Therefore, the correct answer is C.

23.

Two numbers are such that their difference, their sum, and their product are to one another as

1:7:24.

The product of the two numbers is:

6

12

24

48

96

Concepts:ratio and proportion system of equations algebraic manipulation

Difficulty rating: 1210

Small hint:

Represent the difference, sum, and product by

k,

7k,

and

24k

Big hint:

The two numbers are

\frac{7k+k}{2}

and

\frac{7k-k}{2}

Solution:

Let the difference be

k

and the sum be

7k.

The numbers are

4k

and

3k,

so their product is

12k^2.

But the ratio also says the product is

24k.

For distinct numbers

k\ne0,

so

k=2

and the product is

48.

Therefore, the correct answer is D.

24.

Let

y=(x-a)^2+(x-b)^2,

a,

b

constants. For what value of

x

is

y

a minimum?

\dfrac{a+b}{2}

a+b

\sqrt{ab}

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

\dfrac{a+b}{2ab}

Concepts:quadratic completing the square optimization

Difficulty rating: 1140

Small hint:

Expand and collect the terms in

x

Big hint:

Complete the square or use the vertex formula

Solution:

Expanding and completing the square gives

y=2\left(x-\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{2}.

The squared term is minimized at

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

Therefore, the correct answer is A.

25.

The set of values of

m

for which

x^2+3xy+x+my-m

has two factors, with integer coefficients, which are linear in

x

and

y,

is precisely:

0,

12,

-12

0,

12

12,

-12

12

0

Concepts:factoring polynomial system of equations

Difficulty rating: 1870

Small hint:

Write the factors as

(x+ay+b)(x+cy+d)

Big hint:

The missing

y^2

-term and the

3xy

-term force

\{a,c\}=\{0,3\}

Solution:

Write the factorization as

(x+ay+b)(x+cy+d).

Comparing the

y^2

and

xy

coefficients gives

ac=0

and

a+c=3,

so take

a=0,

c=3.

The other coefficients give

\begin{aligned} b+d&=1,\\ bd&=-m,\\ 3b&=m. \end{aligned}

Hence

b(d+3)=0.

If

b=0,

then

m=0;

otherwise

d=-3,

b=4,

and

m=12.

Both values produce valid integer factorizations.

Therefore, the correct answer is B.

26.

In a ten-mile race First beats Second by

2

miles and First beats Third by

4

miles. If the runners maintain constant speeds throughout the race, by how many miles does Second beat Third?

2

2\dfrac14

2\dfrac12

2\dfrac34

3

Concepts:rate ratio and proportion distance rate and time relative speed

Difficulty rating: 1210

Small hint:

When First finishes, Second and Third have run

8

and

6

miles

Big hint:

Compare Third’s distance when Second completes

10

miles

Solution:

Relative to First’s speed, Second’s speed is

\frac{8}{10}=\frac{4}{5}

and Third’s is

\frac{6}{10}=\frac{3}{5}.

Thus Third runs

\frac{3}{4}

as fast as Second. When Second finishes

10

miles, Third has run

10(\frac{3}{4})=7\dfrac12

miles, so Second wins by

2\dfrac12

miles.

Therefore, the correct answer is C.

27.

If

x

is a real number and

|x-4|+|x-3|\lt a,

where

a\gt0,

then:

0\lt a\lt0.01

0.01\lt a\lt1

0\lt a\lt1

0\lt a\leq1

a\gt1

Concepts:absolute value triangle inequality inequality

Difficulty rating: 1260

Small hint:

Interpret the sum as the distances from

x

to

3

and

4

Big hint:

Find its minimum for

3\leq x\leq4

, remembering the inequality is strict

Solution:

By the triangle inequality,

|x-4|+|x-3|\geq|4-3|=1.

Equality holds for every

x

between

3

and

4.

Therefore the strict inequality has a real solution exactly when

a\gt1.

Thus, the correct answer is E.

28.

The sum of

n

terms of an arithmetic progression is

153,

and the common difference is

2.

If the first term is an integer, and

n\gt1,

then the number of possible values for

n

is:

2

3

4

5

6

Concepts:arithmetic sequence divisibility factor counting

Difficulty rating: 1650

Small hint:

If the first term is

a,

the sum is

n(a+n-1)

Big hint:

List the divisors of

153=3^2\cdot17

that exceed

1

Solution:

The arithmetic-series formula simplifies to

153=n(a+n-1).

Thus

n

must divide

153,

and every such

n

gives an integer

a.

The divisors greater than

1

are

3,

9,

17,

51,

and

153,

so there are five possibilities.

Therefore, the correct answer is D.

29.

In this figure

\angle RFS=\angle FDR,

FD=4

inches,

DR=6

inches,

FR=5

inches, and

FS=7\dfrac12

inches. The length of

RS,

in inches, is:

undetermined

4

5\dfrac12

6

6\dfrac14

Difficulty rating: 1710

Small hint:

Compare triangles

RFS

and

FDR

around the marked equal angles

Big hint:

The adjacent side ratios are

\frac{5}{4}=\frac{15}{12}

Solution:

At the equal included angles,

\begin{gathered} \frac{FR}{FD}=\frac54,\\ \frac{FS}{DR}=\frac{15}{12}=\frac54. \end{gathered}

Thus

\triangle RFS\sim\triangle FDR

by SAS. Side

RS

corresponds to

FR=5,

so

RS=(\frac{5}{4})\cdot5=\frac{25}{4}=6\frac14.

Therefore, the correct answer is E.

30.

If

\begin{aligned} (7+4\sqrt3)x^2 &+(2+\sqrt3)x\\ &-2=0, \end{aligned}

the larger root minus the smaller root is:

-2+3\sqrt3

2-\sqrt3

6+3\sqrt3

6-3\sqrt3

3\sqrt3+2

Concepts:quadratic radical algebraic manipulation

Difficulty rating: 1670

Small hint:

Notice that

7+4\sqrt3=(2+\sqrt3)^2

Big hint:

For

Ax^2+Bx+C=0,

the root difference is

\frac{\sqrt{B^2-4AC}}{A}

Solution:

Let

A=7+4\sqrt3=(2+\sqrt3)^2

and

B=2+\sqrt3.

The discriminant is

B^2-4A(-2)=A+8A=9A.

Hence the root difference is

\frac{\sqrt{9A}}A=\frac3{2+\sqrt3}=6-3\sqrt3.

Therefore, the correct answer is D.

31.

Let

\begin{aligned} f(n) &=\frac{5+3\sqrt5}{10} \left(\frac{1+\sqrt5}{2}\right)^n\\ &\quad+\frac{5-3\sqrt5}{10} \left(\frac{1-\sqrt5}{2}\right)^n. \end{aligned}

Then

f(n+1)-f(n-1),

expressed in terms of

f(n),

equals:

\dfrac12f(n)

f(n)

2f(n)+1

f^2(n)

\dfrac12\left(f^2(n)-1\right)

Concepts:recursion Fibonacci exponent

Difficulty rating: 1900

Small hint:

Call the two exponential bases

\alpha

and

\beta

Big hint:

Both roots of

r^2-r-1=0

satisfy

r^{n+1}-r^{n-1}=r^n

Solution:

Each base

r=\frac{1\pm\sqrt5}{2}

satisfies

r^2=r+1,

so

r^2-1=r.

Therefore

r^{n+1}-r^{n-1}=r^{n-1}(r^2-1)=r^n.

Applying this identity term by term to the defining linear combination gives

f(n+1)-f(n-1)=f(n).

Thus, the correct answer is B.

32.

If

\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a},

then:

a

must equal

c

a+b+c+d

must equal zero

either

a=c

or

a+b+c+d=0,

or both

a+b+c+d\ne0

if

a=c

a(b+c+d)=c(a+b+d)

Concepts:rational equation factoring zero product property

Difficulty rating: 1450

Small hint:

Cross-multiply the two fractions

Big hint:

Move all terms to one side and factor out

a-c

Solution:

Cross-multiplication gives

(a+b)(a+d)=(b+c)(c+d).

Subtracting the right side and factoring yields

(a-c)(a+b+c+d)=0.

Hence either

a=c

or the sum is zero, and both may occur.

Therefore, the correct answer is C.

33.

P

is a point interior to rectangle

ABCD

and such that

PA=3

inches,

PD=4

inches, and

PC=5

inches. Then

PB,

in inches, equals:

2\sqrt3

3\sqrt2

3\sqrt3

4\sqrt2

2

Concepts:rectangle British Flag Theorem distance formula

Difficulty rating: 1180

Small hint:

Use the rectangle identity

PA^2+PC^2=PB^2+PD^2

Big hint:

Substitute the three known distances and solve for

PB^2

Solution:

For any point in a rectangle, the British flag theorem gives

PA^2+PC^2=PB^2+PD^2.

Thus

9+25=PB^2+16,

so

PB^2=18

and

PB=3\sqrt2.

Therefore, the correct answer is B.

34.

If

n

is a multiple of

4,

the sum

\begin{aligned} s&=1+2i+3i^2+\cdots\\ &\quad+(n+1)i^n, \end{aligned}

where

i=\sqrt{-1},

equals:

1+i

\dfrac12(n+2)

\dfrac12(n+2-ni)

\dfrac12\left[(n+1)(1-i)+2\right]

\dfrac18(n^2+8-4ni)

Concepts:complex number roots of unity summation

Difficulty rating: 1870

Small hint:

Group the terms in blocks of four powers of

i

Big hint:

Each complete block beginning with coefficient

4j+1

sums to

-2-2i

Solution:

Write

n=4m.

For

j=0,\ldots,m-1,

the four terms with exponents

4j

through

4j+3

sum to

\begin{aligned} &(4j+1)+(4j+2)i\\ &\quad-(4j+3)-(4j+4)i\\ &=-2-2i. \end{aligned}

The final term is

4m+1.

Therefore

\begin{aligned} s&=m(-2-2i)+(4m+1)\\ &=\frac{n+2-ni}{2}. \end{aligned}

Thus, the correct answer is C.

35.

The sides of a triangle are of lengths

13,

14,

and

15.

The altitudes of the triangle meet at point

H.

If

AD

is the altitude to the side of length

14,

what is the ratio

HD:HA?

3:11

5:11

1:2

2:3

25:33

Concepts:altitude coordinate geometry Heron’s Formula

Difficulty rating: 1850

Small hint:

The

13

-

14

-

15

triangle has area

84

, so the altitude to side

14

is

12

Big hint:

The altitude foot divides the side of length

14

into segments

5

and

9

; use coordinates to locate

H

Solution:

Heron’s formula gives area

84,

so

AD=\frac{2(84)}{14}=12.

The adjacent

13

-side has projection

\sqrt{13^2-12^2}=5,

leaving

9

on the base. Put

D=(0,0),\ A=(0,12),

B=(-5,0),\ C=(9,0).

Line

AC

has slope

-\frac{4}{3},

so the altitude through

B

has slope

\frac{3}{4}

and meets

AD

at height

\frac{15}{4}.

Thus

\begin{gathered} HD=\frac{15}{4},\\ HA=12-\frac{15}{4}=\frac{33}{4}, \end{gathered}

giving

HD:HA=5:11.

Therefore, the correct answer is B.

36.

In this figure the radius of the circle is equal to the altitude of the equilateral triangle

ABC.

The circle is made to roll along the side

AB,

remaining tangent to it at a variable point

T

and intersecting lines

AC

and

BC

in variable points

M

and

N,

respectively. Let

n

be the number of degrees in arc

MTN.

Then

n,

for all permissible positions of the circle:

varies from

30^\circ

to

90^\circ

varies from

30^\circ

to

60^\circ

varies from

60^\circ

to

90^\circ

remains constant at

30^\circ

remains constant at

60^\circ

Concepts:circle coordinate geometry equilateral triangle inscribed angle arc

Difficulty rating: 2130

Small hint:

The circle’s center

O

and vertex

C

are the same distance above

AB

, so

CO\parallel AB

Big hint:

Extend

NC

through

C

to meet the circle again at

D

, then use reflection across line

CO

Solution:

Let

O

be the circle’s center. Both

O

and

C

are one triangle altitude above

AB,

so

CO\parallel AB.

Extend

NC

through

C

to meet the circle again at

D.

Since

CD

is opposite to

CN,

\begin{aligned} \angle MCD &=180^\circ-\angle MCN\\ &=120^\circ. \end{aligned}

The line

CO,

parallel to

AB,

bisects this angle. Reflection across

CO

fixes the circle and interchanges rays

CM

and

CD,

so it interchanges

M

and

D.

Hence

CM=CD,

and isosceles triangle

MCD

has base angles

30^\circ.

Because

D,C,N

are collinear,

\angle MDN=30^\circ.

This inscribed angle subtends arc

MTN,

whose measure is therefore

60^\circ

for every permissible circle position.

Thus, the correct answer is E.

37.

Given two positive numbers

a,

b

such that

a\lt b.

Let A.M. be their arithmetic mean and let G.M. be their positive geometric mean. Then A.M. minus G.M. is always less than:

\dfrac{(b+a)^2}{ab}

\dfrac{(b+a)^2}{8b}

\dfrac{(b-a)^2}{ab}

\dfrac{(b-a)^2}{8a}

\dfrac{(b-a)^2}{8b}

Concepts:AM-GM Inequality inequality radical

Difficulty rating: 1670

Small hint:

Rewrite

\frac{a+b}{2}-\sqrt{ab}

using

\sqrt b-\sqrt a

Big hint:

Factor

b-a=(\sqrt b-\sqrt a)(\sqrt b+\sqrt a)

to compare with choice D

Solution:

We have

\frac{a+b}{2}-\sqrt{ab} =\frac{(\sqrt b-\sqrt a)^2}{2}.

After canceling the positive factor

(\sqrt b-\sqrt a)^2,

comparison with choice D reduces to

\frac12 <\frac{(\sqrt b+\sqrt a)^2}{8a}.

This is true because

\sqrt b+\sqrt a\gt2\sqrt a.

Thus A.M. minus G.M. is always less than the expression in choice D.

Therefore, the correct answer is D.

38.

The sides

PQ

and

PR

of triangle

PQR

are respectively of lengths

4

inches and

7

inches. The median

PM

is

3\dfrac12

inches. Then

QR,

in inches, is:

6

7

8

9

10

Concepts:median (geometry)Stewart’s Theorem law of cosines

Difficulty rating: 1210

Small hint:

Use Apollonius’s theorem for the median to

QR

Big hint:

Substitute

PQ=4,\ PR=7,\ PM=\frac{7}{2}

Solution:

Apollonius’s theorem gives

\begin{aligned} QR^2 &=2(4^2+7^2)-4\left(\frac72\right)^2\\ &=81. \end{aligned}

Therefore

QR=9.

Thus, the correct answer is D.

39.

The magnitudes of the sides of triangle

ABC

are

a,

b,

c,

as shown, with

c\leq b\leq a.

Through interior point

P

and the vertices

A,

B,

C,

lines are drawn meeting the opposite sides in

A',

B',

C',

respectively. Let

s=AA'+BB'+CC'.

Then, for all positions of point

P,

s

is less than:

2a+b

2a+c

2b+c

a+2b

a+b+c

Concepts:extremal argument bounding to limit cases inequality

Difficulty rating: 1850

Small hint:

A point on a segment is closer to a fixed vertex than the farther endpoint of that segment

Big hint:

Bound

AA'

,

BB'

, and

CC'

separately using their adjacent side lengths

Solution:

For a fixed vertex, squared distance is a convex function along the opposite side, so its maximum occurs at an endpoint. Since each cevian endpoint is interior to its side,

\begin{gathered} AA'\lt\max(AB,AC)=b,\\ BB'\lt\max(BA,BC)=a,\\ CC'\lt\max(CA,CB)=a. \end{gathered}

Adding gives

s\lt2a+b.

Therefore, the correct answer is A.

40.

A watch loses

2\dfrac12

minutes per day. It is set right at

1

P.M. on March

15.

Let

n

be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows

9

A.M. on March

21,

n

equals:

14\dfrac{14}{23}

14\dfrac1{14}

13\dfrac{101}{115}

13\dfrac{83}{115}

13\dfrac{13}{23}

Concepts:clock date and time rate fraction

Difficulty rating: 1670

Small hint:

The watch advances

\frac{575}{576}

as fast as real time

Big hint:

From the displayed

1

P.M. on March

15

to

9

A.M. on March

21

is

8400

displayed minutes

Solution:

The watch runs at

\frac{575}{576}

of the correct rate. The displayed elapsed time is

5

days

20

hours, or

8400

minutes. Thus the real elapsed time is

8400(\frac{576}{575}),

and the correction is

\begin{aligned} n&=8400\left(\frac{576}{575}-1\right)\\ &=\frac{8400}{575} =\frac{336}{23}\\ &=14\frac{14}{23}. \end{aligned}

Therefore, the correct answer is A.