1984 AIME Solutions | LIVE by Po-Shen Loh

1.

Find the value of

a_2+a_4+a_6+\cdots+a_{98}

if

a_1,

a_2,

a_3,

\ldots

is an arithmetic progression with common difference

1,

and

a_1+a_2+a_3+\cdots+a_{98}=137.

Answer

Concepts:arithmetic sequence summation

Difficulty rating: 1580

Small hint:

Pair the odd-indexed terms with the even-indexed terms

Big hint:

Each even-indexed term is

1

greater than the odd-indexed term immediately before it

Solution:

Let

E=a_2+a_4+\cdots+a_{98}

and

O=a_1+a_3+\cdots+a_{97}.

There are

49

pairs, and

a_{2k}-a_{2k-1}=1,

so

E-O=49.

Also

E+O=137.

Adding the two equations gives

2E=186,

and hence

E=93.

2.

The integer

n

is the smallest positive multiple of

15

such that every digit of

n

is either

8

or

0.

Compute

\frac{n}{15}.

Answer

Concepts:divisibility digits

Difficulty rating: 1890

Small hint:

Divisibility by

5

determines the last digit

Big hint:

Divisibility by

3

restricts the number of digits equal to

8

Solution:

A multiple of

15

must end in

0

and have digit sum divisible by

3.

Because

8\equiv2\pmod3,

the number of

8

’s must be a positive multiple of

3.

The smallest possible number therefore has three

8

’s followed by

0,

namely

n=8880.

Thus

\frac{n}{15}=592.

3.

A point

P

is chosen in the interior of

\triangle ABC

so that when lines are drawn through

P

parallel to the sides of

\triangle ABC,

the resulting smaller triangles,

t_1,

t_2,

and

t_3

in the figure, have areas

4,

9,

and

49,

respectively. Find the area of

\triangle ABC.

Answer

Concepts:similarity area ratio

Difficulty rating: 2260

Small hint:

Each of the three smaller triangles is similar to

\triangle ABC

Big hint:

Convert each area ratio into a linear ratio and add the three linear ratios

Solution:

Let the area of

\triangle ABC

be

K.

The three small triangles are similar to

\triangle ABC,

so their corresponding linear ratios are

\frac{2}{\sqrt K},\qquad \frac{3}{\sqrt K},\qquad \frac{7}{\sqrt K}.

Each such scale factor is also the perpendicular distance from

P

to one side divided by the altitude to that side. These are the three barycentric area ratios

\frac{[PBC]}{K},

\frac{[PCA]}{K},

and

\frac{[PAB]}{K},

which add to

1.

Hence

\frac{2+3+7}{\sqrt K}=1.

Therefore

\sqrt K=12

and

K=144.

4.

Let

S

be a list of positive integers—not necessarily distinct—in which the number

68

appears. The average (arithmetic mean) of the numbers in

S

is

56.

However, if

68

is removed, the average of the remaining numbers drops to

55.

What is the largest number that can appear in

S?

Answer

Concepts:mean optimization

Difficulty rating: 1670

Small hint:

Let

N

be the number of entries in the original list

Big hint:

To maximize one entry, make every unrestricted positive integer as small as possible

Solution:

If the original list has

N

entries, then

56N-68=55(N-1),

so

N=13

and the total of the entries is

56\cdot13=728.

Besides the entry

68,

make eleven entries equal to the least positive integer,

1.

The remaining entry is then

728-68-11=649.

This construction is valid, and no larger entry is possible.

5.

Determine the value of

ab

if

\log_8a+\log_4b^2=5

and

\log_8b+\log_4a^2=7.

Answer

Concepts:logarithm system of equations

Difficulty rating: 1960

Small hint:

Express every logarithm using base

2

Big hint:

Set

A=\log_2a

and

B=\log_2b

to obtain a linear system

Solution:

Let

A=\log_2a

and

B=\log_2b.

The two equations become

\frac A3+B=5,\qquad \frac B3+A=7.

Equivalently,

A+3B=15

and

3A+B=21,

which give

A=6

and

B=3.

Therefore

ab=2^{A+B}=2^9=512.

6.

Three circles, each of radius

3,

are drawn with centers at

(14,92),

(17,76),

and

(19,84).

A line passing through

(17,76)

is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

Answer

Concepts:circle area symmetry slope

Difficulty rating: 2410

Small hint:

The line already bisects the circle centered at

(17,76)

Big hint:

For the other two equal circles, their signed distances from the line must be opposites

Solution:

For a circle of radius

3,

let

G(d)

be the signed difference between the areas on the two sides of a line when the center’s signed distance from the line is

d.

The function

G

is odd and strictly increasing for

-3<d<3,

and is constant only after the line no longer cuts the circle. The circle centered at

(17,76)

contributes zero, so the line must separate the other two centers.

Let

M=(\frac{33}{2},88)

be the midpoint of those two centers. Among the lines through

(17,76)

that separate them, the smaller of their two distances to the line is largest when the distances are equal, namely for the line through

M.

Even then, the distance is

\frac{56}{\sqrt{577}}<3,

so the line must cut at least one of the two circles. Balance then forces it to cut both. The strict increase of

G

therefore forces the two signed distances to be opposites, so the required line passes through

M.

Its slope is

\frac{88-76}{\frac{33}{2}-17}=-24.

Its absolute value is

24.

7.

The function

f

is defined on the set of integers and satisfies

f(n)=n-3

if

n\geq1000,

and

f(n)=f(f(n+5))

if

n<1000.

Find

f(84).

Answer

Concepts:recursion induction parity

Difficulty rating: 2440

Small hint:

Start by evaluating

f(995),

f(996),

\ldots,

f(999)

Big hint:

Use downward induction to find a parity pattern for every integer below

1000

Solution:

Directly from the definition,

\begin{aligned} f(999)&=f(f(1004))\\ &=f(1001)=998,\\ f(998)&=f(f(1003))\\ &=f(1000)=997. \end{aligned}

Continuing gives

f(997)=998,

f(996)=997,

and

f(995)=998.

We now use downward induction. If

n<995

is even, then

n+5

is odd, so the established pattern above

n

gives

f(n)=f(f(n+5)).

Thus

f(n)=f(998)=997.

If

n

is odd, the same argument gives

f(n)=f(997)=998.

Thus every even

n<1000

has value

997.

Since

84

is even,

f(84)=997.

8.

The equation

z^6+z^3+1=0

has one complex root with argument

\theta

between

90^\circ

and

180^\circ

in the complex plane. Determine the degree measure of

\theta.

Answer

Concepts:complex number roots of unity

Difficulty rating: 2210

Small hint:

Substitute

u=z^3

Big hint:

The two possible arguments of

z^3

are

120^\circ

and

240^\circ

Solution:

Let

u=z^3.

Then

u^2+u+1=0,

so

u

has argument

120^\circ

or

240^\circ.

Taking cube roots gives possible arguments

40^\circ,160^\circ,280^\circ,80^\circ,200^\circ,320^\circ.

The only one strictly between

90^\circ

and

180^\circ

is

160^\circ.

9.

In tetrahedron

ABCD,

edge

AB

has length

3

cm. The area of face

ABC

is

15\text{ cm}^2

and the area of face

ABD

is

12\text{ cm}^2.

These two faces meet each other at a

30^\circ

angle. Find the volume of the tetrahedron in

\text{cm}^3.

Answer

Concepts:3D geometry triangle area volume

Difficulty rating: 2650

Small hint:

Find the altitudes from

C

and

D

to the common edge

AB

Big hint:

Express the tetrahedron’s volume using the common edge, the two altitudes, and the sine of the dihedral angle

Solution:

Let

h_C

and

h_D

be the perpendicular distances from

C

and

D

to

AB.

From the two face areas,

\begin{aligned} h_C&=\frac{2\cdot15}{3}=10,\\ h_D&=\frac{2\cdot12}{3}=8. \end{aligned}

The angle between these two perpendicular directions is the

30^\circ

dihedral angle. Hence the scalar triple product gives

\begin{aligned} V&=\frac16(AB)h_Ch_D\sin30^\circ\\ &=\frac16\cdot3\cdot10\cdot8\cdot\frac12\\ &=20. \end{aligned}

10.

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over

80.

From this, John was able to determine the number of problems Mary solved correctly. If Mary’s score had been any lower, but still over

80,

John could not have determined this. What was Mary’s score? (Recall that the AHSME consists of

30

multiple-choice problems and that one’s score,

s,

is computed by the formula

s=30+4c-w,

where

c

is the number correct and

w

is the number wrong; students are not penalized for problems left unanswered.)

Answer

Concepts:Diophantine Equation counting integers in a range

Difficulty rating: 2360

Small hint:

For a fixed score

s,

express the number wrong in terms of

s

and

c

Big hint:

Use

w\geq0

and

c+w\leq30

to bound the possible integer values of

c

Solution:

From

s=30+4c-w,

we have

w=30+4c-s.

The conditions

w\geq0

and

30-c-w\geq0

give

\left\lceil\frac{s-30}{4}\right\rceil \leq c\leq \left\lfloor\frac{s}{5}\right\rfloor.

Evaluating these integer endpoints for scores

81

through

118

always leaves at least two possible values of

c.

At

s=119,

both endpoints equal

23,

so John can determine

c=23.

Thus the first score over

80

with the required property is

119.

11.

A gardener plants three maple trees, four oak trees, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let

\frac{m}{n}

in lowest terms be the probability that no two birch trees are next to one another. Find

m+n.

Answer

Concepts:arrangements with restrictions combinations basic probability

Difficulty rating: 2160

Small hint:

First choose the five positions occupied by birch trees

Big hint:

Place the seven non-birch trees first and use the eight gaps around them

Solution:

The five birch positions form a uniformly chosen

5

-element subset of the

12

positions, so there are

\binom{12}{5}

possibilities. After the seven non-birch trees are placed, there are eight gaps, including the two end gaps. Choosing five distinct gaps gives

\binom85

arrangements with no adjacent birches. Therefore

\frac{m}{n}=\frac{\binom85}{\binom{12}{5}} =\frac{56}{792}=\frac7{99},

and

m+n=106.

12.

A function

f

is defined for all real numbers and satisfies

\begin{aligned} f(2+x)&=f(2-x),\\ f(7+x)&=f(7-x) \end{aligned}

for all real

x.

If

x=0

is a root of

f(x)=0,

what is the least number of roots

f(x)=0

must have in the interval

-1000\leq x\leq1000?

Answer

Concepts:functional equation reflection (geometry)counting integers in a range

Difficulty rating: 2110

Small hint:

Interpret the two identities as reflection symmetries about

2

and

7

Big hint:

Composing the two reflections produces a translation by

10

Solution:

The identities make the root set invariant under reflection about

2

and about

7.

Their composition is translation by

10.

Starting from the root

0,

these operations force every number in

\begin{gathered} \{10k:k\in\mathbb Z\}\\ {}\cup\{4+10k:k\in\mathbb Z\} \end{gathered}

to be a root. In the interval, the first set contributes

201

roots and the second contributes

200,

for a total of

401.

This bound is attainable: define

f

to be

0

on this invariant set and

1

elsewhere. It has both required reflection symmetries. Hence the least possible number is

401.

13.

Find the value of

\begin{aligned} 10\cot\bigl(&\cot^{-1}3+\cot^{-1}7\\ &{}+\cot^{-1}13+\cot^{-1}21\bigr). \end{aligned}

Answer

Concepts:trigonometric identity algebraic manipulation

Difficulty rating: 2280

Small hint:

Use

\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}

Big hint:

Combine the four inverse-cotangent angles two at a time from left to right

Solution:

Let

\alpha=\cot^{-1}3+\cot^{-1}7.

The cotangent addition formula gives

\begin{aligned} \cot\alpha&=\frac{3\cdot7-1}{3+7}\\ &=2. \end{aligned}

Combining the next angle gives

\frac{2\cdot13-1}{2+13}=\frac53,

and combining the last gives

\frac{(\frac{5}{3})\cdot21-1}{\frac{5}{3}+21}=\frac32.

Multiplying by

10

yields

15.

14.

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Answer

Concepts:prime modular arithmetic extremal argument

Difficulty rating: 2650

Small hint:

Test candidate even integers by listing the odd composites no greater than half the candidate

Big hint:

For sufficiently large even

N,

work modulo

6

and try subtracting

9,

25,

or

35

Solution:

In a representation of

38,

the smaller odd composite would be at most

19.

The only possibilities are

9

and

15,

whose complements

29

and

23

are prime. Thus

38

is not representable.

Now let

N>38

be even. If

N\equiv0\pmod6,

write

N=9+(N-9).

If

N\equiv2\pmod6,

write

N=35+(N-35).

If

N\equiv4\pmod6,

write

N=25+(N-25).

In each case the second summand is an odd multiple of

3

greater than

3,

hence is composite; the fixed first summand is also odd and composite. Therefore every even integer greater than

38

is representable, so the largest exception is

38.

15.

Determine

w^2+x^2+y^2+z^2

if

\begin{gathered} \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}\\[-2pt] {}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1,\\[2pt] \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}\\[-2pt] {}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1,\\[2pt] \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}\\[-2pt] {}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1,\\[2pt] \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}\\[-2pt] {}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1. \end{gathered}

Answer

Concepts:rational equation polynomial algebraic manipulation

Difficulty rating: 3060

Small hint:

Regard the four left sides as values of one rational function in

T

Big hint:

Compare the coefficient of

\frac{1}{T}

as

T

tends to infinity

Solution:

Define

\begin{aligned} R(T)&=\frac{x^2}{T-1}+\frac{y^2}{T-9}\\ &\quad{}+\frac{z^2}{T-25}+\frac{w^2}{T-49}\\ &\quad{}-1. \end{aligned}

Put

\begin{aligned} N(T)&=(T-4)(T-16)\\ &\quad{}\cdot(T-36)(T-64),\\ D(T)&=(T-1)(T-9)\\ &\quad{}\cdot(T-25)(T-49). \end{aligned}

With common denominator

D(T),

the numerator of

R(T)

has leading coefficient

-1.

The four equations say its zeros are

4,

16,

36,

and

64.

Therefore

R(T)=-\frac{N(T)}{D(T)}.

Let

S=w^2+x^2+y^2+z^2.

As

T

tends to infinity, the defining expression gives

R(T)=-1+\frac{S}{T}+O(T^{-2}).

On the other hand, the sum of the four numerator roots is

120,

while the sum of the four denominator roots is

84,

so the factored expression gives

\begin{aligned} R(T)&=-1+\frac{120-84}{T}\\ &\quad{}+O(T^{-2}). \end{aligned}

Hence

w^2+x^2+y^2+z^2=36.