1995 AIME Solutions | LIVE by Po-Shen Loh

1.

Square

S_1

is

1\times1.

For

i\geq1,

the lengths of the sides of square

S_{i+1}

are half the lengths of the sides of square

S_i,

two adjacent sides of square

S_i

are perpendicular bisectors of two adjacent sides of square

S_{i+1},

and the other two sides of square

S_{i+1}

are the perpendicular bisectors of two adjacent sides of square

S_{i+2}.

The total area enclosed by at least one of

S_1,

S_2,

S_3,

S_4,

S_5

can be written in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m-n.

Answer

Concepts:geometric sequence area inclusion-exclusion

Difficulty rating: 1900

Small hint:

Each square has one fourth the area of the preceding square

Big hint:

Adjacent squares overlap in one fourth of the smaller square, and nonadjacent interiors do not overlap

Solution:

The sum of the five square areas is

\begin{aligned}1+\frac14+\frac1{16}&+\frac1{64}+\frac1{256}\\&=\frac{1364}{1024}.\end{aligned}

The perpendicular-bisector placement makes the overlap of each adjacent pair one fourth of the smaller square. These four overlaps are disjoint and have total area

\begin{aligned}\frac1{16}+\frac1{64}&+\frac1{256}+\frac1{1024}\\&=\frac{85}{1024}.\end{aligned}

Thus the union has area

\frac{1364-85}{1024}=\frac{1279}{1024},

and

m-n=1279-1024=255.

2.

Find the last three digits of the product of the positive roots of

\sqrt{1995}\,x^{\log_{1995}x}=x^2.

Answer

Concepts:logarithm quadratic modular arithmetic

Difficulty rating: 1780

Small hint:

Set

y=\log_{1995}x,

so

x=1995^y

Big hint:

Compare exponents of

1995

, then use the sum of the two values of

y

Solution:

Put

y=\log_{1995}x,

so

x=1995^y.

The equation becomes

1995^{\frac{1}{2}+y^2}=1995^{2y},

and hence

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

The two values of

y

have sum

2,

so the product of the corresponding positive roots is

1995^2.

Since

1995\equiv-5\pmod {1000},

its last three digits are

025.

The requested AIME answer is

25.

3.

Starting at

(0,0),

an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let

p

be the probability that the object reaches

(2,2)

in six or fewer steps. Given that

p

can be written in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers, find

m+n.

Answer

Concepts:random walk basic counting complementary probability

Difficulty rating: 1850

Small hint:

The target can first be reached only after

4

or

6

steps

Big hint:

From the six-step paths ending at the target, subtract those that already arrived at step

4

Solution:

There are

\binom42=6

four-step paths to

(2,2).

There are

60+60=120

six-step paths ending there: the extra opposite pair is either left-right or down-up. Of these,

6\cdot4=24

first reach the target at step

4

and then make a two-step return. Therefore

p=\frac6{4^4}+\frac{120-24}{4^6}=\frac3{64}.

Thus

m+n=3+64=67.

4.

Circles of radius

3

and

6

are externally tangent to each other and are internally tangent to a circle of radius

9.

The circle of radius

9

has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.

Answer

Concepts:tangent circles chord coordinate geometry

Difficulty rating: 1940

Small hint:

The three circle centers are collinear because their pairwise distances are

6,

3,

and

9

Big hint:

Use signed distances from the two smaller centers to the common tangent to find its distance from the large center

Solution:

Put the radius-

9

circle at the origin and the smaller centers at

(-6,0)

and

(3,0).

Write the common external tangent as

\mathbf n\cdot(x,y)=c,

where

\mathbf n

is a unit normal. Its signed distances from the two centers differ by

6-3=3,

so

9n_x=3

and

n_x=\frac{1}{3}.

Using the distance

3

from

(-6,0)

gives

|c|=5.

Thus the chord lies

5

units from the large center, and its squared length is

4(9^2-5^2)=4\cdot56=224.

5.

For certain real values of

a,

b,

c,

and

d,

the equation

x^4+ax^3+bx^2+cx+d=0

has four non-real roots. The product of two of these roots is

13+i

and the sum of the other two roots is

3+4i,

where

i=\sqrt{-1}.

Find

b.

Answer

Concepts:complex number Vieta’s Formulas polynomial

Difficulty rating: 2110

Small hint:

Because the polynomial has real coefficients, its non-real roots occur in conjugate pairs

Big hint:

Group the six pairwise products into the products within the two groups and the four cross-products

Solution:

Let the first two roots be

\alpha

and

\beta.

Since

\alpha\beta=13+i

is not real, they are not conjugates, so the other roots are

\overline\alpha

and

\overline\beta.

Hence

\alpha+\beta=3-4i

and

\overline\alpha\,\overline\beta=13-i.

By Vieta’s formulas,

\begin{aligned}b&=\alpha\beta+\overline\alpha\,\overline\beta\\&\quad+(3-4i)(3+4i)\\&=26+25=51.\end{aligned}

6.

Let

n=2^{31}3^{19}.

How many positive integer divisors of

n^2

are less than

n

but do not divide

n?

Answer

Concepts:factor counting pairing and grouping prime factorization

Difficulty rating: 2270

Small hint:

Pair each divisor

d

of

n^2

with

\frac{n^2}{d}

Big hint:

Among the divisors below

n

, subtract those that already divide

n

Solution:

The number

n^2=2^{62}3^{38}

has

63\cdot39=2457

divisors. Pairing

d

with

\frac{n^2}{d}

leaves only

n

unpaired, so

\frac{2457-1}{2}=1228

divisors lie below

n.

The number

n

has

32\cdot20=640

divisors, of which

639

are below

n.

Therefore the requested count is

1228-639=589.

7.

Given that

(1+\sin t)(1+\cos t)=\frac54

and

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac mn-\sqrt k,\end{aligned}

where

k,

m,

and

n

are positive integers with

m

and

n

relatively prime, find

k+m+n.

Answer

Concepts:trigonometric identity symmetry (algebra)radical

Difficulty rating: 1900

Small hint:

Set

u=\sin t+\cos t

and express

\sin t\cos t

in terms of

u

Big hint:

Both given products become half of a perfect square in

u

Solution:

Let

u=\sin t+\cos t.

Since

\sin t\cos t=\frac{u^2-1}{2},

\begin{aligned}(1+\sin t)&(1+\cos t)\\&=\frac{(u+1)^2}{2}=\frac54.\end{aligned}

The feasible sign gives

u+1=\sqrt{\frac{5}{2}}.

Therefore

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac{(1-u)^2}{2}\\&=\frac{13}{4}-\sqrt{10}.\end{aligned}

Thus

k+m+n=10+13+4=27.

8.

For how many ordered pairs of positive integers

(x,y),

with

y<x\leq100,

are both

\frac{x}{y}

and

\frac{x+1}{y+1}

integers?

Answer

Concepts:divisibility floor and ceiling functions counting pairs

Difficulty rating: 2060

Small hint:

Write

x=ay

and reduce

x+1

modulo

y+1

Big hint:

The quotient

a

must have the form

1+t(y+1)

; count the possible positive values of

t

Solution:

Write

x=ay.

Modulo

y+1,

we have

y\equiv-1,

so

x+1

is divisible by

y+1

exactly when

a\equiv1\pmod {y+1}.

Since

x>y,

write

a=1+t(y+1)

with

t\geq1.

The bound

x\leq100

becomes

t\leq\left\lfloor\frac{100-y}{y(y+1)}\right\rfloor.

Only

1\leq y\leq9

contribute, giving

\begin{aligned}49+16+8&+4+3\\&+2+1+1+1\\&=85.\end{aligned}

9.

Triangle

ABC

is isosceles, with

AB=AC

and altitude

AM=11.

Suppose that there is a point

D

on

\overline{AM}

with

AD=10

and

\angle BDC=3\angle BAC.

Then the perimeter of

\triangle ABC

may be written in the form

a+\sqrt b,

where

a

and

b

are integers. Find

a+b.

Answer

Concepts:trigonometric identity isosceles triangle trigonometry

Difficulty rating: 2170

Small hint:

Let

BM=x

and let half the apex angle be

\alpha

Big hint:

Use

\tan\alpha=\frac{x}{11},

DM=1,

and the triple-angle formula for tangent

Solution:

Let

BM=x

and

\alpha=\angle BAM,

so

\tan\alpha=\frac{x}{11}.

Because

DM=AM-AD=1,

symmetry gives

\angle BDC=2\arctan x,

while

\angle BAC=2\alpha.

Hence

\arctan x=3\alpha.

Put

t=\tan\alpha=\frac{x}{11}.

Then

\frac{3t-t^3}{1-3t^2}=11t,

so

t^2=\frac{1}{4}

and

x=\frac{11}{2}.

Thus

BC=11

and

AB=\frac{11\sqrt5}{2},

making the perimeter

11+11\sqrt5=11+\sqrt{605}.

Therefore

a+b=11+605=616.

10.

What is the largest positive integer that is not the sum of a positive integral multiple of

42

and a positive composite integer?

Answer

Concepts:modular arithmetic factor extremal argument

Difficulty rating: 2110

Small hint:

For each residue modulo

42

, find the smallest positive composite integer in that residue

Big hint:

Once one number in a residue class is representable, every number

42

larger in that class is representable

Solution:

A number in residue class

r\pmod {42}

is representable once it exceeds the first positive composite in that class by a positive multiple of

42.

Composite residues

r

themselves supply that first value. For the remaining residues, suitable first composites are

\begin{aligned}&0:42,\ 1:85,\ 2:44,\\&3:45,\ 5:215,\ 7:49,\\&11:95,\ 13:55,\ 17:143,\\&19:145,\ 23:65,\ 29:155,\\&31:115,\ 37:121,\ 41:125.\end{aligned}

The largest entry is

215,

and

215-42,

215-84,

215-126,

215-168,

and

215-210

are all prime. Hence

215

is not representable, while every larger integer is.

11.

A right rectangular prism

P

(i.e., a rectangular parallelepiped) has sides of integral length

a,

b,

c,

with

a\leq b\leq c.

A plane parallel to one of the faces of

P

cuts

P

into two prisms, one of which is similar to

P,

and both of which have nonzero volume. Given that

b=1995,

for how many ordered triples

(a,b,c)

does such a plane exist?

Answer

Difficulty rating: 2270

Small hint:

Sort the three side lengths of the smaller prism and compare them in order with

a,

b,

and

c

Big hint:

The two unchanged dimensions force

\frac{a}{1995}=\frac{1995}{c}

Solution:

Let the similar smaller prism have sorted sides

x\leq y\leq z.

It shares two side lengths with

P,

and all three of its sorted sides are smaller than the corresponding sides of

P.

The only possible matching is

y=a

and

z=b=1995.

Similarity then gives

\frac{x}{a}=\frac{a}{1995}=\frac{1995}{c},

so

ac=1995^2.

Conversely every factor pair

a<c

gives a nondegenerate cut. Since

1995=3\cdot5\cdot7\cdot19,

its square has

3^4=81

divisors. Excluding the central pair

a=c=1995

and taking one divisor from each remaining pair gives

\frac{81-1}{2}=40.

12.

Pyramid

OABCD

has square base

ABCD,

congruent edges

\overline{OA},

\overline{OB},

\overline{OC},

and

\overline{OD},

and

\angle AOB=45^\circ.

Let

\theta

be the measure of the dihedral angle formed by faces

OAB

and

OBC.

Given that

\cos\theta=m+\sqrt n,

where

m

and

n

are integers, find

m+n.

Answer

Concepts:3D geometry coordinate geometry vector

Difficulty rating: 2450

Small hint:

Place the square’s vertices at

(\pm1,\pm1,0)

and the apex at

(0,0,h)

Big hint:

Find

h^2

from

\angle AOB,

then take the supplement of the angle between suitable face normals

Solution:

Take adjacent base vertices

A=(1,1,0),

B=(-1,1,0),

C=(-1,-1,0)

and

O=(0,0,h).

From

\angle AOB=45^\circ,

\begin{aligned}\frac{h^2}{h^2+2}&=\frac1{\sqrt2},\\h^2&=2+2\sqrt2.\end{aligned}

Normals to the two faces may be taken as

(0,2h,2)

and

(-2h,0,2).

Their acute angle has cosine

\frac{1}{h^2+1}=3-2\sqrt2.

The interior dihedral angle is its supplement, so

\cos\theta=2\sqrt2-3=-3+\sqrt8.

Thus

m+n=-3+8=5.

13.

Let

f(n)

be the integer closest to

\sqrt[4]{n}.

Find

\sum_{k=1}^{1995}\frac1{f(k)}.

Answer

Concepts:radical counting integers in a range sum of first n squares

Difficulty rating: 1940

Small hint:

Count the integers

n

for which

j-\tfrac12<\sqrt[4]n<j+\tfrac12

Big hint:

The number of occurrences of

f(n)=j

simplifies to

4j^3+j

Solution:

For

j\geq1,

the number of positive integers

n

for which

f(n)=j

is

\begin{aligned}\left(j+\frac12\right)^4&-\left(j-\frac12\right)^4\\&=4j^3+j.\end{aligned}

For

j=1,\ldots,6,

these account for

\sum_{j=1}^6(4j^3+j)=1785

values, and their contribution to the requested sum is

\begin{aligned}\sum_{j=1}^6\frac{4j^3+j}{j}&=4\sum_{j=1}^6j^2+6\\&=370.\end{aligned}

The remaining

1995-1785=210

values have

f(n)=7,

contributing

30.

The total is

400.

14.

In a circle of radius

42,

two chords of length

78

intersect at a point whose distance from the center is

18.

The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form

m\pi-n\sqrt d,

where

m,

n,

and

d

are positive integers and

d

is not divisible by the square of any prime. Find

m+n+d.

Answer

Concepts:sector chord coordinate geometry

Difficulty rating: 2650

Small hint:

Each chord is

9\sqrt3

from the center, so determine the two possible line directions through the intersection point

Big hint:

The unequal chord segments have lengths

30

and

48,

and their endpoints subtend

60^\circ

at the center

Solution:

Put the center at

O=(0,0)

and the intersection at

P=(18,0).

A length-

78

chord is

9\sqrt3

from

O,

so a line through

P

containing such a chord makes angle

60^\circ

or

120^\circ

with

OP.

Solving along either line gives segment lengths

30

and

48.

For either region bordered by unequal segments, the two arc endpoints subtend

60^\circ

at

O.

Its area is the sector minus

\triangle OAB

plus

\triangle PAB:

\begin{aligned}\frac{60}{360}\pi(42)^2&-\frac12(42)^2\sin60^\circ\\&+\frac12(30)(48)\sin60^\circ\\&=294\pi-81\sqrt3.\end{aligned}

Hence

m+n+d=378.

15.

Let

p

be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of

5

heads before one encounters a run of

2

tails. Given that

p

can be written in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers, find

m+n.

Answer

Concepts:recursive probability process simulation basic probability

Difficulty rating: 2210

Small hint:

Use states for

0,

1,

2,

3,

and

4

consecutive heads and one separate state for a single trailing tail

Big hint:

Express every head-run state’s success probability in terms of the trailing-tail state

Solution:

Let

q_i

be the success probability with

i

consecutive heads and no trailing tail, and let

t

be the probability after one tail. Then

\begin{aligned}q_i&=\frac{q_{i+1}+t}{2}\quad(0\leq i<4),\\q_4&=\frac{1+t}{2},\\t&=\frac{q_1}{2}.\end{aligned}

Working backward gives

q_1=\frac{1+15t}{16}.

Since

q_1=2t,

we get

t=\frac{1}{17}

and

q_1=\frac{2}{17}.

Therefore

p=q_0=\frac{q_1+t}{2}=\frac3{34},

so

m+n=3+34=37.