Putnam 2001
A1. Consider a set S and a binary operation ∗, i.e., for each a, b ∈ S, a ∗ b ∈ S. Assume
(a ∗ b) ∗ a = b for all a, b ∈ S. Prove that a ∗ (b ∗ a) = b for all a, b ∈ S.
Solution. We have b = ((b ∗ a) ∗ b) ∗ (b ∗ a) = a ∗ (b ∗ a).
A2. You have coins C1 , C2 , . . . , Cn . For each k, Ck is biased so that, when tossed, it has
probability 1/(2k + 1) of falling heads. If the n coins are tossed, what is the probability that
the number of heads is odd? Express the answer as a rational function of n.
Solution. Let On be the event that n tosses result in an odd number of heads, and let
En be the event that n tosses result in an even number of heads. We have P (O1 ) = 1/3 and
P (On ) = P (Cn lands on T ) P (On−1 ) + P (Cn lands on H) P (En−1 )
µ
¶
1
1
=
1−
P (On−1 ) +
(1 − P (On−1 ))
2n + 1
2n + 1
µ
¶
1
1
1
=
1−
−
P (On−1 ) +
2n + 1 2n + 1
2n + 1
1
2n − 1
(2 (n − 1) + 1) P (On−1 ) + 1
P (On−1 ) +
=
=
2n + 1
2n + 1
³ 2n + 1
´
=
=
=
=
=
(2 (n − 1) + 1)
(2(n−2)+1)P (On−2 )+1
2(n−1)+1
+1
2n + 1
(2 (n − 2) + 1) P (On−2 ) + 2
2n + 1
(2 (n − i) + 1) P (On−i ) + i
··· =
2n + 1 ¡
¢
(2 (n − (n − 1)) + 1) P On−(n−1) + n − 1
2n + 1
1
3· 3 +n−1
n
=
.
2n + 1
2n + 1
A3. For each integer m, consider the polynomial
Pm (x) = x4 − (2m + 4)x2 + (m − 2)2 .
For what values of m is Pm (x) the product of two non-constant polynomials with integer
coefficients?
Solution. We first find the roots r1 , r2 , r3 , r4 of Pm (x). Then
Pm (x) = (x − r1 ) (x − r2 ) (x − r3 ) (x − r4 ) .
1
Thus, one of the nonconstant polynomial factors would be of form x − r1 for some root r1 ,
or (x − r1 ) (x − r2 ) for some roots r1 and r2 . The quadratic formula yields
´
p
1³
2
2
2
r =
(2m + 4) ± (2m + 4) − 4(m − 2)
2
p
√ √
= m + 2 ± (m + 2)2 − (m − 2)2 = (m + 2) ± 2 2 m
³√
√ ´2
=
m± 2 .
So
n√
√ √
√
√
√ o
√
√
m + 2, m − 2, − m + 2, − m − 2
{r1 , r2 , r3 , r4 } =
If one of the nonconstant polynomial factors is of form x − r1 , then the only possibility is
m = 2 and r1 = 0. This possibility is realized, since
¡
¢
P2 (x) = x4 − 8x2 = x2 x2 − 8 .
If one of the nonconstant polynomial factors is of form (x − r1 ) (x − r2 ) , then note that
(x − r1 ) (x − r2 ) = x2 − (r1 + r2 ) x + r1 r2
√
in which√case r1 + r2 and r1 r2 must be integers. The possibilities for r1 + r2 are 0, ±2 m,
and ±2 2 (not an integer). We have
√
√ ª
½ ©√
√
m
+
2,
−
m
−
√
√2ª
√
r1 + r2 = 0 ⇔ {r1 , r2 } = ©√
m − 2, − m + 2
√ √
¢
½ ¡
− ¡m + 2√2 m + 2¢
√
⇒ r1 r2 =
,
− m−2 2 m+2
which is an integer only if m is 2n2 for some integer n ≥ 0. We then have
n√
√ √
√
√
√ o
√
√
{r1 , r2 , r3 , r4 } =
m + 2, m − 2, − m + 2, − m − 2
n√
√ √
√
√
√
√
√ o
=
2n + 2, 2n − 2, − 2n + 2, − 2n − 2
and
³
³√
³ √
√ ´´ ³
√ ´´
2n + 2
x − − 2n − 2 ·
³
³√
³ √
√ ´´ ³
√ ´´
x−
2n − 2
x − − 2n + 2
¡
¢¡
¢
= x2 − 2n2 − 4n − 2 x2 − 2n2 + 4n − 2 = P2n2 (x) .
x−
In the remaining case,
n √
√
√ o
√
√
r1 + r2 = ±2 m ⇔ {r1 , r2 } = ± m + 2, ± m − 2 ⇒ r1 r2 = m − 2
2
and m must either be 2 (as before) or the square of an integer n ≥ 0. If m = n2 , then
n
√
√
√
√ o
{r1 , r2 , r3 , r4 } = n + 2, n − 2, −n + 2, −n − 2 ,
and indeed
³
³
³
³
³
√ ´´ ³
√ ´´ ³
√ ´´ ³
√ ´´
x− n+ 2
x− n− 2
x − −n + 2
x − −n − 2
¡
¢¡
¢
= x2 − 2xn + n2 − 2 x2 + 2xn + n2 − 2
= x4 − (2n2 + 4)x2 + (n2 − 2)2 = Pn2 (x).
Thus, m must be of the form 2n2 or n2 for an integer n ≥ 0.
A4. Triangle ABC has an area 1. Points E, F, G lie, respectively, on sides BC, CA, AB
such that AE bisects BF at point R, BF bisects CG at point S, and CG bisects AE at
point T . Find the area of the triangle RST .
Solution. Choose r, s, t, so that
CE
AF
BG
= r,
= s,
=t
CB
AC
BA
Let µ (U V W ) denote the area of a triangle with vertices U, V, W . Then
µ (AEB) = µ (AEF ) ,
since the triangles have the same base AE and since AE bisects F B the same altitude as
well. Also
µ (ABE) =
BE
BC − EC
µ (ABC) =
· 1 = 1 − r.
BC
BC
Similarly,
CE
µ (ACB)
CB
CF
µ (CEF ) =
µ (CEA) .
CA
µ (ACE) =
Thus,
CF
CF
CF CE
µ (CEA) =
µ (ACE) =
µ (ACB)
CA
CA
CA CB
= (1 − s) rµ (ACB) = (1 − s) r.
µ (CEF ) =
Hence,
1 = µ (ABC) = µ (AEB) + µ (AEF ) + µ (CEF )
= (1 − r) + (1 − r) + (1 − s) r
= 2 − r − rs ⇒ r (1 + s) = 1
3
Similarly, s (1 + t) = 1 and t (1 + r) = 1. Thus,
r (1 + s) = 1, s (1 + t) = 1, t (1 + r) = 1
Now
1
1
1+t
=
and t (1 + r) = 1
1 =
1+s
2+t
1 + 1+t
µ
¶
1+t
⇒ t 1+
= 1 ⇒ t (2 + t + 1 + t) = 2 + t
2+t
⇒ t2 + t ³− 1 = 0 ´
³√
´
√
1
1
⇒ t= 2
5 − 1 or − 2
5 + 1 (negative)
r =
By symmetry,
r=s=t=
1
2
Note that
³√
´
5−1 .
µ (RST ) = µ (ABC) − (µ (ABR) + µ (BCS) + µ (CAT )) .
Also,
BR
AF
µ (ABF ) = 12 µ (ABF ) = 12
µ (ABC)
BF
AC
= 12 sµ (ABC) = 12 t,
µ (ABR) =
and similarly
µ (BCS) = µ (CAT ) = 12 t.
Thus,
³√
´ 7 − 3√5
µ (RST ) = 1 − 2 t + 12 t + 12 t = 1 − 32 · 12
5−1 =
≈ .0729 . . . .
4
¡1
¢
A5. Prove that there are unique positive integers a, n such that an+1 − (a + 1)n = 2001.
Solution. Suppose an+1 − (a + 1)n = 2001. Notice that
2002 = an+1 − (a + 1)n + 1 ≡ 0 mod a
Thus a is a factor of 2002 = 2 × 7 × 11 × 13. Note that a 6= 2, since then
2n+1 − 3n = 2001,
4
which is impossible since 2001 is divisible by 3 unlike 2n+1 . More generally since
an+1 − (a + 1)n = 2001 ≡ 0 mod 3
a ≡ 2 mod 3 is impossible, as is a ≡ 0 mod 3. Thus, a ≡ 1 mod 3, and
(−1)n ≡ (a + 1)n ≡ an+1 ≡ 1 mod 3 ⇒ n even.
Recall that an+1 + 1 has a factor of a + 1 when n is even. Thus, a + 1 is a factor of
an+1 + 1 − (a + 1)n = 2002 = 2 × 7 × 11 × 13
as is a itself. Only two factors differ by 1, namely 13 and 14. Thus, a = 13 and the smallest
value for n is 2. We have
132+1 − 142 = 2001.
It remains to show that no even n > 2 satisfies
13n+1 − 14n = 2001
If n > 2, 14n ≡ 0 mod 8, and 2001 ≡ 1 mod 8. Thus,
13n+1 ≡ 1 mod 8
However, 13 ≡ 5 mod 8 and 132 ≡ 52 ≡ 1 mod 8. Hence, 13n+1 ≡ 5 mod 8 if n is even Thus,
n > 2 and n even is not possible.
A6. Can an arc of a parabola inside a circle of radius 1 have a length greater than 4?
Solution. The answer is yes. Consider the arc of the parabola y = 12 cx2 inside the circle
x2 + (y − 1)2 = 1, where c > 1.
¢
¡
x2 + ( 12 cx2 − 1)2 = 1 ⇒ x2 + 14 c2 x4 − cx2 = 0 ⇒ x2 14 c2 x2 + 1 − c = 0
√
2 c−1
⇒ x = 0 or x = ±
c
The length of the parabolic arc from x = 0 to x =
Z
√
2 c−1
c
√
2 c−1
c
is then
√
1 + c2 x2 dx = (using u = cx)
0
Z √
1 2 c−1 √
=
1 + u2 du = (using the substitution z = sinh u)
c 0
´¯2√c−1
√
1 ³1
¯
1
2
=
arcsinh u + 2 u 1 + u ¯
2
c
0
5
µ
¶
q
√
¢
¡ √
¢2
¡ √
1
=
arcsinh 2 c − 1 + 2 c − 1 1 + 2 c − 1
2c
´
p
√
¢
¡ √
1 ³
=
arcsinh 2 c − 1 + 2 c − 1 1 + 4 (c − 1)
2c
r
¢
¡ √
1
7
3
=
arcsinh 2 c − 1 + 2 1 −
+ 2
2c
4c 4c
We have
2
r
3
7
1−
+ 2 >2
4c 4c
r
µ
µ ¶¶
7
2
7
1
1−
≈2 1− 2
>2− ,
4c
4c
c
for c sufficiently large. Thus, it suffices to prove that
¢ 2
¡ √
1
arcsinh 2 c − 1 >
2c
c
for c sufficiently large, but this is so since arcsinh x increases without limit as x → ∞.
Indeed,
arcsinh x
= lim
x→∞
x→∞
log x
lim
d
dx
arcsinh (x)
1
x
x
= lim √
= 1.
x→∞
1 + x2
Note: Mathematica yields a maximal length of 4.0025901 . . . for c = 199.7944360 . . . .
B1. Let n be an even positive integer. Write the numbers 1, 2, . . . , n2 in the squares of
an n × n grid so that the k-th row, from left to right, is
(k − 1)n + 1, (k − 1)n + 2, . . . , (k − 1)n + n.
Color the squares of the grid so that half of the squares in each row and in each column are
red and the other half are black (a checkerboard coloring is one possibility). Prove that for
each coloring, the sum of the numbers on the red squares is equal to the sum of the numbers
on the black squares.
Solution. Instead of writing the numbers 1, 2, . . . , n2 in the squares, write 0, 1, 2, . . . , n2 −
1. Since half the squares are red and half are black, the assertion to be proven holds iff it
holds with this new numbering. Now rewrite each number in base n in the form d2 d1 , where
we take d2 = 0 in the first row. Note that d2 is the same in each row, while d1 is the same
in each column. Since half of the squares in each row and in each column are red and the
other half are black, the sum of the d2 ’s for the red squares is the same as the sum of the
d2 ’s for the black squares in any row, and hence the sum of all red d2 ’s and all black d2 ’s
is the same. Considering columns, the same equality holds for the sums of d1 ’s for red and
d1 ’s for black squares. Thus, adding base n yields the same sum for red and black squares.
B2. Find all pairs of real numbers (x, y) satisfying the system of equations
1
1
+
= (x2 + 3y 2 )(3x2 + y 2 )
x 2y
1
1
−
= 2(y 4 − x4 ).
x 2y
6
Solution. By adding and subtracting the two given equations, we obtain the equivalent
set of equations
2/x = x4 + 10x2 y 2 + 5y 4
1/y = 5x4 + 10x2 y 2 + y 4 .
Multiplying the first by x and the second by y, we get
2 = x5 + 10x3 y 2 + 5xy 4
1 = 5x4 y + 10x2 y 3 + y 5
Adding these yields
3 = x5 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5 = (x + y)5 .
Subtracting yields
1 = x5 − 5x4 y + 10x3 y 2 − 10x2 y 3 + 5xy 4 − y 5 = (x − y)5
Thus,
x + y = 31/5
x − y = 1,
and so x = (31/5 + 1)/2 and y = (31/5 − 1)/2.
B3. For any positive integer n, let hni denote the closest integer to
∞
X
2hni + 2−hni
n=1
2n
√
n. Evaluate
.
Solution. Since (m − 1/2)2 = m2 − m + 1/4 and (m + 1/2)2 = m2 + m + 1/4, we have
that hni = m if and only if
m2 − m + 1/4 ≤ n ≤ m2 + m + 1/4
However, since n is an integer, this is equivalent to
m2 − m + 1 ≤ n ≤ m2 + m
7
Hence
∞
X
2hni + 2−hni
2n
n=1
=
=
=
=
=
∞
X
m
(2 + 2
−m
m=1
∞
X
∞ X
∞
X
2hni + 2−hni X
=
=
2n
m=1
m=1
hni=m
m2 +m
)
X
=
∞
X
m
(2 + 2
−m
n=m2 −m+1
)2
2m + 2−m
2n
−(m2 −m+1)
m=1
n=m2 −m+1
(2m + 2−m )2−(m
m=1
∞
X
2
−n
2 +m
mX
−2m
) (1 − 2
2 −m+1
1
2
2m−1
X
2−n
n=0
)
=
∞
X
(2m + 2−m )2−(m
m=1
) ¡1 − 2−2m ¢
2 −m
∞ ³
³
´ X
´
2
2
2
2
(2m + 2−m ) 2−m +m − 2−m −m =
2−m +2m − 2−m −2m
m=1
∞
X
m=1
2
X
m=1
¡
−m(m−2)
2
−2
−(m+2)m
¢
=
∞
X
2
−m(m−2)
m=1
−
∞
X
2−m(m−2)
m=3
2−m(m−2) = 2 + 1 = 3.
m=1
B4. Let S denote the set of rational numbers different from {−1, 0, 1}. Define f : S → S
by f (x) = x − 1/x. Prove or disprove that
∞
\
n=1
f (n) (S) = ∅,
where f (n) denotes f composed with itself n times.
Solution. Suppose that
prime factors. Then
n
m
is reduced to lowest terms; i.e., n and m have no common
n
n
m
n2 − m2
(n + m) (n − m)
f( ) =
−
=
=
m
m
n
mn
mn
2
2
−m
is reduced to lowest terms: If a prime p divides mn then it divides m
We claim that n mn
or it divides n. If p divides m, then it cannot divide n, and hence it cannot divide n ± m.
If p divides n, then it cannot divide m, and hence it cannot divide n ± m. In either case, p
cannot divide (n + m) (n − m) , and the claim holds.
¡n¢
n
Define the degree of m
(reduced to lowest terms) by deg m
= |m| + |n|. Thus,
³ n ´ ¯
¯
deg f ( ) = ¯n2 − m2 ¯ + |mn| ≥ 3 + |mn| ,
m
since |n2 − m2 | is smallest (i.e., 3) for {n, m} = {1, 2}. Hence,
³n ´
³ n ´
deg f( ) − deg
) = 3 + |mn| − (|m| + |n|)
m
m
= 2 + 1 + |mn| − (|m| + |n|) = 2 + (|m| − 1) (|n| − 1) ≥ 2
8
Hence the degree
any fraction in f (n) (S) is at least 2n. Since any rational number has
T∞ of (n)
finite degree, n=1 f (S) = ∅.
B5. Let a and b be real numbers in the interval (0, 1/2), and let g be a continuous realvalued function such that g(g(x)) = ag(x) + bx for all real x. Prove that g(x) = cx for some
constant c.
Solution. Pick x0 arbitrary, and define xn recursively by xn+1 = g(xn ) for n ≥ 0.
xn+2 = g(xn+1 ) = g(g(xn )) = ag(xn ) + bxn = axn+1 + bxn
(1)
We can find solutions of this relation of the form xn = rn . To determine r, note that
¡
¢
rn+2 = arn+1 + brn ⇔ 0 = rn+2 − arn+1 − brn = rn r2 − ar − b
³
´
√
1
2
⇔ r = 0 or r = r± := 2 a ± a + 4b .
The general solution of (1) is then
n
n
xn = c+ r+
+ c− r−
.
If we can show that c− = 0 (resp. c+ = 0) for all choices of x0 , then c+ = x0 (resp. c− = 0)
and
g (x0 ) = x0 r+ (resp. g (x0 ) = x0 r− )
and we are done. Note that
− |r+ | < r− < 0 < r+ <
1
2
µ
1
2
+
q
1
4
+
1
2
¶
√
1+ 3
,
=
4
n
n
so that c+ r+
dominates for large positive n, while c− r−
dominates for large negative n.
−1
However, xn is not defined for negative n unless g exists. We prove that g : R → R is 1-1
and onto as follows. Note that
g (u) = g (v)
⇒ ag(u) + bu = g(g(u)) = g(g(v)) = ag(v) + bv = ag(u) + bv
⇒ bu = bv ⇒ u = v.
Thus, g is 1-1, and hence must be strictly increasing or strictly decreasing.
If g is strictly increasing and limx→+∞ g (x) = L+ < +∞, then we have the contradiction
g(L+ ) = lim g(g(x)) = lim (ag(x) + bx) = ∞.
x→+∞
x→+∞
Thus, g is strictly increasing ⇒ limx→+∞ g (x) = +∞. If limx→−∞ g (x) = L− > −∞, then
we have the contradiction
g(L− ) = lim g(g(x)) = lim (ag(x) + bx) = −∞.
x→−∞
x→−∞
9
Thus, under the assumptions, g strictly increasing implies that g : R → R is 1-1 and onto.
If g is strictly decreasing and limx→+∞ g (x) = L+ > −∞, then we have the contradiction
g (L+ ) = lim g(g(x)) = lim (ag(x) + bx) = +∞.
x→+∞
x→+∞
If limx→−∞ g (x) = L− < +∞, then we have the contradiction
g(L− ) = lim g(g(x)) = lim (ag(x) + bx) = −∞.
x→−∞
x→−∞
Hence in either case, g : R → R is 1-1 and onto (i.e., as x goes from −∞ to +∞, g (x)
strictly increases from −∞ to +∞, or g (x) strictly decreases from +∞ to −∞). At any
rate, we may define xn−1 = g −1 (xn ) for n ≤ 0.
Suppose that c+ 6= 0 and c− 6= 0. We will arrive at a contradiction. Note that
µ
µ ¶n ¶
c+ r+
n
n
n
xn = c+ r+ + c− r− = c− r− 1 +
.
c− r−
¯ ³ ´n ¯
¯ ¯
¯
¯
¯ r+ ¯
Since ¯ r− ¯ > 1, ¯ cc−+ rr−+ ¯ < 12 for n sufficiently large and negative, say n < N < 0. Then
for n < N < 0, sign(xn ) = c− (−1)n alternates and
¯
¯
n¯
|xn | > ¯ 12 c− r−
→ ∞ as n → −∞.
Since xn = g (xn−1 ), there are arbitrarily large x > 0 for which g (x) < 0, and so g must be
decreasing. We also have
µ
µ ¶n ¶
c− r−
n
n
n
xn = c+ r+ + c− r− = c+ r+ 1 +
.
c+ r+
¯ ¯
¯ ³ ´n ¯
¯ r− ¯
¯
¯
Since ¯ r+ ¯ < 1, ¯ cc−+ rr−+ ¯ < 12 for n sufficiently large, say n > N > 0. Thus, if c+ > 0, then
for n > N > 0,
n
g (xn−1 ) = xn ≥ 12 c+ r+
and so there are arbitrarily large values of x for which g (x) > 0. This cannot happen if g is
decreasing, as we have just shown is the case. If c+ < 0, then
n
g (xn−1 ) = xn ≤ 12 c+ r+
and there are arbitrarily large negative values of x for which g (x) is negative. This is also impossible for g decreasing. Hence, for each x0 , we have g (x0 ) = r+ x0 or g (x0 ) = r− x0 . Since g
is continuous, both of the sets {x0 ∈ (0, ∞) : g (x0 ) − r+ x0 = 0} and {x0 ∈ (0, ∞) : g (x0 ) − r− x0 = 0}
are closed in (0, ∞). Since (0, ∞) is connected either g (x0 ) = r+ x0 for all x0 > 0, or
g (x0 ) = r− x0 for all x0 > 0. Similarly, this holds for x0 < 0. We must eliminate the possibility g (x) = r± x for x > 0 and g (x0 ) = r∓ x for x < 0, but we know this since g is increasing
or decreasing.
10
B6. Assume that (an )n≥1 is an increasing sequence of positive real numbers such that
lim an /n = 0. Must there exist infinitely many positive integers n such that an−i +an+i < 2an
for i = 1, 2, . . . , n − 1?
Solution. Here, “increasing” must mean strictly increasing, since otherwise a constant
sequence provides an immediate counter-example. Actually, non-decreasing and “not eventually constant” is enough. The condition an−i +¡an+i < 2an means¢ that the line connecting
(n − i, an−i ) and (n + i, an+i ) in R2 has midpoint n, 12 (an−i + an+i ) which lies below (n, an ).
Equivalently, this means that there is a line Ln through (n, an ) such that (n − i, an−i ) and
(n + i, an+i ) are below Ln . Thus, it suffices to show that there are infinitely many n such that
there is a line Ln through (n, an ) such that all of the points (n − i, an−i ) and (n + i, an+i )
for i = 1, 2, . . . , n©− 1 lie
ª below Ln .
−a1
The sequence aii−1
is (eventually) positive for i ≥ 2 and has limit 0. Thus, it has
a positive maximum, say m1 which is attained for some largest i1 > 1. Thus, the line Li1
a −a
through (i1 , ai1 ) with slope m1 := ii11 −1 1 > 0 has the property that all points (i, ai ) with
i < i1 are on or below this line, while all points (i, ai ) for i > i1 are strictly below (since
a −a
a −a
i1 is largest). The ratios ii−i1i1 for i > i1 are then all less than m1 and limi→∞ ii−i1i1 = 0.
a −a
Hence m1 − ii−i1i1 achieves a positive maximum, say 2ε. The line, say Li1 , with slope m1 − ε
through (i1 , ai1 ) then lies above (i, ai ) for all i > i1 . The points (i, ai ) for i < i1 also lie
below this line since they are on or below the line with slope m1 through (i1 , ai1 ). We now
repeat the same argument, starting with i1 instead of 1 to produce a largest i2 , such that
a −a
the maximum of ii−i1i1 for i > i1 is achieved for i = i2 . This maximum, say mi1 , is less than
m1 − ε and so the line with slope mi1 through (i2 , ai2 ) will have the property that all points
(i, ai ) will lie on or below this line. Again, there is some ε > 0, such that the line with slope
mi1 − ε through (i2 , ai2 ) will lie above all other points (i, ai ) , i 6= i2 . In this way, we produce
infinitely many points (ik , aik ) such there is a line through (ik , aik ) which lies above all points
(i, ai ) , i 6= ik .
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