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2010 Stanford Mathematics Tournament
2010 Stanford Mathematics Tournament
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Stanford Mathematics Tournament
Subcontests
(26)
1
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2010 SMT Team Round - Stanford Math Tournament
p1. Compute
lim
x
→
0
tan
x
−
x
−
x
3
3
x
\lim_{x\to 0} \frac{\tan x - x -\frac{x^3}{3}}{x}
lim
x
→
0
x
t
a
n
x
−
x
−
3
x
3
. p2. For how many integers
n
n
n
is
n
20
−
n
\frac{n}{20-n}
20
−
n
n
equal to the square of a positive integer. p3. Find all possible solutions
(
x
1
,
x
2
,
.
.
.
,
x
n
)
(x_1, x_2, ..., x_n)
(
x
1
,
x
2
,
...
,
x
n
)
to the following equations, where
x
1
=
1
2
(
x
n
+
x
n
−
1
2
x
n
)
x_1 =\frac12 \left(x_n +\frac{x^2_{n-1}}{x_n}\right)
x
1
=
2
1
(
x
n
+
x
n
x
n
−
1
2
)
x
2
=
1
2
(
x
1
+
x
n
2
x
1
)
x_2 =\frac12 \left(x_1 +\frac{x^2_{n}}{x_1}\right)
x
2
=
2
1
(
x
1
+
x
1
x
n
2
)
x
3
=
1
2
(
x
2
+
x
1
2
x
2
)
x_3 =\frac12 \left(x_2 +\frac{x^2_{1}}{x_2}\right)
x
3
=
2
1
(
x
2
+
x
2
x
1
2
)
x
4
=
1
2
(
x
3
+
x
2
2
x
3
)
x_4 =\frac12 \left(x_3 +\frac{x^2_{2}}{x_3}\right)
x
4
=
2
1
(
x
3
+
x
3
x
2
2
)
x
5
=
1
2
(
x
4
+
x
3
2
x
4
)
x_5 =\frac12 \left(x_4 +\frac{x^2_{3}}{x_4}\right)
x
5
=
2
1
(
x
4
+
x
4
x
3
2
)
.
.
.
...
...
x
n
=
1
2
(
x
n
−
1
+
x
n
2
x
1
)
=
2010
x_n =\frac12 \left(x_{n-1} +\frac{x^2_{n}}{x_1}\right)= 2010
x
n
=
2
1
(
x
n
−
1
+
x
1
x
n
2
)
=
2010
. p4. Let
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
be a convex hexagon, whose opposite angles are parallel, satisfying
A
F
=
3
AF = 3
A
F
=
3
,
B
C
=
4
BC = 4
BC
=
4
, and
D
E
=
5
DE = 5
D
E
=
5
. Suppose that
A
D
AD
A
D
,
B
E
BE
BE
,
C
F
CF
CF
intersect in a point. Find
C
D
CD
C
D
. p5. Rank the following in decreasing order:
A
=
1
1
+
2
2
+
3
3
1
+
2
+
3
A =\frac{1\sqrt1 + 2\sqrt2 + 3\sqrt3}{\sqrt1 + \sqrt2 + \sqrt3}
A
=
1
+
2
+
3
1
1
+
2
2
+
3
3
,
B
=
1
2
+
2
2
+
3
2
1
+
2
+
3
B =\frac{1^2 + 2^2 + 3^2}{1 + 2 + 3}
B
=
1
+
2
+
3
1
2
+
2
2
+
3
2
,
C
=
1
+
2
+
3
3
C =\frac{1 + 2 + 3}{3}
C
=
3
1
+
2
+
3
,
D
=
1
+
2
+
3
1
1
+
1
2
+
1
3
D =\frac{\sqrt1 + \sqrt2 + \sqrt3}{\frac{1}{\sqrt1}+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}}
D
=
1
1
+
2
1
+
3
1
1
+
2
+
3
. p6. What is the least
m
m
m
such that for any
m
m
m
integers we can choose
6
6
6
integers such that their sum is divisible by
6
6
6
? p7. Find all positive integers
n
n
n
such that
ϕ
(
n
)
=
16
\phi(n) = 16
ϕ
(
n
)
=
16
, where
ϕ
(
n
)
\phi(n)
ϕ
(
n
)
is defined to be the number of positive integers less than or equal to
n
n
n
that are relatively prime to
n
n
n
. p8. Suppose that for an infinitely differentiable function
f
f
f
,
lim
x
→
0
f
(
4
x
)
+
a
f
(
3
x
)
+
b
f
(
2
x
)
+
c
f
(
x
)
+
d
f
(
0
)
x
4
\lim_{x\to 0}\frac{f(4x) + af(3x) + bf(2x) + cf(x) + df(0)}{x^4}
lim
x
→
0
x
4
f
(
4
x
)
+
a
f
(
3
x
)
+
b
f
(
2
x
)
+
c
f
(
x
)
+
df
(
0
)
exists. Find
1000
a
+
100
b
+
10
c
+
d
1000a + 100b + 10c + d
1000
a
+
100
b
+
10
c
+
d
. p9. Minimize
x
3
+
4
y
2
+
9
z
x^3 + 4y^2 + 9z
x
3
+
4
y
2
+
9
z
under the constraints that
x
y
z
=
1
xyz = 1
x
yz
=
1
,
x
,
y
,
z
≥
0
x, y, z \ge 0
x
,
y
,
z
≥
0
. p10. Positive real numbers
x
,
y
x, y
x
,
y
, and
z
z
z
satisfy the equations
x
2
+
y
2
=
9
x^2 + y^2 = 9
x
2
+
y
2
=
9
y
2
+
2
y
z
+
z
2
=
16
y^2 +\sqrt2yz + z^2 = 16
y
2
+
2
yz
+
z
2
=
16
z
2
+
2
z
x
+
x
2
=
25
z^2 +\sqrt2zx + x^2 = 25
z
2
+
2
z
x
+
x
2
=
25
. Compute
2
x
y
+
y
z
+
z
x
\sqrt2xy + yz + zx
2
x
y
+
yz
+
z
x
. p11. Find the volume of the region given by the inequality
∣
x
+
y
+
z
∣
+
∣
x
+
y
−
z
∣
+
∣
x
−
y
+
z
∣
+
∣
−
x
+
y
+
z
∣
≤
4.
|x + y + z| + |x + y - z| + |x - y + z| + | - x + y + z| \le 4.
∣
x
+
y
+
z
∣
+
∣
x
+
y
−
z
∣
+
∣
x
−
y
+
z
∣
+
∣
−
x
+
y
+
z
∣
≤
4.
p12. Suppose we have a polyhedron consisting of triangles and quadrilaterals, and each vertex is shared by exactly
4
4
4
triangles and one quadrilateral. How many vertices are there? p13. Rank the following in increasing order:
A
=
2011
+
2009
2
A =\frac{\sqrt{2011} + \sqrt{2009}}{2}
A
=
2
2011
+
2009
,
B
=
2011
2011
−
2009
2009
3
B =\frac{2011\sqrt{2011} - 2009\sqrt{2009}}{3}
B
=
3
2011
2011
−
2009
2009
,
C
=
2011
+
2
2010
+
2009
4
C =\frac{\sqrt{2011} + 2\sqrt{2010} + \sqrt{2009}}{4}
C
=
4
2011
+
2
2010
+
2009
,
D
=
2010
D =\sqrt{2010}
D
=
2010
,
E
=
2011
−
1
2
2011
E =\sqrt{2011} -\frac{1}{2\sqrt{2011}}
E
=
2011
−
2
2011
1
p14. Suppose
f
f
f
and
g
g
g
are continuously differentiable functions satisfying
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
−
g
(
x
)
g
(
y
)
f(x + y) = f(x)f(y) - g(x)g(y)
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
−
g
(
x
)
g
(
y
)
g
(
x
+
y
)
=
f
(
x
)
g
(
y
)
+
g
(
x
)
f
(
y
)
g(x + y) = f(x)g(y) + g(x)f(y)
g
(
x
+
y
)
=
f
(
x
)
g
(
y
)
+
g
(
x
)
f
(
y
)
Also suppose that
f
′
(
0
)
=
1
f'(0) = 1
f
′
(
0
)
=
1
and
g
′
(
0
)
=
2
g'(0) = 2
g
′
(
0
)
=
2
. Find
f
(
2010
)
2
+
g
(
2010
)
2
f(2010)^2 + g(2010)^2
f
(
2010
)
2
+
g
(
2010
)
2
. p15. Find the number of
n
n
n
-tuples
(
a
1
,
.
.
.
,
a
n
)
(a_1, ..., a_n)
(
a
1
,
...
,
a
n
)
that maximize
a
1
a
2
a
3
+
a
2
a
3
a
4
+
.
.
.
+
a
n
−
2
a
n
−
1
a
n
a_1a_2a_3 + a_2a_3a_4 + ... + a_{n-2}a_{n-1}a_n
a
1
a
2
a
3
+
a
2
a
3
a
4
+
...
+
a
n
−
2
a
n
−
1
a
n
under the constraints that
n
≥
3
n \ge 3
n
≥
3
and
a
1
+
a
2
+
.
.
.
+
a
n
=
3
m
a_1 + a_2 + ...+ a_n = 3m
a
1
+
a
2
+
...
+
a
n
=
3
m
for a fixed integer
m
m
m
, where
a
i
a_i
a
i
are positive integers. Note: 8 problems were common with [url=https://artofproblemsolving.com/community/c4h2765069p24208072]2010 Rice Math Tournament: p1-3, p5, p8, p10-12 PS. You had better use hide for answers.
24
1
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SMT 2010 General #24
We are given a coin of diameter
1
2
\frac{1}{2}
2
1
and a checkerboard of 1\times1 squares of area
2010
×
2010
2010\times2010
2010
×
2010
. We toss the coin such that it lands completely on the checkerboard. If the probability that the coin doesn't touch any of the lattice lines is
a
2
b
2
\frac{a^2}{b^2}
b
2
a
2
where
a
b
\frac{a}{b}
b
a
is a reduced fraction, fi nd
a
+
b
a+b
a
+
b
23
1
Hide problems
SMT 2010 General #23
Let
f
(
X
,
Y
,
Z
)
=
X
5
Y
−
X
Y
5
+
Y
5
Z
−
Y
Z
5
+
Z
5
X
−
Z
X
5
f(X, Y, Z)=X^5Y-XY^5+Y^5Z-YZ^5+Z^5X-ZX^5
f
(
X
,
Y
,
Z
)
=
X
5
Y
−
X
Y
5
+
Y
5
Z
−
Y
Z
5
+
Z
5
X
−
Z
X
5
. Find
f
(
2009
,
2010
,
2011
)
+
f
(
2010
,
2011
,
2009
)
−
f
(
2011
,
2010
,
2009
)
f
(
2009
,
2010
,
2011
)
\frac{f(2009, 2010, 2011)+f(2010, 2011, 2009)-f(2011, 2010, 2009)}{f(2009, 2010, 2011)}
f
(
2009
,
2010
,
2011
)
f
(
2009
,
2010
,
2011
)
+
f
(
2010
,
2011
,
2009
)
−
f
(
2011
,
2010
,
2009
)
25
1
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SMT 2010 General #25
There are balls that look identical, but their weights all di er by a little. We have a balance that can compare only two balls at a time. What is the minimum number of times, in the worst case, we have to use to balance to rank all balls by weight?
22
1
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SMT 2010 General #22
We need not restrict our number system radix to be an integer. Consider the phinary numeral system in which the radix is the golden ratio
ϕ
=
1
+
5
2
\phi = \frac{1+\sqrt{5}}{2}
ϕ
=
2
1
+
5
and the digits
0
0
0
and
1
1
1
are used. Compute
101010
0
ϕ
−
.
01010
1
ϕ
1010100_{\phi}-.010101_{\phi}
101010
0
ϕ
−
.01010
1
ϕ
21
1
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SMT 2010 General #21
How many non-negative integer solutions are there for
x
4
−
2
y
2
=
1
x^4-2y^2=1
x
4
−
2
y
2
=
1
?
20
1
Hide problems
SMT 2010 General #20
Given fi ve circles of radii
1
,
2
,
3
,
4
,
1, 2, 3, 4,
1
,
2
,
3
,
4
,
and
5
5
5
, what is the maximum number of points of intersections possible (every distinct point where two circles intersect counts).
19
1
Hide problems
SMT 2010 General #19
Find the roots of
6
x
4
+
17
x
3
+
7
x
2
−
8
x
−
4
6x^4+17x^3+7x^2-8x-4
6
x
4
+
17
x
3
+
7
x
2
−
8
x
−
4
18
1
Hide problems
SMT 2010 General #18
In an
n
n
n
-by-
m
m
m
grid,
1
1
1
row and
1
1
1
column are colored blue, the rest of the cells are white. If precisely
1
2010
\frac{1}{2010}
2010
1
of the cells in the grid are blue, how many values are possible for the ordered pair
(
n
,
m
)
(n,m)
(
n
,
m
)
17
1
Hide problems
SMT 2010 General #17
An equilateral triangle is inscribed inside of a circle of radius
R
R
R
. Find the side length of the triangle
16
1
Hide problems
SMT 2010 General #16
A wheel is rolled without slipping through
15
15
15
laps on a circular race course with radius
7
7
7
. The wheel is perfectly circular and has radius
5
5
5
. After the three laps, how many revolutions around its axis has the wheel been turned through?
15
1
Hide problems
SMT 2010 General #15
Find the best approximation of
3
\sqrt{3}
3
by a rational number with denominator less than or equal to
15
15
15
14
1
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SMT 2010 General #14
A series of lockers, numbered 1 through 100, are all initially closed. Student 1 goes through and opens every locker. Student 3 goes through and "flips" every 3rd locker ("flipping") a locker means changing its state: if the locker is open he closes it, and if the locker is closed he opens it). Thus, Student 3 will close the third locker, open the sixth, close the ninth. . . . Student 5 then goes through and "flips"every 5th locker. This process continues with all students with odd numbers
n
<
100
n<100
n
<
100
going through and "flipping" every
n
n
n
th locker. How many lockers are open after this process?
13
1
Hide problems
SMT 2010 General #13
Find all the integers
x
x
x
in
[
20
,
50
]
[20, 50]
[
20
,
50
]
such that
6
x
+
5
≡
19
m
o
d
10
6x+5\equiv 19 \mod 10
6
x
+
5
≡
19
mod
10
, that is,
10
10
10
divides
(
6
x
+
15
)
+
19
(6x+15)+19
(
6
x
+
15
)
+
19
.
12
1
Hide problems
SMT 2010 General #12
Consider the sequence
1
,
2
,
1
,
2
,
2
,
1
,
2
,
2
,
2
,
1
,
2
,
2
,
2
,
2
,
1
,
.
.
.
1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1,...
1
,
2
,
1
,
2
,
2
,
1
,
2
,
2
,
2
,
1
,
2
,
2
,
2
,
2
,
1
,
...
Find
n
n
n
such that the fi rst
n
n
n
terms sum up to
2010
2010
2010
.
11
1
Hide problems
SMT 2010 General #11
What is the area of the regular hexagon with perimeter
60
60
60
?
10
3
Hide problems
SMT 2010 Algebra #10
Find the sum of all solutions of the equation
1
x
2
−
1
+
2
x
2
−
2
+
3
x
2
−
3
+
4
x
2
−
4
=
2010
x
−
4
\frac{1}{x^2-1}+\frac{2}{x^2-2}+\frac{3}{x^2-3}+\frac{4}{x^2-4}=2010x-4
x
2
−
1
1
+
x
2
−
2
2
+
x
2
−
3
3
+
x
2
−
4
4
=
2010
x
−
4
SMT 2010 General #10
Compute the base 10 value of
1464
1
99
14641_{99}
1464
1
99
SMT 2010 Geometry #10
A
,
B
,
C
,
D
A, B, C, D
A
,
B
,
C
,
D
are points along a circle, in that order.
A
C
AC
A
C
intersects
B
D
BD
B
D
at
X
X
X
. If
B
C
=
6
BC=6
BC
=
6
,
B
X
=
4
BX=4
BX
=
4
,
X
D
=
5
XD=5
X
D
=
5
, and
A
C
=
11
AC=11
A
C
=
11
, fi nd
A
B
AB
A
B
9
3
Hide problems
SMT 2010 Algebra #9
Suppose
x
y
−
5
x
+
2
y
=
30
xy-5x+2y=30
x
y
−
5
x
+
2
y
=
30
, where
x
x
x
and
y
y
y
are positive integers. Find the sum of all possible values of
x
x
x
SMT 2010 General #9
A straight line connects City A at
(
0
,
0
)
(0, 0)
(
0
,
0
)
to City B, 300 meters away at
(
300
,
0
)
(300, 0)
(
300
,
0
)
. At time
t
=
0
t=0
t
=
0
, a bullet train instantaneously sets out from City A to City B while another bullet train simultaneously leaves from City B to City A going on the same train track. Both trains are traveling at a constant speed of
50
50
50
meters/second. Also, at
t
=
0
t=0
t
=
0
, a super y stationed at
(
150
,
0
)
(150, 0)
(
150
,
0
)
and restricted to move only on the train tracks travels towards City B. The y always travels at 60 meters/second, and any time it hits a train, it instantaneously reverses its direction and travels at the same speed. At the moment the trains collide, what is the total distance that the y will have traveled? Assume each train is a point and that the trains travel at their same respective velocities before and after collisions with the y
SMT 2010 Geometry #9
For an acute triangle
A
B
C
ABC
A
BC
and a point
X
X
X
satisfying
∠
A
B
X
+
∠
A
C
X
=
∠
C
B
X
+
∠
B
C
X
\angle{ABX}+\angle{ACX}=\angle{CBX}+\angle{BCX}
∠
A
BX
+
∠
A
CX
=
∠
CBX
+
∠
BCX
.Fi nd the minimum length of
A
X
AX
A
X
if
A
B
=
13
AB=13
A
B
=
13
,
B
C
=
14
BC=14
BC
=
14
, and
C
A
=
15
CA=15
C
A
=
15
.
8
3
Hide problems
SMT 2010 Algebra #8
Let
P
(
x
)
P(x)
P
(
x
)
be a polynomial of degree
n
n
n
such that
P
(
x
)
=
3
k
P(x)=3^k
P
(
x
)
=
3
k
for
0
≤
k
≤
n
0\le k \le n
0
≤
k
≤
n
. Find
P
(
n
+
1
)
P(n+1)
P
(
n
+
1
)
SMT 2010 General #8
Find all solutions of
a
x
=
x
−
a
a
\frac{a}{x}=\frac{x-a}{a}
x
a
=
a
x
−
a
for
x
x
x
.
SMT 2010 Geometry #8
A sphere of radius
1
1
1
is internally tangent to all four faces of a regular tetrahedron. Find the tetrahedron's volume.
7
1
Hide problems
SMT 2010 Algebra Problem 7
Find all the integers
x
x
x
in
[
20
,
50
]
[20, 50]
[
20
,
50
]
such that
6
x
+
5
≡
−
19
m
o
d
10
,
6x + 5 \equiv -19 \mod 10,
6
x
+
5
≡
−
19
mod
10
,
that is,
10
10
10
divides
(
6
x
+
15
)
+
19.
(6x + 15) + 19.
(
6
x
+
15
)
+
19.
6
2
Hide problems
SMT 2010 Algebra Problem 6
Consider the sequence
1
,
2
,
1
,
2
,
2
,
1
,
2
,
2
,
2
,
1
,
2
,
2
,
2
,
2
,
1
,
.
.
.
1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, ...
1
,
2
,
1
,
2
,
2
,
1
,
2
,
2
,
2
,
1
,
2
,
2
,
2
,
2
,
1
,
...
Find
n
n
n
such that the first
n
n
n
terms sum up to
2010.
2010.
2010.
SMT 2010 General #6
A triangle has side lengths
7
,
9
,
7, 9,
7
,
9
,
and
12
12
12
. What is the area of the triangle?
5
2
Hide problems
SMT 2010 Algebra Problem 5
A series of lockers, numbered 1 through 100, are all initially closed. Student 1 goes through and opens every locker. Student 3 goes through and "flips" every 3rd locker ("fipping") a locker means changing its state: if the locker is open he closes it, and if the locker is closed he opens it. Student 5 then goes through and "flips" every 5th locker. This process continues with all students with odd numbers
n
<
100
n < 100
n
<
100
going through and "flipping" every
n
n
n
th locker. How many lockers are open after this process?
SMT 2010 General #5
Alice sends a secret message to Bob using her RSA public key
n
=
400000001
n=400000001
n
=
400000001
. Eve wants to listen in on their conversation. But to do this, she needs Alice's private key, which is the factorization of
n
n
n
. Eve knows that
n
=
p
q
n=pq
n
=
pq
, a product of two prime factors. Find
p
p
p
and
q
q
q
.
4
3
Hide problems
SMT 2010 Algebra Problem 4
If
x
2
+
1
x
2
=
7
,
x^2+\frac{1}{x^2}=7,
x
2
+
x
2
1
=
7
,
find all possible values of
x
5
+
1
x
5
.
x^5+\frac{1}{x^5}.
x
5
+
x
5
1
.
SMT 2010 General #4
Compute
1
+
1
+
1
+
1
+
1
.
.
.
\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}}}}...}
1
+
1
+
1
+
1
+
1
...
SMT 2010 Geometry #4
Given triangle
A
B
C
ABC
A
BC
.
D
D
D
lies on
B
C
BC
BC
such that
A
D
AD
A
D
bisects
B
A
C
BAC
B
A
C
. Given
A
B
=
3
AB=3
A
B
=
3
,
A
C
=
9
AC=9
A
C
=
9
, and
B
C
=
8
BC=8
BC
=
8
. Find
A
D
AD
A
D
.
3
2
Hide problems
SMT 2010 Algebra Problem 3
Bob sends a secret message to Alice using her RSA public key
n
=
400000001.
n = 400000001.
n
=
400000001.
Eve wants to listen in on their conversation. But to do this, she needs Alice's private key, which is the factorization of
n
.
n.
n
.
Eve knows that
n
=
p
q
,
n = pq,
n
=
pq
,
a product of two prime factors. Find
p
p
p
and
q
.
q.
q
.
SMT 2010 General #3
How many zeros are there at the end of
(
200
124
)
\binom{200}{124}
(
124
200
)
?
2
3
Hide problems
SMT 2010 Algebra Problem 2
Write
0.2010
228
‾
0.2010\overline{228}
0.2010
228
as a fraction.
SMT 2010 General #2
Find the smallest prime
p
p
p
such that the digits of
p
p
p
(in base 10) add up to a prime number greater than
10
10
10
.
SMT 2010 Geometry #2
Find the radius of a circle inscribed in a triangle with side lengths
4
4
4
,
5
5
5
, and
6
6
6
1
3
Hide problems
SMT 2010 Algebra Problem 1
Compute
1
+
1
+
1
+
1
+
1
+
1
+
⋯
\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}}}
1
+
1
+
1
+
1
+
1
+
1
+
⋯
SMT 2010 General #1
Given
8
8
8
coins, at most one of them is counterfeit. A counterfeit coin is lighter than a real coin. You have a free weight balance. What is the minimum number of weighings necessary to determine the identity of the counterfeit coin if it exists
SMT 2010 Geometry #1
Find the reflection of the point
(
11
,
16
,
22
)
(11, 16, 22)
(
11
,
16
,
22
)
across the plane
3
x
+
4
y
+
5
z
=
7
3x+4y+5z=7
3
x
+
4
y
+
5
z
=
7
.