Two is 10% of x and 20% of y. What... 2005 AMC 12 A, Problem #1— “Set up two equations... respectively”

-
Two is 10% of x and 20% of y. What is x − y?
(A) 1
(B) 2
(C) 5
(D) 10
(E) 20
2005 AMC 12 A, Problem #1— “Set up two equations and solve for x and y,
respectively”
- Solution (D) It is given that 0.1x = 2 and 0.2y = 2, so x = 20 and y = 10. Thus x − y = 10.
Difficulty: Easy
NCTM Standard: Number and Operations Standard: Understand meanings of operations and how they relate to one another
Mathworld.com Classification: Number Theory > Arithmetic Fractions > Percent
1
-
The equations 2x + 7 = 3 and bx − 10 = −2 have the same solution. What
is the value of b?
(A) −8
(B) −4
(C) −2
(D) 4
2005 AMC 10 A, Problem #3— “Solve each for x”
- Solution (B) Since 2x + 7 = 3 we have x = −2. Hence
−2 = bx − 10 = −2b − 10,
so
2b = −8, and b = −4.
Difficulty: Easy
NCTM Standard: Algebra Standard: Understand patterns, relations, and functions.
Mathworld.com Classification:
Algebra > General Algebra > Algebra
2
(E) 8
-
A rectangle with a diagonal of length x is twice as long as it is wide. What
is the area of the rectangle?
(A)
1 2
x
4
(B)
2 2
x
5
(C)
1 2
x
2
(D) x2
(E)
3 2
x
2
2005 AMC 10 A, Problem #4— “What is known about the triangle formed in the
rectangle?”
- Solution (B) Let w be the width of the rectangle. Then the length is 2w, and
x2 = w2 + (2w)2 = 5w2 .
The area is consequently w(2w) = 2w2 = 25 x2 .
Difficulty: Medium-hard
NCTM Standard: Geometry Standard: Analyze characteristics and properties of two- and three-dimensional geometric shapes
and develop mathematical arguments about geometric relationships.
Mathworld.com Classification:
Geometry > Plane Geometry > Rectangles > Rectangle
3
-
A store normally sells windows at $100 each. This week the store is offering
one free window for each purchase of four. Dave needs seven windows and
Doug needs eight windows. How much will they save if they purchase the
windows together rather than separately?
(A) 100
(B) 200
(C) 300
(D) 400
(E) 500
2005 AMC 10 A, Problem #5— “Solve for Dave’s, Doug’s, and Dave’s
combined with Doug’s”
- Solution (A)
If Dave buys his windows separately he will purchase six and receive one free, for a cost of $600. If Doug
buys his windows separately, he will purchase seven and receive one free, for a total cost of $700. The total
cost to Dave and Doug purchasing separately will be $1300. If they purchase the windows together, they
will need to purchase only 12 windows, for a cost of $1200, and will receive 3 free. This will result in a
savings of $100.
Difficulty: Medium-easy
NCTM Standard: Problem Solving Standard: Apply and adapt a variety of appropriate strategies to solve problems.
Mathworld.com Classification:
Algebra > General Algebra > Algebra
4
-
The average (mean) of 20 numbers is 30, and the average of 30 other
numbers is 20. What is the average of all 50 numbers?
(A) 23
(B) 24
(C) 25
(D) 26
(E) 27
2005 AMC 10 A, Problem #6— “What is the sum of the 20 numbers? ...of the
30 numbers?”
- Solution (B) The sum of the 50 numbers is 20 · 30 + 30 · 20 = 1200. Their average is 1200/50 = 24.
Difficulty: Medium
NCTM Standard: Number and Operations Standard: Understand meanings of operations and how they relate to one another
Mathworld.com Classification:
Calculus and Analysis > Special Functions > Means > Mean
5
-
Josh and Mike live 13 miles apart. Yesterday Josh started to ride his bicycle
toward Mike’s house. A little later Mike started to ride his bicycle toward
Josh’s house. When they met, Josh had ridden for twice the length of time
as Mike and at four-fifths of Mike’s rate. How many miles had Mike ridden
when they met?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
2005 AMC 10 A, Problem #7— “Use d=rt.”
- Solution (B) Because (rate)(time) = (distance), the distance Josh rode was (4/5)(2) = 8/5 of the distance
that Mike rode. Suppose Mike rode m miles. Then the number of miles between their houses is
8
13
m.
13 = m + m =
5
5
Thus m = 5.
Difficulty: Medium-hard
NCTM Standard: Connections Standard: Recognize and apply mathematics in contexts outside of mathematics.
Mathworld.com Classification:
Calculus and Analysis > Calculus > Differential Calculus > Derivative
6
-
Square EF GH is inside square ABCD so that each side of EF GH can
be extended
to pass through a vertex of ABCD. Square ABCD has side
√
length 50, and BE = 1. What is the area of the inner square EF GH?
A
B
E
H
F
G
D
(A) 25
(B) 32
C
(C) 36
(D) 40
(E) 42
2005 AMC 12 A, Problem #7— “What can one say about triangle BEC?”
- Solution (C) The symmetry of the figure implies that △ABH, △BCE, △CDF , and △DAG are congruent right triangles. So
p
√
BH = CE = BC 2 − BE 2 = 50 − 1 = 7,
and EH = BH − BE = 7 − 1 = 6. Hence the square EF GH has area 62 = 36.
OR
As in the first solution, BH = 7. Now note that △ABH, △BCE, △CDF , and △DAG are congruent
right triangles, so
Area(EF GH) = Area(ABCD) − 4Area(△BCE) = 50 − 4 21 · 1 · 7 = 36.
Difficulty: Medium-easy
NCTM Standard: Geometry Standard: Use visualization, spatial reasoning, and geometric modeling to solve problems.
Mathworld.com Classification:
Geometry > Plane Geometry > Squares > Square
7
-
Let A, M, and C be digits with
(100A + 10M + C)(A + M + C) = 2005.
What is A?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
2005 AMC 12 A, Problem #8— “What is the prime factorization of 2005?”
- Solution (D) Since A, M, and C are digits we have
0 ≤ A + M + C ≤ 9 + 9 + 9 = 27.
The prime factorization of 2005 is 2005 = 5 · 401, so
100A + 10M + C = 401 and A + M + C = 5.
Hence A = 4, M = 0, and C = 1.
Difficulty: Medium-easy
NCTM Standard: Number and Operations Standard: Understand meanings of operations and how they relate to one another
Mathworld.com Classification:
Number Theory > Prime Numbers > Prime Factorization > Prime Factorization Algorithms
8
-
There are two values of a for which the equation 4x2 + ax + 8x + 9 = 0 has
only one solution for x. What is the sum of those values of a?
(A) −16
(B) −8
(C) 0
(D) 8
(E) 20
2005 AMC 10 A, Problem #10— “Try the quadratic equation.”
- Solution (A) The quadratic formula yields
x=
−(a + 8) ±
p
(a + 8)2 − 4 · 4 · 9
.
2·4
The equation has only one solution precisely when the value of the discriminant, (a + 8)2 − 144, is 0. This
implies that a = −20 or a = 4, and the sum is −16.
OR
The
√ equation has one solution if and√only if the polynomial is the square of a binomial with linear term
± 4x2 = ±2x and constant term ± 9 = ±3. Because (2x ± 3)2 has a linear term ±12x, it follows that
a + 8 = ±12. Thus a is either −20 or 4, and the sum of those values is −16.
Difficulty: Medium-hard
NCTM Standard: Connections Standard: Recognize and use connections among mathematical ideas.
Mathworld.com Classification:
Algebra > Algebraic Equations > Quadratic Equation
9
-
A wooden cube n units on a side is painted red on all six faces and then cut
into n3 unit cubes. Exactly one-fourth of the total number of faces of the
unit cubes are red. What is n?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
2005 AMC 10 A, Problem #11— “How many faces are red on the unit cubes?”
- Solution (B) The unit cubes have a total of 6n3 faces, of which 6n2 are red. Therefore
1
6n2
1
= 3 = ,
4
6n
n
so
n = 4.
Difficulty: Medium-hard
NCTM Standard: Geometry Standard: Use visualization, spatial reasoning, and geometric modelling to solve problems.
Mathworld.com Classification:
Geometry > Solid Geometry > Polyhedra > Cubes > Cube
10
-
How many positive integers n satisfy the following condition:
(130n)50 > n100 > 2200 ?
(A) 0
(B) 7
(C) 12
(D) 65
(E) 125
2005 AMC 10 A, Problem #13— “Break it up into two separate inequalities”
- Solution (E) The condition is equivalent to
130n > n2 > 24 = 16,
so 130n > n2 and n2 > 16.
This implies that 130 > n > 4. So n can be any of the 125 integers strictly between 130 and 4.
Difficulty: Hard
NCTM Standard: Number and Operations Standard: Understand numbers, ways of representing numbers, relationships among
numbers, and number systems
Mathworld.com Classification:
Calculus and Analysis > Inequalities > Inequality
11
-
A line passes through A(1, 1) and B(100, 1000). How many other points
with integer coordinates are on the line and strictly between A and B?
(A) 0
(B) 2
(C) 3
(D) 8
(E) 9
2005 AMC 12 A, Problem #12— “How would one represent each point on the
line?”
- Solution (D) The slope of the line is
1000 − 1
111
=
,
100 − 1
11
so all points on the line have the form (1 + 11t, 1 + 111t). Such a point has integer coordinates if and only
if t is an integer, and the point is strictly between A and B if and only if 0 < t < 9. Thus there are 8
points with the required property.
Difficulty: Medium-easy
NCTM Standard: Geometry Standard: Specify locations and describe spatial relationships using coordinate geometry and other
representational systems; apply transformations and use symmetry to analyze mathematical situations
Mathworld.com Classification:
Discrete Mathematics > Point Lattices > Integer Lattice
12
-
In the five-sided star shown, the letters A, B, C, D, and E are replaced by
the numbers 3, 5, 6, 7, and 9, although not necessarily in that order. The
sums of the numbers at the ends of the line segments AB, BC, CD, DE,
and EA form an arithmetic sequence, although not necessarily in this order.
What is the middle term of the arithmetic sequence?
A
D
C
B
E
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
2005 AMC 10 A, Problem #17— “How many times is each letter summed over
the star?”
- Solution (D) Each number appears in two sums, so the sum of the sequence is
2(3 + 5 + 6 + 7 + 9) = 60.
The middle term of a five-term arithmetic sequence is the mean of its terms, so 60/5 = 12 is the middle
term.
The figure shows an arrangement of the five numbers that meets the requirement.
7
5
10
14
9
13
11 12
6
3
Difficulty: Medium-hard
NCTM Standard: Measurement Standard: Understand measurable attributes of objects and the units, systems, and processes
of measurement.
Mathworld.com Classification:
Calculus and Analysis > Series > General Series > Arithmetic Progression
13
-
On a standard die one of the dots is removed at random with each dot
equally likely to be chosen. The die is then rolled. What is the probability
that the top face has an odd number of dots?
(A)
5
11
(B)
10
21
(C)
1
2
(D)
11
21
(E)
6
11
2005 AMC 12 A, Problem #14— “What are the effects on each face?”
- Solution (D) A standard die has a total of 21 dots. For 1 ≤ n ≤ 6, a dot is removed from the face with
n dots with probability n/21. That face is left with an odd number of dots with probability n/21 if n is
even and 1 − n/21 if n is odd. Each face is the top face with probability 1/6. Therefore the top face has
an odd number of dots with probability
1
1
2
3
4
5
6
1
3
1−
+
+ 1−
+
+ 1−
+
=
3+
6
21
21
21
21
21
21
6
21
1 66
11
= ·
=
.
6 21
21
OR
The probability that the top face is odd is 1/3 if a dot is removed from an odd face, and the probability
that the top face is odd is 2/3 if a dot is removed from an even face. Because each dot has the probability
1/21 of being removed, the top face is odd with probability
1
1+3+5
2
2+4+6
33
11
+
=
=
.
3
21
3
21
63
21
Difficulty: Hard
NCTM Standard: Data Analysis and Probability Standard: Understand and apply basic concepts of probability
Mathworld.com Classification:
Probability and Statistics > Probability > Probability
14
-
Let S be the set of the 2005 smallest positive multiples of 4, and let T be
the set of the 2005 smallest positive multiples of 6. How many elements are
common to S and T ?
(A) 166
(B) 333
(C) 500
(D) 668
(E) 1001
2005 AMC 10 A, Problem #22— “What is the gcd of 4 and 6?”
- Solution (D) The sets S and T consist, respectively, of the positive multiples of 4 that doTnot exceed
2005 · 4 = 8020 and the positive multiples of 6 that do not exceed 2005 · 6 = 12,030. Thus S T consists
of the positive multiples of 12 that do not exceed 8020. Let ⌊x⌋
T represent the largest integer that is less
than or equal to x. Then the number of elements in the set S T is
8020
= 668.
12
Difficulty: Medium-hard
NCTM Standard: Number and Operations Standard: Understand numbers, ways of representing numbers, relationships among
numbers, and number systems.
Mathworld.com Classification: Foundations of Mathematics > Set Theory > Sets > Set
15
-
Three circles of radius s are drawn in the first quadrant of the xy-plane. The first circle is
tangent to both axes, the second is tangent to the first circle and the x-axis, and the third
is tangent to the first circle and the y-axis. A circle of radius r > s is tangent to both axes
and to the second and third circles. What is r/s?
r
s
(A) 5
(B) 6
(C) 8
(D) 9
(E) 10
2005 AMC 12 A, Problem #16— “Make the line between the centers of the
circle of radius s and the circle of radius r the hypotenuse of a right triangle.”
- Solution (D) Consider a right triangle as shown. By the Pythagorean Theorem,
(r + s)2 = (r − 3s)2 + (r − s)2
so
and
r2 + 2rs + s2 = r2 − 6rs + 9s2 + r2 − 2rs + s2
0 = r2 − 10rs + 9s2 = (r − 9s)(r − s).
But r 6= s, so r = 9s and r/s = 9.
r−s
r
r − 3s
r+s
3s
s
OR
Because the ratio r/s is independent of the value of s, assume that s = 1 and proceed as in the previous
solution.
Difficulty: Hard
NCTM Standard: Geometry Standard: Specify locations and describe spatial relationships using coordinate geometry and other
representational systems
Mathworld.com Classification:
Geometry > Curves > Plane Curves > Conic Sections > Circle
16
-
A unit cube is cut twice to form three triangular prisms, two of which are
congruent, as shown in Figure 1. The cube is then cut in the same manner
along the dashed lines shown in Figure 2. This creates nine pieces. What is
the volume of the piece that contains vertex W ?
W
Figure 1
(A)
1
12
(B)
Figure 2
1
9
(C)
1
8
(D)
1
6
(E)
1
4
2005 AMC 12 A, Problem #17— “What kind of polyhedra is it?”
- Solution (A) The piece that contains W is shown. It is a pyramid with vertices V, W, X, Y , and Z. Its
base W XY Z is a square with sides of length 1/2 and its altitude V W is 1. Hence the volume of this
pyramid is
2
1
1 1
(1) =
.
3 2
12
Y
W
Z
Y
X
Z
W
X
V
V
Difficulty: Hard
NCTM Standard: Geometry Standard: Use visualization, spatial reasoning, and geometric modeling to solve problems.
Mathworld.com Classification:
Geometry > Solid Geometry > Polyhedra > Pyramids > Pyramid
17
-
Call a number “prime-looking” if it is composite but not divisible by 2,3, or
5. The three smallest prime-looking numbers are 49, 77, and 91. There are
168 prime numbers less than 1000. How many prime-looking numbers are
there less than 1000?
(A) 100
(B) 102
(C) 104
(D) 106
(E) 108
2005 AMC 12 A, Problem #18— “How many numbers less than 1000 are
divisible by 2? By 3? By...?”
- Solution (A) Of the numbers less than 1000, 499 of them are divisible by two, 333 are divisible by 3, and
199 are divisible by 5. There are 166 multiples of 6, 99 multiples of 10, and 66 multiples of 15. And there
are 33 numbers that are divisible by 30. So by the Inclusion-Exclusion Principle there are
499 + 333 + 199 − 166 − 99 − 66 + 33 = 733
numbers that are divisible by at least one of 2,3, or 5. Of the remaining 999 − 733 = 266 numbers, 165
are primes other than 2, 3, or 5. Note that 1 is neither prime nor composite. This leaves exactly 100
prime-looking numbers.
Difficulty: Hard
NCTM Standard: Number and Operations Standard: Understand numbers, ways of representing numbers, relationships among
numbers, and number systems
Mathworld.com Classification:
Number Theory > Prime Numbers > Prime Number Properties > Prime Number
18
-
A faulty car odometer proceeds from digit 3 to digit 5 , always skipping
the digit 4, regardless of position. For example, after traveling one mile
the odometer changed from 000039 to 000050. If the odometer now reads
002005, how many miles has the car actually traveled?
(A) 1404
(B) 1462
(C) 1604
(D) 1605
(E) 1804
2005 AMC 12 A, Problem #19— “Base-9”
- Solution (B) Because the odometer uses only 9 digits, it records mileage in base-9 numerals, except that
its digits 5, 6, 7, 8, and 9 represent the base-9 digits 4, 5, 6, 7, and 8. Therefore the mileage is
2004base 9 = 2 · 93 + 4 = 2 · 729 + 4 = 1462.
OR
The number of miles traveled is the same as the number of integers between 1 and 2005, inclusive, that do
not contain the digit 4. First consider the integers less than 2000. There are two choices for the first digit,
including 0, and 9 choices for each of the other three. Because one combination of choices is 0000, there
are 2 · 93 − 1 = 1457 positive integers less than 2000 that do not contain the digit 4. There are 5 integers
between 2000 and 2005, inclusive, that do not have a 4 as a digit, so the car traveled 1457 + 5 = 1462
miles.
Difficulty: Medium-hard
NCTM Standard: Problem Solving Standard : Apply and adapt a variety of appropriate strategies to solve problems
Mathworld.com Classification:
Number Theory > Arithmetic > Number Bases > Digit
19
-
Let a, b, c, d, e, f, g, and h be distinct elements in the set
{−7, −5, −3, −2, 2, 4, 6, 13}.
What is the minimum possible value of
(a + b + c + d)2 + (e + f + g + h)2 ?
(A) 30
(B) 32
(C) 34
(D) 40
(E) 50
2005 AMC 12 A, Problem #20— “What is the sum of all the elements?”
- Solution (C) Note that the sum of the elements in the set is 8. Let x = a+b+c+d, so e+f +g +h = 8−x.
Then
(a + b + c + d)2 + (e + f + g + h)2 = x2 + (8 − x)2
= 2x2 − 16x + 64 = 2(x − 4)2 + 32 ≥ 32.
The value of 32 can be attained if and only if x = 4. However, it may be assumed without loss of generality
that a = 13, and no choice of b, c, and d gives a total of 4 for x. Thus (x − 4)2 ≥ 1, and
(a + b + c + d)2 + (e + f + g + h)2 = 2(x − 4)2 + 32 ≥ 34.
A total of 34 can be attained by letting a, b, c, and d be distinct elements in the set {−7, −5, 2, 13}.
Difficulty: Hard
NCTM Standard: Algebra Standard: Represent and analyze mathematical situations and structures using algebraic symbols
Mathworld.com Classification:
Foundations of Mathematics > Set Theory > Sets > Element
20
-
How many ordered triples of integers (a, b, c), with a ≥ 2, b ≥ 1, and c ≥ 0,
satisfy both loga b = c2005 and a + b + c = 2005?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
2005 AMC 12 A, Problem #21— “Solve for c.”
- Solution (C) The two equations are equivalent to b = a(c
c = 2005 − a(c
If c > 1, then
2005
b ≥ 2(2
)
2005
2005
)
)
and c = 2005 − b − a, so
− a.
> 2005 > 2005 − a − c = b,
which is a contradiction. For c = 0 and for c = 1, the only solutions are the ordered triples (2004, 1, 0) and
(1002, 1002, 1), respectively. Thus the number of solutions is 2.
Difficulty: Hard
NCTM Standard: Algebra Standard: Represent and analyze mathematical situations and structures using algebraic symbols
Mathworld.com Classification:
History and Terminology > Terminology > Triple
21
-
A rectangular box P is inscribed in a sphere of radius r. The surface area of
P is 384, and the sum of the lengths of its 12 edges is 112. What is r?
(A) 8
(B) 10
(C) 12
(D) 14
(E) 16
2005 AMC 12 A, Problem #22— “Find the radius in terms the surface area and
sum of the lengths.”
- Solution (B) Let the dimensions of P be x, y, and z. The sum of the lengths of the edges of P is
4(x + y + z), and the surface area of P is 2xy + 2yz + 2xz, so
x + y + z = 28 and 2xy + 2yz + 2xz = 384.
Each internal diagonal of P is a diameter of the sphere, so
(2r)2 = (x2 + y 2 + z 2 ) = (x + y + z)2 − (2xy + 2xz + 2yz) = 282 − 384 = 400.
So 2r = 20 and r = 10.
Note: There are infinitely many positive solutions of the system x + y + z = 28, 2xy + 2yz + 2xz = 384,
so there are infinitely many non-congruent boxes meeting the given conditions, but each can be inscribed
in a sphere of radius 10.
Difficulty: Hard
NCTM Standard: Geometry Standard: Use visualization, spatial reasoning, and geometric modeling to solve problems
Mathworld.com Classification:
Geometry > Line Geometry > Concurrence > Inscribed
22
-
Two distinct numbers a and b are chosen randomly from the set
{2, 22, 23, . . . , 225}. What is the probability that loga b is an integer?
(A)
2
25
(B)
31
300
(C)
13
100
(D)
7
50
(E)
1
2
2005 AMC 12 A, Problem #23— “Let a = 2j and b = 2k . . .”
- Solution (B) Let a = 2j and b = 2k . Then
loga b = log2j 2k =
log 2k
k log 2
k
=
= ,
j
log 2
j log 2
j
so loga b is an integer if andj only
k if k is an integer multiple of j. For each j, the number of integer jmultiples
k
25
.
Because
j
=
6
k,
the
number
of
possible
values
of
k
for
each
j
is
− 1.
of j that are at most 25 is 25
j
j
Hence the total number of ordered pairs (a, b) is
25 X
25
j=1
j
−1
= 24 + 11 + 7 + 5 + 4 + 3 + 2(2) + 4(1) = 62.
Since the total number of possibilities for a and b is 25 · 24, the probability that loga b is an integer is
62
31
=
.
25 · 24
300
Difficulty: Hard
NCTM Standard: Data Analysis and Probability Standard: Develop and evaluate inferences and predictions that are based on
data
Mathworld.com Classification:
Probability and Statistics > Probability > Probability
23
-
Let P (x) = (x−1)(x−2)(x−3). For how many polynomials Q(x) does there
exist a polynomial R(x) of degree 3 such that P (Q(x)) = P (x) · R(x) ?
(A) 19
(B) 22
(C) 24
(D) 27
(E) 32
2005 AMC 12 A, Problem #24— “What degree does P (x) · R(x) have?”
- Solution (B) The polynomial P (x) · R(x) has degree 6, so Q(x) must have degree 2. Therefore Q
is uniquely determined by the ordered triple (Q(1), Q(2), Q(3)). When x = 1, 2, or 3, we have 0 =
P (x) · R(x) = P (Q(x)). It follows that (Q(1), Q(2), Q(3)) is one of the 27 ordered triples (i, j, k), where
i, j, and k can be chosen from the set {1, 2, 3}. However, the choices (1, 1, 1), (2, 2, 2), (3, 3, 3), (1, 2, 3),
and (3, 2, 1) lead to polynomials Q(x) defined by Q(x) = 1, 2, 3, x, and 4 − x, respectively, all of which
have degree less than 2. The other 22 choices for (Q(1), Q(2), Q(3)) yield non-collinear points, so in each
case Q(x) is a quadratic polynomial.
Difficulty: Hard
NCTM Standard: Algebra Standard: Analyze change in various contexts
Mathworld.com Classification:
Algebra > Polynomials > Polynomial
24
-
Let S be the set of all points with coordinates (x, y, z), where x, y, and z
are each chosen from the set {0, 1, 2}. How many equilateral triangles have
all their vertices in S ?
(A) 72
(B) 76
(C) 80
(D) 84
(E) 88
2005 AMC 12 A, Problem #25— “Good Luck”
- Solution (C) Let A(x1 , y1 , z1 ), B(x2 , y2 , z2 ), and C(x3 , y3 , z3 ) be the vertices of such a triangle. Let
(∆xk , ∆yk , ∆zk ) = (xk+1 − xk , yk+1 − yk , zk+1 − zk ), for 1 ≤ k ≤ 3,
where (x4 , y4 , z4 ) = (x1 , y1 , z1 ). Then (|∆xk | , |∆yk | , |∆zk |) is a permutation of one of the ordered triples
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2), (0, 2, 2), (1, 1, 1), (1, 1, 2), (1, 2, 2), or (2, 2, 2). Since △ABC is
equilateral, AB, BC, and CA correspond to permutations of the same ordered triple (a, b, c). Because
3
X
∆xk =
k=1
the sums
3
X
k=1
|∆xk |,
3
X
∆yk =
k=1
3
X
k=1
|∆yk |,
3
X
∆zk = 0,
and
3
X
k=1
k=1
|∆zk |
are all even. Therefore (|∆xk | , |∆yk | , |∆zk |) is a permutation of one of the triples (0, 0, 2), (0, 1, 1),
(0, 2, 2), (1, 1, 2), or (2, 2, 2).
If (a, b, c) = (0, 0, 2), each side of △ABC is parallel to one of the coordinate axes, which is impossible.
If (a, b, c) = (2, 2, 2), each side of △ABC is an interior diagonal of the 2 × 2 × 2 cube that contains S,
which is also impossible.
If (a, b, c) = (0, 2, 2), each side of △ABC is a face diagonal of the 2 × 2 × 2 cube that contains S. The
three faces that join at any vertex determine such a triangle, so the triple (0, 2, 2) produces a total of 8
triangles.
If (a, b, c) = (0, 1, 1), each side of △ABC is a face diagonal of a unit cube within the larger cube that
contains S. There are 8 such unit cubes producing a total of 8 · 8 = 64 triangles.
There are two types of line segments for which (a, b, c) = (1, 1, 2). One type joins the center of the face
of the 2 × 2 × 2 cube to a vertex on the opposite face. The other type joins the midpoint of one edge
of the cube to the midpoint of another edge. Only the second type of segment can be a side of △ABC.
The midpoint of each of the 12 edges is a vertex of two suitable triangles, so there are 12 · 2/3 = 8 such
triangles.
The total number of triangles is 8 + 64 + 8 = 80.
Difficulty: Hard
NCTM Standard: Geometry Standard: Specify locations and describe spatial relationships using coordinate geometry and other
representational systems
Mathworld.com Classification:
Geometry > Coordinate Geometry > Cartesian Coordinates
25