the numbers on the two top faces are added. What is
the probability that the sum is 7 ?
(A)19(B)18(C)16(D)211(E)15
2009 AMC 10 A, Problem #22—“Theprocessofrandomlypickingthenumbers,randomly axing them to the dice, rolling the dice, andadding the top numbers is equivalent to picking two ofthe twelve numbers at random and adding them.”
Solution
Answer (D):
Suppose that the two dice originally had the numbers 1, 2,
3, 4, 5, 6 and
1
0
,
2
0
,
3
0
,
4
0
,
5
0
,
6
0
, respectively. The process of randomly
picking the numbers, randomly affixing them to the dice, rolling the dice,
and adding the top numbers is equivalent to picking two of the twelve
numbers at random and adding them.
There are
(
12
2
)
= 66
sets of two
elements taken from
S
=
{
1
,
1
0
,
2
,
2
0
,
3
,
3
0
,
4
,
4
0
,
5
,
5
0
,
6
,
6
0
}
.
There are 4
ways to use a 1 and 6 to obtain 7, namely,
{
1
,
6
}
,
{
1
,
6
0
}
,
{
1
0
,
6
}
, and
{
1
0
,
6
0
}
. Similarly there are 4 ways to obtain the sum of 7 using a 2 and 5,
and 4 ways using a 3 and 4. Hence there are 12 pairs taken from
S
whose
sum is 7. Therefore the requested probability is
12
66
=
2
11
.
Difficulty:
Hard
NCTM Standard:
Data Analysis and Probability Standard: understand and apply basic concepts
of probability.
Mathworld.com Classification:
Probability and Statistics
>
Probability
>
Probability
Convex quadrilateral
ABCD
has
AB
= 9
and
CD
=
12
. Diagonals
AC
and
BD
intersect at
E
,
AC
= 14
,
and
4
AED
and
4
BEC
have equal areas. What is
AE
?
2009 AMC 10 A, Problem #23—2009 AMC 12 A, Problem #20—“ABkCD.”
Solution
Answer (E):
Because
4
AED
and
4
BEC
have equal areas, so do
4
ACD
and
4
BCD
. Side
CD
is common to
4
ACD
and
4
BCD
, so
the altitudes from
A
and
B
to
CD
have the same length. Thus
AB
k
CD
,
so
4
ABE
is similar to
4
CDE
with similarity ratio
AE
EC
=
AB
CD
=
9
12
=
3
4
.
Let
AE
= 3
x
and
EC
= 4
x
.
Then
7
x
=
AE
+
EC
=
AC
= 14
, so
x
= 2
, and
AE
= 3
x
= 6
.
A
B
C
D
E
9
12
Difficulty:
Hard
NCTM Standard:
Geometry Standard: explore relationships (including congruence and similarity)
among classes of two- and three-dimensional geometric objects, make and test conjectures about
them, and solve problems involving them.
Mathworld.com Classification:
Geometry
>
Plane Geometry
>
Quadrilaterals
>
Quadrilateral
Three distinct vertices of a cube are chosen at random.
What is the probability that the plane determined by
these three vertices contains points inside the cube?
2009 AMC 10 A, Problem #24—“A plane that intersects at least three vertices of acube either cuts into the cube or is coplanar with acube face.”
Solution
Answer (C):
A plane that intersects at least three vertices of a cube either
cuts into the cube or is coplanar with a cube face.
Therefore the three
randomly chosen vertices result in a plane that does not contain points
inside the cube if and only if the three vertices come from the same face of
the cube. There are 6 cube faces, so the number of ways to choose three
vertices on the same cube face is
6
·
(
4
3
)
= 24
. The total number of ways
to choose the distinct vertices without restriction is
(
8
3
)
= 56
. Hence the
probability is
1
-
24
56
=
4
7
.
Difficulty:
Hard
NCTM Standard:
Geometry Standard: analyze properties and determine attributes of two- and
three-dimensional objects.
Mathworld.com Classification:
Geometry
>
Solid Geometry
>
Polyhedra
>
Cubes
>
Cube
For
k >
0
, let
I
k
= 10
. . .
064
, where there are
k
zeros
between the 1 and the 6. Let
N
(
k
)
be the number of
factors of 2 in the prime factorization of
I
k
. What is
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