2017 AIME II Problems
Find the number of subsets of
that are subsets of neither
nor
.
The number of subsets of a set with
elements is
. The total number of subsets of
is equal to
. The number of sets that are subsets of at least one of
or
can be found using complimentary counting. There are
subsets of
and
subsets of
. It is easy to make the mistake of assuming there are
sets that are subsets of at least one of
or
, but the
subsets of
are overcounted. There are
sets that are subsets of at least one of
or
, so there are
subsets of
that are subsets of neither
nor
.
.
Theams
,
,
, and
are in the playoffs. In the semifinal matches,
plays
, and
plays
. The winners of those two matches will play each other in the final match to determine the champion. When
plays
, the probability that
wins is
, and the outcomes of all the matches are independent. The probability that
will be the champion is
, where
and
are relatively prime positive integers. Find
.
There are two scenarios in which
wins. The first scenario is where
beats
,
beats
, and
beats
, and the second scenario is where
beats
,
beats
, and
beats
. Consider the first scenario. The probability
beats
is
, the probability
beats
is
, and the probability
beats
is
. Therefore the first scenario happens with probability
. Consider the second scenario. The probability
beats
is
, the probability
beats
is
, and the probability
beats
is
. Therefore the second scenario happens with probability
. By summing these two probabilities, the probability that
wins is
. Because this expression is equal to
, the answer is
.
A triangle has vertices
,
, and
. The probability that a randomly chosen point inside the triangle is closer to vertex
than to either vertex
or vertex
can be written as
, where
and
are relatively prime positive integers. Find
.
[asy]
pair A,B,C,D,X,Z,P;
A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4);
fill(B--X--P--Z--cycle,lightgray);
draw(A--B--C--cycle);
dot(A);
label("
",A,SW);
dot(B);
label("
",B,SE);
dot(C);
label("
",C,N);
draw(X--P,dashed);
draw(Z--P,dashed);
dot(X);
label("
",X,NE);
dot(Z);
label("
",Z,S);
dot(P);
label("
",P,NW);
[/asy]
A diagram is above. The set of all points closer to point
than to point
lie to the right of the perpendicular bisector of
(line
in the diagram), and the set of all points closer to point
than to point
lie below the perpendicular bisector of
(line
in the diagram). Therefore, the set of points inside the triangle that are closer to
than to either vertex
or vertex
is bounded by quadrilateral
. Because
is the midpoint of
and
is the midpoint of
,
and
. The coordinates of point
is the solution to the system of equations defined by lines
and
. Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope
is
, the equation for line
is
and the equation for line
is
. The solution of this system is
. Using the shoelace formula on quadrilateral
and triangle
, the area of quadrilateral
is
and the area of triangle
is
. Finally, the probability that a randomly chosen point inside the triangle is closer to vertex
than to vertex
or vertex
is the ratio of the area of quadrilateral
to the area of
, which is
. The answer is
.
Find the number of positive integers less than or equal to
whose base-three representation contains no digit equal to
.
The base-
representation of
is
. Because any
-digit base-
number that starts with
and has no digit equal to
must be greater than
, all
-digit numbers that have no digit equal to
must start with
or
in base
. Of the base-
numbers that have no digit equal to
, there are
-digit numbers that start with
,
-digit numbers that start with
,
-digit numbers,
-digit numbers,
-digit numbers,
-digit numbers,
-digit numbers, and
-digit numbers. Summing these up, the answer is
.
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are
,
,
,
,
, and
. Find the greatest possible value of
.
Let these four numbers be
,
,
, and
, where
.
needs to be maximized, so let
and
because these are the two largest pairwise sums. Now
needs to be maximized. Notice
. No matter how the numbers
,
,
, and
are assigned to the values
,
,
, and
, the sum
will always be
. Therefore we need to maximize
. The maximum value of
is achieved when we let
and
be
and
because these are the two largest pairwise sums besides
and
. Therefore, the maximum possible value of
.
Find the sum of all positive integers
such that
is an integer.
Manipulating the given expression,
. The expression under the radical must be an square number for the entire expression to be an integer, so
. Rearranging,
. By difference of squares,
. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately,
is found to be
and
. The two values of
that satisfy one of the equations are
and
. Summing these together, the answer is
.
Find the number of integer values of
in the closed interval
for which the equation
has exactly one real solution.
Find the number of positive integers
less than
such that
is an integer.
A special deck of cards contains
cards, each labeled with a number from
to
and colored with one of seven solors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one cardf with each number, the probability that Sharon can discard one of her cards and
have at least one card of each color and at least one card with each number if
, where
and
are relatively prime positive integers. Find
.
Rectangle
has side lengths
and
. Point
is the midpoint of
, point
is the trisection point of
closer to
, and point
is the intersection of
and
. Point
lies on the quadrilateral
, and
bisects the area of
. Find the area of
.
Five towns are connected by a system of raods. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).
Circle
has radius
, and the point
is a point on the circle. Circle
has radius
and is internally tangent to
at point
. Point
lies on circle
so that
is located
counterclockwise from
on
. Circle
has radius
and is internally tangent to
at point
. In this way a sequence of circles
and a sequence of points on the circles
are constructed, where circle
has radius
and is internally tangent to circle
at point
, and point
lies on
counterclockwise from point
, as shown in the figure below. There is one point
inside all of these circles. When
, the distance from the center
to
is
, where
and
are relatively prime positive integers. Find
.
[asy]
draw(Circle((0,0),125));
draw(Circle((25,0),100));
draw(Circle((25,20),80));
draw(Circle((9,20),64));
dot((125,0));
label("
",(125,0),E);
dot((25,100));
label("
",(25,100),SE);
dot((-55,20));
label("
",(-55,20),E);
[/asy]
For each integer
, let
be the number of
-element subsets of the vertices of the regular
-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of
such that
.
A
grid of points consists of all points in space of the form
, where
,
, and
are integers between
and
, inclusive. Find the number of different lines that contain exactly
of these points.
Tetrahedron
has
,
, and
. For any point
in space, define
. The least possible value of
can be expressed as
, where
and
are positive integers, and
is not divisible by the square of any prime. Find
.