From the bar graph,
students chose chocolate ice cream, students chose strawberry ice cream,
and students chose vanilla ice
cream.
The total number of students is , so the answer is .
Answer:
Since , is equilateral so .
Since is parallel to , .
Since is parallel to and is perpendicular to , is also perpendicular to , so .
Therefore, .
Answer:
Let the real number
satisfying .
From we get that .
From we get that .
This means .
We are given that , so or .
Therefore, .
Answer:
The terms being added and subtracted are the integers , , , and so on up to .
This means there are terms in
total, which is an odd number of terms.
We will now insert parentheses to group the terms in pairs starting with
the first two terms, then the second two terms, and so on.
Since the total number of terms is odd, will not be grouped with another
term: Each of
the parenthetical expressions is equal to , and the number of parenthetical
expressions is .
Therefore,
Answer:
Suppose for some
integer . Then is even and so must be even, which means is an integer.
Dividing both sides of by
gives , and
since is an integer,
this means is a perfect
square.
We are given that ,
which implies .
The perfect squares between and
are , , and , among which is the only multiple of , so or .
This value of satisfies , and .
We have shown that is the
only that satisfies the
conditions, so it must be the largest.
Answer:
Substituting into gives .
Substituting and into gives .
Since is a two-digit integer,
we must have , so
.
Since is an integer, this implies
.
We also know that is a two-digit
integer, so , which means
that , , or .
Using the equations from earlier, we have , so the larger
is, the larger is.
With , we get , , and , which are all two-digit
integers.
Therefore, the conditions are satisfied when , its largest possible value, which
means corresponds to the
largest possible value of .
Therefore, the answer is .
Answer:
We can rewrite the given expression as follows: The
integer is the
-digit integer with leading
digits followed by ’s.
Thus, the digits of the given integer, from left to right, are , , and followed by ’s, and the final four digits are , , , and .
Therefore, the sum of the digits of the integer is
Answer:
By rearranging, we get The
quantity is an integer if
is an integer, so the given
expression in is an integer
exactly when is an
integer.
Suppose that and consider
the functions , , and .
The graph of is a parabola
with a positive coefficient of ,
which means that there are finitely many integers with the property that (there could be no such
integers).
Since , we have shown
that there are only finitely many integers for which is false.
The graph of is a line that is
neither vertical nor horizontal since , so there must be infinitely
many integers for which .
Since infinitely many integers satisfy and only finitely many integers
fail , there must exist
an integer with the property that
.
This means there exists an integer for which but this
means
is not an integer. Therefore, we have shown that if , then there is an integer for which the expression in the problem
is not an integer, so we conclude that or .
We now have that the expression in the question is equivalent to which is an
integer exactly when is an integer.
If is positive, then can be chosen so that is larger than , which would mean that the positive
expression is
not an integer.
Similarly, if is negative, then
there are integers for which
is not an
integer.
This shows that if , then
there are integers for which the
expression in the question is not an integer.
We can conclude that , or
.
Therefore, we have that and
, so the expression given in the
problem is equivalent to
Thus, the answer to the question is .
Answer:
We will use the fact that if two circles are externally tangent,
then the line connecting their centres passes through the point of
tangency.
The side lengths of
are ,
,
and .
Notice that and so the sides of satisfy . By the
converse of the Pythagorean theorem, has .
Therefore, the area of is .
Since , the
unshaded sector that lies inside the triangle has area
of the area of the circle centred at .
The area of the circle centred at
is .
The area of the shaded region is therefore
which means , , and , so .
Answer:
We will count the integers with the desired property by
considering four cases, one for each possible leading digit.
The leading digit is : In
this situation, we are looking for integers of the form where is an integer from to inclusive so that is a multiple of .
For to be a multiple of
, there must be some integer so that , which can be rearranged to get
.
Thus, will have the desired
property exactly when is a
positive factor of that is
less than .
The positive factors of that
are less than are
There are integers in the list
above, so we get integers in
this case.
The leading digit is :
Similar to case , we need to find
all integers for which and is a multiple of .
The latter condition means there is an integer with the property that which can be rearranged to get
.
The positive factors of that
are less than are
of which there are . Therefore,
there are integers in this
case.
The leading digit is :
As in cases and , we need to find all integers for which and is a multiple of .
Again, this means we need to count the positive factors of that are less than .
If is a factor of , then it is either a multiple of
or it is not.
If it is not a multiple of , then
it is one of the factors of
from case .
Otherwise, it is times one of the
factors of from case .
Among the factors from case , all
except remain less than when multiplied by .
Therefore, there are
integers in this case.
The leading digit is : In
this case, we need to count the positive factors of that are smaller than .
The prime factorization of is
.
Each factor of is either a
power of , times a power of , times a power of , or times a power of .
Each of , , , , , and is less than , so of the factors are powers of .
Multiplying each of the powers of
in the previous line by gives
, , , , , and , each of which is less than .
Multiplying by each of the
factors in the previous line gives , , , , , and , each of which is less than .
Multiplying by each of the
factors in the previous line gives , , , , , , three of which are less than .
This gives a total of
factors of that are less than
, so we get integers in this case.
Thus, the number of integers with the desired property is .
Answer:
Substituting into gives so .
Substituting into gives .
Answer:
The smallest two positive two-digit multiples of are and .
Since is a multiple of and is not, the answer is .
Answer:
Since the first digit of a four-digit integer cannot be , there are three integers with the
given property: , , and .
The largest is and the
smallest is , so the answer is
.
Answer:
Let denote the number of
people who answered “yes” and let
denote the number of people who answered “no”.
The number of people is , so . As well, since of respondents said “yes”, we have
that .
The equation implies
which can be
rearranged to .
Dividing both sides by gives
.
Since and are numbers of people, they are
non-negative integers. Furthermore, since a non-zero percentage of
people responded with each of “yes” and “no”, they must be positive
integers.
Therefore, the equation
implies that is a positive
multiple of since and do not have any positive factors in
common other than .
The smallest that can be is , and if , then .
If and , then , which means and
.
Therefore, and satisfy the conditions, and we have
already argued that cannot be any
smaller.
If is greater than , then is greater than since must be a positive multiple of . Therefore, the answer is .
Answer:
In the diagram below, there are squares shaded in such a way that no
two shaded squares share an edge, which shows that the answer is at
least .
We will now argue that it is impossible to shade more than square in such a way that no two shaded
squares share an edge, which will show that the answer to the question
is .
If four or five squares in the bottom row are shaded, then at least two
shaded squares will share an edge.
Therefore, at most squares can be
shaded in the bottom row.
If all three squares in the middle row are shaded, then there will be
shaded squares that share an edge.
Therefore, at most squares can be
shaded in the middle row.
There is only one square in the top row, so at most square can be shaded in the top
row.
Thus, the number of squares that can be shaded so that no two shaded
squares share an edge is at most .
Answer:
Multiplying both sides of the equation by gives , so .
Answer:
An isosceles triangle must have at least two sides of equal
length.
If an isosceles triangle has at least one side of length and at least one side of length , then its three side lengths are
either , , and or , , and .
In any triangle, the sum of the lengths of any two sides must be larger
than the length of the other side.
Since , a triangle
cannot have sides of length ,
, and .
Therefore, the triangle’s sides have lengths , , and , so the answer is .
Answer:
The area of interest is a triangle with vertices at the origin,
the -intercept of , and the point of intersection
of the lines with equations
and .
Setting in gives which implies .
Therefore, the base of the triangle has length .
Setting leads to , and substituting this
into gives , which is the height
of the triangle.
Using that the area of the triangle is , we have that or .
Dividing both sides of this equation by gives , which can be
rearranged to get , which
implies that .
Answer:
The side length of a cube with volume is .
Since and , the side length of the original
cube is and the side
length of the smaller cube is .
The new figure has nine faces: three
squares, three squares, and three faces obtained by removing a
square from the corner of a
square.
Thus, the surface area of the new figure in is [Note
that this is exactly the surface area of the original cube. Can you see
a simple reason why this is true?]
The shaded region is a square, so the ratio we seek is or .
Therefore, .
Answer:
The expression is undefined
when . For every integer , the expression is equal to .
Thus, if has the property that
is also
an integer, then must be an
integer different from with the
property that is a factor of
.
The factors of are The following table
has these values of in the
first column, the corresponding values of in the second column, and the
corresponding values of in the third
column.
Other than when , we get an
integer value of .
The sum of the integer values of is .
Answer:
Let be the amount in
dollars initially invested in Account A, let be the amount in dollars initially
invested in Account B, and let be
the amount in dollars initially invested in Account C.
The given information leads to the equations , , and .
The second and third equations can be solved for and , respectively, to get and .
Substituting into
the equation , we
get .
Substituting and into the equation gives
Therefore, the amount in dollars initially invested in Account C is
.
Answer:
Solution 1
The original line contains the points and .
Translating these points units up
gives the points and .
Translating these points units to
the left gives and .
Reflecting these points in the line gives and .
The slope of the line through these two points is .
Substituting and into gives or .
Therefore, the equation of the resulting line is , so the answer is .
Solution 2
The equation of the line obtained by translating up by units is or .
The equation of the line obtained by translating to the left by units is or .
To find the equation of the line obtained by reflecting the line in
, we interchange and to get , then solve for to get . Therefore, the answer
is .
Answer:
The sum of the integers in the list is The number of integers in the list is
The
average of the integers in the list is , so this means
The above equation is equivalent to , or , which can be solved for
to get .
Answer:
Suppose Devi was given the bill for Bohan’s meal. From the second
bullet point, this would mean that Bohan and Devi were given each
other’s bills.
This means that Ann’s, Che’s, and Eden’s bills were distributed among
each-other in some way.
Che was given Ann’s bill by the fourth bullet point, so if Ann was given
Che’s bill, then Eden would have been given her own bill.
Since nobody was given their own bill, Ann must have been given Eden’s
bill, so the only bill remaining for Eden to have been given is
Che’s.
Thus, assuming that Devi was given Bohan’s bill, we have deduced that
Ann was given Eden’s bill, Eden was given Che’s bill, and Che was given
Ann’s bill.
By the first bullet point, Che and Eden ordered meals with the same
cost.
Since Eden was given Che’s bill, the third bullet point is
violated.
This means that Devi was not given Bohan’s bill.
Because Che was given Ann’s bill, Devi was not given Ann’s bill.
As well, Devi could not have been given Che’s bill because this would
imply, by the second bullet point, that Che was given Devi’s bill, which
would violate the fourth bullet point.
By eliminating all other possibilities, we have deduced that Devi’s
and Eden’s bills were interchanged.
This means that the remaining three bills, belonging to Ann, Bohan, and
Che, were distributed among the same three people.
Similar to the reasoning from before, since nobody was given their own
bill and Che was given Ann’s bill, the only possibility is that Bohan
was given Che’s bill and Ann was given Bohan’s bill.
Answer: Ann
Solution 1
The line segment has slope
.
Since , the
slope of is the negative
reciprocal of , which
is .
The line segment has slope , so
.
This implies , so or .
Solution 2
In any right-angled triangle, the three vertices are equidistant from
the midpoint of the hypotenuse.
The right angle is at , so this
means that the hypotenuse of is .
The coordinates of are and the coordinates of are , so the coordinates of the midpoint
of is at .
The length of is The
distance from to the midpoint of
is half of , which leads to the
equations It can be checked that if the
coordinates of are , that the side lengths of are , , and , which are indeed the lengths
of the sides of a right-angled triangle.
Answer:
Using that is a solution
to we have that or .
This means is , the roots of which
are and , which means .
The polynomial factors
as , so upon expanding
the latter expression, we have .
Therefore, .
Answer:
For positive , A similar calculation shows that Therefore, so .
Answer:
If , then . This means we have .
The expression on the left can be factored to get .
Therefore, the given equation is true exactly when or .
Since , the only value of with is .
The only values of with
are and .
Therefore, the answer is .
Answer:
The distance from point to
each of points , , and is .
Since is a square of side
length , the Pythagorean theorem
implies that the distance from to
is .
A similar calculation shows that the distance from to is and the distance from to is .
To compute the distance from to
, observe that is right-angled at with , so the distance from to is .
The distance from to is which is greater than .
Notice that is the closest point
to among the points , , , , , , , and , so none of these points can be a
distance of from .
Therefore, the points that are at a distance of from are , , and , for a total of .
By symmetry, there will be a total of points at a distance of from each of , , , , , , and .
Next consider point . By reasoning
similar to that which is above, each of , , , and has a distance of at least from .
Applications of the Pythagorean theorem can be used to show that the
points that are away from
are , , , , and , for a total of points.
By symmetry, there are a total of
points at a distance of
from each of , , , , , , and .
We have counted points at a
distance of from each of
the points through , and points at a distance of from each of the points from through .
This is a total of .
The lengths of the line segments were counted once for each end point,
which means the total of counts
every distance of
twice.
Therefore, the answer is .
Answer:
Label the centre of the circle by , the point of tangency of with the circle by , and the point of intersection of with by :
Let . Then , so , which means because is a rectangle.
Using circle properties, is
perpendicular to , and since
is parallel to , we also have that , and hence, , is perpendicular to .
We now have that is
right-angled at , and is also right-angled at
.
Since and are radii of the same circle, they
must be equal.
Triangles and also share side , so they are congruent by
hypotenuse-side similarity.
Therefore, is the midpoint of
, so .
It is given that the radius of the circle is , which implies that and .
By the Pythagorean theorem, , and after substituting
the lengths computed earlier, this leads to the following equivalent
equations: this implies that either
or .
We cannot have since is the side-length of a rectangle.
Therefore, or .
The area of is .
Answer:
Solution 1
Suppose has side length
and let be on so that is perpendicular to . Set .
When the paper is folded,
lands on and does not move.
This means that . As well, by construction.
Since and have two angles in common,
they must have three angles in common.
They also share side , so they
must be congruent.
The area of is the sum of the
areas of and .
By symmetry, is
congruent to , which
means the area of is The length of is since it is a diagonal of a
square with side length .
Therefore, the area of is
.
The area of the shaded region is twice the area of , and since is congruent to , we have , so the shaded region has area
Therefore, the fraction we seek is .
Solution 2
Let the side length of the square be , set , and define as in Solution 1.
Using the argument from Solution 1, we get that as well.
Since is the diagonal of a
square, is isosceles
with , which
means .
Since by
construction, is an
isosceles right triangle, so .
We also have that and
, so this means .
Solving for , we get .
Using the argument from Solution 1, , , , and are all congruent.
This means the area of the shaded region is equal to twice the area of
, which is The area of is the area of the square minus the
combined area of and
.
This means the area of is
minus twice the area of , or .
Therefore, the ratio is
Answer:
Factoring , we
have or .
Therefore, the integers , , , and satisfy or .
Because is a prime number and
and are integers, it must be that is a factor pair of .
Observe that , so
must be equal to the sum of the
integers in a factor pair of .
Since is a prime number, there
are only two ways to express as
the product of two integers, and they are and .
The sum of the factors in these products are and , respectively.
Therefore, , so .
Answer:
We can categorize the
small cubes into four groups:
cubes that are completely in the interior of the larger cube (and are
completely invisible),
“face cubes” that have exactly one of their faces showing, “edge cubes” that have
exactly two faces showing, and
“corner cubes” that have faces
showing.
Notice that , so this
indeed accounts for all small
cubes.
The diagram below has all visible faces of the smaller cubes labelled so
that faces of edge cubes are labelled with an
E, faces of face cubes are labelled with
an F, and faces of corner cubes are
labelled with a C:
Each of the corner cubes has
of its faces exposed.
Because of the way the faces are labelled on the small cubes, each of
the corner cubes will always
contribute a total of to
the sum of the numbers on the outside.
Regardless of how the small cubes are arranged, the corner cubes
contribute a total of
to the total of all numbers on the outside of the larger cube.
The edge cubes can each show either a and a or a and a , for a total of either or .
Thus, each edge cube contributes a total of either or to the total on the outside of the
larger cube.
This means the smallest possible total that the edge cubes contribute is
, and the largest
possible total that they contribute is .
Notice that it is possible to arrange the edge cubes to show any total
from to inclusive.
This is because if we want the total to be where is any integer from to inclusive, we simply need to arrange
exactly of the edge cubes to show
a total of .
Each of the face cubes shows a total of or a total of . There are of them, so the total showing on the
face cubes is at least and at
most .
Similar to the reasoning for the edge cubes, every integer from to inclusive is a possibility for the
total showing on the face cubes.
Therefore, the smallest possible total is , and the largest possible
total is .
By earlier reasoning, every integer between and inclusive is a possibility, so the
number of possibilities is .
Answer:
The sum of the positive integers from to inclusive is . Since each row sum must be the same,
this means each row must be one third of which is .
In the second row, the first two cells contain the integers and , so the third must contain .
The integers that still need to be placed are , , , , , and .
Notice that and that and are the only two integers in this list
that have a sum of .
Therefore, the integers , , and must be in one of the remaining rows
and the integers , , and must be in the other.
In fact, if the integers are placed in such a way that all conditions
are satisfied and the top and bottom rows are exchanged, all properties
will still be satisfied and the column products will not have
changed.
Therefore, we can assume that ,
, and are in the top row.
This gives six possible ways to fill in the top two rows. They are shown
below.
The digits that remain to be placed are , , and .
We will first focus on the first partially filled in grid in which the
first row contains , , and from left to right.
The third column already contains
and . The integer to be placed in
the third column in the bottom row is at least , so this means the third column product
is at least .
By similar reasoning, the first column product is at most .
Therefore, no matter how the remaining integers are placed, the
difference between the largest and smallest column product will be at
least , which is greater
than .
Therefore, the , , and cannot be placed in that order.
In the second grid, the second column product is at least and the first
column product is at most , so the difference between the largest and smallest column
product is at least
which is greater than , so the
, , and cannot be placed in the order , , , either.
By similar reasoning, the third and fourth possibilities above can be
eliminated.
Therefore, the top row must contain, from left to right, either , , and , or , , and .
If the is placed in the third
column of the grid below,
then the third column product will be .
This would mean that the largest the first column product can be is
.
This would lead to the largest and smallest column products having a
difference greater than , so
cannot be placed in the third
column.
If is placed in the third column,
the difference between the largest and smallest column products would be
even larger, so the must be
placed in the third column.
By similar reasoning, the must be
placed in the second column of the last grid.
We have now reduced to four possible ways to place the integers, and
they are shown below with the column products given below each
column.
The column products in the first grid are , , and . Since and differ by more than , this grid does not satisfy the
conditions.
The column products in the second grid are , , and .
The difference between the largest and smallest column product is which is less than , so this grid satisfies all of the
conditions.
Its largest column product is ,
so is one of the
possibilities.
The column products in the third grid are , , and . Since and differ by more than , this grid does not satisfy the
conditions.
The column products in the fourth grid are , , and .
The difference between the largest and smallest column product is which is less than , so this grid satisfies all of the
conditions.
Its largest column product is ,
so is another possibility.
We have shown that there are exactly two ways to arrange the integers so
that the conditions are satisfied, and the largest column products in
these two arrangements are and
.
Answer: ,
The solution will use two facts. The first is a general fact
about triangles.
Fact 1: Suppose has on and on so that and . If is the point of intersection of and and is the point where the extension of
intersects , then .
The second fact that we will use is specific to the question.
Fact 2: The point is on the line segment connecting to so that .
(Note: this justifies the fact that and intersect as implied by the problem
statement.)
We will answer the question before proving the two facts.
To start, we will show that the altitude of tetrahedron is times that of .
Let be the point where the line
through intersects and let be the point in so that is perpendicular to , and let be the point on so that is perpendicular to .
By construction, is the altitude
of from to and is the altitude of from to .
By Fact 2, is on so that and is on so that by assumption.
By Fact 1, .
Since and share an angle at and both have a right angle, the
triangles are similar.
This means .
Thus, the length of the altitude of is five times that of .
Since is the midpoint of , the area of is half that of .
Thus, the base area of tetrahedron is half that of tetrahedron , while the heights are in ratio
, which means that the volumes
are in the ratio .
Now for the proofs of the facts.
Proof of Fact 1
We will first show that the height of from is one fifth the height of from .
Suppose has height
from and has height from . The first goal is to show that .
The area of is and the area of is .
Thus, if we take the ratio of their areas we get
and so the ratio of their heights is the same as the ratio of their
areas.
This means we can show that by showing
that the area of is
times the area of .
To do this, draw a line from to
and label the areas of , , , , and by , , , and , respectively.
We will now make several observations about the areas through .
Since and have the same altitude from
to and their bases satisfy , we get that By similar reasoning, we
also get the equations Substituting Equation (1) into
Equation (3) gives which simplifies to
.
Substituting Equation (2) into Equation (4) gives which simplifies
to .
Adding and gives .
Adding to both sides gives
, which
exactly says that the area of is five times that of .
We will now let be the point on
so that is perpendicular to and let be on so that is perpendicular to .
Note that is the height of
from and is the height of from , so we know that .
Since and , is similar to .
Since ,
we get that , which proves
the fact.
Proof of Fact 2
Consider and . It is given that ,
and since the triangles have a common angle at , they are similar by side-angle-side
similarity.
Let be the point at which intersects .
Since is similar
to , .
We also have that since they are the same angle, which implies that is similar to since if two triangles have
two angles in common, they must have three angles in common.
Since ,
similarity implies .
By similar reasoning, .
Rearranging these two equations leads to and , but is the midpoint of , so , and hence, .
This means is the midpoint of
, so , which shows that is on since is on .
Finally, since and
are similar, that
,
which means that is on so that .
Answer:
(Note: Where possible, the solutions to parts (b) and (c) of each
Relay are written as if the value of is not initially known, and then is substituted at the end.)