amc 2015 10a
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AMC 10 năm 2015TRANSCRIPT
http://www.artofproblemsolving.com/wiki/index.php?title=AMC_10_Problems_and_Solutions
2015: AMC 10A
Problem 1
What is the value of
Problem 2
A box contains a collection of triangular and square tiles. There are tiles in the box, containing edges total. How many square tiles are there in the box?
Problem 3
Ann made a 3-step staircase using 18 toothpicks as shown in the figure. How many toothpicks does she need to add to complete a 5-step staircase?
Problem 4
Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?
Problem 5
Mr. Patrick teaches math to students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was . After he graded Payton's test, the test average became . What was Payton's score on the test?
Problem 6
The sum of two positive numbers is times their difference. What is the ratio of the larger number to the smaller number?
Problem 7
How many terms are there in the arithmetic sequence , , , . . ., , ?
Problem 8
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be : ?
Problem 9
Two right circular cylinders have the same volume. The radius of the second cylinder is more than the radius of the first. What is the relationship between the heights of the two cylinders?
Problem 10
How many rearrangements of are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either or .
Problem 11
The ratio of the length to the width of a rectangle is : . If the rectangle has diagonal of length , then the area may be expressed as for some constant . What is ?
Problem 12
Points and are distinct points on the graph of . What
is ?
Problem 13
Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
Problem 14
The diagram below shows the circular face of a clock with radius cm and a circular disk with radius cm externally tangent to the clock face at o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
Problem 15
Consider the set of all fractions where and are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by , the value of the fraction is increased by ?
Problem 16
If , and , what is the value of ?
Problem 17
A line that passes through the origin intersects both the line and the line . The three lines create an equilateral triangle. What is the perimeter of the triangle?
Problem 18
Hexadecimal (base-16) numbers are written using numeric digits through as well as the letters through to represent through . Among the first positive integers, there are whose hexadecimal representation contains only numeric digits. What is the sum of the digits of ?
Problem 19
The isosceles right triangle has right angle at and area . The rays trisecting intersect at and . What is the area of ?
Problem 20
A rectangle with positive integer side lengths in has area and perimeter . Which of the following numbers cannot equal ?
NOTE: As it originally appeared in the AMC 10, this problem was stated incorrectly and had no answer; it has been modified here to be solvable.
Problem 21
Tetrahedron has , , , , , and . What is the volume of the tetrahedron?
Problem 22
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
Problem 23
The zeroes of the function are integers. What is the sum of the possible values of ?
Problem 24
For some positive integers , there is a quadrilateral with positive integer side lengths, perimeter , right angles at and , , and . How many different values of are possible?
Problem 25
Let be a square of side length . Two points are chosen independently at random on the sides of
. The probability that the straight-line distance between the points is at least is , where , ,
and are positive integers with . What is ?
2015 AMC 10A Answer Key
1. C
2. D
3. D
4. B
5. E
6. B
7. B
8. B
9. D
10. C
11. C
12. C
13. C
14. C
15. B
16. B
17. D
18. E
19. D
20. B (Note: This problem was originally stated incorrectly, and all contestants received full credit regardless of their answer.)
21. C 22. A 23. C 24. B 25. A
Solution
Problem 1
.Problem 2
Let be the amount of triangular tiles and be the amount of square tiles.Triangles have edges and squares have edges, so we have a system of equations.We have tiles total, so .We have edges total, so .
Solving gives, and , so the answer is .
Alternate Solution
If all of the tiles were triangles, there would be edges. This is not enough, so there need to be some squares. Trading a triangle for a square results in one additional edge each time, so we must
trade out triangles for squares. Answer:
.Problem 3
We can see that a -step staircase requires toothpicks and a -step staircase requires toothpicks. Thus, to go from a -step to -step staircase, additional toothpicks are needed and to go from a -step to -step staircase, additional toothpicks are needed. Applying this pattern, to go from a -step to -step staircase, additional toothpicks are needed and to go from a -step to -step
staircase, additional toothpicks are needed. Our answer is
.Problem 4
Assign a variable to the number of eggs Mia has, say . Then, because we are given that Sofia has twice the number of eggs Mia has, Sofia has eggs, and Pablo, having three times the number of eggs as Sofia, has eggs.
For them to all have the same number of eggs, they must each have eggs. This means Pablo must give eggs to Mia and a eggs to Sofia, so the answer
is
.Problem 5
If the average of the first peoples' scores was , then the sum of all of their tests is . When Payton's score was added, the sum of all of the scores
became . So, Payton's score must be
Alternate Solution
The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first students each scored . If Payton also scored an , the average would
still be . In order to increase the overall average to , we need to add one more point to all of the scores, including Payton's. This means we need to add a total of more points, so Payton
needs
.Problem 6
Let be the bigger number and be the smaller.
.
Solving gives , so the answer is .
.Problem 7
Solution
, so the amount of terms in the sequence , , , , , is the same as in the sequence , , , , , .In this sequence, the terms are the multiples of going up to , and there are multiples of in .
However, one more must be added to include the first term. So, the answer is .
Solution 2
.
Solution 3
Using the formula for arithmetic sequence's nth term, we see
that .
.Problem 8
This problem can be converted to a system of equations. Let be Pete's current age and be Claire's current age.
The first statement can be written as . The second statement can be written
as To solve the system of equations:
Let be the number of years until Pete is twice as old as Claire.
The answer is .
.Problem 9
Let the radius of the first cylinder be and the radius of the second cylinder be . Also, let the height
of the first cylinder be and the height of the second cylinder be . We are told
Substituting the first equation into the second and dividing both sides by , we get
Therefore,
.Problem 10
The first thing one would want to do is see a possible value that works and then stem off of it. For example, if we start with an , we can only place a or next to it. Unfortunately, after that step, we can't do too much, since:
is not allowed because of the , and is not allowed because of the .We get the same problem if we start with a , since a will have to end up in the middle, causing it to be adjacent to an or .If we start with a , the next letter would have to be a , and since we can put an next to it and then a after that, this configuration works. The same approach applies if we start with a .So the solution must be the two solutions that were allowed, one starting from a and the other with a , giving us:
.Problem 11
Let the rectangle have length and width . Then by triangles (or the Pythagorean
Theorem), we have , and so . Hence, the area of the rectangle
is , so the answer is
.Problem 12
Since points on the graph make the equation true, substitute in to the equation and then solve to find and .
There are only two solutions to the equation, so one of them is the value of and the other is . The order does not matter because of the absolute value sign.
The answer is
.Problem 13
Solution #1
Let Claudia have 5-cent coins and 10-cent coins. It is easily observed that any multiple
of between and inclusive can be obtained by a combination of coins. Thus, combinations can be made, so . But the answer is not because
we are asked for the number of 10-cent coins, which is
Solution #2
Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of To have exactly different multiples of we will need to make up to cents. If all twelve coins were 5-cent coins, we will have cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain
cents, and as we need to gain cents, the answer is
.Problem 14
Solution 1
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk
made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4.
The answer is .
Solution 2
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 =
120°, which is 4 o' clock.
Solution 3
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is , so that is traveled on a arrow path. This is a ratio of 1/3, so the angle it carves is 120
degrees, which leads us to the correct answer of 4 o' clock.
.Problem 15
Solution 1
You can create the equation Cross multiplying and combining like terms gives .
This can be factored into . and must be positive, so and , so and .
This leaves the factor pairs: and But we can't stop here because and must be relatively prime.
gives and . and are not relatively prime, so this doesn't work.
gives and . This doesn't work.
gives and . This does work.
We found one valid solution so the answer is
Solution 2
The condition required is .
Observe that so is at most
By multiplying by and simplifying we can rewrite the condition as . Since and
are integer, this only has solutions for . However, only the first yields a that is relative prime to .
There is only one valid solution so the answer is
.Problem 16
Solution 1
Note that we can add the two equations to yield the equation
Moving terms gives the equation
We can also subtract the two equations to yield the equation
Moving terms gives the equation
Because we can divide both sides of the equation by to yield the equation
Substituting this into the equation for that we derived earlier gives
Solution 2 (Algebraic)
Subtract from the LHS of both equations, and use difference of squares to yield the equations
and .
It may save some time to find two solutions, and , at this point. However, in these solutions.
Substitute into .
This gives the equation
which can be simplified to
.
Knowing and are solutions now is helpful, as you divide both sides by . This can also be done using polynomial division to find as a factor. This gives
.
Because the two equations and are symmetric, the and values are
the roots of the equation, which are and .Squaring these and adding the together gives
.
.Problem 17
Solution 1
Since the triangle is equilateral and one of the sides is a vertical line, the other two sides will have
opposite slopes. The slope of the other given line is so the third must be . Since this third
line passes through the origin, its equation is simply . To find two vertices of the triangle, plug in to both the other equations.
We now have the coordinates of two vertices, and . The length of one
side is the distance between the y-coordinates, or .
The perimeter of the triangle is thus , so the answer is
Solution 2
Draw a line from the y-intercept of the equation perpendicular to the line x=1. There is a square of side length 1 inscribed in the equilateral triangle. The problems becomes reduced to
finding the perimeter of a equilateral triangle with a square of side length 1 inscribed in it. The side
length is 2 + 1. After multiplying the side length by 3 and rationalizing, you
get .
.Problem 18
Notice that is in hexadecimal. We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.The first digit could be or and the second two could be any digit , giving combinations. However, this includes so this number must be diminished by Therefore, there are valid corresponding to those positive integers less than that consist of only numeric digits. (Notice that in hexadecimal.) Finally, our
answer is
.Problem 19
can be split into a right triangle and a right triangle by dropping a perpendicular from to side . Let be where that perpendicular intersects .
Because the side lengths of a right triangle are in ratio , .
Because the side lengths of a right triangle are in ratio and
+ , .
Setting the two equations for equal to each other, .
Solving gives .
The area of . is congruent to , so their areas are equal.
A triangle's area can be written as the sum of the figures that make it up,
so .
.
Solving gives , so the answer is
Solution 2
The area of is , and so the leg length of is Thus, the altitude
to hypotenuse , , has length by right triangles. Now, it is clear that , and so by the Exterior Angle Theorem, is an
isosceles triangle. Thus, by the Half-Angle
formula, and so the area of is . The answer is
thus
.Problem 20
Let the rectangle's length and width be and . Its area is and the perimeter is .
Then . Factoring, this is .Looking at the answer choices, only cannot be written this way, because then either or would be .
So the answer is .
.Problem 21
Solution 1
Let the midpoint of be . We have , and so by the Pythagorean
Theorem and . Because the altitude from of tetrahedron passes touches plane on , it is also an altitude of triangle . The area of triangle is, by Heron's Formula, given by
Substituting and performing huge (but manageable) computations yield ,
so . Thus, if is the length of the altitude from of the
tetrahedron, . Our answer is thus
and so our answer is
Solution 2
Drop altitudes of triangle and triangle down from and , respectively. Both will hit the same point; let this point be . Because both triangle and triangle are 3-4-5
triangles, . Because
, it follows that the is a right triangle,
which means , which means that planes and are perpendicular to each other. Now, we can treat as the base of the tetrahedron and as the height. Thus, the
desired volume is which is answer
.Problem 22
Solution 1
We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by at the end. We casework on how many people are standing.
Case people are standing. This yields arrangement.
Case person is standing. This yields arrangements.
Case people are standing. This yields arrangements, because the two people cannot be next to each other.
Case people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding possible arrangements.
More difficult is:
Case people are standing. First, choose the location of the first person standing ( choices). Next, choose of the remaining people in the remaining legal seats to stand, amounting to arrangements considering that these two people cannot stand next to each other. However, we have
to divide by because there are ways to choose the first person given any three. This
yields arrangements for Case
Summing gives and so our probability is .
Solution 2
We will count how many valid standing arrangements there are counting rotations as distinct and divide by at the end. Line up all people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka ball and urn.
If there are standing, there are ways to place them. For there are ways. etc. Summing, we
get ways.
Now we consider that the far right person can be standing as well, so we
have ways
Together we have , and so our probability is .
Solution 3
We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only people. If they stand, we count the arrangements with instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with person there are two ways and with people there are three ways. Carrying out the Fibonacci recursion until we get to people, we find there are standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for people to
stand in a line, which is from our sequence. Therefore our probability is
.Problem 23
Solution 1
By Vieta's Formula, is the sum of the integral zeros of the function, and so is integral.Because the zeros are integral, the discriminant of the function, , is a perfect square, say .
Then adding 16 to both sides and completing the square yields
Hence and Let an
d ; then, and so . Listing all possible pairs (not counting transpositions because this does not affect
), , yields . These sum to , so our answer
is .
Solution 2
Let and be the integer zeroes of the quadratic.
Since the coefficent of the term is , the quadratic can be written as
or .By comparing this with , and .
Plugging the first equation in the second, . Rearranging gives .
This can be factored as .
These factors can be: .We want the number of distinct , and these factors gives .
So the answer is .
.Problem 24
Let and be positive integers. Drop a perpendicular from to to
show that, using the Pythagorean Theorem, that Simplifying
yields , so . Thus, is one more than a perfect square.
The perimeter must be less than 2015. Simple
calculations demonstrate that is valid, but is not. On the
lower side, does not work (because ), but does work. Hence, there are 31
valid (all such that for ), and so our answer is
.Problem 25
Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point be in the bottom-left segment. Then, it is easy to see that any point in the 5
segments not bordering the bottom-left segment will be distance at least apart from . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance
at least 0.5 apart from is because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)If the second point is on the left-bottom segment, then if is distance away from the left-bottom
vertex, then must be at least away from that same vertex. Thus, using an averaging argument we find that the probability in this case is
(Alternatively, one can equate the problem to finding all valid with such
that , i.e. is outside the unit circle with radius )
Thus, averaging the probabilities gives
Our answer is .
Solution 2
Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the
same side is , on an adjacent side is , and on the the opposite side is . We discuss these three cases.
Case 1: Two points are on the same side. Let the first point be and the second point be in the -
axis with . Consider a point on the unit square on the -plane.
The region has the area of . Therefore, the probability
that is .
Case 2: Two points are on two adjacent sides. Let the two sides be on the x-axis and on
the y-axis and let one point be and the other point be . Then and the
distance between the two points is . As in Case 1, is a point on the unit
square . The area of the region is and the area of its complementary set inside the square
(i.e. ) is . . Therefore, the probability that the
distance between and is at least is .Case 3: Two points are on two opposite sides. In this case, the probability that the distance between
the two points is at least is obviously .
Thus the probability that the probability that the distance between the two points is at least is
given by Therefore , , and . Thus, and the answer is (a).