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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).