美国中学数学竞赛(真题加详解)2019 AMC-12B

资源大全更新时间:2025-05-01 14:11:47

2024年4月11日发(作者：高三模拟考数学试卷)

2019 AMC 10B Problems/Problem 1

The following problem is from both the 2019 AMC 10B #1 and 2019 AMC 12B #1, so both problems redirect to this page.

Contents

1

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Problem

Solution

Solution 2

See Also

Problem

Alicia had two containers. The ﬁrst was full of water and the second was empty. She poured all the water from the ﬁrst container

into the second container, at which point the second container was full of water. What is the ratio of the volume of the ﬁrst

container to the volume of the second container?

Solution

Let the ﬁrst jar\'s volume be and the second\'s be . It is given that . We ﬁnd that

We already know that this is the ratio of smaller to larger volume because it is less than

--mguempel

Solution 2

We can set up a ratio to solve this problem. If x is the volume of the ﬁrst container, and y is the volume of the second container,

then:

Cross Multiplying allows us to get . Thus the ratio of the volume of the ﬁrst container to the

second container is

An alternate solution is to plug in some maximum volume for the ﬁrst container - let\'s say

ﬁrst container, and then the second container also has a volume of

iron

, so you get

, so there was a volume of 60 in the

. Thus, .

See Also

2019 AMC 10B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=43&year=2019))

Preceded by

First Problem

Followed by

Problem 2

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 10 Problems and Solutions

2019 AMC 10B Problems/Problem 2

The following problem is from both the 2019 AMC 10B #2 and 2019 AMC 12B #2, so both problems redirect to this page.

Problem

Consider the statement, \"If is not prime, then

statement?

is prime.\" Which of the following values of is a counterexample to this

Solution

Since a counterexample must be when n is not prime, n must be composite, so we eliminate A and C. Now we subtract 2 from the

remaining answer choices, and we see that the only time

iron

minor edit (the inclusion of not) by AlcBoy1729

is prime is when , which is .

See Also

2019 AMC 10B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=43&year=2019))

Preceded by

Problem 1

Followed by

Problem 3

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 1

Followed by

Problem 3

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America ()\'s American

Mathematics Competitions ().

Retrieved from \"/wiki/?title=2019_AMC_10B_Problems/Problem_2&oldid=102779\"

2019 AMC 12B Problems/Problem 3

Problem

Which of the following rigid transformations (isometries) maps the line segment

image of is and the image of is

reﬂection in the -axis

counterclockwise rotation around the origin by

translation by 3 units to the right and 5 units down

reﬂection in the -axis

clockwise rotation about the origin by

onto the line segment

?

so that the

Solution

We can simply graph or use coordinate rules to realize that both

rotated , so -Dodgers66

and are rotated about the origin, therefore is

See Also

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 2

Followed by

Problem 4

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America ()\'s American

Mathematics Competitions ().

Retrieved from \"/wiki/?title=2019_AMC_12B_Problems/Problem_3&oldid=102393\"

2019 AMC 10B Problems/Problem 6

The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.

Contents

1

2

3

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5

6

Problem

Solution 1

Solution 2

Solution 3

Solution 4

See Also

Problem

There is a real such that . What is the sum of the digits of ?

Solution 1

.

iron

Solution 2

Dividing both sides by gives

Since is positive, . The answer is

Solution 3

Divide both sides by

:

factor out :

prime factorization of

is

and a bit of experimentation gives us and , so , so the answer

Solution 4

Obviously n must be very close to . By quick inspection, works.

See Also

2019 AMC 10B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=43&year=2019))

Preceded by

Problem 5

Followed by

Problem 7

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 3

Followed by

Problem 5

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America ()\'s American

Mathematics Competitions ().

Retrieved from \"/wiki/?title=2019_AMC_10B_Problems/Problem_6&oldid=102789\"

2019 AMC 10B Problems/Problem 7

The following problem is from both the 2019 AMC 10B #7 and 2019 AMC 12B #5, so both problems redirect to this page.

Contents

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2

3

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5

Problem

Solution 1

Solution 2

Solution 3

See Also

Problem

Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red

candy, 14 pieces of green candy, 15 pieces of blue candy, or pieces of purple candy. A piece of purple candy costs 20 cents.

What is the smallest possible value of ?

Solution 1

If he has enough money to buy 12 pieces of red candy, 14 pieces of green candy, and 15 pieces of blue candy, then the least money

he can have is = 420. Since a piece of purple candy costs 20 cents, the least value of n can be

iron (don\'t take credit away from other people @wwt7534)

Solution 2

We simply need to ﬁnd a value of that divides 12, 14, and 15. divides 12 and 15, but not 14. successfully

divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy, so the minimum

value of .

Solution 3

This problem is equivalent to ﬁnding the LCM of 12, 14, 15, and 20 (and then dividing it by 20). It is easy to see that the prime

factorization of said LCM must be . We can divide by 20 now, before we ever multiply it out, leaving us with

. However, in this case multiplying it out nets us

- Robin\'s solution

, which is worth the time it takes all on its own.

See Also

2019 AMC 10B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=43&year=2019))

Preceded by

Problem 6

Followed by

Problem 8

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 10 Problems and Solutions

2019 AMC 10B Problems/Problem 10

The following problem is from both the 2019 AMC 10B #10 and 2019 AMC 12B #6, so both problems redirect to this page.

Contents

1

2

3

4

Problem

Solution

Solution 2

See Also

Problem

In a given plane, points and are units apart. How many points

is units and the area of is square units?

are there in the plane such that the perimeter of

Solution

Notice that whatever point we pick for C, AB will be the base of the triangle. WLOG, points A and B are (0,0) and (0,10) [notice that

for any other combination of points, we can just rotate the plane to be the same thing]. When we pick point C, we have to make

sure the y value of C is 20, because that\'s the only way the area of the triangle can be 100.

We ﬁgure that the one thing we need to test to see if there is such a triangle is when the perimeter is minimized, and the value of C

is (x, 20). Thus, we put C in the middle, so point C is (5, 20). We can easily see that AC and BC will both be

. The perimeter of this minimized triangle is , which is larger than 50. Since the

.minimized perimeter is greater than 50, there is no triangle that satisﬁes the condition, giving us

iron

Solution 2

WLOG let be a horizontal segment of length 10. Now, realize has to lie on one of the lines parallel to and vertically 20

units away from it. But 10+20+20 is already 50, and this doesn\'t form a triangle. Otherwise, WLOG . Dropping altitude

we have a right triangle with hypotenuse and leg which is clearly impossible, giving us

.

See Also

2019 AMC 10B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=43&year=2019))

Preceded by

Problem 9

Followed by

Problem 11

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 5

Followed by

Problem 7

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

2019 AMC 10B Problems/Problem 13

The following problem is from both the 2019 AMC 10B #13 and 2019 AMC 12B #7, so both problems redirect to this page.

Contents

1

2

3

4

Problem

Solution 1

Solution 2

See Also

Problem

What is the sum of all real numbers for which the median of the numbers

numbers?

and is equal to the mean of those ﬁve

Solution 1

There are cases: is the median, is the median, and is the median. In all cases, the mean is

For case 1,

For case 2,

For case 3,

. This allows 6 to be the median because the set is

. This is impossible because the set is .

. This is impossible because the set is

, so the answer is .

.

Only case 1 yields a solution,

Solution by a1b2

Solution 2

The mean is

There are 3 possibilities: either the median is 6, 8, or x.

Let\'s start with 6.

when

Now let the mean=8

when

Finally we let the mean=x

and the sequence is 4, 6, 8, 8.75, 17 which has median

8 so no go.

So the only option for x is

--mguempel

and the sequence is 4, 5, 6, 8, 17 which has median 6 so no go.

and the sequence is -5, 4, 6, 8, 17 which has 6 as the median so we\'re good.

.

See Also

2019 AMC 12B Problems/Problem 8

Contents

1

2

3

4

Problem

Solution 1

Solution 2

See Also

Problem

Let . What is the value of the sum

?

Solution 1

Note that . We can see from this that the terms cancel and the answer is .

Solution 2

We can ﬁrst plug in a few numbers and see what happens. We get and so on.

Then, we can skip to the end and see that the last term and the ﬁrst term are equal, and cancel each other out because they have

different signs. Therefore we see that every number cancels out. It might seem that there is some term in the middle, but if we use

a smaller example to check, we see that that is not the case. Therefore, the answer is

-- clara32356 (Claire)

See Also

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 7

Followed by

Problem 9

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America ()\'s American

Mathematics Competitions ().

Retrieved from \"/wiki/?title=2019_AMC_12B_Problems/Problem_8&oldid=102787\"

2019 AMC 12B Problems/Problem 9

Contents

1

2

3

4

Problem

Solution

Solution 2

See Also

Problem

For how many integral values of can a triangle of positive area be formed having side lengths ?

Solution

Note is a lower bound for , corresponding to a triangle with side lengths

, violating the triangle inequality.

. If ,

Note also that is an upper bound for , corresponding to a triangle with side lengths

, again violating the triangle inequality.

and

is .

. If ,

It is easy to verify all satisfy

satisﬁed trivially). The number of integers strictly between and

-DrJoyo

(the third inequality is

Solution 2

Note that ,

as it\'s less restrictive in all cases than the last.

Let\'s raise the ﬁrst to the power of .

Doing the same for the second nets us:

Thus, x is an integer strictly between

- Robin\'s solution

and : .

, and . The second one is redundant,

. Thus, .

.

See Also

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 8

Followed by

Problem 10

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America ()\'s American

Mathematics Competitions ().

Retrieved from \"/wiki/?title=2019_AMC_12B_Problems/Problem_9&oldid=102445\"

2019 AMC 12B Problems/Problem 10

Contents

1

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3

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5

Problem

Solution (with some graph theory)

Solution 2

Solution 3

See Also

Problem

The ﬁgure below is a map showing cities and roads connecting certain pairs of cities. Paula wishes to travel along exactly

of those roads, starting at city and ending at city , without traveling along any portion of a road more than once. (Paula is

allowed to visit a city more than once.)

How many different routes can Paula take?

Solution (with some graph theory)

Let the bottom right vertex be (0,0) and let each of the edges have length 1 so that all of the vertices are at lattice points. First,

notice that for any vertex excluding and on the graph, we can visit it at most where deg is

the number of edges connected to . This is because to visit any of these vertices, we would have to visit and exit on two

different edges (This is a pretty standard principle in graph theory). We will label each vertex with this number.

Additionally, notice that if we visit vertices (not necessarily distinct; i.e. if we visit a vertex twice we count it as two vertices)

along our path, we must traverse edges (this is also pretty easy to prove). Thus, if we want to traverse 13 edges total, we

have to visit 14 vertices. We will visit and for sure, leaving us with 12 vertices to visit from the rest of the graph. If we add up

the maximum number of visits for each vertex, we ﬁnd that this sum is equal to exactly 12. This means that we have to visit each

vertex times and we have to traverse edges connected to each vertex. Speciﬁcally, this tells us that we must

traverse all four of the edges connected to the two center vertices ((1,1) and (1,2)), both edges connected to the two corner

vertices (excluding and ), and any two edges connected to the other vertices along the outside edge of the rectangle. With

this information, we can proceed with just a few cases.

In this case, we see that we must immediately move right to in order to traverse the edge from to since we

can never revisit . However, by the same logic, we must immediately move to . This is a contradiction, so there are no

paths in this case.

Similar to the last case, we see that we must immediately move to . By symmetry, we conclude that our last two moves

must be . With this information, we have reduced the problem to traveling from to

with the same constraints as before. We redraw the graph, removing the edges we have already used and updating our labels (Note

that and are still labeled with since we will pass through them twice, including the start and end). Then, we

remove the vertices with label 0 and the edges we know we can never traverse, giving us:

Now, it is clear that we must complete a cycle in the lower left square, return to , go to , and complete a cycle in the

top right square, returning to . There are two ways to traverse each cycle (clockwise and anti-clockwise), giving us a total of

paths of length 13 from

(Solution by vedadehhc)

to .

Solution 2

Note the following route, which isn\'t that hard to discover:

!! Someone with good Latex/Asymptote skills please help !!

Look at the two square \"loop\"s. Each one can be oriented in one of two directions (lower left: either go down or left ﬁrst; upper

right: either go right or up ﬁrst). Therefore, the answer is 1 route * 2 * 2 = . Note that no choice is larger than it.

Solution 3

Note that of the 12 cities, 6 of them (2 on the top and bottom, 1 on the sides) have 3 edges connecting to them. Therefore, at least

1 edge connecting to them cannot be used. Additionally, the same goes for the start and end point as we don\'t want to return to

them. so we have 8 points that we know have 1 unused edge, and we have a total of 4 unused edges to work with (17-13), so we

easily ﬁnd there is only one conﬁguration that satisﬁes this:

X\'s represent unused edges, by necessity, lines are ﬁlled in for the paths.

Now, we ﬁnd that at each of the 2 cities marked with an O, we have 2 possibilities to follow the path, we can either continue straight

and cross back over it later, or make a left turn and turn right when we approach the junction again/ This gives us

See Also

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 9

Followed by

Problem 11

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America ()\'s American

Mathematics Competitions ().

Retrieved from \"/wiki/?title=2019_AMC_12B_Problems/Problem_10&oldid=102517\"

2019 AMC 12B Problems/Problem 11

Contents

1

2

3

4

Problem

Solution 1

Solution 2

See Also

Problem

How many unordered pairs of edges of a given cube determine a plane?

Solution 1

WLOG, pick one of the 12 edges of the cube to be among the two selected. We seek the answer by computing the probability that a

random choice of second edge satisﬁes the problem statement.

For two line segments in space to correspond to a common plane, they must correspond to lines that either intersect or are

parallel. If all 12 line segments are extended to lines, our ﬁrst edge\'s line intersects 4 lines and is parallel to another 3. Thus 7 of the

11 line segments satisfy the problem statement.

We compute:

//Can somebody better with latex than me put a picture in here?

Solution 2

Case 1: The two edges are on the same face. There are possibilities.

Case 2: The two edges are parallel but not on the same face. There are possibilities.

See Also

2019 AMC 12B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=44&year=2019))

Preceded by

Problem 10

Followed by

Problem 12

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America ()\'s American

Mathematics Competitions ().

Retrieved from \"/wiki/?title=2019_AMC_12B_Problems/Problem_11&oldid=102714\"

2019 AMC 12B Problems/Problem 12

Contents

1

2

3

4

5

Problem

Solution 1

Solution 1.5 (Little bit of coordinate bash)

Solution 2

See Also

Problem

Right triangle with right angle at is constructed outwards on the hypotenuse

with leg length , as shown, so that the two triangles have equal perimeters. What is

of isosceles right triangle

?

Solution 1

Observe that the \"equal perimeter\" part implies that . A quick Pythagorean chase gives

. (Note: You could set up variables such as X, 2-X to denote the side lengths). Use the sine addition

formula on angles and

on angle

Feel free to elaborate if necessary.

(which requires ﬁnding their cosines as well), and this gives the sine of

to get .

. Now, use

Solution 1.5 (Little bit of coordinate bash)

After using Pythagorean to ﬁnd and , we can instead notice that the angle between the y-coordinate and is

degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point , we can then proceed to ﬁnd the

height and base of this new triangle (deﬁned by where is the intersection of the altitude and ) by coordinate-

bashing, which turns out to be and respectively.

By double angle formula and difference of squares, it\'s easy to see that our answer is

~Solution by MagentaCobra

Solution 2

Let and , so .

. To write this in terms of and , we

.

By the double-angle formula,

can say that we are looking for

Using trigonometric addition and subtraction formulas, we know that

and

. .

So

Now we just need to ﬁgure out what the numerical answer is.

From the given information about the triangles\' perimeters, we can deduce that

theorem tell us that . These two equations allow us to write

redundancy:

Plugging these into

and

, we\'ll get

.

. Also, the Pythagorean

in terms of without

and

.

If we set these equal to each other, now there is an algebraic equation that can be easily solved:

Now that we know what is equal to, we can also ﬁgure out .

Thus,

See Also

2019 AMC 12B Problems/Problem 13

The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page.

Contents

1

2

3

4

5

6

7

Problem

Solution 1

Solution 1 Variant

Solution 2 (variant)

Solution 3 (inﬁnite geometric series)

Solution 4 (quickest)

See Also

Problem

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each

ball, the probability that it is tossed into bin is

higher-numbered bin than the green ball?

for What is the probability that the red ball is tossed into a

Solution 1

The probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher

numbered bin. The probability of both landing in the same bin is . The sum is equal to . Therefore the other two

probabilities have to both be

Solution by a1b2

.

Solution 1 Variant

We solve for the probability by doing .

We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we

have the probability of equality being equal to

The summation of this expression is equal to

. Using the geometric sum formula, we obtain the summation of this expression to be or .

Solution 2 (variant)

Suppose the green ball goes in bin , for some

the red ball goes in a higher-numbered bin is

, and the red ball goes in a bin greater than , is

. The probability of this occurring is . Given this occurs, the probability that

. Thus the probability that the green ball goes in bin

. Summing from to inﬁnity, we get

(Note: to ﬁnd this sum, we use the formula . Since in this case , the answer is . If you don\'t

,know this formula, you may instead note that if you multiply the sum by , it is equivalent to adding . Thus:

which clearly simpliﬁes to .

- scrabbler94 (explanation of inﬁnite sum provided by Robin)

Solution 3 (inﬁnite geometric series)

The probability that the two balls will go into adjacent bins is

. The probability that the two balls will go into bins that

have a distance of 2 from each other is

each time we add a bin between the two balls, the probability halves. Thus, our answer is

converges into .

. We can see that

, which

Solution 4 (quickest)

Deﬁne a win as a ball appearing in higher numbered box.

Start from the ﬁrst box.

There are 4 possible results in the box: Red, Green, Red and Green, none, with probability of for each. Red win, Green win, Tie all

have the same probability of . If none of the balls is in the ﬁrst box, the game restarts at the second box with the same kind of

probability distribution.

So ﬁnally, Red win, Green win and Tie all have a probability of

The answer is

-- Solution by mathsuper(丹神)

See Also

2019 AMC 10B (Problems • Answer Key • Resources (/Forum/re

?c=182&cid=43&year=2019))

Preceded by

Problem 16

Followed by

Problem 18

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AMC 10 Problems and Solutions

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