## 2021 AMC 12B 真题与答案

文末答案

## Problem 1

How many integer values of satisfy

## Problem 2

At a math contest, students are wearing blue shirts, and another students are wearing yellow shirts. The students are assigned into pairs. In exactly of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

## Problem 3

SupposeWhat is the value of

## Problem 4

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is , and the afternoon class’s mean score is . The ratio of the number of students in the morning class to the number of students in the afternoon class is . What is the mean of the score of all the students?

## Problem 5

The point in the -plane is first rotated counterclockwise by around the point and then reflected about the line . The image of after these two transformations is at . What is

## Problem 6

An inverted cone with base radius and height is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of . What is the height in centimeters of the water in the cylinder?

## Problem 7

Let What is the ratio of the sum of the odd divisors of to the sum of the even divisors of

## Problem 8

Three equally spaced parallel lines intersect a circle, creating three chords of lengths and . What is the distance between two adjacent parallel lines?

## Problem 9

What is the value of

## Problem 10

Two distinct numbers are selected from the set so that the sum of the remaining numbers is the product of these two numbers. What is the difference of these two numbers?

## Problem 11

Triangle has and . Let be the point on such that . There are exactly two points and on line such that quadrilaterals and are trapezoids. What is the distance

## Problem 12

Suppose that is a finite set of positive integers. If the greatest integer in is removed from , then the average value (arithmetic mean) of the integers remaining is . If the least integer in is also removed, then the average value of the integers remaining is . If the greatest integer is then returned to the set, the average value of the integers rises to The greatest integer in the original set is greater than the least integer in . What is the average value of all the integers in the set

## Problem 13

How many values of in the interval satisfy

## Problem 14

Let be a rectangle and let be a segment perpendicular to the plane of . Suppose that has integer length, and the lengths of and are consecutive odd positive integers (in this order). What is the volume of pyramid

## Problem 15

The figure is constructed from line segments, each of which has length . The area of pentagon can be written is , where and are positive integers. What is

## Problem 16

Let be a polynomial with leading coefficient whose three roots are the reciprocals of the three roots of where What is in terms of and

## Problem 17

Let be an isosceles trapezoid having parallel bases and with Line segments from a point inside to the vertices divide the trapezoid into four triangles whose areas are and starting with the triangle with base and moving clockwise as shown in the diagram below. What is the ratio

## Problem 18

Let be a complex number satisfying What is the value of

## Problem 19

Two fair dice, each with at least faces are rolled. On each face of each dice is printed a distinct integer from to the number of faces on that die, inclusive. The probability of rolling a sum if is of the probability of rolling a sum of and the probability of rolling a sum of is . What is the least possible number of faces on the two dice combined?

## Problem 20

Let and be the unique polynomials such thatand the degree of is less than What is

## Problem 21

Let be the sum of all positive real numbers for whichWhich of the following statements is true?

## Problem 22

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one “wall” among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes and can be changed into any of the following by one move: or

Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

## Problem 23

Three balls 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 for More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is where and are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins and ) What is

## Problem 24

Let be a parallelogram with area . Points and are the projections of and respectively, onto the line and points and are the projections of and respectively, onto the line See the figure, which also shows the relative locations of these points.

Suppose and and let denote the length of the longer diagonal of Then can be written in the form where and are positive integers and is not divisible by the square of any prime. What is

## Problem 25

Let be the set of lattice points in the coordinate plane, both of whose coordinates are integers between and inclusive. Exactly points in lie on or below a line with equation The possible values of lie in an interval of length where and are relatively prime positive integers. What is

## Solution 1

1、Since is about , we multiply 9 by 2 and add 1 to get

2、 . Since is approximately , is approximately . We are trying to solve for , where . Hence, , for . The number of integer values of is . Therefore, the answer is .

## Solution2

There are students paired with a blue partner. The other students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are students remaining. Therefore the requested number of pairs is

## Solution3

Subtracting from both sides and taking reciprocals gives . Subtracting from both sides and taking reciprocals again gives . Subtracting from both sides and taking reciprocals for the final time gives or .

## Solution 4

1、WLOG, assume there are students in the morning class and in the afternoon class. Then the average is

2、Let there be students in the morning class and students in the afternoon class. The total number of students is . The average is . Therefore, the answer is .

3、Suppose the morning class has students and the afternoon class has students. We have the following chart:

We are also given that which rearranges as

The mean of the scores of all the students is

## Solution5

The final image of is . We know the reflection rule for reflecting over is . So before the reflection and after rotation the point is .

By definition of rotation, the slope between and must be perpendicular to the slope between and . The first slope is . This means the slope of and is .

Rotations also preserve distance to the center of rotation, and since we only “travelled” up and down by the slope once to get from to it follows we shall only use the slope once to travel from to .

Therefore point is located at . The answer is .

## Solution 6

1、The volume of a cone is where is the base radius and is the height. The water completely fills up the cone so the volume of the water is .

The volume of a cylinder is so the volume of the water in the cylinder would be .

We can equate these two expressions because the water volume stays the same like this . We get and .

So the answer is

2、The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has of the volume of the cylinder, and so the height is divided by . Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since ).

Therefore, the height is divided by and divided by , which is

## Solution 7

1、Prime factorize to get . For each odd divisor of , there exist even divisors of , therefore the ratio is

2、Prime factorizing , we see . The sum of ‘s odd divisors are the sum of the factors of without , and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors. The sum of odd divisors is given byand the total sum of divisors is. Thus, our ratio is

## Solution 8

1、

Since two parallel chords have the same length (), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be . Thus, the distance from the center of the circle to the chord of length is

and the distance between each of the chords is just . Let the radius of the circle be . Drawing radii to the points where the lines intersect the circle, we create two different right triangles:

– One with base , height , and hypotenuse ( on the diagram)

– Another with base , height , and hypotenuse ( on the diagram)

By the Pythagorean theorem, we can create the following system of equations:

Solving, we find , so .

2、Because we know that the equation of a circle is where the center of the circle is and the radius is , we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is . Now, we can set the distance between the chords as so the distance from the chord with length 38 to the diameter is .

Therefore, the following points are on the circle as the y-axis splits the chord in half, that is where we get our x value:

Now, we can plug one of the first two value in as well as the last one to get the following equations:

Subtracting these two equations, we get – therefore, we get . We want to find because that’s the distance between two chords. So, our answer is .

## Solution9

Note that , and similarly

Expanding,

All the log terms cancel, so the answer is .

## Solution10

The sum of the first integers is given by , so .

Therefore,

Rearranging,

Looking at the possible divisors of , and are within the constraints of so we try those:

Therefore, the difference , choice E).

## Solution 11

1、Toss on the Cartesian plane with and . Then by the trapezoid condition, where . Since , point is of the way from to and is located at . Thus line has equation . Since and is parallel to the ground, we know has the same -coordinate as , except it’ll also lie on the line . Therefore,

To find the location of point , we need to find the intersection of with a line parallel to passing through . The slope of this line is the same as the slope of , or , and has equation . The intersection of this line with is . Therefore point is located at

The distance is equal to the distance between and , which is

2、Using Stewart’s Theorem we find . From the similar triangles and we haveSo

3、Let be the length . From the similar triangles and we haveTherefore . Now extend line to the point on , forming parallelogram . As we also have so .

We now use the Law of Cosines to find (the length of ):As , we have (by Law of Cosines on triangle )ThereforeAnd . The answer is then

4、Let the brackets denote areas. By Heron’s Formula, we haveIt follows that the height of is

Next, we drop the altitudes and of By the Pythagorean Theorem on we get By the AA Similarity, with the ratio of similitude It follows that Since is a rectangle, By the Pythagorean Theorem on we get

By again, we have and Also, by the AA Similarity, with the ratio of similitude It follows that

Finally,

## Solution 12

1、Let be the greatest integer, be the smallest, be the sum of the numbers in S excluding and , and be the number of elements in S.

Then,

Firstly, when the greatest integer is removed,

When the smallest integer is also removed,

When the greatest integer is added back,

We are given that

After you substitute , you have 3 equations with 3 unknowns , and .

This can be easily solved to yield , , .

average value of all integers in the set , D)

2、We should plug in and assume everything is true except the part. We then calculate that part and end up with . We also see with the formulas we used with the plug in that when you increase by the part decreases by . The answer is then .

## Solution 13

1、First, move terms to get . After graphing, we find that there are solutions (two in each period of ). -dstanz5

2、We can graph two functions in this case: and .Using transformation of functions, we know that is just a cos function with amplitude 5 and period . Similarly, is just a sin function with amplitude 3 and shifted 1 unit downwards. So:We have

## Solution 14

1、This question is just about pythagorean theoremWith these calculation, we find out answer to be ~Lopkiloinm

2、Let be , be , be , , , be , , respectively.

We have three equations:

Subbing in the first and third equation into the second equation, we get:Therefore,,Solving for other values, we get , . The volume is then

## Solution 15

1、Let be the midpoint of . Noting that and are triangles because of the equilateral triangles, . Also, and so .

2、

Draw diagonals and to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles and , they each have area . For triangle , we can see that and . Using Pythagorean Theorem, the altitude for this triangle is , so the area is . Adding each part up, we get .

## Solution 16

1、Note that has the same roots as , if it is multiplied by some monomial so that the leading term is they will be equal. We haveso we can see thatTherefore

2、Let the three roots of be , , and . (Here e does NOT mean 2.7182818…) We know that , , and , and that (Vieta’s). This is equal to , which equals . -dstanz5

3、Because the problem doesn’t specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let’s take . Then has a triple root of . Then has a triple root of , and it’s monic, so . We can see that this is , which is answer choice .

-Darren Yao

4、If we let and be the roots of , and . The requested value, , is thenThe numerator is (using the product form of ) and the denominator is , so the answer is

## Solution17

1、Without loss let have vertices , , , and , with and . Also denote by the point in the interior of .

Let and be the feet of the perpendiculars from to and , respectively. Observe that and . Now using the formula for the area of a trapezoid yieldsThus, the ratio satisfies ; solving yields .

2、Let be the bottom base, be the top base, be the height of the bottom triangle, be the height of the top triangle. Thus, so Let so we get This gives us a quadratic in ie. so

## Solution 18

1、Using the fact , the equation rewrites itself as

As the two quantities in the parentheses are real, both quantities must equal so

2、The answer being in the form means that there are two solutions, some complex number and its complex conjugate.We should then be able to test out some ordered pairs of . After testing it out, we get the ordered pairs of and its conjugate . Plugging this into answer format gives us ~Lopkiloinm

3、Let . Then . From the answer choices， we know that is real and , so . Then we havePlugging the above back to the original equation, we haveSo .

~Sequoia

4、There are actually several ways to see that I present two troll ways of seeing it, and a legitimate way of checking.

Rewrite using

Symmetric in and so if is a sol, then so is

TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, which means they must be conjugates and so

TROLL OBSERVATION #2: Note that because either solution must give the same answer! which means that

Alternatively, you can check: Let and Thus, we have and the discriminant of this must be nonnegative as is real. Thus, or which forces as claimed.

Thus, we plug in and get: ie. or which means and that’s our answer since we know

– ccx09

5、Observe that all the answer choices are real. Therefore, and must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product () to be real. Thus . We will test all the answer choices, starting with . Suppose the answer is . If then and . Note that if works, then so does . It is relatively easy to see that if , then and . Thus the conditionis satisfied for , and the answer is .

## Solution19

Suppose the dice have and faces, and WLOG . Since each die has at least faces, there will always be ways to sum to . As a result, there must be ways to sum to . There are at most nine distinct ways to get a sum of , which are possible whenever . To achieve exactly eight ways, must have faces, and . Let be the number of ways to obtain a sum of , then . Since , . In addition to , we only have to test , of which both work. Taking the smaller one, our answer becomes .

## Solution 20

1、Note thatso if is the remainder when dividing by ,Now,So , andThe answer is

2、Instead of dealing with a nasty , we can instead deal with the nice , as is a factor of . Then, we try to see what is. Of course, we will need a , getting . Then, we’ve gotta get rid of the term, so we add a , to get . This pattern continues, until we add a to get rid of , and end up with . We can’t add anything more to get rid of the , so our factor is . Then, to get rid of the , we must have a remainder of , and to get the we have to also have a in the remainder. So, our product isThen, our remainder is . The remainder when dividing by must be the same when dividing by , modulo . So, we have that , or . This corresponds to answer choice . ~rocketsri

3、One thing to note is that takes the form of for some constants A and B. Note that the roots of are part of the solutions of They can be easily solved with roots of unity:Obviously the right two solutions are the roots of We substitute into the original equation, and becomes 0. Using De Moivre’s theorem, we get:Expanding into rectangular complex number form:Comparing the real and imaginary parts, we get:The answer is . ~Jamess2022(burntTacos;-;)

4、Note that the equation above is in the form of polynomial division, with being the dividend, being the divisor, and and being the quotient and remainder respectively. Since the degree of the dividend is and the degree of the divisor is , that means the degree of the quotient is . Note that R(x) can’t influence the degree of the right hand side of this equation since its degree is either or . Since the coefficients of the leading term in the dividend and the divisor are both , that means the coefficient of the leading term of the quotient is also . Thus, the leading term of the quotient is . Multiplying by the divisor gives . We have our term but we have these unnecessary terms like . We can get rid of these terms by adding to the quotient to cancel out these terms, but this then gives us . Our first instinct will probably be to add , but we can’t do this as although this will eliminate the term, it will produce a term. Since no other term of the form where is an integer less than will produce a term when multiplied by the divisor, we can’t add to the quotient. Instead, we can add to the coefficient to get rid of the term. Continuing this pattern, we get the quotient asThe last term when multiplied with the divisor gives . This will get rid of the term but will produce the expression , giving us the dividend as . Note that the dividend we want is of the form . Therefore, our remainder will have to be in order to get rid of the term in the expression and give us , which is what we want. Therefore, the remainder is

## Solution 21

1、Note that this solution is not recommended unless you’re running out of time.

Upon pure observation, it is obvious that one solution to this equality is . From this, we can deduce that this equality has two solutions, since grows faster than (for greater values of ) and is greater than for and less than for , where is the second solution. Thus, the answer cannot be or . We then start plugging in numbers to roughly approximate the answer. When , , thus the answer cannot be . Then, when , . Therefore, , so the answer is . ~Baolan

2、

(At this point we see by inspection that is a solution.)

LHS is a line. RHS is a concave curve that looks like a logarithm and has intercept at There are at most two solutions, one of which is But note that at we have meaning that the log log curve is above the line, so it must intersect the line again at a point Now we check and see that which means at the line is already above the log log curve. Thus, the second solution lies in the interval The answer is

## Solution22

1、First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. is always winning for the first player. Furthermore, is losing and so is We look at all the positions created from as is obviously winning by playing There are several different positions that can be played by the first player from They are Now we list refutations for each of these moves:

This proves that is losing for the first player.

-Note: In general, this game is very complicated. For example is winning for the first player but good luck showing that.

2、 can be turned into by Arjun, which is symmetric, so Beth will lose.

can be turned into by Arjun, which is symmetric, so Beth will lose.

can be turned into by Arjun, which is symmetric, so Beth will lose.

can be turned into by Arjun, which is symmetric, so Beth will lose.

That leaves or .

3、Let the nim-value of the ending game state, where someone has just removed the final brick, be . Then, any game state with a nim-value of is losing. It is well-known that the nim-value of a supergame (a combination of two or more individual games) is the binary xor function on the nim-values of the individual games that compose the supergame. Therefore, we calculate the nim-values of the states with a single wall up to bricks long (since the answer choices only go up to ).

First, the game with brick has a nim-value of .

Similarly, the game with bricks has a nim-value of .

Next, we consider a brick wall. After the next move, the possible resulting game states are brick, a brick wall, or separate bricks. The first two options have nim-values of and . The final option has a nim-value of , so the nim-value of this game state is .

Next, the brick wall. The possible states are a brick wall, a brick wall, a brick wall and a brick wall, or a brick wall and a brick wall. The nim-values of these states are , , , and , respectively, and hence the nim-value of this game state is . (Wait why is the nim-value of it ? – awesomediabrine)

The possible game states after the brick wall are the following: a brick wall, a brick wall, a brick wall and a brick wall, a two brick walls, and a brick wall plus a brick wall. The nim-values of these are , , , , and , respectively, meaning the nim-value of a brick wall is .

Finally, we find the nim-value of a brick wall. The possible states are a brick wall, a brick wall and a brick wall, a brick wall and a brick wall, a brick wall, a brick wall and a brick wall, and finally two brick walls. The nim-values of these game states are , , , , , and , respectively. This means the brick wall has a nim-value of .

The problem is asking which of the answer choices is losing, or has a nim-value of . We see that option has a nim-value of , option has a nim-value of , option has a nim-value of , option has a nim-value of , and option has a nim-value of , so the answer is .

This method can also be extended to solve the note after the first solution. The nim-values of the brick wall and the brick wall are and , using the same method as above. The nim-value of is therefore , which is winning.

## Solution23

1、”Evenly spaced” just means the bins form an arithmetic sequence.

Suppose the middle bin in the sequence is . There are different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these bins are chosen is , so the probability is the middle bin is . Then, we want the sumThe answer is

2、As in solution 1, note that “evenly spaced” means the bins are in arithmetic sequence. We let the first bin be and the common difference be . Further note that each pair uniquely determines a set of 3 bins.

We have because the leftmost bin in the sequence can be any bin, and , because the bins must be distinct.

This gives us the following sum for the probability:Therefore the answer is , which is choice (A).

-Darren Yao

3、This is a slightly messier variant of solution 2. If the first ball is in bin and the second ball is in bin , then the third ball is in bin . Thus the probability is

## Solution24

1、Let denote the intersection point of the diagonals and . Remark that by symmetry is the midpoint of both and , so and . Now note that since , quadrilateral is cyclic, and sowhich implies .

Thus let be such that and . Then Pythagorean Theorem on yields , and soSolving this for yields , and soThe requested answer is .

2、Let denote the intersection point of the diagonals and and let . Then, by the given conditions, . So,Combining the above 3 equations, we getSince we want to find we let ThenSolving this, we get so .

3、Let be the intersection of diagonals and . By symmetry , and , so now we have reduced all of the conditions one quadrant. Let . , by similar triangles and using the area condition we get . Note that it suffices to find because we can double and square it to get . Solving for in the above equation, and then using .

4、Again, Let be the intersection of diagonals and . Note that triangles and are similar because they are right triangles and share . First, call the length of . By the definition of an area of a parallelogram, , so . Using similar triangles on and , . Therefore, finding , . Now, applying the Pythagorean theorem once, we find + = . Solving this equation for , we find .

## Solution 25

1、First, we find a numerical representation for the number of lattice points in that are under the line For any value of the highest lattice point under is Because every lattice point from to is under the line, the total number of lattice points under the line is

Now, we proceed by finding lower and upper bounds for To find the lower bound, we start with an approximation. If lattice points are below the line, then around of the area formed by is under the line. By using the formula for a triangle’s area, we find that when Solving for assuming that is a point on the line, we get Plugging in to we get:

We have a repeat every values (every time goes through a lattice point). Thus, we can use arithmetic sequences to calculate the value above:

This means that is a possible value of Furthermore, it is the lower bound for This is because goes through many points (such as ). If was lower, would no longer go through some of these points, and there would be less than lattice points under it.

Now, we find an upper bound for Imagine increasing slowly and rotating the line starting from the lower bound of The upper bound for occurs when intersects a lattice point again

In other words, we are looking for the first that is expressible as a ratio of positive integers with For each , the smallest multiple of which exceeds is respectively, and the smallest of these is Note: start listing the multiples of from and observe that they get further and further away from Alternatively, see the method of finding upper bounds in solution 2.

The lower bound is and the upper bound is Their difference is so the answer is

~JimY

2、I know that I want about of the box of integer coordinates above my line. There are a total of 30 integer coordinates in the desired range for each axis which gives a total of 900 lattice points. I estimate that the slope, m, is . Now, although there is probably an easier solution, I would try to count the number of points above the line to see if there are 600 points above the line. The line separates the area inside the box so that of the are is above the line.

I find that the number of coordinates with above the line is 30, and the number of coordinates with above the line is 29. Every time the line hits a y-value with an integer coordinate, the number of points above the line decreases by one. I wrote out the sum of 30 terms in hopes of finding a pattern. I graphed the first couple positive integer x-coordinates, and found that the sum of the integers above the line is . The even integer repeats itself every third term in the sum. I found that the average of each of the terms is 20, and there are 30 of them which means that exactly 600 above the line as desired. This give a lower bound because if the slope decreases a little bit, then the points that the line goes through will be above the line.

To find the upper bound, notice that each point with an integer-valued x-coordinate is either or above the line. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates. We turn our attention to which the line intersects at . The point (30,20) is already counted below the line, and we can clearly see that if we slowly increase the slope of the line, we will hit the point (28,19) since (28, ) is closer to the lattice point. The slope of the line which goes through both the origin and (28,19) is . This gives an upper bound of .

Taking the upper bound of m and subtracting the lower bound yields . This is answer .

~theAJL

Diagram

## AMC8/AMC10/AMC12/AIME

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