## 2021 AMC 12A 真题与答案

文末答案

## Problem1

What is the value of

## Problem2

Under what conditions does hold, where and are real numbers?

It is never true.

It is true if and only if .

It is true if and only if .

It is true if and only if and .

It is always true.

## Problem3

The sum of two natural numbers is . One of the two numbers is divisible by . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

## Problem4

Tom has a collection of snakes, of which are purple and of which are happy. He observes that

all of his happy snakes can add,

none of his purple snakes can subtract, and

all of his snakes that can't subtract also can't add.

Which of these conclusions can be drawn about Tom's snakes?

Purple snakes can add.

Purple snakes are happy.

Snakes that can add are purple.

Happy snakes are not purple.

Happy snakes can't subtract.

## Problem5

When a student multiplied the number by the repeating decimalwhere and are digits, he did not notice the notation and just multiplied times . Later he found that his answer is less than the correct answer. What is the -digit number

## Problem6

A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is . When black cards are added to the deck, the probability of choosing red becomes . How many cards were in the deck originally?

## Problem7

What is the least possible value of for all real numbers and

## Problem8

A sequence of numbers is defined by and for . What are the parities (evenness or oddness) of the triple of numbers , where denotes even and denotes odd?

## Problem9

Which of the following is equivalent to

## Problem10

Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are cm and cm. Into each cone is dropped a spherical marble of radius cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

## Problem11

A laser is placed at the point . The laser bean travels in a straight line. Larry wants the beam to hit and bounce off the -axis, then hit and bounce off the -axis, then hit the point . What is the total distance the beam will travel along this path?

## Problem12

All the roots of the polynomial are positive integers, possibly repeated. What is the value of ?

## Problem13

Of the following complex numbers , which one has the property that has the greatest real part?

## Problem14

What is the value of

## Problem15

A choir direction must select a group of singers from among his tenors and basses. The only requirements are that the difference between the numbers of tenors and basses must be a multiple of , and the group must have at least one singer. Let be the number of different groups that could be selected. What is the remainder when is divided by ?

## Problem16

In the following list of numbers, the integer appears times in the list for .What is the median of the numbers in this list?

## Problem17

Trapezoid has , and . Let be the intersection of the diagonals and , and let be the midpoint of . Given that , the length of can be written in the form , where and are positive integers and is not divisible by the square of any prime. What is ?

## Problem18

Let be a function defined on the set of positive rational numbers with the property that for all positive rational numbers and . Furthermore, suppose that also has the property that for every prime number . For which of the following numbers is ?

## Problem19

How many solutions does the equation have in the closed interval ?

## Problem20

Suppose that on a parabola with vertex and a focus there exists a point such that and . What is the sum of all possible values of the length

## Problem21

The five solutions to the equationmay be written in the form for where and are real. Let be the unique ellipse that passes through the points and . The eccentricity of can be written in the form where and are relatively prime positive integers. What is ? (Recall that the eccentricity of an ellipse is the ratio , where is the length of the major axis of and is the is the distance between its two foci.)

## Problem22

Suppose that the roots of the polynomial are and , where angles are in radians. What is ?

## Problem23

Frieda the frog begins a sequence of hops on a grid of squares, moving one square on each hop and choosing at random the direction of each hop up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

## Problem24

Semicircle has diameter of length . Circle lies tangent to at a point and intersects at points and . If and , then the area of equals , where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. What is ?

## Problem25

Let denote the number of positive integers that divide , including and . For example, and . (This function is known as the divisor function.) LetThere is a unique positive integer such that for all positive integers . What is the sum of the digits of

## Solution1

We evaluate the given expression to get that

## Solution 2

方法1、Square both sides to get . Then, . Also, it is clear that both sides of the equation must be nonnegative. The answer is .

方法2、The left side of the original equation is the **arithmetic square root**, which is always nonnegative. So, we need which eliminates and Next, picking reveals that is incorrect, and picking reveals that is incorrect. By POE (Process of Elimination), the answer is

~MRENTHUSIASM

方法3、If we graph then we get the positive -axis and the positive -axis, plus the origin. Therefore, the answer is

## Solution 3

方法1、The units digit of a multiple of will always be . We add a whenever we multiply by . So, removing the units digit is equal to dividing by .

Let the smaller number (the one we get after removing the units digit) be . This means the bigger number would be .

We know the sum is so . So . The difference is . So, the answer is .

方法2、Since the ones place of a multiple of is , this implies the other integer has to end with a since both integers sum up to a number that ends with a . Thus, the ones place of the difference has to be , and the only answer choice that ends with an is

## Solution 4

方法1、We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is .

方法2、We are given that

Combining and into below, we have

Clearly, the answer is

## Solution5

It is known that and . Let . We have that . Solving gives that so . ~aop2014

## Solution 6

方法1、If the probability of choosing a red card is , the red and black cards are in ratio . This means at the beginning there are red cards and black cards.

After black cards are added, there are black cards. This time, the probability of choosing a red card is so the ratio of red to black cards is . This means in the new deck the number of black cards is also for the same red cards.

So, and meaning there are red cards in the deck at the start and black cards.

So the answer is .

方法2、For the number of cards, the final deck is of the original deck. Adding cards to the original deck is the same as increasing the original deck by of itself. So, the original deck has cards.

~MRENTHUSIASM

方法3、Suppose there were cards in the deck originally. Now, the deck has cards, which must be a multiple of

Only is a multiple of So, the answer is

## Solution 7

方法1、Expanding, we get that the expression is or . By the trivial inequality(all squares are nonnegative) the minimum value for this is , which can be achieved at . ~aop2014

方法2、Like solution 1, expand and simplify the original equation to and let . To find local extrema, find where . First, find the first partial derivative with respect to x and y and find where they are :

Thus, there is a local extreme at . Because this is the only extreme, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning is the minimum of . Plugging into , we find 1

## Solution8

Making a small chart, we have

This starts repeating every 7 terms, so , , and . Thus, the answer is ~JHawk0224

## Solution 9

方法1、All you need to do is multiply the entire equation by . Then all the terms will easily simplify by difference of squares and you will get or as your final answer. Notice you don't need to worry about because that's equal to .

方法2、If you weren't able to come up with the insight, then you could just notice that the answer is divisible by , and . We can then use Fermat's Little Theorem for on the answer choices to determine which of the answer choices are divisible by both and . This is .

方法3、After expanding the first few terms, the result after each term appears to be where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by would give , and all the previous terms multiplied by would give . Their sum is equal to , so the proof is complete. Since is equal to , the answer is .

## Solution 10(Use Tables to Organize Information)

__方法1、Initial Scenario__

By similar triangles:

For the narrow cone, the ratio of base radius to height is which remains constant.

For the wide cone, the ratio of base radius to height is which remains constant.

Equating the initial volumes gives which simplifies to

__Final Scenario (Two solutions follow from here.)__

### 1.1 (Fraction Trick)

Let the base radii of the narrow cone and the wide cone be and respectively, where We have the following table:

Equating the final volumes gives which simplifies to or

Lastly, the requested ratio is

**PS:**

1. This problem uses the following fraction trick:

For unequal positive numbers and if then

__Quick Proof__

From we know that and . Therefore,

2. The work above shows that, regardless of the shape or the volume of the solid dropped in, as long as the solid sinks to the bottom and is completely submerged without spilling any liquid, the answer will remain unchanged.

~MRENTHUSIASM

### 1.2 (Bash)

Let the base radii of the narrow cone and the wide cone be and respectively.

Let the rises of the liquid levels of the narrow cone and the wide cone be and respectively. We have the following table:

By similar triangles discussed above, we have

The volume of the marble dropped in is

Now, we set up an equation for the volume of the narrow cone and solve for

Next, we set up an equation for the volume of the wide cone Using the exact same process from above (but with different numbers), we getRecall that Therefore, the requested ratio is

方法2、The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii 3 and 6 and infinitely large height. Then the base area of the wide cylinder is 4 times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise times as much.

## Solution 11

方法1、Every time the laser bounces off a wall, instead we can imagine it going straight by reflecting it about the wall. Thus, the laser starts at and ends at , so the path's length is

方法2、Let be the point where the beam hits the -axis, and be the point where the beam hits the -axis.

Reflecting about the -axis gives Then, reflecting over the -axis gives Finally, reflecting about the -axis gives as shown below.

It follows that The total distance that the beam will travel isGraph in

方法3、Define points and as Solution 2 does.

When a line segment hits and bounces off a coordinate axis at point the ray entering and the ray leaving have negative slopes. **Geometrically, the rays coincide when reflected about the line perpendicular to that coordinate axis, creating a line symmetry.** Let the slope of be It follows that the slope of is and the slope of is Here, we conclude that

Next, we locate on such that thus is a parallelogram, as shown below.

Let By the property of slopes, we get By symmetry, we obtain

Applying the slope formula on and givesEquating the last two expressions gives

By the Distance Formula, and The total distance that the beam will travel is

方法4、Define points and as Solution 2 does.

Since choices and all involve we suspect that one of them is the correct answer. We take a guess in faith that and form angles with the coordinate axes, then we get that and This result verifies our guess. Following the penultimate paragraph of Solution 3 gives the answer

## Solution 12

方法1、By Vieta's formulae, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, so the function is . Therefore, .

方法2、Using the same method as Solution 1, we find that the roots are and . Note that is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the products we obtain

## Solution 13

方法1、First, .

Taking the real part of the 5th power of each we have:

,

which is negative

which is zero

Thus, the answer is .

方法2、For every complex number where and are real numbers and its magnitude is For each choice, we get that the magnitude is

Rewriting each choice to the polar form we know that by the **De Moivre's Theorem**, the real part of isWe make a table as follows:Clearly, the answer is

## Solution 14

方法1、This equals

方法2、We use the following property of logarithms:

We can prove it quickly using the Change of Base Formula:Now, we simplify the expressions inside the summations:andUsing these results, we evaluate the original expression:

方法3、First, we can get rid of the exponents using properties of logarithms:

(Leaving the single in the exponent will come in handy later). Similarly,

Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms:

To evaluate the exponent of the in the first logarithm, we use the triangular numbers equation:

Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify:

Thus,

方法4、In note that the addends are greater than for all

In note that the addends are greater than for all

By a rough approximation,from which we eliminate choices and We get the answer by either an educated guess or continued estimation: Since it follows that By a (very) rough approximation,From here, it should be safe to guess that the answer is

As an extra guaranty, note that Therefore, we must have

## Solution 15

方法1、We know the choose function and we know the pair multiplication so we do the multiplications and additions.

方法2、The problem can be done using a roots of unity filter. Let . By expanding the binomials and distributing, is the generating function for different groups of basses and tenors. That is,where is the number of groups of basses and tenors. What we want to do is sum up all values of for which except for . To do this, define a new functionNow we just need to sum all coefficients of for which . Consider a monomial . If ,otherwise, is a sum of these monomials so this gives us a method to determine the sum we're looking for:(since and it can be checked that ). Hence, the answer is with the for which gives . ~lawliet163

方法3、We will use the Vandermonde's Identity to find the requested sum:

**方法4、We claim that if the empty group is allowed, then there are ways to choose the singers satisfying the requirements.**

First, we set one tenor and one bass aside. We argue that each group from the remaining singers (of any size, including ) corresponds to exactly one desired group from the original singers.

The remaining singers can formgroups. The left side counts directly, while the right side uses casework (selecting tenors and basses for each group). Now, we map each group from the to a group from the

By casework:

Clearly, the mapping is satisfied. For each group from the we can obtain a desired group from the by adding one tenor or one bass accordingly.

Since we can select basses instead of basses, without changing the number of groups. Therefore, we have the absolute difference Since we conclude that and the mapping is satisfied by case

By the same reasoning as Case we select basses instead of basses. The absolute difference also is Since is even, it follows that is also even, and The mapping is satisfied by case

## Solution 16

方法1、There are numbers in total. Let the median be . We want to find the median such thatorNote that . Plugging this value in as gives, so is the nd and rd numbers, and hence, our desired answer. .

Note that we can derive through the formulawhere is a perfect square less than or equal to . We set to , so , and . We then have .

方法2、The th number of this sequence is via the quadratic formula. We can see that if we halve we end up getting . This is approximately the number divided by . and since looks like the only number close to it, it is answer

方法3、We can look at answer choice , which is first. That means that the number of numbers from to is roughly the number of numbers from to .

The number of numbers from to is which is approximately The number of numbers from to is which is approximately as well. Therefore, we can be relatively sure the answer choice is

## Solution 17

方法1、Angle chasing reveals that , thereforeAdditional angle chasing shows that , thereforeSince is right, the Pythagorean theorem implies that

方法2、Since is isosceles with legs and it follows that the median is also an altitude of Let and We have

Since by AA, we have

Let the brackets denote areas. Notice that (By the same base/height, Subtracting from both sides gives ). Doubling both sides, we have

In we haveandFinally,

方法3、Let and is perpendicular bisector of Let so

(1) so we get or

(2) pythag on gives

(3) with ratio so

Thus, or And so and the answer is

方法4、Observe that is congruent to ; both are similar to . Let's extend and past points and respectively, such that they intersect at a point . Observe that is degrees, and that . Thus, by ASA, we know that , thus, , meaning is the midpoint of . Let be the midpoint of . Note that is congruent to , thus , meaning is the midpoint of

Therefore, and are both medians of . This means that is the centroid of ; therefore, because the centroid divides the median in a 2:1 ratio, . Recall that is the midpoint of ; . The question tells us that ; ; we can write this in terms of ; .

We are almost finished. Each side length of is twice as long as the corresponding side length or , since those triangles are similar; this means that . Now, by Pythagorean theorem on , .

## Solution 18

方法1、Looking through the solutions we can see that can be expressed as so using the prime numbers to piece together what we have we can get , so or .

-Lemonie

方法2、We know that . By transitive, we haveSubtracting from both sides gives AlsoIn we have .

In we have .

In we have .

In we have .

In we have .

Thus, our answer is

方法3、Consider the rational , for integers. We have . So . Let be a prime. Notice that . And . So if , . We simply need this to be greater than what we have for . Notice that for answer choices and , the numerator has less prime factors than the denominator, and so they are less likely to work. We check first, and it works, therefore the answer is .

方法4、We have the following important results:

for all positive integers

for all positive rational numbers

for all positive rational numbers

__Proofs__

Result can be shown by induction.

Result Since powers are just repeated multiplication, we will use result to prove result

Result For all positive rational numbers we haveTherefore, we get So, result is true.

Result For all positive rational numbers we haveIt follows that and result is true.

For all positive integers and suppose and are their prime factorizations, respectively, we have

We apply function on each fraction in the choices:

Therefore, the answer is

## Solution 19

方法1、

The ranges of and are both , which is included in the range of , so we can use it with no issues.

This only happens at on the interval , because one of and must be and the other . Therefore, the answer is

方法2、By the cofunction identity for all we simplify the given equation:for some integer We keep simplifying:By rough constraints, we know that so that The only possibility is From here, we getfor some integer

The **possible** solutions in are but only check the original equation (Note that is an extraneous solution formed by squaring above.). Therefore, the answer is

方法3、Let and This problem is equivalent to counting the intersections of the graphs of and in the closed interval We make a table of values, as shown below:The graph of in (from left to right) is the same as the graph of in (from right to left). The output is from to (from left to right), inclusive, and strictly decreasing.

The graph of in (from left to right) has two parts:

in (from left to right). The output is from to (from left to right), inclusive, and strictly decreasing.

in (from right to left). The output is from to (from left to right), inclusive, and strictly increasing.

If then and So, their graphs do not intersect.

If then Clearly, the graphs intersect at and (at points and respectively), but we will prove/disprove that they are the **only** points of intersection:

Let and It follows that Since we know that by the cofunction identity:

Applying Solution 2's argument (starts from its last block of equations) to deduce that and are the only points of intersection. So, the answer is

## Solution20

Let be the directrix of ; recall that is the set of points such that the distance from to is equal to . Let and be the orthogonal projections of and onto , and further let and be the orthogonal projections of and onto . Because , there are two possible configurations which may arise, and they are shown below.

Set , which by the definition of a parabola also equals . Then as , we have and . Since is a rectangle, , so by Pythagorean Theorem on triangles and ,This equation simplifies to , which has solutions . Both values of work - the smaller solution with the right configuration and the larger solution with the left configuration - and so the requested answer is .

## Solution 21

方法1、The solutions to this equation are , , and . Consider the five points , , and ; these are the five points which lie on . Note that since these five points are symmetric about the -axis, so must .

Now let denote the ratio of the length of the minor axis of to the length of its major axis. Remark that if we perform a transformation of the plane which scales every -coordinate by a factor of , is sent to a circle . Thus, the problem is equivalent to finding the value of such that , , and all lie on a common circle; equivalently, it suffices to determine the value of such that the circumcenter of the triangle formed by the points , , and lies on the -axis.

Recall that the circumcenter of a triangle is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments and arerespectively. These two lines have different slopes for , so indeed they will intersect at some point ; we want . Plugging into the first equation yields , and so plugging into the second equation and simplifying yieldsSolving yields .

Finally, recall that the lengths , , and (where is the distance between the foci of ) satisfy . Thus the eccentricity of is and the requested answer is .

方法2、Completing the square in the original equation, we getfrom which Now, we will find the equation of an ellipse that passes through and in the -plane. By symmetry, the center of must be on the -axis.

The formula of iswith the center at and the axes' lengths and Plugging the points and in, respectively, we get the following system of three equations:Clearing fractions givesSince for all real numbers we rewrite the system asApplying the Transitive Property in and we getApplying the results of and on we getSubstituting this into we get

Substituting the current results into we get

Finally, we haveandOur answer is

## Solution 22

方法1、Part 1: solving for a

is the negation of the sum of roots

The real values of the 7th roots of unity are: and they sum to .

If we subtract 1, and condense identical terms, we get:

Therefore, we have

Part 2: solving for b

is the sum of roots two at a time by Vieta's

We know that

By plugging all the parts in we get:

Which ends up being:

Which was shown in the first part to equal , so

Part 3: solving for c

Notice that

is the negation of the product of roots by Vieta's formulas

Multiply by

Then use sine addition formula backwards:

Finally multiply or .

方法2、Letting the roots be , , and , Vietas givesWe use the Taylor series for ,to approximate the roots. Taking the sum up to yields a close approximation, so we haveNote that these approximations get worse as gets larger, but they will be fine for the purposes of this problem. We then haveWe further approximate these values to , , and (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have . ~ciceronii

Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.

方法3、Note sum of roots of unity equal zero, sum of real parts equal zero, and thus which means

By product to sum, so

By product to sum, so

方法4、Using geometric series, we can show that

Desmos graph of where (the th roots of unity): https://www.desmos.com/calculator/kjnnkhgq6u

Graphically, the imaginary parts of these complex numbers sum to Using the above result, the real parts of these complex numbers sum to too. It follows thatfrom whichas it contributes half the real part of Two solutions follow from here:

4.1 We know that are solutions ofas they can be verified graphically. Now, let It follows thatRewriting from above in terms of we haveIt follows that and

4.2 Let Since is a th root of unity, Graphically, it follows that

Recall that (so that ), and let By Vieta's Formulas and the results above, the answer is

## Solution 23(complementary counting)

We will use complementary counting. First, the frog can go left with probability . We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since , we will ignore the leading probability.

From the left, she either goes left to another edge () or back to the center (). Time for some casework.

She goes back to the center.

Now, she can go in any 4 directions, and then has 2 options from that edge. This gives . --End case 1

She goes to another edge (rightmost).

Subcase 1: She goes back to the left edge. She now has 2 places to go, giving

Subcase 2: She goes to the center. Now any move works.

for this case. --End case 2

She goes back to the center in Case 1 with probability , and to the right edge with probability

So, our answer is

But, don't forget complementary counting. So, we get .

## Solution24

Let be the center of the semicircle, be the center of the circle, and be the midpoint of By the Perpendicular Chord Theorem Converse, we have and Together, points and must be collinear.

Applying the Extended Law of Sines on we havein which the radius of is

By the SAS Congruence, we have both of which are -- triangles. By the side-length ratios, and By the Pythagorean Theorem in we get and By the Pythagorean Theorem on we obtain

As shown above, we construct an altitude of Since and we know that We construct on such that Clearly, is a rectangle. Since by alternate interior angles, we have by the AA Similarity, with ratio of similitude Therefore, we get that and

The area of isand the answer is

## Solution 25

方法1、Consider the prime factorizationBy the Multiplication Principle,Now, we rewrite asAs for all positive integers it follows that for all positive integers and , if and only if So, is maximized if and only ifis maximized.

For every factor with a fixed where the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime we look for the for which is a relative maximum:

Finally, the number we seek is The sum of its digits is

Actually, once we get that is a factor of we know that the sum of the digits of must be a multiple of Only choice is possible.

方法2、Using the answer choices to our advantage, we can show that must be divisible by 9 without explicitly computing , by exploiting the following fact:

**Claim**: If is not divisible by 3, then .

**Proof**: Since is a multiplicative function, we have and . Then

Note that the values and do not have to be explicitly computed; we only need the fact that which is easy to show by hand.

The above claim automatically implies is a multiple of 9: if was not divisible by 9, then which is a contradiction, and if was divisible by 3 and not 9, then , also a contradiction. Then the sum of digits of must be a multiple of 9, so only choice works.

## AMC8/AMC10/AMC12/AIME

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