## 2021 AMC 10A 真题及答案

参考答案见文末（仅供参考）

## Problem1

What is the value of

## Problem2

Portia’s high school has times as many students as Lara’s high school. The two high schools have a total of students. How many students does Portia’s high school have?

## 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

A cart rolls down a hill, travelling inches the first second and accelerating so that during each successive -second time interval, it travels inches more than during the previous -second interval. The cart takes seconds to reach the bottom of the hill. How far, in inches, does it travel?

## Problem5

The quiz scores of a class with students have a mean of . The mean of a collection of of these quiz scores is . What is the mean of the remaining quiz scores of terms of ?

## Problem6

Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at miles per hour. She meets Jean at the halfway point. What was Jean’s average speed, in miles per hour, until they meet?

## Problem7

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.

## 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

Which of the following is equivalent to

## Problem11

For which of the following integers is the base- number not divisible by ?

## Problem12

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?

## Problem13

What is the volume of tetrahedron with edge lengths , , , , , and ?

## Problem14

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

## Problem15

Values for and are to be selected from without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves and intersect? (The order in which the curves are listed does not matter; for example, the choices is considered the same as the choices )

## 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 ?

## Diagram

## 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

The area of the region bounded by the graph ofis , where and are integers. What is ?

## Problem20

In how many ways can the sequence be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

## Problem21

Let be an equiangular hexagon. The lines and determine a triangle with area , and the lines and determine a triangle with area . The perimeter of hexagon can be expressed as , where and are positive integers and is not divisible by the square of any prime. What is ?

## Problem22

Hiram’s algebra notes are pages long and are printed on sheets of paper; the first sheet contains pages and , the second sheet contains pages and , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly . How many sheets were borrowed?

## 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

The interior of a quadrilateral is bounded by the graphs of and , where a positive real number. What is the area of this region in terms of , valid for all ?

## Problem25

How many ways are there to place indistinguishable red chips, indistinguishable blue chips, and indistinguishable green chips in the squares of a grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally.

## 1

## Solution 1

Call the different colors A,B,C. There are ways to rearrange these colors to these three letters, so must be multiplied after the letters are permuted in the grid. WLOG assume that A is in the center.In this configuration, there are two cases, either all the A’s lie on the same diagonal:or all the other two A’s are on adjacent corners:In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.

In each case there is only one way to put the three B’s and the three C’s as shown in the diagrams.This means that there are ways to arrange A,B, and C in the grid, and there are 6 ways to rearrange the colors. Therefore, there are ways in total, which is .

-happykeeper

## Solution 2 (Casework)

Without the loss of generality, we place a red ball in the top-left square. There are two cases:

**Case (1):** The two balls adjacent to the top-left red ball have different colors.Each placement has permutations, as there are ways to permute RBG.

There are three sub-cases for Case (1):So, Case (1) has ways.

**Case (2)**: The two balls adjacent to the top-left red ball have the same color.Each placement has permutations, as there are ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB). There are three sub-cases for Case (2):So, Case (2) has ways.

Together, the answer is

~MRENTHUSIASM

## Solution 3 (Casework and Derangements)

**Case (1)**: We have a permutation of R, B, and G as all of the rows. There are ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, , so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are possible permutations for the last row. Thus, there are possibilities.

**Case (2)**: All of the rows have two balls that are the same color and one that is different. There are obviously possible configurations for the first row, for the second, and for the third. .

Therefore, our answer is

2

## Solution 1

The following system of equations can be formed with representing the number of students in Portia’s high school and representing the number of students in Lara’s high school.Substituting with we get . Solving for , we get . Since we need to find we multiply by 3 to get , which is

-happykeeper

## Solution 2 (One Variable)

Suppose Lara’s high school has students. It follows that Portia’s high school has students. We know that or Our answer is

~MRENTHUSIASM

## Solution 3 (Arithmetics)

Clearly, students is times as many students as Lara’s high school. Therefore, Lara’s high school has students, and Portia’s high school has students.

~MRENTHUSIASM

## Solution 4 (Answer Choices)

### Solution 4.1 (Quick Inspection)

The number of students in Portia’s high school must be a multiple of This eliminates and . Since is too small (as is clearly true), we are left with

~MRENTHUSIASM

### Solution 4.2 (Plug in the Answer Choices)

For we have So, is incorrect.

For we have So, is incorrect.

For we have So, is correct. For completeness, we will check choices and

For we have So, is incorrect.

For we have So, is incorrect.

3.

## Solution 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 .

–abhinavg0627

## Solution 2 (Lazy Speed)

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

~CoolJupiter 2021

## Solution 3 (Vertical Addition and Logic)

Let the larger number be It follows that the smaller number is Adding vertically, we haveWorking from right to left, we haveThe larger number is and the smaller number is Their difference is

4.

## Solution 1 (Arithmetic Series)

Sincewe seek the sumin which there are 30 addends. The last addend is Therefore, the requested sum isRecall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely~MRENTHUSIASM

## Solution 2 (Answer Choices and Modular Arithmetic)

From the -term sumin the previous solution, taking modulo givesThe only answer choices that are are and By a quick estimate, is too small, leaving us with

5.

## Solution 1 (Generalized)

The total score in the class is The total score on the quizzes is Therefore, for the remaining quizzes ( of them), the total score is Their mean score is

~MRENTHUSIASM

## Solution 2 (Convenient Values and Observations)

Set The answer is the same as the last student’s quiz score, which is From the answer choices, only yields a negative value for

6.

## Solution 1 (Generalized Distance)

Let miles be the distance from the start to the fire tower. When Chantal meets Jean, she has traveled forhours. Jean also has traveled for hours, and he travels for miles. So, his average speed ismiles per hour.

~MRENTHUSIASM

## Solution 2 (Convenient Distance)

**We use the same template as shown in Solution 1, except that we replace with a concrete number.**

Let miles be the distance from the start to the fire tower. When Chantal meets Jean, she travels forhours. Jean also has traveled for hours, and he travels for miles. So, his average speed is miles per hour.

7.

## Solution 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 .

–abhinavg0627

## Solution 2 (Explains Solution 1 Using Arrows)

We are given that

Combining and into below, we have

Clearly, the answer is

8.

## Solution

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

9.

## Solution 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

## Solution 2 (Beyond Overkill)

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

10.

## Solution 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 .

-Lemonie

## Solution 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 .

-MEWTO

## Solution 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 .

-SmileKat32

## Solution 4 (Engineer’s Induction)

We can compute some of the first few partial products, and notice that . As we don’t have to prove this, we get the product is , and smugly click . ~rocketsri

10.

## Solution 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 .

-Lemonie

## Solution 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 .

-MEWTO

## Solution 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 .

-SmileKat32

## Solution 4 (Engineer’s Induction)

We can compute some of the first few partial products, and notice that . As we don’t have to prove this, we get the product is , and smugly click . ~rocketsri

11.

## Solution

We haveThis expression is divisible by **unless** The only choice congruent to modulo is

12.

## Solution 1 (Use Tables to Organize Information)

__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.)__

### Solution 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

### Solution 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

~MRENTHUSIASM

## Solution 2 (Quick and dirty)

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.

13.

## Solution

Drawing the tetrahedron out and testing side lengths, we realize that the triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid: , so we have an answer of . ~IceWolf10

14.

## Solution 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, . ~JHawk0224

## Solution 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

15.

## Solution 1 (Intuition):

Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn’t matter. Then and . Therefore the number of ways to choose the four integers is , and the answer is . ~IceWolf10

## Solution 2 (Algebra):

Setting , we find that , so by the trivial inequality. This implies that and must both be positive or negative. If two distinct values are chosen for and respectively, there are ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by at the end, however, since the curves aren’t considered distinct. Calculating, we get

16.

## Solution 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 . ~approximation by ciceronii

## Solution 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 ~Lopkiloinm

## Solution 3 (answer choices)

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

17.

## Solution 1

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

~mn28407

## Solution 2 (One Pair of Similar Triangles, then Areas)

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,

~MRENTHUSIASM

## Solution 3 (short)

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

~ ccx09

## Solution 4 – Extending the line

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 , .

18.

## Solution 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

– awesomediabrine

## Solution 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

~JHawk0224 ~awesomediabrine

## Solution 3 (Deeper)

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 .

~yofro

## Solution 4 (Most Comprehensive, Similar to Solution 3)

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

~MRENTHUSIASM

## Solution 5

The problem gives us that f(p)=p. If we let a=p and b=1, we get f(p)=f(p)+f(1), which implies f(1)=0. Notice that the answer choices are all fractions, which means we will have to multiply an integer by a fraction to be able to solve it. Therefore, let’s try plugging in fractions and try to solve them. Note that if we plug in a=p and b=1/p, we get f(1)=f(p)+f(1/p). We can solve for f(1/p) as -f(p)! This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of E.

19.

## Solution 1

In order to attack this problem, we need to consider casework:

Case 1:

Substituting and simplifying, we have , i.e. , which gives us a circle of radius centered at .

Case 2:

Substituting and simplifying again, we have , i.e. . This gives us a circle of radius centered at .

Case 3:

Doing the same process as before, we have , i.e. . This gives us a circle of radius centered at .

Case 4:

One last time: we have , i.e. . This gives us a circle of radius centered at .

After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:

Now, the area of the shaded region is just a square with side length with four semicircles of radius . The area is . The answer is which is

20.

## Solution 1 (Bashing)

We write out the cases. These cases are the ones that work: We count these out and get permutations that work. ~contactbibliophile

## Solution 2 (Casework)

Reading the terms from left to right, we have two cases:

( stands for increase and stands for decrease.)

For note that for the second and fourth terms, one of which must be a and the other one must be a or We have four sub-cases:

For the first two blanks must be and in some order, and the last blank must be a for a total of possibilities. Similarly, also has possibilities.

For there are no restrictions for the numbers and So, we have possibilities. Similarly, also has possibilities.

Together, has possibilities. By symmetry, also has possibilities. Together, the answer is

21.

## Solution (Misplaced problem?)

Note that the extensions of the given lines will determine an equilateral triangle because the hexagon is equiangular. The area of the first triangle is , so the side length is . The area of the second triangle is , so the side length is . We can set the first value equal to and the second equal to by substituting some lengths in with different sides of the same equilateral triangle. The perimeter of the hexagon is just the sum of these two, which is and

22.

## Solution

Suppose the roommate took pages through , or equivalently, page numbers through . Because there are numbers taken,The first possible solution that comes to mind is if , which indeed works, giving and . The answer is

~Lcz

## Solution 2 (Different Variable Choice, Similar Logic)

Suppose the smallest page number removed is and pages are removed. It follows that the largest page number removed is

**Remarks:**

1. pages are removed means that sheets are removed, from which must be even.

2. must be odd, as the smallest page number removed is on the right side (odd-numbered).

3.

4. The sum of the page numbers removed is

Together, we haveThe factors of areSince is even, we only have a few cases to consider:

Since only are possible:

If then the notebook will run out if we take pages starting from page

If then the average page number of the remaining pages will be undefined, as there is no page remaining (after taking pages starting from page ).

So, the only possibility is from which pages are taken out, which is sheets.

23.

## Solution 1 (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 . ~ firebolt360

Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360

## Solution 2 (direct counting and probability states)

We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have , so the answer is . ~IceWolf10

## Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)

**Denominator**

There are ways to make hops without restrictions.

**Numerator (Casework)**

Suppose Frieda makes hops without stopping. We perform casework on which hop reaches a corner for the first time.

Hop (Hops and have no restrictions)

The independent hops have options, respectively. So, this case has ways.

Hop (Hop has no restriction)

No matter which direction the first hop takes, the second hop must “wrap around”.

The independent hops have options, respectively. So, this case has ways.

Hop

Two sub-cases:

The second hop “wraps around”. It follows that the third hop also “wraps around”.

The independent hops have options, respectively. So, this sub-case has ways.

The second hop backs to the center.

The independent hops have options, respectively. So, this sub-case has ways.

Together, Case has ways.

The numerator is

**Probability**

This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3

~MRENTHUSIASM

## Solution 4

Let be the probability that Frieda is on the central square after n moves, be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in specific direction out of total directions from the middle of an edge, so . The ways to reach the middle of an edge are by moving in any direction from the center or by moving in specific direction from the middle of an edge, so . The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in specific directions from the middle of an edge, so . Since Frieda always start from the center, , , and . We use the previous formulas to work out and find it to be .

## Solution 5

Imagine an infinite grid of by squares such that there is a by square centered at for all ordered pairs of integers

It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at . (minus the teleportations) Since counting the complement set is easier, we’ll count the number of -step paths such that Frieda never reaches a corner point.

In other words, since the reachable corner points are and Frieda can only travel along the collection of points included in , where is all points on and such that and , respectively, plus all points on the big square with side length centered at We then can proceed with casework:

Case : Frieda never reaches nor

When Frieda only moves horizontally or vertically for her four moves, she can do so in ways for each case . Thus, total paths for the subcase of staying in one direction. (For instance, all length combinations of and except , , , and for the horizontal direction.)

There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are , , , and Thus, a total of ways for this subcase.

Total for Case :

Case : Frieda reaches or .

Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of paths for this case.

Our total number of paths never reaching coroners is thus making for an answer of

-fidgetboss_4000

## Solution 6 (Casework)

We take cases on the number of hops needed to reach a corner. For simplicity, denote as a move that takes Frieda to an edge, as wrap-around move and as a corner move. Also, denote as a move that takes us to the center.

**2 Hops**

Then, Frieda will have to as her set of moves. There are ways to move to an edge, and corners to move to, for a total of cases here. Then, there are choices for each move, for a probability of .

**3 Hops**

In this case, Frieda must wrap-around. There’s only one possible combination, just . There are ways to move to an edge, way to wrap-around (you must continue in the same direction) and corners, for a total of cases here. Then, there are choices for each move, for a probability of .

**4 Hops**

Lastly, there are two cases we must consider here. The first case is , and the second is . For the first case, there are ways to move to an edge, way to return to the center, ways to move to an edge once again, and ways to move to a corner. Hence, there is a total of cases here. Then, for the second case, there are ways to move to a corner, way to wrap-around, way to wrap-around again, and ways to move to a corner. This implies there are cases here. Then, there is a total of cases, out of a total of cases, for a probability of .

Then, the total probability that Frieda ends up on a corner is , corresponding to choice . ~rocketsri

## Solution 7

I denote 3×3 grid by

– HOME square (x1)

– CORN squares (x4)

– SIDE squares (x4)

Transitions:

– HOME always move to SIDE

– CORN is DONE

– SIDE move to CORN with move to SIDE with and move to CORN with

After one move, will be on square

After two moves, will be

After three moves, will be

After four moves, probability on CORN will be

24.

## Solution 1

The conditions and give and or and . The slopes here are perpendicular, so the quadrilateral is a rectangle. Plug in and graph it. We quickly see that the area is , so the answer can’t be or by testing the values they give (test it!). Now plug in . We see using a ruler that the sides of the rectangle are about and . So the area is about . Testing we get which is clearly less than , so it is out. Testing we get which is near our answer, so we leave it. Testing we get , way less than , so it is out. So, the only plausible answer is ~firebolt360

## Solution 2 (Casework)

For the equation the cases are

This is a line with -intercept -intercept and slope

This is a line with -intercept -intercept and slope

For the equation the cases are

This is a line with -intercept -intercept and slope

This is a line with -intercept -intercept and slope

Plugging into the choices gives

Plugging into the four above equations and solving systems of equations for intersecting lines [ and and and and ], we get the respective solutions

### Solution 2.1 (Rectangle)

Since the slopes of the intersecting lines (from the four above equations) are negative reciprocals, the quadrilateral is a rectangle. Finally, by the Distance Formula, the length and width of the rectangle are and The area we seek is

The answer is

~MRENTHUSIASM

### Solution 2.2 (Shoelace Formula)

Even if we do not recognize that the solutions form the vertices of a rectangle, we can apply the Shoelace Formula on ** consecutive** vertices

The area formula isTherefore, the answer is

Suggested Reading for the Shoelace Formula: https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem

~MRENTHUSIASM

## Solution 3 (Geometry)

Similar to Solution 2, we will use the equations of the four cases:

(1) This is a line with -intercept , -intercept , and slope

(2) This is a line with -intercept , -intercept , and slope

(3)* This is a line with -intercept , -intercept , and slope

(4)* This is a line with -intercept , -intercept , and slope

The area of the rectangle created by the four equations can be written as

=

=

=

(Note: slope )

-fnothing4994

## Solution 4 (bruh moment solution)

Trying narrows down the choices to options , and . Trying and eliminates and , to obtain as our answer. -¢

25.

## Solution 1

Call the different colors A,B,C. There are ways to rearrange these colors to these three letters, so must be multiplied after the letters are permuted in the grid. WLOG assume that A is in the center.In this configuration, there are two cases, either all the A’s lie on the same diagonal:or all the other two A’s are on adjacent corners:In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.

In each case there is only one way to put the three B’s and the three C’s as shown in the diagrams.This means that there are ways to arrange A,B, and C in the grid, and there are 6 ways to rearrange the colors. Therefore, there are ways in total, which is .

-happykeeper

## Solution 2 (Casework)

Without the loss of generality, we place a red ball in the top-left square. There are two cases:

**Case (1):** The two balls adjacent to the top-left red ball have different colors.Each placement has permutations, as there are ways to permute RBG.

There are three sub-cases for Case (1):So, Case (1) has ways.

**Case (2)**: The two balls adjacent to the top-left red ball have the same color.Each placement has permutations, as there are ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB). There are three sub-cases for Case (2):So, Case (2) has ways.

Together, the answer is

~MRENTHUSIASM

## Solution 3 (Casework and Derangements)

**Case (1)**: We have a permutation of R, B, and G as all of the rows. There are ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, , so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are possible permutations for the last row. Thus, there are possibilities.

**Case (2)**: All of the rows have two balls that are the same color and one that is different. There are obviously possible configurations for the first row, for the second, and for the third. .

Therefore, our answer is