【全网首发】2021AIME II 真题及答案
文末答案
Problem 1
Find the arithmetic mean of all the threedigit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as or .)
Problem 2
Equilateral triangle has side length . Point lies on the same side of line as such that . The line through parallel to line intersects sides and at points and , respectively. Point lies on such that is between and , is isosceles, and the ratio of the area of to the area of is . Find .
Diagram
Problem 3
Find the number of permutations of numbers such that the sum of five products
Problem 4
There are real numbers and such that is a root of and is a root of These two polynomials share a complex root where and are positive integers and Find
Problem 5
For positive real numbers , let denote the set of all obtuse triangles that have area and two sides with lengths and . The set of all for which is nonempty, but all triangles in are congruent, is an interval . Find .
Problem 6
For any finite set , let denote the number of elements in . Find the number of ordered pairs such that and are (not necessarily distinct) subsets of that satisfy
Problem 7
Let and be real numbers that satisfy the system of equationsThere exist relatively prime positive integers and such thatFind .
Problem 8
An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly 8 moves that ant is at a vertex of the top face on the cube is , where and are relatively prime positive integers. Find
Problem 9
Find the number of ordered pairs such that and are positive integers in the set and the greatest common divisor of and is not .
Problem 10
Two spheres with radii and one sphere with radius are each externally tangent to the other two spheres and to two different planes and . The intersection of planes and is the line . The distance from line to the point where the sphere with radius is tangent to plane is , where and are relatively prime positive integers. Find .
Diagram
Remarks

 Let be the plane that is determined by the centers of the spheres, as shown in the black points. Clearly, the sidelengths of the black dashed triangle are and

 Plane is tangent to the spheres at the green points. Therefore, the blue dashed line segments are the radii of the spheres.
 By symmetry, since planes and are reflections of each other about plane it follows that the three planes are concurrent to line From here, we can conclude all of the following:

 The four black dashed line segments all lie in plane

 The four green solid line segments all lie in plane
 The red point (the foot of the perpendicular from the smallest sphere's center to line ), as well as the other points on line all lie in planes and

Problem 11
A teacher was leading a class of four perfectly logical students. The teacher chose a set of four integers and gave a different number in to each student. Then the teacher announced to the class that the numbers in were four consecutive twodigit positive integers, that some number in was divisible by , and a different number in was divisible by . The teacher then asked if any of the students could deduce what is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of . Find the sum of all possible values of the greatest element of .
Problem 12
A convex quadrilateral has area and side lengths and in that order. Denote by the measure of the acute angle formed by the diagonals of the quadrilateral. Then can be written in the form , where and are relatively prime positive integers. Find .
Problem 13
Find the least positive integer for which is a multiple of .
Problem 14
Let be an acute triangle with circumcenter and centroid . Let be the intersection of the line tangent to the circumcircle of at and the line perpendicular to at . Let be the intersection of lines and . Given that the measures of and are in the ratio the degree measure of can be written as where and are relatively prime positive integers. Find .
Problem 15
Let and be functions satisfyingandfor positive integers . Find the least positive integer such that .
Problem 1
Solution 1
Recall that the arithmetic mean of all the digit palindromes is just the average of the largest and smallest digit palindromes, and in this case the palindromes are and and and is the final answer.
~ math31415926535
Solution 2
For any palindrome , note that is 100A + 10B + A, which is also 101A + 10B. The average for A is 5 since A can be any of 1, 2, 3, 4, 5, 6, 7, 8, or 9. The average for B is 4.5 since B is either 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, the answer is 505 + 45 = .
 ARCTICTURN
Solution 3 (Symmetry and Generalization)
For every threedigit palindrome with and note that must be another palindrome by symmetry. Therefore, we can pair each threedigit palindrome uniquely with another threedigit palindrome so that they sum toFor instances:and so on.
From this symmetry, the arithmetic mean of all the threedigit palindromes is
Remark
By the Multiplication Principle, there are threedigit palindromes in total. Their sum is as we can match them into pairs such that the sum of each pair is
~MRENTHUSIASM
Solution 4 (Very, Very Easy and Quick)
We notice that a threedigit palindrome looks like this
And we know a can be any number from 19, and b can be any number from 09, so there are threedigit palindromes
We want to find the sum of these 90 palindromes and divide it by 90 to find the arithmetic mean
How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: 101a+10b
Thus, all of these 90 palindromes can be broken into this form
Thus, the sum of these 90 palindromes will be , because each a will be in 10 different palindromes (since for each a, there are 10 choices for b). The same logic explains why there is a times 9 when computing the total sum of b.
We get a sum of
But don't compute this! There's no need. Divide this by 90 and you will get
~
Solution 5 (Extremely Fast Solution)
The average values of the first and last digits are each and the average value of the middle digit is so the average of all threedigit palindromes is
Problem 2
Solution 1 (Area Formulas for Triangles)
By anglechasing, is a triangle, and is a triangle.
Let It follows that and By the sidelength ratios in we have and
Let the brackets denote areas. We haveand
We set up and solve an equation for Since it is clear that Therefore, we take the positive square root for both sides:
~MRENTHUSIASM
Solution 2
We express the areas of and in terms of in order to solve for
We let Because is isosceles and is equilateral,
Let the height of be and the height of be Then we have that and
Now we can find and in terms of Because we are given that This allows us to use the sin formula for triangle area: the area of is Similarly, because the area of
Now we can make an equation:
To make further calculations easier, we scale everything down by (while keeping the same variable names, so keep that in mind).
Thus Because we scaled down everything by the actual value of is
~JimY
Solution 3 (Pretty Straightforward)
So, If is isosceles, it means that .
Let
So,
In , , Hence (because )
Therefore,
So,
Now, as we know that the ratio of the areas of and is
Substituting the values, we get
Hence, . Solving this, we easily get
We have taken , Hence,
Arnav Nigam
Solution 4 (Similar Triangles)
Since is isosceles, , and since is equilateral, . Thus, , and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from and to line ; call the meeting points and , respectively. is clearly congruent to both and , and thus each of these new triangles has the same area as . But we can "slide" over to make it adjacent to , thus creating an equilateral triangle whose area has a ratio of 18:8 when compared to (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio 18:8 reduces to 9:4, the ratio of their sides must be 3:2. So, because and represent sides of these triangles, and they add to 840, must equal twofifths of 840, or .
Problem 3
Solution 1
Since is one of the numbers, a product with a in it is automatically divisible by , so WLOG , we will multiply by afterward since any of would be , after some cancelation we see that now all we need to find is the number of ways that is divisible by , since is never divisible by , now we just need to find the number of ways is divisible by , after some calculation you will see that there are ways to choose and in this way. So the desired answer is .
~ math31415926535
Solution 2 (Cyclic Symmetry and Casework)
The expression has cyclic symmetry. Without the loss of generality, let It follows that We have
 are congruent to in some order.
We construct the following table for the case with all values in modulo For Row 1, can be either or and can be either or By the Multiplication Principle, Row 1 produces permutations. Similarly, Rows 2, 5, and 6 each produce permutations.
Together, we get permutations for the case By the cyclic symmetry, the cases and all have the same count. Therefore, the total number of permutations is
~MRENTHUSIASM
Solution 3
WLOG, let So, the terms are divisible by .
We are left with and . We need The only way is when They are or (mod 3)
The numbers left with us are which are (mod ) respectively.
(of or ) (of or ) = .
(of or ) (of or ) =
But, as we have just two and two , Hence, We will have to take and Among these two, we have a and in common, i.e. (because and are common in and )
So, i.e. values.
For each value of we get values for Hence, in total, we have ways.
But any of the can be So,
Problem 4
Solution 1 (Complex Conjugate Root Theorem)
By the Complex Conjugate Root Theorem, the imaginary roots for each of and are a pair of complex conjugates. Let and It follows that the roots of are and the roots of are
We know that
Applying Vieta's Formulas to we have from which
Applying Vieta's Formulas to we have from whichFinally, we get by and
~MRENTHUSIASM
Solution 2 (Somewhat Bashy)
, hence
Also, , hence
satisfies both we can put it in both equations and equate to 0.
In the first equation, we get Simplifying this further, we get
Hence, and
In the second equation, we get Simplifying this further, we get
Hence, and
Comparing (1) and (2),
and
;
Substituting these in gives,
This simplifies to
Hence,
Consider case of :
Also,
(because c = 1) Also, Also, Equation (2) gives
Solving (4) and (5) simultaneously gives
[AIME can not have more than one answer, so we can stop here also 😁... Not suitable for Subjective exam]
Hence,
Arnav Nigam
Solution 3 (Heavy Calculation Solution)
start off by applying vieta's and you will find that and . After that, we have to use the fact that and are roots of and , respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore and and you can set these two equations equal to each other while also substituting the values of , , , and above to give you , then you can rearrange the equation into . With this property, we know that is divisible by therefore that means which results in which finally gives us m=10 mod 21. We can test the first obvious value of which is and we see that this works as we get and . That means your answer will be
Jske25
Solution 4 (Synthetic Division)
We note that and for some polynomials and . Through synthetic division (ignoring the remainder as we can set and to constant values such that the remainder is zero), , and By the complex conjugate root theorem, we know that and share the same roots, and they share the same leading coefficient, so . Therefore, and . Solving the system equations, we get and , so . Finally, by the quadratic formula, we have roots of , so our final answer is
faefeyfa
Solution 5 (Fast and Easy)
We plug 20 into the equation obtaining , likewise, plugging 21 into the second equation gets .
Both equations must have 3 solutions exactly, so the other two solutions must be and .
By Vieta's, the sum of the roots in the first equation is , so must be .
Next, using Vieta's theorem on the second equation, you get: However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with 21, so
Given that is , then is equal to .
Therefore, the answer to the equation is
Problem 5
Solution 1
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given 4 and 10 as the sides, so we know that the 3rd side is between 6 and 14, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles sides are between 6 and exclusive, and the larger bound is between and 14, exclusive. The area of these triangles are from 0 (straight line) to on the first "small bound" and the larger bound is between 0 and 20. is our first equation, and is our 2nd equation. Therefore, the area is between and , so our final answer is .
~ARCTICTURN
Solution 2 (Casework: Detailed Explanation of Solution 1)
If and are the sidelengths of an obtuse triangle with then both of the following must be satisfied:
 Triangle Inequality Theorem:
 Pythagorean Inequality Theorem:
For one such obtuse triangle, let and be its sidelengths and be its area. By the Pythagorean Inequality Theorem, it is clear that We apply casework to its longest side:
Case (1): The longest side has length so
By the Triangle Inequality Theorem, we have from which
By the Pythagorean Inequality Theorem, we have from which
Taking the intersection of these three inequalities produces for this case.
At the obtuse triangle degenerates into a straight line with area at the obtuse triangle degenerates into a right triangle with area Together, we obtain or
Case (2): The longest side has length so
By the Triangle Inequality Theorem, we have from which
By the Pythagorean Inequality Theorem, we have from which
Taking the intersection of these three inequalities produces for this case.
At the obtuse triangle degenerates into a straight line with area at the obtuse triangle degenerates into a right triangle with area Together, we obtain or
Answer
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers are in exactly one of or Taking the exclusive disjunction, the set of all such isfrom which
~MRENTHUSIASM
Solution 3
We have the diagram below.
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for that they yield .
If angle is obtuse, then we have that . This is because is attained at , and the area of the triangle is strictly decreasing as increases beyond . This can be observed fromby noting that is decreasing in .
Then, we note that if is obtuse, we have . This is because we get when , yileding . Then, is decreasing as increases by the same argument as before.
cannot be obtuse since .
Now we have the intervals and for the cases where and are obtuse, respectively. We are looking for the that are in exactly one of these intervals, and because , the desired range isgiving
Solution 4
Note: Archimedes15 Solution which I added an answer here are two cases. Either the and are around an obtuse angle or the and are around an acute triangle. If they are around the obtuse angle, the area of that triangle is as we have and is at most . Note that for the other case, the side lengths around the obtuse angle must be and where we have . Using the same logic as the other case, the area is at most . Square and add and to get the right answer
Problem 6
Solution 1
By PIE, , and after some algebra you see that we need or . WLOG , then for each element there are possibilities, either it is in both and , it is in but not , or it is in neither nor . This gives us possibilities, and we multiply by since it could have also been the other way around. Now we need to subtract the overlaps where , and this case has ways that could happen. It is because each number could be in the subset or it could not be in the subset. So the final answer is .
~ math31415926535
Solution 2
We denote . We denote , , , .
Therefore, and the intersection of any two out of sets , , , is an empty set. Therefore, is a partition of .
Following from our definition of , , , we have .
Therefore, the equation
can be equivalently written as
This equality can be simplified as
Therefore, we have the following three cases: (1) and , (2) and , (3) . Next, we analyze each of these cases, separately.
Case 1: and .
In this case, to count the number of solutions, we do the complementary counting.
First, we count the number of solutions that satisfy .
Hence, each number in falls into exactly one out of these three sets: , , . Following from the rule of product, the number of solutions is .
Second, we count the number of solutions that satisfy and .
Hence, each number in falls into exactly one out of these two sets: , . Following from the rule of product, the number of solutions is .
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy minus the number of solutions that satisfy and , i.e., .
Case 2: and .
This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., .
Case 3: and .
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is .
By putting all cases together, following from the rule of sum, the total number of solutions is equal to
~ Steven Chen (www.professorchenedu.com)
Solution 3 (Casework)
By the Principle of InclusionExclusion (abbreviated as PIE), we have from which we rewrite the given equation asRearranging and applying Simon's Favorite Factoring Trick givefrom which at least one of the following is true:
Let For each value of we will use PIE to count the ordered pairs
Suppose There are ways to choose the elements for These elements must also appear in Next, there are ways to add any number of the remaining elements to (Each element has options: in or not in ). There are ordered pairs for Similarly, there are ordered pairs for
To fix the overcount, we subtract the number of ordered pairs that are counted twice, in which There are such ordered pairs.
Therefore, there areordered pairs for
Two solutions follow from here:
Solution 3.1 (Binomial Theorem)
The answer is
Problem 7
Solution 1
From the fourth equation we get substitute this into the third equation and you get . Hence . Solving we get or . From the first and second equation we get , if , substituting we get . If you try solving this you see that this does not have real solutions in , so must be . So . Since , or . If , then the system and does not give you real solutions. So . Since you already know and , so you can solve for and pretty easily and see that . So the answer is .
~ math31415926535
Solution 2 (Easy Algebra)
We can factor out of the last two equations. Therefore, it becomes . Notice this is just , since . We now have and . We then find in terms of , so . We solve for and find that it is either or . We can now try for these two values, and plug the rest into the equation. Thus, we have . We have and we're done.
~Arcticturn
Solution 3 (Easy and Straightforward Algebra)
can be rewritten as . Hence,
Rewriting , we get . Substitute and solving, we get, call this Equation 1
gives . So, , which implies or call this equation 2.
Substituting Eq 2 in Eq 1 gives,
Solving this quadratic yields that
Now we just try these 2 cases.
For substituting in Equation 1 gives a quadratic in which has roots
Again trying cases, by letting , we get , Hence We know that , Solving these we get or (doesn't matter due to symmetry in a,b)
So, this case yields solutions
Similarly trying other three cases, we get no more solutions, Hence this is the solution for
Finally,
So,
 Arnav Nigam
Solution 4 (Bash: Two Variables, Two Equations)
Number the given equations and in this order. Rearranging and solving for we haveSubstituting into and solving for we getSubstituting and into and simplifying, we rewrite the left side of in terms of and only:Let from whichMultiplying both sides by rearranging, and factoring give Substituting back and completing the squares produceIf then combining this with we know that and are the solutions of the quadratic Since the discriminant is negative, neither nor is a real number.
If then combining this with we know that and are the solutions of the quadratic or from which Substituting into and we obtain and respectively. Together, we haveso the answer is
Problem 8
Solution 1 (Recursion)
For all positive integers let
 be the number of ways to make a sequence of exactly moves, where the last move is from the bottom face to the bottom face.
 be the number of ways to make a sequence of exactly moves, where the last move is from the bottom face to the top face.
 be the number of ways to make a sequence of exactly moves, where the last move is from the top face to the bottom face.
 be the number of ways to make a sequence of exactly moves, where the last move is from the top face to the top face.
The base case occurs at from which
Suppose the ant makes exactly moves. We will perform casework on its last move:

 If its last move is from the bottom face to the bottom face, then its next move has


 way to move from the bottom face to the bottom face.



 way to move from the bottom face to the top face.


 If its last move is from the bottom face to the top face, then its next move has ways to move from the top face to the top face.

 If its last move is from the top face to the bottom face, then its next move has ways to move from the bottom face to the bottom face.
 If its last move is from the top face to the top face, then its next move has

 way to move from the top face to the bottom face.

 way to move from the top face to the top face.

Alternatively, this recursion argument is illustrated below, where each dashed arrow indicates way, and each solid arrow indicates ways:Therefore, we have the following relationships:Using these equations, we recursively fill out the table below:Clearly, there are ways to make exactly moves. To guarantee correctness, note that for each value of the four counts in that column must total up to
Finally, the requested probability isfrom which the answer is
~Arcticturn (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (Casework)
On each move, we can either stay on the level we previously were (stay on the bottom/top) or switch levels (go from top to bottom and vise versa). Since we start on the bottom, ending on the top means that we will have to switch an odd number of times; since we cannot switch twice in a row, over an eightmove period we can either make one or three switches. Furthermore, once we switch to a level we can choose one of two directions of traveling on that level: clockwise or counterclockwise (since we can't go back to our previous move, our first move on the level after switching determines our direction).
 Case 1: one switch. Our one switch can either happen at the start/end of our moves, or in the middle. There are ways to do this, outlined below.
 Subcase 1: switch happens at ends. If our first move is a switch, then there are two ways to determine the direction we travel along the top layer. Multiply by 2 to count for symmetry (last move is a switch) so this case yields possibilities.
 Subcase 2: switch happens in the middle. There are six places for the switch to happen; the switch breaks the sequences of moves into two chains, with each having 2 ways to choose their direction of travel. This case yields possibilities.
 Case 2: three switches. Either two, one, or none of our switches occur at the start/end of our moves. There are ways to do this, outlined below. (Keep in mind we can't have two switches in a row.)
 Subcase 1: start and end with a switch. Since our third switch can't be in moves 2 or 7, there are four ways to place our switch, breaking our sequence into two chains. This case yields possibilities.
 Subcase 2: one of our switches is at the start/end. WLOG our first move is a switch; moves 2 and 8 cannot be switches. We can choose 2 from any of the remaining 5 moves to be switches, but we have to subtract the 4 illegal cases where the two switches are in a row (34, 45, 56, 67). These three switches break our sequence into three chains; accounting for symmetry this case yields possibilities.
 Subcase 3: all our switches are in the middle. We choose 3 from any of the 6 middle moves to be our switches, but have to subtract the cases where at least two of them are in a row. If at least two switches are in a row, there are five places for the group of 2 and four places for the third switch; however this overcounts the case where all three are in a row, which has 4 possibilities. These three switches break our sequence into four chains, so this case yields possibilities.
Our probability is then .
Solution 3 (Faster Recursion)
Define to be the probability that after moves, the ant ends up on the level it started on (assuming the first move is a normal move where the ant can stay or move to the opposite level with half chance each). Note and .
Consider when the ant has moves left. It can either stay on its current level with chance and moves left, or travel to the opposite level with chance and moves left (it must spend another move as it cannot travel back immediately). We then have the recurrence
On the first move, the ant can stay on the bottom level with chance and moves left. Or, it can move to the top level with chance and moves left (it has to spend one on the top as it can not return immediately). So the requested probability is .
Computing we get and , resulting in .
Problem 9
Solution 1
We make use of the (olympiad number theory) lemma that .
Noting , we have (by difference of squares)It is now easy to calculate the answer (with casework on ) as .
~Lcz
Solution 2 (Comprehensive and Generalized)
Claim (Solution 1's Lemma)
If and are positive integers with then
There are two proofs to this claim, as shown below.
~MRENTHUSIASM
Proof 1 (Euclidean Algorithm)
If then from which the claim is clearly true.
Otherwise, let without the loss of generality. Note that for all integers the Euclidean Algorithm states thatWe apply this result repeatedly to reduce the larger number:Continuing, we havefrom which the proof is complete.
~MRENTHUSIASM
Proof 2 (Bézout's Identity)
Let It follows that and
By Bézout's Identity, there exist integers and for which sofrom which We know that
Next, we notice thatSince is a common divisor of and we conclude that from which the proof is complete.
~MRENTHUSIASM
Solution (Detailed Explanation of Solution 1)
By the Euclidean Algorithm, we haveWe are given that Multiplying both sides by giveswhich implies that must have more factors of than does.
We construct the following table for the first positive integers:To count the ordered pairs we perform casework on the number of factors of that has:

 If has factors of then has options and has options. So, this case has ordered pairs.

 If has factor of then has options and has options. So, this case has ordered pairs.

 If has factors of then has options and has options. So, this case has ordered pairs.
 If has factors of then has options and has option. So, this case has ordered pairs.
Together, the answer is
~MRENTHUSIASM
Remark (GCD Property)
In of the solution, we use the following property of the greatest common divisor:
For positive integers and if then
As and are relatively prime (have no prime divisors in common), this property is intuitive.
Problem 10
Solution 1
The centers of the three spheres form a 494972 triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the 72 side of this triangle. Take its midpoint , which is 36 away from the midpoint of the 72 side , and connect these two midpoints.
Now consider the point at which the plane is tangent to the small sphere, and connect with the small sphere's tangent point . Extend through B until it hits the ray from through the center of the small sphere (convince yourself that these two intersect). Call this intersection , the center of the small sphere , we want to find .
By Pythagorus AC= , and we know . We know that must be parallel, using ratios we realize that . Apply Pythagorean theorem on triangle BCD; , so 312 + 23 =
Ross Gao
Solution 2 (Coord Bash)
Let's try to see some symmetry. We can use a coordinate plane to plot where the circles are. The 2 large spheres are externally tangent, so we'll make them at 0, 36, 0 and 0, 36, 0. The center of the little sphere would be x, 0, and 23 since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that x is 24 (since 24 wouldn't make sense). Now, we draw a line through the little sphere and the origin. It also intersects because of the symmetry we created.
lies on the plane too, so these 2 lines must intersect. The point at where it intersects is 24a, 0, and 23a. We can use the distance formula again to find that a = . Therefore, they intersect at . Since the little circle's x coordinate is 24 and the intersection point's x coordinate is , we get  24 = . Therefore, our answer to this problem is 312 + 23 = .
~Arcticturn
Solution 3 (Illustration of Solution 1)
This solution refers to the Diagram section.
As shown below, let be the centers of the spheres (where sphere is the smallest) and be their respective points of tangency to plane Suppose is the foot of the perpendicular from to line so is the perpendicular bisector of We wish to find
As the intersection of planes and is line we know that both and must intersect line Furthermore, since and it follows that from which and are coplanar.
We will focus on crosssections and

 In the threedimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.Clearly, crosssection intersects line at exactly one point. Let the intersection of and line be which must also be the intersection of and line
 In crosssection let be the foot of the perpendicular from to line and be the foot of the perpendicular from to
We have the following diagram:
In crosssection since as discussed, we obtain by AA, with the ratio of similitude Therefore, we get or
In crosssection note that and Applying the Pythagorean Theorem to right we have Moreover, since and we obtain so that by AA, with the ratio of similitude Therefore, we get or
Finally, note that and Since is a rectangle, we have Applying the Pythagorean Theorem to right gives from which the answer is
Problem 11
Solution 1
Let's start by listing the multiples of 6 and 7: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96; 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, and 98. First of all, the multiples of 6 and 7 wouldn't work, so we can cross out 42 and 84. Since the students said they didn't know the answer, it must be that the set has no distinct integers from the sets. Let's list out the possible sets. Since they are in lists of 4, we need multiples of 6 or 7. Let's list them out. 11, 12, 13, 14 12, 13, 14, 15 But if a student has 11 or 15, they would know. So that wouldn't work. The second set would be this: They couldn't know since the multiples of 7 is 35, and if you list it out, they wouldn't know about it since there are other sets such as 35, 36, 37, 38, and 33, 34, 35, 36. The people with these sets would say no, so the second set would WORK. The other sets are only 2 numbers, and since 2 numbers are sufficiently smaller than our 2nd set, we have concluded the answer is 37+50+79+92 = . ~Arcticturn
Solution 2 (Illustrations)
Note that It is clear that and otherwise the three other elements in are divisible by neither nor
In the table below, the multiples of are colored in yellow, and the multiples of are colored in green. By the least common multiple, we obtain cycles: If is a possible maximum value of then must be another possible maximum value of and vice versa. By observations, we circle all possible maximum values of From the second row of the table above, we perform casework on the possible maximum value of Finally, all possibilities for are and from which the answer is
Remarks
 Alternatively, we can reconstruct the second table in this solution as follows, where Y and N denote the replies of "yes" and "no", respectively. Notice that this table has some kind of symmetry!
 As a confirmation, we can verify that each student will be able to deduce what is upon hearing the four replies of "no" in unison. For example, if then all students will know that no one gets or otherwise that student will reply yes (as discussed). Therefore, all students will conclude that has only one possibility.
Problem 12
Solution 1
We denote by , , and four vertices of this quadrilateral, such that , , , . We denote by the point that two diagonals and meet at. To simplify the notation, we denote , , , .
We denote . Hence, and .
First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral .
We havewhere the second equality follows from the formula to use the sine function to compute a triangle area, the the fourth equality follows from the property that .
Because , we have.
Second, we use the law of cosines to establish four equations for four sides of the quadrilateral .
In , following from the law of cosines, we have
Because and , we have
In , following from the law of cosines, we have
Because and , we have
In , following from the law of cosines, we have
Because and , we have
In , following from the law of cosines, we have
Because and , we have
By taking , we get
By taking , we get
Therefore, by writing this answer in the form of , we have and . Therefore, the answer to this question is .
~ Steven Chen (www.professorchenedu.com)
Solution 2
Since we are asked to find , we can find and separately and use their values to get . We can start by drawing a diagram. Let the vertices of the quadrilateral be , , , and . Let , , , and . Let , , , and . We know that is the acute angle formed between the intersection of the diagonals and .
We are given that the area of quadrilateral is . We can express this area using the areas of triangles , , , and . Since we want to find and , we can represent these areas using as follows:
We know that . Therefore it follows that:
From here we see that . Now we need to find . Using the Law of Cosines on each of the four smaller triangles, we get following equations:
We know that . We can substitute this value into our equations to get:
If we subtract the sum of the first and third equation from the sum of the second and fourth equation, the squared terms cancel, leaving us with:
From here we see that .
Since we have figured out and , we can calculate :
Therefore our answer is .
Problem 13
Solution 1
1000 divides this expression iff 8, 125 both divide it. It should be fairly obvious that ; so we may break up the initial condition into two subconditions.
(1) . Notice that the square of any odd integer is 1 modulo 8 (proof by plugging in , , , into modulo 8), so the LHS of this expression goes "5,1,5,1, ...", while the RHS goes "1,2,3,4,5,6,7,8,1, ...". The cycle length of the LHS is 2, RHS is 8, so the cycle length of the solution is . Indeed, the n that solve this equation are exactly those such that .
(2) . This is extremely computationally intensive if you try to calculate all , so instead, we take a divideandconquer approach. In order for this expression to be true is necessary; it shouldn't take too long for you to go through the 20 possible LHSRHS combinations and convince yourself that or .
With this in mind we consider modulo 25. By the generalized Fermat's little theorem, , but we already have n modulo 20! Our calculation is greatly simplified. The LHS cycle length is 20, RHS cycle length is 25, the lcm is 100, in this step we need to test all the numbers between 1 to 100 that or . In the case that , , and RHS goes "3,23,43,63,83"; clearly . In the case that , by a similar argument, .
In the final step, we need to calculate modulo 125. The former is simply ; because modulo 125, . is .This time, LHS cycle is 100, RHS cycle is 125, so we need to figure out what is n modulo 500. It should be .
Put everything together. By the second subcondition, the only candidates < 100 are . Apply the first subcondition, n=797 is the desired answer.
Ross Gao
Solution 2
We have that , or and by CRT. It is easy to check don't work, so we have that . Then, and , so we just have and . Let us consider both of these congruences separately.
First, we look at . By Euler's Totient Theorem (ETT), we have , so . On the RHS of the congruence, the possible values of are all nonnegative integers less than and on the RHS the only possible values are and . However, for to be we must have , a contradiction. So, the only possible values of are when .
Now we look at . Plugging in , we get . Note, for to be satisfied, we must have and . Since as , we have . Then, . Now, we get . Using the fact that , we get . The inverse of modulo is obviously , so , so . Plugging in , we get .
Now, we are finally ready to plug into the congruence modulo . Plugging in, we get . By ETT, we get , so . Then, . Plugging this in, we get , implying the smallest value of is simply . ~rocketsri
Solution 3 (Chinese Remainder Theorem and Binomial Theorem)
We wish to find the least positive integer for which Rearranging givesApplying the Chinese Remainder Theorem, we get the following systems of linear congruences:It is clear that from which we simplify to

 For quick inspections produce that are congruent to modulo respectively. More generally, if is odd, and if is even. As is always odd (so is ), we must have That is, for some nonnegative integer

 For we substitute the result from and simplify:
Note that and so we apply the Binomial Theorem to the left side:Since it follows that
That is, for some nonnegative integer
Substituting this back into we getAs is a multiple of it has at least three factors of Since contributes one factor, it follows that contributes at least two factors, or must be a multiple of Therefore, the least such nonnegative integer is
Finally, combining the two results from above (bolded) generates the least such positive integer at
Problem 14
Solution 1
Let be the midpoint of . Because , and are cyclic, so is the center of the spiral similarity sending to , and . Because , it's easy to get from here.
~Lcz
Solution 2
Let be the midpoint of . Because we have cyclic and so ; likewise since we have cyclic and so . Now note that are collinear since is a median, so . But . Now letting we have and so .
Guessing Solution for last 3 minutes (unreliable)
Notice that looks isosceles, so we assume it's isosceles. Then, let and Taking the sum of the angles in the triangle gives so so the answer is
Problem 15
Solution
Consider what happens when we try to calculate where n is not a square. If for (positive) integer k, recursively calculating the value of the function gives us . Note that this formula also returns the correct value when , but not when . Thus for .
If , returns the same value as . This is because the recursion once again stops at . We seek a case in which , so obviously this is not what we want. We want to have a different parity, or have the same parity. When this is the case, instead returns .
Write , which simplifies to . Notice that we want the expression to be divisible by 3; as a result, . We also want n to be strictly greater than , so . The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is , giving .
Indeed  if we check our answer, it works. Therefore, the answer is
AMC8/AMC10/AMC12/AIME
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