**2022AIME I 真题及答案**

## Problem1

Quadratic polynomials and have leading coefficients and respectively. The graphs of both polynomials pass through the two points and Find

## Solution 1 (Linear Polynomials)

Let Since the -terms of and cancel, we conclude that is a linear polynomial.

Note thatso the slope of is

It follows that the equation of isfor some constant and we wish to find

We substitute into this equation to get from which

~MRENTHUSIASM

## Solution 2 (Quadratic Polynomials)

Letfor some constants and

We are given thatand we wish to findWe need to cancel and Since we subtract from to get~MRENTHUSIASM

## Solution 3 (Pure Brute Force)

Let

By substitutes and into these equations, we can get:Hence, and .

Similarly,Hence, and .

Notice that and . Therefore

~Littlemouse

## Problem2

Find the three-digit positive integer whose representation in base nine is where and are (not necessarily distinct) digits.

## Solution 1

We are given thatwhich rearranges toTaking both sides modulo we haveThe only solution occurs at from which

Therefore, the requested three-digit positive integer is

~MRENTHUSIASM

## Solution 2

As shown in Solution 1, we get .

Note that and are large numbers comparatively to , so we hypothesize that and are equal and fills the gap between them. The difference between and is , which is a multiple of . So, if we multiply this by , it will be a multiple of and thus the gap can be filled. Therefore, the only solution is , and the answer is .

~KingRavi

## Solution 3

As shown in Solution 1, we get

We list a few multiples of out:Of course, can't be made of just 's. If we use one , we get a remainder of , which can't be made of 's either. So doesn't work. can't be made up of just 's. If we use one , we get a remainder of , which can't be made of 's. If we use two 's, we get a remainder of , which can be made of 's. Therefore we get so and . Plugging this back into the original problem shows that this answer is indeed correct. Therefore,

~Technodoggo

## Solution 4

As shown in Solution 1, we get .

We can see that is larger than , and we have an . We can clearly see that is a multiple of , and any larger than would result in being larger than . Therefore, our only solution is . Our answer is .

~Arcticturn

## Problem3

In isosceles trapezoid , parallel bases and have lengths and , respectively, and . The angle bisectors of and meet at , and the angle bisectors of and meet at . Find .

## Diagram

~MRENTHUSIASM ~ihatemath123

## Solution 1

We have the following diagram:

Let and be the points where and extend to meet , and be the height of . As proven in Solution 2, triangles and are congruent right triangles. Therefore, . We can apply this logic to triangles and as well, giving us . Since , .

Additionally, we can see that is similar to and . We know that . So, we can say that the height of the triangle is while the height of the triangle is . After that, we can figure out the distance from to and the height of triangle .

Finally, since the ratio between the height of to the height of is and is ,

~Cytronical

## Solution 2

Extend line to meet at and at . The diagram looks like this:Because the trapezoid is isosceles, by symmetry is parallel to and . Therefore, by interior angles and by the problem statement. Thus, is isosceles with . By symmetry, is also isosceles, and thus . Similarly, the same thing is happening on the right side of the trapezoid, and thus is the midline of the trapezoid. Then, .

Since and , we have . The length of the midline of a trapezoid is the average of their bases, so . Finally, .

~KingRavi

## Solution 3

We have the following diagram:Extend lines and to meet line at points and , respectively, and extend lines and to meet at points and , respectively.

Claim: quadrilaterals and are rhombuses.

Proof: Since , . Therefore, triangles , , and are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, is congruent to the other three. Therefore, , so is a rhombus. By symmetry, is also a rhombus.

Extend line to meet and at and , respectively. Because of rhombus properties, . Also, by rhombus properties, and are the midpoints of segments and , respectively; therefore, by trapezoid properties, . Finally, .

~ihatemath123

## Solution 4

Let and be the feet of the altitudes from and , respectively, to , and let and be the feet of the altitudes from and , respectively, to . Side is parallel to side , so is a rectangle with width . Furthermore, because and trapezoid is isosceles, .

Also because is isosceles, is half the total sum of angles in , or . Since and bisect and , respectively, we have , so .

Letting , applying Pythagoras to yields . We then proceed using similar triangles: and , so by AA similarity . Likewise, and , so by AA similarity . Thus .

Adding our two equations for and gives . Therefore, the answer is .

~Orange_Quail_9

## Solution 5

This will be my first solution on AoPS. My apologies in advance for any errors.

Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that is equidistant from and and is equidistant from and If we let the feet of the altitudes from to and be called and respectively, we can say that Analogously, we let the feet of the altitudes from to and be and respectively. Thus, Because is an isosceles trapezoid, we can say that all of the altitudes are equal to each other.

By SA as well as SS congruence for right triangles, we find that triangles and are congruent. Similarly, and by the same reasoning. Additionally, since and are congruent rectangles.

If we then let let and let we can create the following system of equations with the given side length information:Adding the first two equations, subtracting by twice the second, and dividing by yields

~regular

## Problem4

Let and where Find the number of ordered pairs of positive integers not exceeding that satisfy the equation

## Solution 1

We rewrite and in polar form:The equation becomesfor some integer

Since and we conclude thatNote that the values for and the values for have one-to-one correspondence.

We apply casework to the values for

There are values for so there are values for It follows that so there are values for

There are ordered pairs in this case.

There are values for so there are values for It follows that so there are values for

There are ordered pairs in this case.

There are values for so there are values for It follows that so there are values for

There are ordered pairs in this case.

Together, the answer is

~MRENTHUSIASM

## Solution 2

First we recognize that and because the cosine and sine sums of those angles give the values of and , respectively. By Demoivre's theorem, . When you multiply by , we can think of that as rotating the complex number counterclockwise in the complex plane. Therefore, by the equation we know that and land on the same angle.

This means thatwhich we can simplify toNotice that this means that cycles by for every value of . This is because once hits , we get an angle of and the angle laps onto itself again. By a similar reasoning, laps itself every times, which is much easier to count. By listing the possible values out, we get the pairs :We have columns in total: values for the first column, for the second, for the third, and then for the fourth, for the fifth, for the sixth, etc. Therefore, this cycle repeats every columns and our total sum is .

~KingRavi

## Problem5

A straight river that is meters wide flows from west to east at a rate of meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of meters downstream from Sherry. Relative to the water, Melanie swims at meters per minute, and Sherry swims at meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find .

## Solution 1 (Euclidean)

Define as the number of minutes they swim for.

Let their meeting point be . Melanie is swimming against the current, so she must aim upstream from point , to compensate for this; in particular, since she is swimming for minutes, the current will push her meters downstream in that time, so she must aim for a point that is meters upstream from point . Similarly, Sherry is swimming downstream for minutes, so she must also aim at point to compensate for the flow of the current.

If Melanie and Sherry were to both aim at point in a *currentless* river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are and meters. We can draw out this new scenario, with the dimensions that we have:(While it is indeed true that the triangle above with side lengths , and is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)

By Pythagorean, we have

Subtracting the first equation from the second gives us , so . Substituting this into our first equation, we have that

So .

~ihatemath123

## Solution 2 (Vectors)

We have the following diagram:Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.

Let and be some positive numbers. We have the following table:Recall that soWe subtract from to get from which Substituting this into either equation, we have

It follows that Melanie and Sherry both swim for minutes. Therefore, the answer is~MRENTHUSIASM

## Problem5

Find the number of ordered pairs of integers such that the sequenceis strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.

## Solution 1

Since and cannot be an arithmetic progression, or can never be . Since and cannot be an arithmetic progression, can never be . Since , there are ways to choose and with these two restrictions in mind.

However, there are still specific invalid cases counted in these pairs . Sincecannot form an arithmetic progression, .cannot be an arithmetic progression, so ; however, since this pair was not counted in our , we do not need to subtract it off.cannot form an arithmetic progression, so .cannot form an arithmetic progression, so .cannot form an arithmetic progression, ; however, since this pair was not counted in our (since we disallowed or to be ), we do not to subtract it off.

Also, the sequences , , , , and will never be arithmetic, since that would require and to be non-integers.

So, we need to subtract off progressions from the we counted, to get our final answer of .

~ ihatemath123

## Solution 2 (Rigorous)

We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.

We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of , which is impossible.)

If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.

Finally, we count the sequences that are something like (one of 3,4,5,), , (one of 30, 40, 50). If this is to be the case, then let be the starting value in the sequence. The sequence will be ; We see that if we subtract the largest term by the smallest term we have , so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are . Of these, only the last is invalid because it gives , larger than our bounds . Therefore, we subtract from this case.

Our final answer is

~KingRavi

## Solution 3

Denote .

Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .

Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .

Hence, is a subset of , such that there exists an arithmetic sequence that has 4 terms and includes both and .

Hence, this problem asks us to compute

First, we compute .

We have .

Second, we compute .

: .

We have . Thus, the number of solutions is 21.

: .

We have . Thus, the number of solutions is 9.

Thus, .

Third, we compute .

In , we have . However, because , we have . Thus, .

This implies . Thus, .

Fourth, we compute .

: In the arithmetic sequence, the two numbers beyond and are on the same side of and .

Hence, . Therefore, the number solutions in this case is 3.

: In the arithmetic sequence, the two numbers beyond and are on the opposite sides of and .

: The arithmetic sequence is .

Hence, .

: The arithmetic sequence is .

Hence, .

: The arithmetic sequence is .

Hence, .

Putting two cases together, .

Therefore,

~Steven Chen (www.professorchenedu.com)

## Solution 4

divide cases into .(Notice that can't be equal to , that's why I divide them into two parts. There are three cases that arithmetic sequence forms: .(NOTICE that IS NOT A VALID SEQUENCE!) So when , there are possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)

When , there are ways.

In all, there are possible sequences.

~bluesoul

## Problem6

Find the number of ordered pairs of integers such that the sequenceis strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.

## Solution 1

Since and cannot be an arithmetic progression, or can never be . Since and cannot be an arithmetic progression, can never be . Since , there are ways to choose and with these two restrictions in mind.

However, there are still specific invalid cases counted in these pairs . Sincecannot form an arithmetic progression, .cannot be an arithmetic progression, so ; however, since this pair was not counted in our , we do not need to subtract it off.cannot form an arithmetic progression, so .cannot form an arithmetic progression, so .cannot form an arithmetic progression, ; however, since this pair was not counted in our (since we disallowed or to be ), we do not to subtract it off.

Also, the sequences , , , , and will never be arithmetic, since that would require and to be non-integers.

So, we need to subtract off progressions from the we counted, to get our final answer of .

~ ihatemath123

## Solution 2 (Rigorous)

We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.

We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of , which is impossible.)

If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.

Finally, we count the sequences that are something like (one of 3,4,5,), , (one of 30, 40, 50). If this is to be the case, then let be the starting value in the sequence. The sequence will be ; We see that if we subtract the largest term by the smallest term we have , so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are . Of these, only the last is invalid because it gives , larger than our bounds . Therefore, we subtract from this case.

Our final answer is

~KingRavi

## Solution 3

Denote .

Denote by a subset of , such that there exists an arithmetic sequence that has 4 terms and includes but not .

Hence, is a subset of , such that there exists an arithmetic sequence that has 4 terms and includes both and .

Hence, this problem asks us to compute

First, we compute .

We have .

Second, we compute .

: .

We have . Thus, the number of solutions is 21.

: .

We have . Thus, the number of solutions is 9.

Thus, .

Third, we compute .

In , we have . However, because , we have . Thus, .

This implies . Thus, .

Fourth, we compute .

: In the arithmetic sequence, the two numbers beyond and are on the same side of and .

Hence, . Therefore, the number solutions in this case is 3.

: In the arithmetic sequence, the two numbers beyond and are on the opposite sides of and .

: The arithmetic sequence is .

Hence, .

: The arithmetic sequence is .

Hence, .

: The arithmetic sequence is .

Hence, .

Putting two cases together, .

Therefore,

~Steven Chen (www.professorchenedu.com)

## Solution 4

divide cases into .(Notice that can't be equal to , that's why I divide them into two parts. There are three cases that arithmetic sequence forms: .(NOTICE that IS NOT A VALID SEQUENCE!) So when , there are possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)

When , there are ways.

In all, there are possible sequences.

~bluesoul

## Problem7

Let be distinct integers from to The minimum possible positive value ofcan be written as where and are relatively prime positive integers. Find

## Solution

To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that

If we minimize the numerator, then Note that so It follows that and are consecutive composites with prime factors no other than and The smallest values for and are and respectively. So, we have and from which

If we do not minimize the numerator, then Note that

Together, we conclude that the minimum possible positive value of is Therefore, the answer is

~MRENTHUSIASM ~jgplay

## Problem8

Equilateral triangle is inscribed in circle with radius Circle is tangent to sides and and is internally tangent to Circles and are defined analogously. Circles and meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of are the vertices of a large equilateral triangle in the interior of and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of The side length of the smaller equilateral triangle can be written as where and are positive integers. Find

## Diagram

~MRENTHUSIASM ~ihatemath123

## Solution 1 (Coordinate Geometry)

We can extend and to and respectively such that circle is the incircle of .Since the diameter of the circle is the height of this triangle, the height of this triangle is . We can use inradius or equilateral triangle properties to get the inradius of this triangle is (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is .

Let be the center of the largest circle. We will set up a coordinate system with as the origin. The center of will be at because it is directly beneath and is the length of the larger radius minus the smaller radius, or . By rotating this point around , we get the center of . This means that the magnitude of vector is and is at a degree angle from the horizontal. Therefore, the coordinates of this point are and by symmetry the coordinates of the center of is .

The upper left and right circles intersect at two points, the lower of which is . The equations of these two circles are:We solve this system by subtracting to get . Plugging back in to the first equation, we have . Since we know is the lower solution, we take the negative value to get .

We can solve the problem two ways from here. We can find by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find as they lie on the same vertical, is degrees so we can make use of triangles, and because is the center of triangle . We can draw the diagram as such:Note that . It follows thatFinally, the answer is .

~KingRavi

## Solution 2 (Euclidean Geometry)

For equilateral triangle with side length , height , and circumradius , there are relationships: , , and .

There is a lot of symmetry in the figure. The radius of the big circle is , let the radius of the small circles , , be .

We are going to solve this problem in steps:

We have is a triangle, and , ( and are tangent), and . So, we get and .

Since and are tangent, we get .

Note that is an equilateral triangle, and is its center, so .

Note that is an isosceles triangle, so

In , Power of a Point gives and .

It follows that . We solve this quadratic equation: .

Since is the circumradius of equilateral , we have .

Therefore, the answer is .

~isabelchen

## Problem9

Ellina has twelve blocks, two each of red (), blue (), yellow (), green (), orange (), and purple (). Call an arrangement of blocks if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangementis even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is where and are relatively prime positive integers. Find

## Solution 1

Consider this position chart:Since there has to be an even number of spaces between each ball of the same color, spots , , , , , and contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of possible arrangements, so the probability is:which is in simplest form. So, .

~Oxymoronic15

## Solution 2

We can simply use constructive counting. First, let us place the red balls; choose the first slot in ways, and the second in ways, because the number is cut in half due to the condition in the problem. This gives ways to place the blue balls. Similarly, there are ways to place the blue balls, and so on, until there are ways to place the purple balls. Thus, the probability isand the desired answer extraction is .

~A1001

## Problem10

Three spheres with radii , , and are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at , , and , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that . Find .

## Diagrams

## Solution 1

We let be the plane that passes through the spheres and and be the centers of the spheres with radii and . We take a cross-section that contains and , which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information, . Since is a trapezoid, we can drop an altitude from to to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is and let the distance from to be . Then we have .

We have because of the rectangle, so . Squaring, we have . Subtracting, we get . We also notice that since we had means that and since we know that , .

We now look at our second diagram.

. Since , we have . Using Pythagorean theorem, . Therefore,

~KingRavi

## Solution 2

Let the distance between the center of the sphere to the center of those circular intersections as separately. . According to the problem, we have . After solving we have , plug this back to

The desired value is

~bluesoul

## Solution 3

Denote by the radius of three congruent circles formed by the cutting plane. Denote by , , the centers of three spheres that intersect the plane to get circles centered at , , , respectively.

Because three spheres are mutually tangent, , .

We have , , .

Because and are perpendicular to the plane, is a right trapezoid, with .

Hence,

Recall that

Hence, taking , we get

Solving (1) and (3), we get and .

Thus, .

Thus, .

Because and are perpendicular to the plane, is a right trapezoid, with .

Therefore,

In our solution, we do not use the conditio that spheres and are externally tangent. This condition is redundant in solving this problem.

~Steven Chen (www.professorcheneeu.com)

## Problem10

Three spheres with radii , , and are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at , , and , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that . Find .

## Diagrams

## Solution 1

We let be the plane that passes through the spheres and and be the centers of the spheres with radii and . We take a cross-section that contains and , which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information, . Since is a trapezoid, we can drop an altitude from to to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is and let the distance from to be . Then we have .

We have because of the rectangle, so . Squaring, we have . Subtracting, we get . We also notice that since we had means that and since we know that , .

We now look at our second diagram.

. Since , we have . Using Pythagorean theorem, . Therefore,

~KingRavi

## Solution 2

Let the distance between the center of the sphere to the center of those circular intersections as separately. . According to the problem, we have . After solving we have , plug this back to

The desired value is

~bluesoul

## Solution 3

Denote by the radius of three congruent circles formed by the cutting plane. Denote by , , the centers of three spheres that intersect the plane to get circles centered at , , , respectively.

Because three spheres are mutually tangent, , .

We have , , .

Because and are perpendicular to the plane, is a right trapezoid, with .

Hence,

Recall that

Hence, taking , we get

Solving (1) and (3), we get and .

Thus, .

Thus, .

Because and are perpendicular to the plane, is a right trapezoid, with .

Therefore,

In our solution, we do not use the conditio that spheres and are externally tangent. This condition is redundant in solving this problem.

## Problem11

Let be a parallelogram with . A circle tangent to sides , , and intersects diagonal at points and with , as shown. Suppose that , , and . Then the area of can be expressed in the form , where and are positive integers, and is not divisible by the square of any prime. Find .

## Solution 1 (No trig)

Let's redraw the diagram, but extend some helpful lines.

We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of . Then . Similarly, the power of and . We let and label the diagram accordingly.

Notice that because . Let be the center of the circle. Since and intersect and , respectively, at right angles, we have is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from to and to , and both are equal to . Since , . Since and . We can now use Pythagorean theorem on ; we have and .

We know that because is a parallelogram. Using Pythagorean theorem on , . Therefore, base . Thus the area of the parallelogram is the base times the height, which is and the answer is

~KingRavi

## Solution 2

Let the circle tangent to at separately, denote that

Using POP, it is very clear that , let , using LOC in ,, similarly, use LOC in , getting that . We use the second equation to minus the first equation, getting that , we can get .

Now applying LOC in , getting , solving this equation to get , then , , the area is leads to

~bluesoul

## Solution 3

Denote by the center of the circle. Denote by the radius of the circle. Denote by , , the points that the circle meets , , at, respectively.

Because the circle is tangent to , , , , , , .

Because , , , are collinear.

Following from the power of a point, . Hence, .

Following from the power of a point, . Hence, .

Denote . Because and are tangents to the circle, .

Because is a right trapezoid, . Hence, . This can be simplified as \[ 6 x = r^2 . \hspace{1cm} (1) \]

In , by applying the law of cosines, we have \begin{align*} AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\ & = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\ & = AB^2 + CB^2 + 2 AB \cdot CB \cdot \frac{AE - BF}{AB} \\ & = AB^2 + CB^2 + 2 CB \left( AE - BF \right) \\ & = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\ & = 24 x + 676 . \end{align*}

Because , we get . Plugging this into Equation (1), we get .

Therefore, \begin{align*} {\rm Area} \ ABCD & = CB \cdot EF \\ & = \left( 20 + x \right) \cdot 2r \\ & = 147 \sqrt{3} . \end{align*}

Therefore, the answer is .

~Steven Chen (www.professorchenedu.com)

## Solution 4

Let be the circle, let be the radius of , and let the points at which is tangent to , , and be , , and , respectively. Note that PoP on and with respect to yields and . We can compute the area of in two ways:

1. By the half-base-height formula, .

2. We can drop altitudes from the center of to , , and , which have lengths , , and . Thus, .

Equating the two expressions for and solving for yields .

Let . By the Parallelogram Law, . Solving for yields . Thus, , for a final answer of .

~ Leo.Euler

## Problem12

For any finite set , let denote the number of elements in . Definewhere the sum is taken over all ordered pairs such that and are subsets of with . For example, because the sum is taken over the pairs of subsetsgiving . Let , where and are relatively prime positive integers. Find the remainder when is divided by 1000.

## Solution 1 (Easy to Understand)

Let's try out for small values of to get a feel for the problem. When is obviously . The problem states that for is . Let's try it out for .

Let's perform casework on the number of elements in .

In this case, the only possible equivalencies will be if they are the exact same element, which happens times.

In this case, if they share both elements, which happens times, we will get for each time, and if they share only one element, which also happens times, we will get for each time, for a total of for this case.

In this case, the only possible scenario is that they both are the set , and we have for this case.

In total, .

Now notice, the number of intersections by each element , or in general, is equal for each element because of symmetry - each element when adds to the answer. Notice that - let's prove that (note that you can assume this and answer the problem if you're running short on time in the real test).

Let's analyze the element - to find a general solution, we must count the number of these subsets that appears in. For to be in both and , we need and (Basically, both sets contain and another subset of through not including ).

For any that is the size of both and , the number of ways to choose the subsets and is for both subsets, so the total number of ways to choose the subsets are . Now we sum this over all possible 's to find the total number of ways to form sets and that contain . This is equal to . This is a simplification of Vandermonde's identity, which states that . Here, , and are all , so this sum is equal to . Finally, since we are iterating over all 's for values of , we have , proving our claim.

We now plug in to the expression we want to find. This turns out to be . Expanding produces .

After cancellation, we have

and don't have any common factors with , so we're done with the simplification. We want to find

~KingRavi

## Linearity of Expectation Solution

We take cases based on the number of values in each of the subsets in the pair. Suppose we have elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be by linearity of expectation because for each of the elements, there is a probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by . Summing, we getNotice that we can rewrite this asWe can simplify this using Vandermonde's identity to get . Evaluating this for and givesEvaluating the numerators and denominators mod gives

- pi_is_3.14

## Solution 2 (Rigorous)

For each element , denote , where (resp. ).

Denote .

Denote .

Hence,

Therefore,

This is in the lowest term. Therefore, modulo 1000,

~Steven Chen (www.professorchenedu.com

## Problem13

Let be the set of all rational numbers that can be expressed as a repeating decimal in the form where at least one of the digits or is nonzero. Let be the number of distinct numerators obtained when numbers in are written as fractions in lowest terms. For example, both and are counted among the distinct numerators for numbers in because and Find the remainder when is divided by

## Solution 1

, .

Then we need to find the number of positive integers less than 10000 can meet the requirement.Suppose the number is x.

Case 1: (9999, x)=1. Clearly x satisfies.

Case 2: 3|x but x is not a multiple of 11 or 101. Then the least value of abcd is 9x, so that , 334 values from 3 to 1110.

Case 3: 11|x but x is not a multiple of 3 or 101. Then the least value of abcd is 11x, so that , 55 values from 11 to 902.

Case 4: 101|x. None.

Case 5: 3, 11|x. Then the least value of abcd is 11x, 3 values from 33 to 99.

To sum up, the answer is

## Problem

Let and be positive real numbers satisfying the system of equations:Then can be written as where and are relatively prime positive integers. Find

## Solution 1 (geometric interpretation)

First, we note that we can let a triangle exist with side lengths , , and opposite altitude . This shows that the third side, which is the nasty square-rooted sum, is going to have the length equal to the sum on the right - let this be for symmetry purposes. So, we note that if the angle opposite the side with length has a value of , then the altitude has length and thus so and the triangle side with length is equal to .

We can symmetrically apply this to the two other triangles, and since by law of sines, we have is the circumradius of that triangle. Hence. we calculate that with , and , the angles from the third side with respect to the circumcenter are , and . This means that by half angle arcs, we see that we have in some order, , , and (not necessarily this order, but here it does not matter due to symmetry), satisfying that , , and . Solving, we get , , and .

We notice that

- kevinmathz

## Solution 2 (pure algebraic trig, easy to follow)

(This eventually whittles down to the same concept as Solution 1)

Note that in each equation in this system, it is possible to factor , , or from each term (on the left sides), since each of , , and are positive real numbers. After factoring out accordingly from each terms one of , , or , the system should look like this:This should give off tons of trigonometry vibes. To make the connection clear, , , and is a helpful substitution:From each equation can be factored out, and when every equation is divided by 2, we get:which simplifies to (using the Pythagorean identity ):which further simplifies to (using sine addition formula ):Without loss of generality, taking the inverse sine of each equation yields a simple system:giving solutions , , . Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: , , and . When plugging into the expression , noting that helps to simplify this expression into:

Now, all the cosines in here are fairly standard: , , and . With some final calculations:This is our answer in simplest form , so

-Oxymoronic15