Rank. Name: Score
1. danzhi: 14
2. dinoboy: 13
2. awesomeusername: 13
4. CRICKET229: 12
5. esque: 11
5. ksun48: 11
7. SteinsChaos: 10
7. exmath89: 10
9. DanielKang: 9
9. mathisawesome: 9
11. mymathboy: 8
12. ABCDE: 7
13. bluehall90: 6
13. andrewjjiang97: 6
15. Faustus: 5
15. ahaanomegas: 5
17. kad2361: 3
17. cjyq7575: 3
17. fireonice: 3

1) 196
2) 079
3) 004
4) 011
5) 172
6) 132
7) 065
8) 245
9) 262
10) 000
11) 006
12) 037
13) 600
14) 339
15) 028

Solutions to the Problems from the 2012 Mock AIME II

1)
Problem: Given that where and are positive relatively prime integers, find the remainder when is divided by .
Answer: 196
Solution:
Consider . We note that , thus we have a telescoping sequence and we need only consider the first numerator and last denominator.


Note that however . Also, note that however . Since , we know that . Now note that we want , therefore we use and to give us .



Problem author: djmathman, solution authors: sjaelee/Binomial-theorem


2)
Problem: Let , and be defined as where , and is an integer. If for being a positive integer greater than and being a positive integer greater than 2, find the smallest possible value of .
Answer: 079
Solution:
The sequence goes as follows:

Note that this sequence repeats, because once we get to , we subtract from the number under the square root sign and take the square root of that, then when you take the difference of the squares of these two numbers, they have a difference of , so you get back to .

The first perfect square less than is . Therefore .

We present two ways to finding :
Way 1: Now notice that, for , we have . Also, the terms under the radical counts down one for every such that . The number of before a number can be seen to be . Therefore, we can see that . Setting this equal to , we see that . We can see that both and satisfy this, however, since , we disregard this, and we find that , so .


Way 2: Look at the sequence by grouping terms as follows:
. Note that the number of these groups of three that we have is equal to the number of terms in the sequence . This has terms in it. Before the last group, we have terms, and is therefore the th term, and .

Both ways lead to and , so the answer is .

To show that no smaller sums can be created, note that the next perfect square will be and will take more than as the value of , and hence all other perfect squares will take more than , and we have as our sum which is greater than since .

Problem authors: Binomial-theorem/admin25, solution authors: Binomial-theorem/admin25
3)
Problem: The \textit{digital root} of a number is defined as the result obtained by repeatedly adding the digits of the number until a single digit remains. For example, the \textit{digital root} of is (). Find the \textit{digital root} of .

Answer: 004

Solution: The sum of the digits when we get this down to a one digit number is the same thing as the number modulo 9. This is because let’s say we start with a number . When we take the sum of the digits modulo 9, this is the same thing as taking the number modulo 9. However, the sum of the digits modulo 9 is the same thing as the sum of the sum of the digits modulo 9, and this pattern repeats.

We are left with finding . Note that , therefore we get . By Euler’s Totient Theorem, or . We now need to find , which is equal to . Note that , and , so this pattern repeats in the exponents. Since , we get . From this, we get , so we need to find . Since , we get an answer of .

Problem author: tc1729, solution authors: tc1729/Binomial-theorem

4)
Problem: Let be a triangle, and let , , and be the points where the angle bisectors of , , and , respectfully, intersect the sides opposite them. Given that , , and , then the ratio can be written in the form where and are positive relatively prime integers. Find .
Answer: 011
Solution:
Let and . We use the Angle Bisector Theorem twice to get two different equations relating and . From the angle bisector , we have the proportion , or . From the angle bisector , we have the proportion , or . Multiply the second equation by to get , then plug this expression for into the first equation to get . Solving this equation gives us . Finally, the ratio is, by the Angle Bisector Theorem, equivalent to the ratio , which is equal to so the answer is .
Problem author: djmathman, solution author: djmathman


5)
Problem: A die with sides numbered through inclusive is rolled times. The probability that the sum of the rolls is is nonzero and is equivalent to the probability that a sum of is rolled. Find the last 3 digits of the minimum value of .
Answer: 172
Solution: Consider the generating function for a 12 sided die. When rolled n times, the generating function is . This polynomial is clearly symmetric, and the coefficient of is thus the same as the coefficient of .

Thus, the coefficient of is the same as the coefficient of . Note that , and thus the minimum value for n is 168. The minimum value of is thus , so the answer is .


Problem author: BOGTRO, solution author: BOGTRO

6)
Problem: A circle with radius and center in the first quadrant is placed so that it is tangent to the -axis. If the line passing through the origin that is tangent to the circle has slope , then the -coordinate of the center of the circle can be written in the form where , , and are positive integers, and . Find .
Answer: 132
Solution:



SOLUTION 1:
Since the circle has a radius of , is tangent to the y-axis, and has its center in the first quadrant, its center has coordinates for some positive . Also, since the circle is tangent to the line , the distance from the center to that line must be . The equation for the line can be rewritten as . Using the point-to-line formula, we must have . Simplifying this, we have . This gives two solutions for . We have or . We reject the negative value, and so we have , and .


SOLUTION 2:
Let be the angle between the line and the -axis. Then we have , so and . Drawing in the line from the center of the circle to the origin, we see that this line bisects the angle between the -axis and the line , so its tangent is equal to . Now using the upper right triangle, we see that also equals , where is the -coordinate of the center of the circle. Thus, we have , so and . Our answer is thus .

Problem author: djmathman, solution 1 author: admin25, solution 2 author: djmathman

7)
Problem: Given that are positive real numbers such that , then the value can be expressed as , where and are relatively prime positive integers. Find .
Answer: 065
Solution 1:

Lemma:
Proof: We can rearrange the given equation to get . Notice that the RHS is almost a perfect square. . Notice now that the RHS is always positive, and so we must have . Since is positive, we can divide by to get .

Seeing that we can make perfect squares with the linear terms and square root terms, we try to force perfect squares. Since the coefficient of is already a perfect square, we begin with that. We have . Now we divide both sides of the equation by to get . We recognize from our original equation as , and we substitute that back in to get . We again recognize a template for a perfect square in the right hand fraction, so we have . We can also right the other terms in the numerator of the right hand fraction as perfect squares in terms of and . . We now see that . We can take the square root of both sides to see the full picture. . This equation is just the equality case for the RMS-AM inequality! Therefore, from the equality case, we must have , from which we find that and . We see that these satisfy the original equality, and so their product is .

Note: We had to prove that so that , one of the terms in the RMS-AM inequality, is positive, otherwise the inequality is false.

Solution 2:
Let and . We have . Moving this to one side, we have . Completing the square, we have


The above simplifies into , and since squares are nonnegative, we have both squares equal to , which means and . Thus, and , so and .

Problem author: admin25, Solution 1 author: admin25, Solution 2 author: sjaelee



8)
Problem: Let be a point outside circle with center and radius such that the tangents from to , and , form . Let first intersect the circle at , and extend the parallel to from to meet the circle at . The length , where ,, and are positive integers and is not divisible by the square of any prime. Find .
Answer: 245
Solution: Let intersect at .
Note that , thus . Then we have . We can use the law of cosines to find , .
Thus, we have .

Problem author: sjaelee, solution author: sjaelee


9)
Problem: In , , , and . and lie on , and , respectively. If , and , the area of can be expressed in the form where are all positive integers, and and are not divisible by the square of any prime. Find .
Answer: 266
Solution:
Here is a diagram (note that D should be on AC and F should be on BC):

.
Start out by finding . Remark that by the law of sines, . Therefore . We know because in , therefore is in the first quadrant. Use the Pythagorean Identity to give us .

Now, note that is the same thing as , from , therefore we have .

Next, use the law of sines to give us . This gives us . This gives us .

Now, we use coordinates to find that the coordinates of , , , and . is going to be the point from adding point to create triangle as shown in this diagram (where is the right angle, is the angle, and is the angle):


Now, note that we have to find the coordinates of and . Assume WLOG that is and be . is obviously , is going to be and the coordinates of is going to be . Since and have the same value, we can find by using . The base is going to be equal to the distance from to , which is the same thing as , and the height is the change in the coordinate from to . This is the same thing as . Hence, plugging these into give us . The answer is thus .

Solution 2 (sketch):
Note that , thus we need to find .

Problem author: Binomial-theorem, solution 1 author: Binomial-theorem, solution 2 author: sjaelee

10)
Problem: Call a set of positive integers \textit{lucky} if it can be split into two nonempty disjoint subsets and with such that the product of the elements in and the product of the elements in sum up to the cardinality of . Find the number of \textit{lucky} sets such that the largest element is less than .
Answer: 000
Solution: Let the cardinality of be . We therefore want the sum of the products of the elements of and to be , and this is constant no matter what the elements of are. The set with the smallest such sum of products is the set , since if one of the elements is not in this set, then we can replace it with one of the elements in this set and get a smaller product. However, notice that whichever subset or contains the element , then the product of the elements of that set will be greater than or equal to , and the product of the elements of the other set will be positive, and so the sum of the products must be greater than . Therefore, no lucky set of positive integers is possible and the answer is .
Problem author: tc1729 Solution author: admin25


11)
Problem: There exist values of and such that , , and for some value of . Let be the sum of all possible values of . Find .
Answer: 006
Solution: First, if , then . We now assume that .
Now, note that .
Also, we have .

Next, . But we know , so .

Since the only possible values of are and , our final answer is .

(It is easy to check that there exists satisfying the equations.)

Problem author: djmathman, solution author: Anderson Wang, helpful advice: Anderson Wang




12)
Problem: Let . Assume the value of has three real solutions . If , where and are relatively prime positive integers, find .
Answer: 037
Solution: Let . Then and . From this, we have the system


Substituting the first equation into the second, we obtain


Plugging this into the third equation yields .

Thus, . Note that our three real roots multiply to .
However, since , we need to multiply by , so our is

We need . Using vieta’s and making sure we count for each factor of we divided off, we have .

Our answer is , thus .

Problem author: sjaelee, solution author: sjaelee

13)
Problem: Regular octahedron (such that points , , , and are coplanar and form the vertices of a square) is divided along plane , parallel to line , into two polyhedra of equal volume. The cosine of the acute angle plane makes with plane is . Given that , find the area of the cross section made by plane with octahedron .
Answer: 600
Solution:
Problem author: admin25

14)
Problem: Call a number a \textit{near Carmichael number} if for all prime divisors of the \textit{near Carmichael number}, , divides and has a perfect square as a divisor. Find the sum of all two digit \textit{near Carmichael numbers}.
Answer: 339
Solution:
Remark that all numbers of the form for are \textit{near Carmichael numbers}. The proof of this is noting that , which is proven by noting that , and therefore we get . Therefore, some two digit \textit{near Carmichael numbers} that we get are .

Now, we look for \textit{near Carmichael numbers} of the form . Note that this means that , which is impossible, so we can’t have and together. Also, this implies that and , so we can’t have times any other prime to give us a \textit{near Carmichael number}. We now look at numbers of the form . We can’t have and together because this would imply that divides , but , so this is impossible. Hence, we look for numbers of the form and . The first case gives us or . Therefore , which gives us . However, gives us no values for , therefore we must have and this gives us to give another \textit{near Carmichael number}:

Now, looking at the form gives us or or . Therefore , but this gives us no possibilities for that are greater than (which don’t fall into the first case), therefore there are none of this form.

Lastly, look at the form where is a one digit prime and is a two digit prime. Start out by looking at . For reasons explained before, this cannot be a \textit{near Carmichael number} if is a two digit prime number. Now, look at numbers of the form . Since , we are going to need for . Therefore . gives which isn’t a two digit prime number, gives which gives a largest value of as . gives us , therefore or which aren’t prime. gives us no two digit \textit{near Carmichael numbers}, since for is greater than .

Now, look at the form to note that either or . The first case gives us and the second gives us , which implies that which isn’t a two digit prime number. Therefore, we look at the first case. We need for or . We also need or . However, this gives us , which gives us no two digit numbers that work.

Lastly, look at numbers of the form . Since implies that which isn’t a two digit prime number, we must have , and therefore we get numbers of the form . Note that therefore or . The first one gives and both false, and the second one gives and , which the second one isn’t true.

Therefore, we have found our \textit{near Carmichael numbers} to be which gives us a sum of .
Problem author: Binomial-theorem, solution author: Binomial-theorem


15)
Problem: Define for and . Given that is a polynomial, and is an arithmetic sequence, find the smallest positive integer value of such that .
Answer: 028
Solution:

Lemma: is a quadratic
Proof: Note that using the method of finite differences, once we get to a constant term, the rest of the terms in the polynomial that have not been eliminated are going to be .
Since is an arithmetic sequence, difference(let this be h), we get in general . Therefore . We now have equations with our polynomials. Subtract all consecutive equations to give us , , . Let . Note that subtracting two equations eliminates , and we are going to be taking two more differences to get the equations equal to . These two more differences subtract the term and the term, because by method of finite differences, you only have to take and differences respectively to eliminate the linear/quadratic term. Therefore is quadratic.

Now, let . Since , we have or . Since , we get and . Subtract the LHS from the RHS of the equations to give us and . Subtract these two equations to give us or . Now, substitute this into to give us or . Therefore .

Since , we get , we are going to have this being true for . Since we want being positive, we use for to give us . The RHS is the same as . Our goal now is to find approximately. Note that (where , and are digits of ), therefore , and substitute this into our equation for to give a smallest possible value of being to give us and hence the smallest possible positive integer value for is .
Problem author: Binomial-theorem, solution author: Binomial-theorem,
helpful advice: BOGTRO, pythag011