## Olympiad Number Theory+AIME Number Theory

1993 AIME #9: Two thousand points are given on a circle. Label one of the points 1. From this point, count 2 points in the clockwise direction and label this point 2. From the point labeled 2, count 3 points in the clockwise direction and label this point 3. (See figure.) Continue this process until the labels  are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as 1993?

Solution

From the point , the distance to each must be the same .Therefore,  letting the smallest be .

Therefore,

Therefore,  or .

Note that  and  can’t divide  using basic Euclid’s, therefore,  and  or  and  (unless one of them is  which will be obviously not true after doing computations).

The first gives  and . The second gives .

The first gives  (after doing some simple cases). For the second, we get a cycle of: which mod 16 gives us  therefore we are done and the answer is .

Two positive integers differ by  The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

Solution

and . We therefore get . Let  and . We therefore get  when  and  are perfect squares.Therefore,  and . Therefore, .

We want to maximize . We have either  is a multiple of , or odd (theorem on perfect squares). In the first case,  or .

However,  gives  which gives  which definitely isn’t maximal and same with . Also,  gives  being a perfect square and  so that’s absurd.

Therefore, . The first case gives us  which gives  for maximum, or . Therefore,  in this case.

gives  or  as maximum therefore .

gives  or  as maximum therefore

gives  or  as maximum therefore

Therefore, maximum value of  is  in all cases. Therefore, we get  therefore the sum is .

ISL 1995 S3:

For an integer  let  be the least prime that does not divide  and define  to be the product of all primes less than  In particular,  For  having  define  Consider the sequence  defined by  and

for  Find all  such that

Solution

Okay so note that . Then . We let this be . Then,  let this be . I will prove in general the correspondence between binary numbers.Assume via induction that this binary pattern holds. The base case is proved above. Now, assume it holds for the set  to  and I prove that it works for  to .

Note that the numbers without the factor of the th prime (this is the leading factor) are going to create the group  to  by induction hypothesis. Also, we get that for  to  , we get an added factor of the th prime by all these numbers, therefore we are done by induction.

. Therefore, putting this in this form of induction, we get  (where we only include digits which are in the prime factorization). Therefore, we get . This is .

For a positive integer , let  denote the sum of the remainders when  is divided by respectively. Prove that  for infinitely many positive integers . (1981 K¨ursch´ak Competition)

Solution

Okay so note that  and  therefore I desire to prove  always works.

For , we compare remainders with . Starting with  we are going to get a sum starting with  and going down, which is the same with  and going down. Also,  gives , therefore we are going to prove that from  down they are the same.

We are going to have the remainder when  divides a number always to be  more than unless the number we divide by divides . This happens only for  and in this case we get  remainder being  and  remainder being . Therefore, summing this up we get .

When it doesn’t divide this, we get  incidences (subtracting all the ones that worked above), which is the same as above, and hence we are done.

## Complex Numbers and some Number Theory

Let  be the three sides of a triangle, and let , be the angles opposite them. If , find  (1989 AIME #10)

Solution

Use the hint in the book to turn all the  into  and . Then, use Law of cosines to give us  or therefore . Next, we are going to put all the sin’s in term of . We get . Therefore, we get .

Next, use Law of Cosines to give us . Therefore, . Also, . Hence, .

Lastly, . Therefore, we get .

Now, . After using , we get .

For how many positive integers  less than or equal to  is  true for all real t?  (2005 AIME II Problem 9)

Solution

.
.
and . The second equation gives  which is the same as the first.
Therefore  or … however, testing gives  and  gives  therefore,  which gives  solutions.
The equation  has one complex root with argument  between  and  in the complex plane. Determine the degree measure of . (1984 AIME Problem 8)

Solution

We get that  must be a 9th root of unity but not a 3rd root of unity. Therefore, we find  so the answer is .
Given that  is a complex number such that  find the least integer that is greater than  (2000 AIME Problem 9)

Solution

Use the hint to solve for  to give  or therefore . Hence, .

(after simplifying) so an answer of .

a. Prove that  is never an integer for .
b. Prove that  is never an integer for .

a.

Lemma (from dinoboy): If  then .
Proof: This is pretty simple, let , then  and . Also, . Therefore,  and since , we have  .

We now have .

Now, using the Lemma , and then continue on in this manner we are going to get the highest power of  in the denominator, therefore we have . We need, however, for this to be true we need  however since  this is false.

As an aside: The reason we choose  is because this way we always have  or  and can never have equality (due to parity).

b

Let  (I think this is the right notation).

We therefore get

. Now,  because .

Continue on in this manner of grouping to give us . We need this to be greater than or equal to  or therefore,  which is obviously false because .

Prove that  divides  if and only if  for some positive integer . (Canada 1985 #4)
We look at the power of  in both numbers.  and . We have  from the given statement.

Let .

We have  (with equality when  for some non-negative integer . We have  (with equality again when . Therefore, as a conclusion of this long inequality chain, with equality when . Therefore, at most we will have equality which holds when (as we need for it to be greater).

This completes both sides of our if and only if statement, so we are done.

Given the triangle  we consider the points  such that the triangles  are equilateral, and they don’t have intersection with . Let  be the midpoint of  the midpoint of , and  the midpoints of , respectively. Prove that .

Solution

Hints from AoPS

127) Let  be the origin. What must be true about  if and only if ?
85) If you’re stuck with a ratio that you’d like to show is inamginary, here’s a hint. If , then what does  equal?

Let  be the origin

We desire:  being imaginary.

We know:

Also,  and .

Now we have a HIGHLY computationally intense problem so let’s try it.

. Next, . Also,  and .

which we need to be imaginary.

This is the same thing as

This is also the same as

Therefore, we get

We now prove .

This is crucial in the next part of the problem.

or  Therefore, we get  which is true by the definition of .

Yay, lemma proven.

Find  where .

We note that letting this value be , we get .

Therefore, we have .

Therefore, we must now prove that  is imaginary.

Since  is a primitive 6th root of unity, we get  or .

Therefore,  or .

In the first case, we get  or therefore it is the same as

We then get
.

In the second case, we get .
Also, we get .

We are finally done. .

## A bunch of inequalities

Let . Prove that . (2004 USAMO #5)

Solution

because  because . (This was the first hint from the AoPS book)Now, because , we have . Therefore, .Now, we desire  which is directly true by Holder’s.

Prove that for all positive real numbers ,(IMO 2001 #2)Solution

We use the technique found on page 3 of the online math circle lecture: http://onlinemathcircle.com/wp-content/ … older1.pdfFrom Holder’s, .Now, we have .

We have our desired from above.

Therefore, we now have to prove . We desire .

Therefore, we now want  which is true by AM-GM.

Let  be the side lengths of a triangle. Prove that . (Samin Riasat)Solution

Use Ravi’s transformations with . Therefore, we now desire  or . We now have .Therefore, we now desire . We can now just apply Titu’s:, and we are done.
Prove that  for . (Source: WOOT)Solution

By Power Mean or Holder’s, we have  (try to find the way that both inequalities are applied!) Now, .Now, we want .. This is true iff  which is true by  (or power mean if you like).

Let  be positive real numbers such that . Prove that . (Source: Mathematical Database)Solution

There is probably a method using Holder’s, but I want to use the finding equality method.I find one case of equality is when . So use AM-GM, with the equality case to give .We now want . Therefore, we need  or . Therefore, we need  which can be proven many ways.

et  and  be positive real numbers. Prove that, . (Vasile Cirtoaje)Solution

We use the same method on page 3 of the onlinemathcircle Holder document.By Holder’s, . Therefore, , and we are done.

Prove that  (Online Math Circle)

Solution

Use Holder’s to give us  and we are done.

Let  be positive real numbers. Prove that, . (Online Math Circle)

Solution

We think of what terms we want on the greater side to produce the respective terms on the less than side in Holder’s.We get .All the terms match up nicely when using Holder’s, besides for . We note that by AM-GM, so now we try to put this into the equation above, and we find that it works out nicely.

Therefore,  (AM-GM)
(Holder’s)

Therefore, now by taking the cube root of this inequality, we are done.

For any real numbers , prove that . (Russia 1992)Solution

By Titu’s, .Therefore, we want to prove which is true.
Let  be real numbers such that  and  are non-negative. Prove that, . (Online Math Circle)Solution

The constant on the LHS is weird. Try letting , then we want .We know perfectly how to proceed. If  for all positive , then we are done. We must find the positive max of . Let , giving us . Now, we must maximize  which has max at  giving the value of . Therefore,  as desired, and we are done.

Let  be real numbers such that . Prove that . (Centro American Math Olympiad, 2009)

Solution

Rewrite the RHS as .Now, expand both sides to give us (using symmetric sum notation).Use the well known  (from http://www.math.ust.hk/excalibur/v11_n1.pdf page 2, fact) to give us  as desired.

Let  and  be the altitudes of triangle  and let  be its inradius. Prove that, . (Puerto Rico TST 2009)Solution

and similarly,  and .Also, .Therefore,

This is the same as  which is true by Cauchy (Titu’s).

Let  be positive reals such that . Prove . (Source: Online Math Circle)

Solution

Using Cauchy (or QM-AM which is derived from cauchy), to give
.Now, we need , which is true by AM-GM, so we are done.

Let  be positive reals such that . Prove that . (Belarus IMO TST 1999)

Solution

by Cauchy (Titu’s). We now need This is the same as needing  which is true by .

For arbitrary positive numbers , prove that  (1999 Czech and Slovak Republic)

Solution

This was fun.By Cauchy, . Therefore, (doing some simplification). Also, , therefore,  as desired.

Let  be positive real numbers with . Prove that, . (Ireland 1999).

Solution

Use Cauchy (or Titu’s) to give .

Let  be positive reals. Prove that . (Online Math Circle)

Solution

.

Let  be positive reals such that . Prove that . When is there equality?
(French TST 2006 Problem 2)

Solution

First off, we expand the LHS and put it under a common denominator.We get

.Which is true by AM-GM.

Equality holds in AM-GM when  is given, so and  gives , and similarly  is where equality holds.

Let  be positive numbers with . Prove that . (1991 APMO)Solution

.
If  satisfy that , prove that .(Source: Inequalities: A mathematical olympiad approach)Solution

We get , which is true by AM-GM and  (on the denominator).

## Some geometry AIME problems and other problems

1984 AIME Problem #3:
A point  is chosen in the interior of  so that when lines are drawn through  parallel to the sides of , the resulting smaller triangles, , and  in the figure, have areas 4, 9, and 49, respectively. Find the area of .

Solution

Note that if we have , and , then . We use this to get the following ratios:

Using this, we get . Therefore, . Similarly, we get , and . Sum up all these areas in the picture to give us .

1984 AIME Problem #9:

In tetrahedron , edge  has length 3 cm. The area of face  is 15  and the area of face  is 12 . These two faces meet each other at a  angle. Find the volume of the tetrahedron in
Solution

I’ve seen this problem way too many times.

So, we find the height of  to be . Therefore, the height of the whole figure is , and the total area of the figure is going to be .

1985 AIME Problem #4:

A small square is constructed inside a square of area 1 by dividing each side of the unit square into equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of  if the the area of the small square is exactly 1/1985.

Solution

We find after doing a lot of algebra that the two intersection points on one side of the square are . The distance between these two points is going to be . This is the side length of the square, so the area of the square is . Therefore, . This gives us after dividing by  and factoring, , so .

As shown in the figure, triangle  is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle .

(1985 AIME Problem 6)

Solution

Denote the point on  to be , the point on  to be  and the point on  to be .

Let . Note that  because of same altitudes. Using Ceva’s on these lines intersecting at  gives us . Using similar logic as before, we have  and . Therefore, we now have . Denote  and  now as well. Therefore, we get .

Now,  or therefore .

Hence, now . Therefore, .

Substituting this back into  gives us . This gives us . We list out the factors, which are . Therefore, we find . Next, .

Therefore, the sum of all the areas is .

1988 AIME Problem #7:
In triangle , and the altitude from  divides  into segments of length 3 and 17. What is the area of triangle ?

Solution

We get letting the altitude be , that  (using sum formula), so after expanding and solving (I did this via hand don’t want to show work though), . We want .

1990 AIME Problem #7:

A triangle has vertices , and . The equation of the bisector of  can be written in the form . Find .

Solution

First off, using the distance formula, we find  (yes, it is a right triangle, but that’s irrelevant to this problem). Use the angle bisector theorem now, to give us letting the angle bisector be  and . Therefore, solving these, we get . Therefore, the point  is  of the points along the line segment , hence it is the point  (using basic algebra).

Therefore, the line we want is from . The slope of this line is , so the equation of the line is . This gives us . Therefore,  and our answer is .

1994 AIME Problem #2:

A circle with diameter  of length 10 is internally tangent at  to a circle of radius 20. Square  is constructed with  and  on the larger circle,  tangent at  to the smaller circle, and the smaller circle outside . The length of  can be written in the form , where  and  are integers. Find .

Solution

Use coordinates. the center of the largest circle is (0,0). Now, therefore, letting CD be vertical, we get the coordinates of C being (10,m) and of D being (10, -m). Now, by using the circle , we get the two other points  and  being  and . Therefore, setting the square to have equal side lengths, we get . We desire  so
Let . Consider the sequence  where  for . Compute the number of values  such that , and  are distinct, but  for all .  (ARML)

Solution

Note that  gives the desired  (induction). Therefore,  or .

gives us . Since  giving  and completing this case.

gives us  or  (since ). Therefore,  giving us  giving two values of .

Find all functions  such that for all real  and , we have . (Saudi Arabia)

Solution

We use separation, by dividing by  to give:

.

We continue with putting everything involving  on the LHS to give us .

Therefore,  for some constant .

Therefore,  or therefore .

Substituting this into our original equation gives which is true.

Therefore our general form is true, and  for all constants . Since  is any constant, we can also write this in the form .

. If  is a positive integer, and  is the sum of the digits of the number of possible values of , find the sum of the digits of . (ARML)

Solution

Okay, so after doing some algebra, we get that .

Therefore, there are  total ways.

First sum of the digits is going to be  looking at number of ‘s.

Then, this is the same as , with  ‘s. Therefore, the sum is

Let  and  be positive real numbers such that . Prove . (1971 CMO #2)

Solution

By Cauchy, .

Now, by AM-GM, . Therefore  and we are done .

.

.If  and  are integers such that  is a factor of , then find . (AHSME)

Solution

Note that .

Now we get .

Comparing  coefficients, we have  or .

Therefore, we get .

Lastly, .

Therefore, we get the  term (which is ), to be .

If  is the product of two linear factors with integral coefficients, find the value of . (CMO)

Solution

These are the equations we get. From the first equation, and because we have integral coefficients, we WLOG let  and .

The equations are now:

From the first two equations, . Therefore, letting  and , we have  and . Therefore, .

Case 1:

We go through cases on  gives us  absurd.  gives  absurd.  gives  absurd.  gives  absurd. gives  which is possible.  gives  absurd. Therefore, . Furthermore, . But  so this isn’t possible.

Case 2:

We go through cases on  again.  gives us  absurd.  gives us absurd.  gives us  gives us .

gives  and  gives . But  so again absurd.

gives us . These satisfy all the equations.

Therefore .

Let  be a fourth degree polynomial with leading coefficient  such that , and . Find . (Intermediate Algebra)

Solution

Note that letting , we have  giving us after subtracting out equations, . Also,  so after subtracting out equations, we have . Therefore, , and . Also  is given.

Now it’s a matter of just solving equations.

gives us  or
gives us  or .

Therefore,  and therefore . Also, .

Therefore, .

Show that the polynomial  does not have a pair of real roots with product . (Intermediate Algebra).

Solution

We use a method similar to the one in the book by assuming there is. Hence

Therefore, . Next,  and . Therefore,  or . Therefore, . However, this doesn’t check with the last equation, so we are done .

Solve for all complex numbers  such that . (HMMT)

Solution

After doubting any rational roots, we are going to try to put this in terms of two quadratic polynomials.

We have  looking at  terms. Therefore, . Hence, we have .

Next, we have  or

is next  and lastly .

From , and , we have . WLOG, we let the pair be

The first case gives us . We now have  as desired.

The second case gives us . We have , so therefore .

Therefore, the polynomial is hence .

Therefore,

Prove that  is the only integer-valued function defined on the integers that satisfies the following conditions: (i)  for all integers , (ii)  for all integers , (iii)  (Putnam 1992 A1)

Solution

Note that from (ii), . Now use (i) to give us .

Now, first off, , therefore . Assume that  holds for all integers . Therefore,  or therefore  and our induction is completed. For negative values, assume that  holds. Therefore,  completing induction as well.

Now, for odd integers, we get .
Now, use similar induction by assuming  for all integers . Therefore,  completing our induction. For negative values, assume  holds. Then,  completing our induction.

Therefore, we are finished with both cases and we are done .

.

## Logarithmic, Polynomial, Sequences and Series Problems

Let  be a positive integer, and let  be a sequence of reals such that , and  for . Find . (2005 AIME II Problem 11)

Solution

Let . Therefore, using the recurrence, . Now, use the recurrence on  to give us  or . Using similar logic, we obtain the equation . We desire to have  and  giving us . Therefore  and so  doing simple calculations.

Evaluate the infinite product . (Source: Putnam)

Solution

We get . This is equal to ..

Find a closed form for . (AoPS)

Solution

. We now use partial fraction decomposition to give us . We now note that the sum cancels for each  and , because the denominators are going to be the same for the plus term and the minus term. Therefore, we end up with .

A sequence of numbers  satisfies  and  for all . Determine the value of  for . (CMO)

Solution

Using the given equations, we have  and .Subtracting the two given equations gives us . We now get  or .

Therefore, using telescoping series. Therefore, we get .

Find the last three digits of the product of the positive roots of(1995 AIME #2)

Solution

Let  so . Therefore, we have , and , so we get . Take  of both sides to give us  or therefore . Therefore we want the product of all ‘s. There are two values of , therefore we have . We want this  or
Find  if .
Solution
Okay so let . Then we get after doing some equation stuff with the given equation that .Then we get  so  so .
Problem:
Let  and  be positive numbers such that  and . (Bay Area Math Meet)

Solution:
Let . Therefore, we get:.Therefore, we have

Therefore we get  so  or therefore .

The system of equations
has two solutions  and . Find . (2001 AIME I Problem 9)Solution

First equation gives .Wait first off let’s let .Therefore  or therefore . This is the same as .

The second equation gives . Therefore,  or .

Lastly, the third equation gives  or therefore .

Multiply all the equations to give us .

We care most about the variable , so we go from teh equation  to give us . First off, simplify this, to give us . Therefore . Therefore, substitute  to give us .

The first gives us  and second gives us  so their sum is .

Determine the value of . (USAMTS)Solution

First off, each term is of the form  using simple algebraic manipulation. Now, the desired sum is the same thing as  which using W|A gives us .

Find the value of . (HMMT)

Solution

First off, each term is of the form . Take the sum of this from to infinity to give us .This is the same as .
Let  be the Fibonacci sequence:  (), and let  be the polynomial of degree 990 satisfying  for , 993, , 1982. Prove that . (ISL 1983)Solution

Row  contains the elements  using finite differences (and simple fibonacci formulas). Therefore, the last row  contains the element . Therefore, the next element in this row is  and we have to add this to all of the last elements in the other rows to give us the next element in row . Therefore, we need to prove . This is true by a well known fibonacci identity (proven via induction).

If you don’t know this identity, here’s the proof: Assume it holds for . Therefore, . We need . Therefore, we must have  which is true by definition. The last thing we need is for the base case  to work, which it does, since .

Let  be a polynomial of degree 100, such that  for , 1, , 100. Find .  (Duke Math Meet)
Solution

I claim that the first number in all of the rows is going to be  for row . Here is my proof: By the handout, . From this, we re-write this as . Therefore, our desired sum is  (note that we ignore the first term since it’s ). We now use this same formula, but with .
Let  be the curve  and  three distinct collinear points on . Show that

(Sweden, 1992)
Change the points into:We desire to prove . This is true if , so we’ll be looking at the coefficient of  in our third degree polynomial.

Note that for the points to be on a line, we need for them to satisfy  for all of them. Let  and  be constants as they are already, and we have . Therefore, the roots of the polynomial  are , so hence  and we are done .

## USAMO, AIME, Inequalities

The sum of the following seven numbers is exactly 19: . It is desired to replace each  by an integer approximation , so that the sum of the ‘s is also 19 and so that , the maximum of the “errors” , the maximum absolute value of the difference, is as small as possible. For this minimum , what is ? (1985 AIME Problem 8)

Solution

So essentially, if a number is more than 4 or less than 1, then we get the largest difference at least  but I can make it less than . We have  solve to give . Therefore we get  going to  rest going to , therefore largest is  for answer of .
Let  be a list of positive integers – not necessarily distinct – in which the number 68 appears. The average (arithmetic mean) of the numbers in  is 56. However, if 68 is removed, the average of the remaining numbers drops to 55. What is the largest number that can appear in ?
Solution
Let the set have  ‘s and   (essentially being greedy as ‘s bring up the average. Therefore, letting largest element in set be , we have . We also have . Therefore,  and . We therefore get , or therefore  or . We therefore get  or therefore .
Let  be positive real numbers. Prove that . (AoPS)Solution

Expand gives .This is the equivalent of .

After expanding this out, the rest is obvious by Muirheads  majorizes .

Problem:  Prove that if  for all  then .  (AoPS)15 second method

All  terms cancel, rest is Muirhead’s .

5 minute method

Expand as before, and then you desire:.

This is the equivalent of proving  or .

We get this being the equivalent of proving .

The rest is simple, we note that if  then both terms are greater than  so product is greater than . If , then both are negative, so the product is greater than  so all products are greater than  so their sum is as well .

Problem:  If  and  are two of the roots of , prove that  is a root of . (1977 USAMO Problem 3)
Solution:
Let  be the roots of the original equation. Then  are the roots of the next equation is what we’re trying to prove. Note that:.

. Note that , and  (both from previous equation), so we get
so we get the coefficient is .
so the coefficient of this is
so the coefficient of this is .
so the coefficient of this is .

Therefore, the polynomial with roots  is  so we are done

## Polynomials and Catan

Show that  where  is a positive integer.

Solution

We induct on  are easy to check, first gives  second gives .

Now, assume it works for , therefore we have . We must prove .

We have by hypothesis

Again, by hypothesis, , so  has remainder  when divided by

Therefore .

Since , we can add  to it to further simplify it (with a wishful eye!), to give us  or , so we have proven

We wanted to prove  or  which is exactly what I got above, so we are done .

The roots of  are in geometric progression for some constant . Find . (AoPS)

Solution

Let the roots be . Therefore, , or , so .

Next, , or .

Lastly, , so . This gives that . We could use this to find , but that’s terribly un-interesting, instead we have  or .

How many -digit numbers are such that the digits, as read left-to-right, are non decreasing, and that theth digit from the left is at most ? (For example, 12235 is such a number.) (AoPS)

Solution

We have  numbers, which we let each difference be , and we set up a one to one correspondence with a similar problem. For every spot before a sequence of ‘s (which sum up to ), we use a ( and for every spot where we have another number added to the difference, we use a ). Therefore, we start out with a (, and the rest gives catalan numbers in the sense that it starts with a (, we must have more (‘s than )’s at all points, because this is by definition of catalan numbers, and we must have the difference total starting at 0 being less than i. Lastly, we use the non-decreasing fact by that all differences are positive, so we have . We use  to give us .
Brazil defeats Germany in a wild World Cup final by the score of  to . Assuming the  goals were equally likely to be scored in any order, find the probability that the score was never tied (expect at ). (AoPS)

Solution

Okay, so the total number of outcomes is going to be . We draw a huge diagram, and get that we can never have Germany beating Brazil because if so then Brazil and Germany will be tied at a point with the final score.

Okay, so Brazil wins the first game and we get  to  without going through any points on  on this graph. We make a one to one correspondence with this and  to . We also make another one to one correspondence between going through  and . If it goes through  it must have gone through  so to count to , we just move up (we can’t move right since it’s a    grid), so we get  or .

Prove that the number of rooted binary trees with  internal nodes is the th Catalan number. (AoPS)

Solution(using hint)

Denote  to be the number of rooted binary trees with  internal nodes. We assume using strong induction that the result holds for all .

So now, looking at the casework that we have  internal nodes on the left node to start out with, then , then so forth until we get . This gives us . By our inductive hypothesis, we have all of these , therefore we get  which by the recursive definition of catalan numbers is true .

The expression  is equal to  for some integer . Find .  (ARML)

Solution

Letting the expression be , we want .

Due to ugliness, let  for now. Therefore .

Now, letting , we get  which is .

A sequence of positive integers with  and  is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all , the terms  are in geometric progression, and the terms , and  are in arithmetic progression. Let the greatest term in the sequence less than  be . Find . (2004 AIME II #9)

Solution

I used the book hint to experiment a bit on the second term. This hint worked quite nice!

For , we get:

Letting the second term be , I note that the third term is going to be , the fifth term is going to be , the seventh term is going to be , and the ninth term is going to be . My proof of the lemma that it works is that noting that this means the fourth term is going to be d(2d-1), then d(3d-2), and so forth and this satisfies arithmetic sequence condition. The eight term is hence . Therefore, we have:

Eight term=(3d-2)(4d-3)
Ninth term=(4d-3)(4d-3)
Tenth term=(5d-4)(4d-3)

Therefore, we get (4d-3)(9d-7) which we want equal to 646. d=5 works (646=2*17*19 using W|A).

Therefore, we get, ninth term of (17)^2, tenth term of 21*17, eleventh term of 21^2, twelve term of 21*25, thirteenth term of 25^2, fourteenth term of 25*29, fifteenth term of 29^2, sixteenth term of 29*33, seventeenth term of 33^2.

Since 33^2>1000, and 29*33<1000, , and since this is the sixteenth term, our answer is .

Set  consists of  consecutive integers whose sum is , and set  consists of consecutive integers whose sum is . The absolute value between the greatest element of  and the greatest element of  is . Find . (2004 AIME I Problem 2)

Solution(outline)

Let the smallest element in  be , and the smallest in  be . Therefore, we get  and .

Now, subtract those two, we get .

We have either  or .

For the first case, we get , so we get  absurd.

For the second case, we get , so , or .

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 30. Find the sum of the four terms. (2003 AIME I Problem 8)

Solution

Let the terms be .

Then the sum is  or .

Also, , so , or therefore .

This also gives .

Therefore, our desired answer is the same as .

Now, since  is positive, and when multiplying by  and taking , we get , we do casework on  gives  or  contradiction. gives  or , but  since the first term is positive, so contradiction. Therefore,  giving  or .

This gives the last term (to check for all conditions) to be equal to  which is clearly an integer, so hence the answer is .

## A bunch of polynomial problems and one induction

Consider the polynomials  and  Given that  and  are the roots of find  (2003 AIME II)

Solution

Note that this is .

By Vieta’s, . Now,  or therefore , so  or .

Next, , therefore  so .

Next, . So , so . Lastly, , so  therefore .

points (where  is a positive integer) are given in space, such that no  of them are on a line. We draw  line segments connecting pairs of these points. Prove that there must exist a triangle whose vertices are  of the points and whose sides are  of the drawn segments.  (AoPS)

Solution

First off, WLOG let the figure be a regular figure. If it’s not, then we can just rotate the regular figure into whatever form we want and still have the triangle in it.

I’m going to use strong induction here, and start out with the base case , which gives  points and  line segments. Since there are  total line segments we can draw, and we must remove one we are still left with a triangle no matter what (removing one line doesn’t take away all triangles, this can be proven with taking away a diagonal giving a triangle with another diagonal and a side giving a triangle with 2 other sides and a diagonal).

I assume that it works for all values of  up to . I’ll prove it works for  now. First off, I claim that WLOG there is two points next to each other that are connected in the regular figure. For the cases there are not, move one of the line segments so that the two points are next to each other. Everything else in the diagram stays the same, so if we prove it for the two points next to each other case, we are done.

Start out with drawing in this line in the diagram and singling out this one line segment. We note that if for the other  points, that there are at least  lines in this region of  points, then we are done by inductive hypothesis. So assume not, and therefore there are more that  points with one of the two “starting” points. Therefore, we must have at least  lines, and since we already have one, we have  more lines to draw. There are  points that are not these starting points, and we have  lines connecting through either of the  points or both. Therefore, we have  objects, and we have  boxes, and objects, so one object must have at least  objects in it by pigeonhole, so one point has a line through both of these starting points. These starting points are also connected, so we have formed a triangle and we are done .

Let  and  be real numbers that differ in absolute value and satisfy the equations  Find . (Dropped AIME)

Solution

Start with multiplying the first equation by  second by  then subtracting then flip-flopping to give us:

and .

Now, multiply the first by  and second by  to give us  and . Subtracting the two gives us .

This is the same as  so we get (*)

We desire to find . Manipulate (*) to give us  or . Therefore, now (**).

Now, go back to the original equation, to give us  and  so therefore we get  and .

Therefore we get . Apply (*) to this to give  or therefore .

Substitute this into (**) gives us .

Show that  for all distinct  .  (AoPS)

Solution

Multiply by  to give us the numerator equal to which is  so done.

Factor  into a product of linear terms. (AoPS)

Solution

Let it be  then  (left for the reader to check). Therefore divides the expression(multivariable polynomial factor theorem), and by symettry so does  and . Also,  (again, left to reader) so  divides the expression, and since , we get .

Substitute  to give , so we get , so  and our factorization is .

Factor . (AoPS)

Solution Outline

Hmmm so use multivariable factor theorem to get  are all factors.

Using some grouping, we get desired=(c-b).

So now, inside the parenthesis, we get .

Therefore beginning expression is the same as .

We now need an  as well…

We end up getting ugly starting thing=.

Note that to homogenize we take  giving us .

Since the degree of that is , and I don’t see any factorizations of it, we’re done.

Let  and . If , then find . (AoPS)

Solution

So divide by  to give  (also  because if so then contradiction). Subtract these two to give , so since . Therefore, we multiply our two original equations by  to give us  and , so  and . Therefore,  and , therefore we get  (keeping in mind  since ). In both cases, .

## Polynomial problems

Determine  if , and  are the three real roots of the polynomial . (Mandelbrot)

Solution

.Since , we get
using Vieta’s.

Show that if the roots of  are rational numbers , and , then the roots of  are also rational. (Intermediate Algebra)

Solution

We have  from Vieta’s on the first equation. Now, we desire to have  being rational, so since  and  are rational, we need being rational. From , we have . Therefore , or  which is rational so we are done .

Let , and  be the distinct roots of . Find . (ARML)

Solution

We get
.

.

Consider all lines that meet the graph of  in four distinct points, , and . Show that  is independent of the line and find it’s value. (Putnam)

Solution

Let the line it’s intersecting be . Therefore the roots of the equation  are . By Vieta’s, , so  which is independent of the line since our final answer doesn’t have any ‘s or ‘s in it.
Let  be the  zeros of the cubic polynomial . Then, the expression  equals a rational number. Find this number.Solution

Essentially, it is the same thing as

So now, we get the right term is the same as .

Since by Vieta’s, , we get the term in parenthesis equals , so our answer is . Since , we get an answer of .

Let  and  be the roots of the equation . Show that  are the roots of the equation . (Source: Hungary)My Solution

First off, , so therefore  matching the constant term of the desired polynomial. We now have to prove that . We now that , so .The given equation is true

.

Which is true, so reversing our steps we are done .

## USAMO 1997 #2

Let  be a triangle. Take points  on the perpendicular bisectors of respectively. Show that the lines through  perpendicular to  respectively are concurrent.

Solution:

Consider the circles, centered at  passing through . These circles exist since  lie on the perpendicular bisectors and their radical axis are the perpendiculars through  to  as the radical axis is perpendicular to the line connecting the centers and the points lie on both circles. Thus they are concurrent at the radical centers.