# Problems Collection

This is a page where you can share the problems you made (try not to use past exams).

If you have problems or solutions to problems already on this page to contribute, please put it below. In the former case, please give at least one solution to that problem. Problems/solutions in the following section shall be inspected by Ddk001 and will be put in the section after that section. If you would like, put your user name to the list of contributors of this page (at the near-end).

## Contents

- 1 Contributed Problems and Solutions
- 2 Problems
- 3 Solutions
- 3.1 Problem 1
- 3.2 Solution 1
- 3.3 Problem 2
- 3.4 Solution 1 (Slow, probably official MAA)
- 3.5 Solution 2 (Fast)
- 3.6 Solution 3 (Faster)
- 3.7 Problem 3
- 3.8 Solution 1(Probably official MAA, lots of proofs)
- 3.9 Solution 2 (Fast, risky, no proofs)
- 3.10 Problem 4
- 3.11 Solution 1
- 3.12 Problem 5
- 3.13 Solution 1 (Euler's Totient Theorem)
- 3.14 Problem 6
- 3.15 Solution 1 (Recursion)
- 3.16 Problem 7
- 3.17 Solution 1 (Tedious Casework)
- 3.18 Solution 2 (Official)
- 3.19 Solution 3 (Official and Fastest)
- 3.20 Problem 8
- 3.21 Solution 1
- 3.22 Problem 9
- 3.23 Solution 1
- 3.24 Problem 10
- 3.25 Solution 1(Wordless endless bash)
- 3.26 Problem 11
- 3.27 Solution 1 (Analytic geo)
- 3.28 Solution 2 (Hard vector bash)
- 3.29 Problem 12
- 3.30 Solution 1 (Plain old chapter 3 analysis bash)
- 3.31 Problem 13
- 3.32 Solution 1 (Plain high school contest)
- 3.33 Solution 2 (Lagrange Multipliers)
- 3.34 Solution 4 (AM-GM bash)
- 3.35 Problem 14
- 3.36 Solution 1 (Lagrange Multipliers)
- 3.37 Solution 2 (Relative extrema by calculus)

- 4 Contributors
- 5 See also

## Contributed Problems and Solutions

Please contribute whatever problems you have.

## Problems

### AMC styled

Nothing yet to show

### AIME styled

1.There is one and only one perfect square in the form

where and is prime. Find that perfect square.

2. and are positive integers. If , find .

3.The fraction,

where and are side lengths of a triangle, lies in the interval , where and are rational numbers. Then, can be expressed as , where and are relatively prime positive integers. Find .

4.Suppose there are complex values and that satisfy

Find .

5.Suppose

Find the remainder when is divided by 1000.

6.Suppose that there is rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are other pegs positioned sufficiently apart. A is made if

(i) ring changed position (i.e., that ring is transferred from one peg to another)

(ii) No bigger rings are on top of smaller rings.

Then, let be the minimum possible number that can transfer all rings onto the second peg. Find the remainder when is divided by .

7.Let be a 2-digit positive integer satisfying . Find the sum of all possible values of .

8.Suppose is a -degrees polynomial. The Fundamental Theorem of Algebra tells us that there are roots, say . Suppose all integers ranging from to satisfies . Also, suppose that

for an integer . If is the minimum possible positive integral value of

.

Find the number of factors of the prime in .

9. is an isosceles triangle where . Let the circumcircle of be . Then, there is a point and a point on circle such that and trisects and , and point lies on minor arc . Point is chosen on segment such that is one of the altitudes of . Ray intersects at point (not ) and is extended past to point , and . Point is also on and . Let the perpendicular bisector of and intersect at . Let be a point such that is both equal to (in length) and is perpendicular to and is on the same side of as . Let be the reflection of point over line . There exist a circle centered at and tangent to at point . intersect at . Now suppose intersects at one distinct point, and , and are collinear. If , then can be expressed in the form , where and are not divisible by the squares of any prime. Find .

### Olympiad styled

### Putnam styled (Calculus version of AMC, AIME, and Olympiad)

1.Suppose where and are relatively prime positive integers. Find .

2. Let and be real numbers. Show that

3. Common blood types are determined genetically by 3 alleles: ,, and . A person whose blood genotype is , , , , , or is heterozygous while a person whose blood genotype is , , or is homozygous. The Hardy-Weinberg Law states that the proportion P of heterozygous individuals in a certain population (in genetic equilibrium) is given by

where is the percent of allele , the percent of allele , and the percent allele within the population. Find the maximum possible proportion of heterozygous individuals in a population (in genetic equilibrium).

### Others (proofs & ect.)

1.In with , is the foot of the perpendicular from to . is the foot of the perpendicular from to . is the midpoint of . Prove that .

2.Show that the series

where p and m are real numbers converge if or but and diverge otherwise.

## Solutions

### Problem 1

There is one and only one perfect square in the form

where and is prime. Find that perfect square.

### Solution 1

. Suppose . Then,

, so since , so is less than both and and thus we have and . Adding them gives so by Simon's Favorite Factoring Trick, in some order. Hence, .~Ddk001

### Problem 2

and are positive integers. If , find .

### Solution 1 (Slow, probably official MAA)

Let and . Then,

### Solution 2 (Fast)

Recall that a perfect square can be written as . Since is a perfect square, the RHS must be in this form. We substitute for to get that . To make the middle term have an exponent of , we must have . Then and , so .

~ cxsmi

### Solution 3 (Faster)

Calculating the terms on the RHS, we find that . We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that works, and it produces . Then .

~ (also) cxsmi

### Problem 3

The fraction,

where and are side lengths of a triangle, lies in the interval , where and are rational numbers. Then, can be expressed as , where and are relatively prime positive integers. Find .

### Solution 1(Probably official MAA, lots of proofs)

**Lemma 1: **

*Proof:* Since the sides of triangles have positive length, . Hence,

, so now we just need to find .

Since by the Trivial Inequality, we have

as desired.

To show that the minimum value is achievable, we see that if , , so the minimum is thus achievable.

Thus, .

**Lemma 2: **

*Proof:* By the Triangle Inequality, we have

.

Since , we have

.

Add them together gives

Even though unallowed, if , then , so

.

Hence, , since by taking and close , we can get to be as close to as we wish.

### Solution 2 (Fast, risky, no proofs)

By the Principle of Insufficient Reason, taking we get either the max or the min. Testing other values yields that we got the max, so . Another extrema must occur when one of (WLOG, ) is . Again, using the logic of solution 1 we see so so our answer is . ~Ddk001

### Problem 4

Suppose there are complex values and that satisfy

Find .

### Solution 1

To make things easier, instead of saying , we say .

Now, we have . Expanding gives

.

To make things even simpler, let

, so that .

Then, if , Newton's Sums gives

Therefore,

Now, we plug in

.

We substitute to get

.

Note: If you don't know Newton's Sums, you can also use Vieta's Formulas to bash.~Ddk001

### Problem 5

Suppose

Find the remainder when is divided by 1000.

### Solution 1 (Euler's Totient Theorem)

We first simplify

so

.

where the last step of all 3 congruences hold by the Euler's Totient Theorem. Hence,

Now, you can bash through solving linear congruences, but there is a smarter way. Notice that , and . Hence, , so . With this in mind, we proceed with finding .

Notice that and that . Therefore, we obtain the system of congruences :

.

Solving yields , and we're done. ~Ddk001

### Problem 6

Suppose that there is rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are other pegs positioned sufficiently apart. A is made if

(i) ring changed position (i.e., that ring is transferred from one peg to another)

(ii) No bigger rings are on top of smaller rings.

Then, let be the minimum possible number that can transfer all rings onto the second peg. Find the remainder when is divided by .

### Solution 1 (Recursion)

Let be the minimum possible number that can transfer rings onto the second peg. To build the recursion, we consider what is the minimum possible number that can transfer rings onto the second peg. If we use only legal , then will be smaller on the top, bigger on the bottom. Hence, the largest ring have to be at the bottom of the second peg, or the largest peg will have nowhere to go. In order for the largest ring to be at the bottom, we must first move the top rings to the third peg using , then place the largest ring onto the bottom of the second peg using , and then get all the rings from the third peg on top of the largest ring using another . This gives a total of , hence we have . Obviously, . We claim that . This is definitely the case for . If this is true for , then

so this is true for . Therefore, by induction, is true for all . Now, . Therefore, we see that

.

But the part is trickier. Notice that by the Euler's Totient Theorem,

so is equivalent to the inverse of in , which is equivalent to the inverse of in , which, by inspection, is simply . Hence,

, so by the Chinese Remainder Theorem, .

### Problem 7

Let be a 2-digit positive integer satisfying . Find the sum of all possible values of .

### Solution 1 (Tedious Casework)

**Case 1: **

In this case, we have

.

If , we must have

, but this contradicts the original assumption of , so hence we must have .

With this in mind, we consider the unit digit of .

**Subcase 1.1: **

In this case, we have that

.

There is no apparent contradiction here, so we leave this as it is.

**Subcase 1.2: **

In this case, we have that

.

This contradicts with the fact that , so this is impossible.

**Subcase 1.3: **

In this case, we have that

.

However, this is impossible for all .

**Subcase 1.4: **

In this case, we have that

.

Again, this yields , which, again, contradicts .

Hence, we must have .

Now, with determined by modular arithmetic, we actually plug in the values.

To simplify future calculations, note that

.

For , this does not hold.

For , this does not hold.

For , this does hold. Hence, is a solution.

For , this does not hold.

For , this does not hold.

Hence, there is no positive integers and between and inclusive such that .

**Case 2: **

For this case, we must have

which is impossible if a is a integer and .

**Case 3: **

In this case, we have

.

If , we must have

which is impossible since and .

Hence, .

**Subcase 3.1: **

Testing cases, we can see that there is no such .

**Subcase 3.2: **

Testing cases, we can see that there is no such .

**Subcase 3.3: **

Testing cases, we can see that there is no such .

**Subcase 3.4: **

Testing cases, we can see that there is no such .

We see that . ~Ddk001

### Solution 2 (Official)

cleary square of a positive double-digit integer is either 3-digit or 4-digit as it ranges between and .

This means both and are less than or equal to 7 as any greater than 7! exceeds 9999.

Now at least one of or must be greater than or equal to 5 or else the maximum possible value of LHS is 4!+4!=48<100 also, both and can't be greater than or equal to 5 as if they were so. The unit digit of LHS would be zero but the unit digit of RHS would be either 5,6 or 9.

Hence, one of and is greater than or equal to 5 and the other is less than 5. this means the unit digit of RHS is the unit digit of a factorial which is less than 5.

Hence unit digit of RHS is 0,1,2, 6 or 4. 0,2,4 and 6 are rejected as follows:-

**1**. 2 can't be the unit digit of a perfect square.

**2**. If 6 is the unit digit of OF LHS this means one of the numbers is 3( as only 3! has the unit digit 6) and also this would mean that is either 4 or 6. This directly means that 36 and 34 are the only cases to be tested. 34 can't be as both the digits are less than 5 and 36 clearly doesn't satisfy

**3**. If 0 is the unit digit of LHS then 50 60 70 are the only cases (as one of the digits is greater than or equal to 5) that don't satisfy

**4**. If 4 is the unit digit of LHS this means one of the numbers is 4( as only 4! has the unit digit 4) and also this would mean that is either 2 or 8. This directly means that 42 and 48 are the only cases to be tested. 42 can't be as both the digits are less than 5 and 48 can't also as one of the digits is 8

Hence, the unit digit of LHS must be 1 for which =1 or 9(rejected). This means 51 61 and 71 are the only cases to be tried now which is really very easy to calculate and get 71 as the only possible solution to the problem and

thus, our answer is 7+1=.~ SANSGANKRSNGUPTA

### Solution 3 (Official and Fastest)

cleary square of a positive double-digit integer is either 3-digit or 4-digit as it ranges between and .

This means both and are less than or equal to 7 as any greater than 7! exceeds 9999.

Now at least one of or must be greater than or equal to 5 or else the maximum possible value of LHS is 4!+4!=48<100 also, both and can't be greater than or equal to 5 as if they were so. The unit digit of LHS would be zero but the unit digit of RHS would be either 5,6 or 9.

Hence, one of and is greater than or equal to 5 and the other is less than 5. Also, RHS is a perfect square, so it must be 0 or 1

Hence the number less than 5 can be either 1 or 4 because 2! and 3! are 2

If it is 4, then the unit digit of RHS is 4 meaning is either 2 or 8 both of which aren't possible as 8>7 and the number less than 5 is 4 not 2.

If it is 1 then the unit digit of RHS is 1 implying is either 9(rejected 9>7) or 1. This means is 1, this means 51 61 and 71 are the only cases to be tried

which is very easy to calculate and get 71 as the only possible solution to the problem and

thus, our answer is 7+1=.~ SANSGANKRSNGUPTA

### Problem 8

Suppose is a -degrees polynomial. The Fundamental Theorem of Algebra tells us that there are roots, say . Suppose all integers ranging from to satisfies . Also, suppose that

for an integer . If is the minimum possible positive integral value of

.

Find the number of factors of the prime in .

### Solution 1

Since all integers ranging from to satisfies , we have that all integers ranging from to satisfies , so by the Factor Theorem,

.

since is a -degrees polynomial, and we let to be the leading coefficient of .

Also note that since is the roots of ,

Now, notice that

Similarly, we have

To minimize this, we minimize . The minimum can get is when , in which case

, so there is factors of . ~Ddk001

### Problem 9

is an isosceles triangle where . Let the circumcircle of be . Then, there is a point and a point on circle such that and trisects and , and point lies on minor arc . Point is chosen on segment such that is one of the altitudes of . Ray intersects at point (not ) and is extended past to point , and . Point is also on and . Let the perpendicular bisector of and intersect at . Let be a point such that is both equal to (in length) and is perpendicular to and is on the same side of as . Let be the reflection of point over line . There exist a circle centered at and tangent to at point . intersect at . Now suppose intersects at one distinct point, and , and are collinear. If , then can be expressed in the form , where and are not divisible by the squares of any prime. Find .

Someone mind making a diagram for this?

### Solution 1

Line is tangent to with point of tangency point because and is perpendicular to so this is true by the definition of tangent lines. Both and are on and line , so intersects at both and , and since we’re given intersects at one distinct point, and are not distinct, hence they are the same point.

Now, if the center of tangent circles are connected, the line segment will pass through the point of tangency. In this case, if we connect the center of tangent circles, and ( and respectively), it is going to pass through the point of tangency, namely, , which is the same point as , so , , and are collinear. Hence, and are on both lines and , so passes through point , making a diameter of .

Now we state a few claims :

**Claim 1: is equilateral. **

*Proof:*

where the last equality holds by the Power of a Point Theorem.

Taking the square root of each side yields .

Since, by the definition of point , is on . Hence, , so

, and since is the reflection of point over line , , and since , by the Pythagorean Theorem we have

Since is the perpendicular bisector of , and we have hence is equilateral.

With this in mind, we see that

Here, we state another claim :

**Claim 2 : is a diameter of **

*Proof:* Since , we have

and the same reasoning with gives since .

Now, apply Ptolemy’s Theorem gives

so is a diameter.

From that, we see that , so . Now,

, so

, so

, and we’re done. ~Ddk001

### Problem 10

Suppose where and are relatively prime positive integers. Find .

### Solution 1(Wordless endless bash)

### Problem 11

In with , is the foot of the perpendicular from to . is the foot of the perpendicular from to . is the midpoint of . Prove that .

### Solution 1 (Analytic geo)

Let

We set it this way to simplify calculation when we calculate the coordinates of and (Notice to find , you just need to take the x coordinate of and let the y coordinate be ).

Obviously,

Now, we see that

, so , as desired. ~Ddk001

### Solution 2 (Hard vector bash)

#### Solution 2a (Hard)

Hence, . ~Ddk001

#### Solution 2b (Harder)

Since is the midpoint of ,

Now come the coordinates. Let

so that

.

Therefore,

Hence, we have that is perpendicular to . ~Ddk001

### Problem 12

Show that the series

where p and m are real numbers converge if or but and diverge otherwise.

### Solution 1 (Plain old chapter 3 analysis bash)

Before we even get started, let's state a few definitions first.

**Definition 1:** A set , whose elements we shall call *points*, is said to be a *metric space* if with any two points and of there is associated a real number , called the *distance* from to , such that

if

, for any .

Any function with these properties is called a *distance function* or a *metric*

**Definition 2:** Let be a metric space. All points and sets mentioned below are understood to be elements and subsets of .

(a) A *neighborhood* of a point is a set consisting of all points such that . The number is said to be the *radius* of .

(b) A point is a *limit point* of a set if every neighborhood of contains a point such that .

(c) If and is not a limit point of , then is called a *isolated point* of .

(d) is *closed* if every limit point of is a point of .

(e) A point is an *interior point* of if there is a neighborhood of such that

(f) is *open* if every point of is a interior point of .

(g) The *complement* of (denoted by ) is the set of points such that but .

(h) is bounded if there is a ream number and a point such that for all .

**Definition 3:** Let be a metric space. All points and sets mentioned below are understood to be elements and subsets of .

(a) By a *open cover* of a set in we mean a collection of open subsets of such that .

(b) A subset of is said to be *compact* if every open cover of contain a finite subcover.

**Definition 4:** Let be a metric space and . Let denote the set of limit points of . The *closure* of is said to be .

**Lemma 1:** Closed subsets of compact sets are compact.

*Proof:* Suppose , is closed and is compact. Let be a open cover of . We wish to show that have no finite subcover of . Let be adjoined to . Then we obtain an open cover of . (Note that complements of closed sets are open since if is closed, all the limit points of is a point of so if , . Every neighborhood of contain a point of that is not itself so no neighborhood of is completely in so is not a interior point of , so is open. Thus is open.) Since is compact, there is a finite subcollection of that cover , and hence . If , we can exclude it from and can thus still have a open cover of . The obtained open cover is thus a finite subcollection of that cover , so is compact.

**Definition 5:** A subset of is said to be a *k-cell* if is the set of such that , where the 's and 's are real numbers.

**Lemma 2:** Every k-cell is compact.

*Proof:* Let be a k-cell, consisting of the points such that . Put

Thus would imply .

Suppose, for the sake of contradiction, that there exist an open cover of which contain no finite subcover. Put . The intervals and then determine k-cells whose union is . Thus at least one of these sets, call it , cannot be covered by any finite subcollection of (otherwise could be so covered). We divide like we did , and obtain as we did . As the process continue we thus obtain a sequence of 's such that

(a) ;

(b) is not covered by any subcollection of ;

(c) If , then .

We claim that there is a point in every . In fact, we can even prove the stronger statement

"Let be a positive interger. If is a sequence of k-cells such that , then ."

To prove that, it suffices to prove it with . The rest follows easily from induction. The case can be rephrased as follows:

"If is a sequence of intervals such that , then

To prove this, let . Then let denote the set of all the 's. Obviously is non-empty and is bounded above (by , for one). Then the least upper bound of exist, call it . If and are any positive integers, then

so for each . Since , obviously .

The result follow.

Thus there is a point in every . Now, since covers it follows that for some . Since is open there is a such that implies . If n is so large that then (c) implies that , which contradicts (b). A contradiction is obtained and the result must follow.

**Corollary 1 (Heine-Borel Theorem):** Suppose . Then the following two statements are equivalent:

(a) is closed and bounded

(b) is compact

*Proof:* It suffices to show that . We shall only prove that since we will only use that part. However, a proof of the converse is given in a remark after the solution.

Since is closed and bounded, is a closed subset of a k-cell, a compact set. Thus is compact by Lemma 1.

Thus, in , every single bounded set have a compact closure.

**Definition 6:** Let be a subset of a metric space , and let be the set such that

.

Then the smallest upper bound of is called the *diameter* of and is often written as .

**Lemma 3:** (a) Let be the closure of in a metric space . Then

(b) If is a sequence of compact sets in such that and if

, then consist of exactly one point.

*Proof:* (a) Since , it is obvious that

Thus it suffice to show that

To do so, choose . Then for each there exist points and such that and that . Hence,

It follows that

. Since is ambitarily chosen, (a) is proved.

(b) Put . It suffice to show that has at most 1 point and that it exist. The former is easy. If have 1 or more points, which contradict the fact that .

It remains to show that is non-empty.

*Sublemma 3.1:* If is a collection of compact subsets of a metric space such that the intersection of every finite subcollection of is non-empty, then is non-empty.

*Proof:* Fix a member of and let for each . Suppose, for the sake of contradiction that no point of belong to every single member of . Note that every is open by the parenthetical remark in the proof of lemma 1. Thus the sets form a open cover of ; however, since is compact there exist finitely many indices such that

But this would mean

, contradictory to the hypothesis.

A contradiction is obtained so we are done. .

The result follow from the sublemma.

Now, we show state another important lemma that will use the above lemma (lemma 3). Part (a) and (c) of the following lemma is called the *Cauchy Criterion*.

Before starting the proof, however, we have one more definition to make.

**Definition 7:** A sequence of a metric space is said to be a *Cauchy Sequence* if for every there exist a integer such that if .

Obviously, we can state definition 7 in terms of definition 6:

" A sequence of a metric space is said to be a *Cauchy Sequence* if

where is the set consisting of the points . "

In fact, its converse is also true (and its proof is trivial):

"Let be a sequence of sets such that

.

Then if a sequence is chosen with

*More to come*

### Problem 13

Let and be real numbers. Show that

### Solution 1 (Plain high school contest)

It might seem insane, but we are going to divide this problem into 2 cases:

and

**Case 1: **

By the Trivial Inequality, we have

. Add these inequalities gives

and expanding yields

Since , we can square both sides and obtain

so

so the result follow if .

**Case 2: **

By the Trivial Inequality, we have

. Add these inequalities gives

and expanding yields

Since , we can square both sides and obtain

and the rest proceeds as in case 1.

The result follows. ~Ddk001

### Solution 2 (Lagrange Multipliers)

Put and let $g(

Under construction

===Solution 3 (Finding relative extrema with calculus)=== Let <math>f(x, y, z) = (x^2 + z^2)^2 + y^4 - 4xzy^2$ (Error compiling LaTeX. Unknown error_msg). We will prove that is never negative (equivalent to the problem statement) by finding the critical values of . (Note that is a polynomial of , , and . Therefore, we do not need to consider discontinuities.) We will begin by setting partial derivatives of to . (Note that if , , or , the inequality in the problem statement reduces to , which is true by the Trivial Inequality. Hereon, we assume ).

If is a critical point of , then it must satisfy equations (1), (2), and (3). Substituting (3) into (1), we find:

If , then and by the Trivial Inequality. We now only need to consider the case where . Let be the function obtained when we substitute into function .

.

We must now prove that is never negative (proving this statement finishes the last case of the proof). We similarly find the critical values of by taking partial derivatives of and setting them to .

If is a critical point of , then . If then . Therefore, is never negative. QED

### Solution 4 (AM-GM bash)

Under construction (I am not very sure if this could work)

### Problem 14

Common blood types are determined genetically by 3 alleles: ,, and . A person whose blood type is , , or is heterozygous while , , or is homozygous. The Hardy-Weinberg Law states that the proportion P of heterozygous individuals in a certain population is given by

where is the percent of allele , the percent of allele , and the percent allele . Find the maximum possible proportion of heterozygous individuals in a certain population.

### Solution 1 (Lagrange Multipliers)

Note that allele , , and make up all the alleles, so therefore,

The problem reduces to determining the maximum value of given that . Now, by the Method of Lagrange multipliers, we find all values of , , and and such that the gradient of is times the gradient of , or

or

Solving yields

Now, since , we have

so . Therefore, the maximum of given is

.

### Solution 2 (Relative extrema by calculus)

Just as in solution 1, we see that . To turn this into a 2-D problem, substitute

## Contributors

## See also

Here's the source for the problems:

1,2,3,4,5,6,8,9,10,11: Ddk001, credits given to Ddk001

7: SANSKAR'S OG PROBLEMS, credits given to SANSGANKRSNGUPTA

- Note: Problem 6 is based on the Tower of Hanoi Problem

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*