# User:Ddk001

## Contents

- 1 Introduction
- 2 User Counts
- 3 Cool asyptote graphs
- 4 Problems Sharing Contest
- 5 Contributions
- 6 Problems I made
- 7 Answer key
- 8 Solutions
- 8.1 Problem 1
- 8.2 Solution 1
- 8.3 Problem 2
- 8.4 Solution 1 (Slow, probably official MAA)
- 8.5 Solution 2 (Fast)
- 8.6 Solution 3 (Faster)
- 8.7 Problem 3
- 8.8 Solution 1(Probably official MAA, lots of proofs)
- 8.9 Solution 2 (Fast, risky, no proofs)
- 8.10 Solution 3
- 8.11 Problem 4
- 8.12 Solution 1
- 8.13 Problem 5
- 8.14 Solution 1 (Euler's Totient Theorem)
- 8.15 Problem 6
- 8.16 Solution 1 (Recursion)
- 8.17 Problem 7
- 8.18 Solution 1
- 8.19 Problem 8
- 8.20 Solution 1
- 8.21 Problem 9
- 8.22 Solution 1(Wordless endless bash)
- 8.23 Problem 10
- 8.24 Solution 1 (Analytic geo)
- 8.25 Solution 2 (Hard vector bash)

- 9 Vandalism area
- 10 See also

## Introduction

I am a 5th grader who likes making and doing problems, doing math, and redirecting pages (see Principle of Insufficient Reasons). I like geometry and don't like counting and probability. My number theory skill are also not bad.

## User Counts

If this is your first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)

For those of you who want more boxes, me too. However, this is the max number of boxes. Also, I check the pages history so I know if someone edited something.

(Please don't mess with the user count)

Doesn't that look like a number on a pyramid

## Cool asyptote graphs

Asymptote is fun!

## Problems Sharing Contest

Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start:

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

where and are prime. Find that perfect square.

1. We can expand the product in the expression. . Suppose this equals for some positive integer . We rewrite using the square of a binomial pattern to find that . Through trial and error on small values of and , we find that and must equal and in some order. The perfect square formed using these numbers is .

Note: I will be the first to admit that this solution is somewhat lucky.

2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle intersects itself. has leg length . The perimeter of this diamond is expressible as , where , , and are integers, and is not divisible by the square of any prime. What is the remainder when is divided by ?

## Contributions

2005 AMC 8 Problems/Problem 21 Solution 2

2022 AMC 12B Problems/Problem 25 Solution 5 (Now it's solution 6)

2023 AMC 12B Problems/Problem 20 Solution 3

2016 AIME I Problems/Problem 10 Solution 3

2017 AIME I Problems/Problem 14 Solution 2

2019 AIME I Problems/Problem 15 Solution 6

2022 AIME II Problems/Problem 3 Solution 3

Restored diagram for 1994 AIME Problems/Problem 7

Principle of Insufficient Reasons

## Problems I made

### Aime styled

#### Introductory

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

where and are prime. Find that perfect square.

2. and are positive integers. If , find .

#### Intermediate

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 is complex values and that satisfy

Find .

5. Suppose

Find the remainder when is divided by .

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 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. 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 .

#### Olympiad

8. (Much harder) 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?

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

### Proofs

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

I will leave a big gap below this sentence so you won't see the answers accidentally.

## Answer key

1. 049

2. 019

3. 092

4. 170

5. 736

6. 895

7. 011

8. 054

9. 077

## Solutions

- Note: Most the solutions so far have been made by me :)

I like your solutions.~Ddk001

### 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, .

### 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 .

### Solution 3

Expand the denominator. We now have . Consider its reciprocal; if this expression takes values on the interval , then its reciprocal will take values on the interval . This is important because we can now write the reciprocal of the expression as . We attempt to maximize and minimize . To maximize the expression, we consider the triangle inequality. From it, we find the following. We rewrite. Add all of the inequalities. We find the following. Considering the equality case and plugging into the expression, we find that the maximum value of the expression is . However, since this "equality case" cannot actually happen, this part of the interval must be open. Now, we minimize the inequality by using the Power Mean Inequality (specifically, the QM-AM part of the inequality). Considering the terms , , and , we find the following. Square both sides. Rewrite as follows. Considering the equality case and plugging into the expression, we find that the minimum value of the expression is . Since the expression (which we said was the reciprocal of the original expression) takes values on the interval , the original expression must take values on the interval . Then , so our final answer is .

~ cxsmi

### 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.

### 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.

### 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

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 .

### Problem 8

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.

### Problem 9

Suppose where and are relatively prime positive integers. Find .

### Solution 1(Wordless endless bash)

### Problem 10

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.

### Solution 2 (Hard vector bash)

#### Solution 2a (Hard)

Hence, .

#### Solution 2b (Harder)

Since is the midpoint of ,

Now come the coordinates. Let

so that

.

Therefore,

Hence, we have that is perpendicular to .

## Vandalism area

Here, you can add anything, delete anything, and do anything! (Don't delete this line since it's instruction and don't be inappropriate) Do not delete the see also. However, do NOT vandalize before this word (Feel free to delete this and the period that follows).

(ok :) :) this page is so cool!)

## See also

The problems on this page are NOT copyrighted by the Mathematical Association of America's American Mathematics Competitions.