Difference between revisions of "User:Ddk001"

Revision as of 19:51, 6 August 2024

If you have a problem or solution to contribute, please go to this page.

I am a aops user 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 Count

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27

Credits given to Firebolt360 for inventing the box above.

Cool asyptote graphs

Asymptote is fun! $[asy]draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));[/asy]$

$[asy]draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));[/asy]$

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

$(p^2+1)(q^2+1)-((pq)^2-pq+1)$

where $p$ and $q$ are prime. Find that perfect square. ~Ddk001

$\textbf{Solution by cxsmi}$

1. We can expand the product in the expression. $(p^2+1)(q^2+1)-((pq)^2-pq+1) = p^2q^2+p^2+q^2+1-((pq)^2-pq+1) = p^2 + q^2 + pq$ . Suppose this equals $m^2$ for some positive integer $m$ . We rewrite using the square of a binomial pattern to find that $m^2 = (p + q)^2 - pq$ . Through trial and error on small values of $p$ and $q$ , we find that $p$ and $q$ must equal $3$ and $5$ in some order. The perfect square formed using these numbers is $\boxed{49}$ .

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 $\triangle ABC$ intersects $\triangle ABC$ itself. $\triangle ABC$ has leg length $2024$ . The perimeter of this diamond is expressible as $a\sqrt{b}-c$ , where $a$ , $b$ , and $c$ are integers, and $c$ is not divisible by the square of any prime. What is the remainder when $a + b + c$ is divided by $1000$ ?

$[asy] unitsize(1inch); draw((0,0)--(0,2)); draw((0,2)--(2,0)); draw((2,0)--(0,0)); draw(circle((0.586,0.586),0.586)); draw((0,0)--(0,1.172),red); draw((0,1.172)--(1.172,1.172)); draw((1.172,1.172)--(1.172,0)); draw((1.172,0)--(0,0),red); draw((0,1.172)--(0.828,1.172),red); draw((0.828,1.172)--(1.172,0.828),red); draw((1.172,0.828)--(1.172,0),red); draw((0,0.1)--(0.1,0.1)); draw((0.1,0.1)--(0.1,0)); label("$A$",(0,2.1)); label("$B$",(0,-0.1)); label("$C$",(2,-0.1)); label("$2024$",(-0.2,1)); label("$2024$",(1,-0.2)); [/asy]$

$\textbf{Solution by Ddk001}$

Solution 1

The inradius of $\Delta ABC$ , $r$ , can be calculated as

$r=\frac{\textbf{Area}_{ABC}}{\textbf{Semiperimeter}} \implies r=\frac{2024^2/2}{(2024+2024+2024 \sqrt{2})/2}=\frac{2024}{2+\sqrt{2}}=2024-1012\sqrt{2}$

so the square have side length $4048-2024 \sqrt{2}$ . Let the $D$ be the vertex of the square $D \ne B$ on side $BC$ . Then $DC= 2024 (\sqrt{2} -1)$ . Let the sides of the square intersect $AC$ at $E$ and $F$ , with $AE<AF$ . Then $AE=CF=2024(2-\sqrt{2})$ so $EF=2024 (3 \sqrt{2} -4)$ . Let $G$ be the vertex of the square across from $B$ . Then $EG=FG=2024 (3-2\sqrt{2})$ . Thus the perimeter of the diamond is

$4(4048-2024 \sqrt{2})-2 \cdot 2024 (3-2\sqrt{2})+2024 (3 \sqrt{2} -4)=2024 (8-4 \sqrt{2}-6+4\sqrt{2}+3\sqrt{2}-4)=2024(3\sqrt{2}-2)=7072\sqrt{2}-4048$

The desired sum is $7072+2+4048=\boxed{11122}$ . $\blacksquare$

Ddk001 Presents

THE FOLLOWING PROBLEM

Note: This is one of my favorite problems. Very well designed and actually used two of my best tricks without looking weird.

Suppose

$x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}$

Find the remainder when $\min{x}$ is divided by 1000.

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

1978 USAMO Problems/Problem 1 Solution 4

Restored diagram for 1994 AIME Problems/Problem 7

Divergence Theorem

Stokes' Theorem

Principle of Insufficient Reasons

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!)

honestly i think your user page is very cool. :)

Hi Ddk001 User:zhenghua (Taking Oly Geo)

Zhenghua I havent seen you since forever!!! I'm not focusing on compitition right now so you probably won't see me in any of your classes.

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 <div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div>
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 [[2022 AIME II Problems/Problem 3]] Solution 3
+[[1978 USAMO Problems/Problem 1]] Solution 4
 Restored diagram for [[1994 AIME Problems/Problem 7]]
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 [[Principle of Insufficient Reasons]]
 ==Vandalism area==

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