User:Ddk001

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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38

Credits given to Firebolt360 for inventing the box above.

Cool asyptote graphs

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

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

Contributions

AMC

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

2023 AMC 12B Problems/Problem 20 Solution 3

AIME

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

USAMO

1978 USAMO Problems/Problem 1 Solution 4

IMO

1995 IMO Problems/Problem 6 Solution 1 (I got the solution from a forum discussion)

1986 IMO Problems/Problem 3 Solution 2 (I may or may not had gotten the solution from another source)

Restored solution for 2022 USAMO Problems/Problem 6 Solution 1 See https://artofproblemsolving.com/wiki/index.php?title=2022_USAMO_Problems/Problem_6&action=history

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.

Vandalism area

Congratulations! You had reached the end of this boring user page. Time for vandalism! Write anything under 45000 bytes.

Hi Ddk001 ~ zhenghua

Today my math teacher at school thought that a triangle had to be equilateral just because an altitude could be drawn in it and graded my homework 90% because the wrong answer was correct.... (Wait can't any triangle have an altitude?)

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User:Ddk001

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Problems Sharing Contest

Solution 1

Vandalism area

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