Difference between revisions of "User:Ddk001"
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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> | <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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− | <math>\textbf{ | + | <math>\textbf{Solution by Ddk001}</math> |
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+ | ===Solution 1=== | ||
+ | The inradius of <math>\Delta ABC</math>, <math>r</math>, can be calculated as | ||
+ | |||
+ | <cmath>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}</cmath> | ||
+ | |||
+ | so the square have side length <math>4048-2024 \sqrt{2}</math>. Let the <math>D</math> be the vertex of the square <math>D \ne B</math> on side <math>BC</math>. Then <math>DC= 2024 (\sqrt{2} -1)</math>. Let the sides of the square intersect <math>AC</math> at <math>E</math> and <math>F</math>, with <math>AE<AF</math>. Then <math>AE=CF=2024(2-\sqrt{2})</math> so <math>EF=2024 (3 \sqrt{2} -4)</math>. Let <math>G</math> be the vertex of the square across from <math>B</math>. Then <math>EG=FG=2024 (3-2\sqrt{2})</math>. Thus the perimeter of the diamond is | ||
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+ | <cmath>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</cmath> | ||
+ | |||
+ | The desired sum is <math>7072+2+4048=\boxed{11122}</math>. <math>\blacksquare</math> | ||
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+ | |||
+ | [[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 | ||
+ | |||
+ | <cmath>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</cmath> | ||
+ | |||
+ | Find the remainder when <math>\min{x}</math> is divided by 1000. | ||
==Contributions== | ==Contributions== | ||
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[[2022 AIME II Problems/Problem 3]] Solution 3 | [[2022 AIME II Problems/Problem 3]] Solution 3 | ||
+ | |||
+ | [[1978 USAMO Problems/Problem 1]] Solution 4 | ||
Restored diagram for [[1994 AIME Problems/Problem 7]] | Restored diagram for [[1994 AIME Problems/Problem 7]] | ||
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[[Principle of Insufficient Reasons]] | [[Principle of Insufficient Reasons]] | ||
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==Vandalism area== | ==Vandalism area== |
Latest revision as of 20:51, 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
Credits given to Firebolt360 for inventing the box above.
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. ~Ddk001
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
?
Solution 1
The inradius of ,
, can be calculated as
so the square have side length . Let the
be the vertex of the square
on side
. Then
. Let the sides of the square intersect
at
and
, with
. Then
so
. Let
be the vertex of the square across from
. Then
. Thus the perimeter of the diamond is
The desired sum is .
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
Find the remainder when 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
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.
See also
The problems on this page are NOT copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Can someone help me clear out this page?