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Difference between revisions of "User:Ddk001"

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==Problems (I made it, not copied)==
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If you have a problem or solution to contribute, please go to [[Problems Collection|this page]].
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See if you can solve these:  
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 +
 
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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
 +
 
 +
<br>
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__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3">
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==
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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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<center><font size="100px">User Count</font></center>
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</div>
 +
 
 +
Credits given to [[User:Firebolt360|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>
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 +
 
 +
 
 +
==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
 
1. There is one and only one perfect square in the form
  
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<math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>
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<cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath>
 +
 
 +
where <math>p</math> and <math>q</math> are prime. Find that perfect square. ~[[Ddk001]]
 +
 
 +
<math>\textbf{Solution by cxsmi}</math>
 +
 
 +
1. We can expand the product in the expression. <math>(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</math>. Suppose this equals <math>m^2</math> for some positive integer <math>m</math>. We rewrite using the square of a binomial pattern to find that <math>m^2 = (p + q)^2 - pq</math>. Through trial and error on small values of <math>p</math> and <math>q</math>, we find that <math>p</math> and <math>q</math> must equal <math>3</math> and <math>5</math> in some order. The perfect square formed using these numbers is <math>\boxed{49}</math>.
 +
 
 +
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 <math>\triangle ABC</math> intersects <math>\triangle ABC</math> itself. <math>\triangle ABC</math> has leg length <math>2024</math>. The perimeter of this diamond is expressible as <math>a\sqrt{b}-c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers, and <math>c</math> is not divisible by the square of any prime. What is the remainder when <math>a + b + c</math> is divided by <math>1000</math>?
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 +
<asy>
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unitsize(1inch);
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draw((0,0)--(0,2));
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draw((0,2)--(2,0));
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draw((2,0)--(0,0));
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draw(circle((0.586,0.586),0.586));
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draw((0,0)--(0,1.172),red);
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draw((0,1.172)--(1.172,1.172));
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draw((1.172,1.172)--(1.172,0));
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draw((1.172,0)--(0,0),red);
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draw((0,1.172)--(0.828,1.172),red);
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draw((0.828,1.172)--(1.172,0.828),red);
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draw((1.172,0.828)--(1.172,0),red);
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draw((0,0.1)--(0.1,0.1));
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draw((0.1,0.1)--(0.1,0));
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label("$A$",(0,2.1));
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label("$B$",(0,-0.1));
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label("$C$",(2,-0.1));
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label("$2024$",(-0.2,1));
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label("$2024$",(1,-0.2));
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</asy>
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 +
 
 +
<math>\textbf{Solution by Ddk001}</math>
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 +
 
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===Solution 1===
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The inradius of <math>\Delta ABC</math>, <math>r</math>, can be calculated as
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<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>
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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>
 +
 
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The desired sum is <math>7072+2+4048=\boxed{11122}</math>. <math>\blacksquare</math>
 +
 
  
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Find that perfect square.
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[[Ddk001]] Presents
  
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2. Suppose there is complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
 
  
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<math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</math>
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THE FOLLOWING PROBLEM
  
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Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
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Note: This is one of my favorite problems. Very well designed and actually used two of my best tricks without looking weird.
  
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3. Suppose  
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Suppose
  
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<math>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</math>
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<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.
 
Find the remainder when <math>\min{x}</math> is divided by 1000.
  
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4. Suppose <math>f(x)</math> is a <math>10000000010</math>-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
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==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
  
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<math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
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[[2017 AIME I Problems/Problem 14]] Solution 2
  
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for an integer <math>m</math>. If <math>p</math> is the minimum possible value of
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[[2019 AIME I Problems/Problem 15]] Solution 6
  
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<math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
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[[2022 AIME II Problems/Problem 3]] Solution 3
  
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Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
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[[1978 USAMO Problems/Problem 1]] Solution 4
  
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5. (Much harder) <math>\Delta ABC</math> is an isosceles triangle where <math>CB=CA</math>. Let the circumcircle of <math>\Delta ABC</math> be <math>\Omega</math>. Then, there is a point <math>E</math> and a point <math>D</math> on circle <math>\Omega</math> such that <math>AD</math> and <math>AB</math> trisects <math>\angle CAE</math> and <math>BE<AE</math>, and point <math>D</math> lies on minor arc <math>BC</math>. Point <math>F</math> is chosen on segment <math>AD</math> such that <math>CF</math> is one of the altitudes of <math>\Delta ACD</math>. Ray <math>CF</math> intersects <math>\Omega</math> at point <math>G</math> (not <math>C</math>) and is extended past <math>G</math> to point <math>I</math>, and <math>IG=AC</math>. Point <math>H</math> is also on <math>\Omega</math> and <math>AH=GI<HB</math>. Let the perpendicular bisector of <math>BC</math> and <math>AC</math> intersect at <math>O</math>. Let <math>J</math> be a point such that <math>OJ</math> is both equal to <math>OA</math> (in length) and is perpendicular to <math>IJ</math> and <math>J</math> is on the same side of <math>CI</math> as <math>A</math>. Let <math>O’</math> be the reflection of point <math>O</math> over line <math>IJ</math>. There exist a circle <math>\Omega_1</math> centered at <math>I</math> and tangent to <math>\Omega</math> at point <math>K</math>. <math>IO’</math> intersect <math>\Omega_1</math> at <math>L</math>. Now suppose <math>O’G</math> intersects <math>\Omega</math> at one distinct point, and <math>O’, G</math>, and <math>K</math> are collinear. If <math>IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2</math>, then <math>\frac{EH}{BH}</math> can be expressed in the form <math>\frac{\sqrt{b}}{a} (\sqrt{c} + d)</math>, where <math>b</math> and <math>c</math> are not divisible by the squares of any prime. Find <math>a^2+b^2+c^2+d^2+abcd</math>.
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Restored diagram for [[1994 AIME Problems/Problem 7]]
  
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==User Counts==
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[[Divergence Theorem]]
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==
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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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[[Stokes' Theorem]]
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<center><font size="100px">0</font></center>
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</div>
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[[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==
 +
* My [[User talk:Ddk001|talk page]]
 +
* [[Problems Collection|My problems collection]]
 +
 
 +
The problems on this page are NOT copyrighted by the [http://www.maa.org Mathematical Association of America]'s [http://amc.maa.org American Mathematics Competitions]. [[File:AMC_logo.png|middle]]
 +
<div style="clear:both;">
 +
 
 +
Can someone help me clear out [[Problems Collection|this page]]?

Latest revision as of 20:57, 24 November 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

If this is your first time visiting this page, edit it by incrementing the user count below by one.
User Count

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.

See also

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

Can someone help me clear out this page?

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=User:Ddk001&oldid=235533"