Difference between revisions of "User:Ddk001"

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.

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Difference between revisions of "User:Ddk001"

Latest revision as of 20:57, 24 November 2024

User Count

Cool asyptote graphs

Problems Sharing Contest

Solution 1

Contributions

Vandalism area

See also

@@ Line 1: / Line 1: @@
-==User Counts==
+If you have a problem or solution to contribute, please go to [[Problems Collection|this page]].
-If this is you first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)
-<math>\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{1}}}}}}</math>
+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
-(Please don't mess with the user count)
+<br>
+__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3">
-Doesn't that look like a number on a pyramid?
+==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==
+<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>
+<center><font size="100px">User Count</font></center>
+</div>
+Credits given to [[User:Firebolt360|Firebolt360]] for inventing the box above.
 ==Cool asyptote graphs==
@@ Line 16: / Line 21: @@
 <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 I made==
-See if you can solve these:
+==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:
-. (Much easier) There is one and only one perfect square in the form
+. There is one and only one perfect square in the form
-<math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>
+<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.
+where <math>p</math> and <math>q</math> are prime. Find that perfect square. ~[[Ddk001]]
-. Suppose there is complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
+<math>\textbf{Solution by cxsmi}</math>
-<math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</math>
+. 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>.
-Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
+Note: I will be the first to admit that this solution is somewhat lucky.
-. Suppose
-<math>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</math>
+. 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>?
-Find the remainder when <math>\min{x}</math> 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>
-. 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
-<math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
+<math>\textbf{Solution by Ddk001}</math>
-for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
-<math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
+===Solution 1===
+The inradius of <math>\Delta ABC</math>, <math>r</math>, can be calculated as
-Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
+<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>
-. (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>.
+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
-Someone mind making a diagram for this?
+<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>
-==Answer key & solution to the problems==
+The desired sum is <math>7072+2+4048=\boxed{11122}</math>. <math>\blacksquare</math>
-I will leave a big gap below this sentence so you won't see the answers accidentally.
+[[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.
-dsf
-fsd
-===Answer key===
-. 049
-. 170
-. 736
-. 011
-. 054
-===Solutions===
-====Problem 1====
-There is one and only one perfect square in the form
-<math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>
-where <math>p</math> and <math>q</math> is prime. Find that perfect square.
-====Solution 1====
-<math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2 \cdot q^2 +p^2+q^2+1-p^2 \cdot q^2 +pq-1=p^2+q^2+pq</math>.
-Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>.
-Then, <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)</math>, so since <math>n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}</math>, <math>n>p,n>q</math> so <math>p+q-n</math> is less than both <math>p</math> and <math>q</math> and thus we have <math>p+q-n=1</math> and <math>p+q+n=pq</math>. Adding them gives <math>2p+2q=pq+1</math> so by [[Simon's Favorite Factoring Trick]], <math>(p-2)(q-2)=3 \implies (p,q)=(3,5)</math> in some order. Hence, <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}</math>.<math>\square</math>
-====Problem 2====
-Suppose there are complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
-<math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</math>
-Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
-====Solution 1====
-To make things easier, instead of saying <math>x_i</math>, we say <math>x</math>.
-Now, we have
-<math>(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}</math>.
-Expanding gives
-<math>x^3-(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}) \cdot x^2+(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})x-(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})=0</math>.
-To make things even simpler, let
-<math>a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}</math>, so that <math>x^3-ax^2+bx-c=0</math>.
-Then, if <math>P_n=x_{1}^n+x_{2}^n+x_{3}^n</math>, [[Newton's Sums]] gives
-<math>P_1+(-a)=0</math>  <math>(1)</math>
-<math>P_2+(-a) \cdot P_1+2 \cdot b=0</math>  <math>(2)</math>
-<math>P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0</math>  <math>(3)</math>
-Therefore,
-<math>P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))</math>
-<math>=a \cdot P_2-b \cdot P_1+3 \cdot c</math>
-<math>=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c</math>
-<math>=a(a^2-2b)-ab+3c</math>
-<math>=a^3-3ab+3c</math>
-Now, we plug in <math>a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}:</math>
-<math>P_3=(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})^3-3(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})+3(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})</math>.
-As we have done many times before, we substitute <math>x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}</math> to get
-<math>P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})</math>
-<math>=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3(x^2y+y^2x+x^2z+z^2x+z^2y+y^2z+3xyz)+3xyz+1</math>
-<math>=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3x^2y-3y^2x-3x^2z-3z^2x-3z^2y-3y^2z-9xyz+3xyz+1</math>
-<math>=x^3+y^3+z^3+1</math>
-<math>=13+53+103+1</math>
-<math>=\boxed{170}</math>. <math>\square</math>
-Note: If you don't know [[Newton's Sums]], you can also use [[Vieta's Formulas]] to bash.
-====Problem 3====
 Suppose
-<math>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</math>
+<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.
-====Solution 1 (Euler's Totient Theorem)====
+==Contributions==
-We first simplify <math>\cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:</math>
+[[2005 AMC 8 Problems/Problem 21]] Solution 2
-<math>2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6=42^4+6 \cdot 30^6=(\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)}</math>
-so
-<math>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)} \equiv 1 \pmod{5}</math>
-<math>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \pmod{6}</math>
-<math>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 6 \cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)} \equiv 6 \pmod{7}</math>.
-where the last step of all 3 congruences hold by the [[Euler's Totient Theorem]].
-Hence,
-<math>x \equiv 1 \pmod{5}</math>
-<math>x \equiv 0 \pmod{6}</math>
-<math>x \equiv 6 \pmod{7}</math>
-Now, you can bash through solving linear congruences, but there is a smarter way. Notice that <math>5|x-6,6|x-6</math>, and <math>7|x-6</math>. Hence, <math>210|x-6</math>, so <math>x \equiv 6 \pmod{210}</math>. With this in mind, we proceed with finding <math>x \pmod{7!}</math>.
-Notice that <math>7!=5040= \text{lcm}(144,210)</math> and that <math>x \equiv 0 \pmod{144}</math>. Therefore, we obtain the system of congruences :
-<math>x \equiv 6 \pmod{210}</math>
-<math>x \equiv 0 \pmod{144}</math>.
-Solving yields <math>x \equiv 2\boxed{736} \pmod{7!}</math>, and we're done. <math>\square</math>
-====Problem 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
-<math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
-for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
-<math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
-Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
-====Solution 1====
-Since all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>, we have that all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)-n=0</math>, so by the [[Factor Theorem]],
-<math>n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n</math>
-<math>\implies (n+1)n \dots (n-10000000008)|f(n)-n</math>.
-<math>\implies f(n)=a(n+1)n \dots (n-10000000008)+n</math>
-since <math>f(n)</math> is a <math>10000000010</math>-degrees polynomial, and we let <math>a</math> to be the leading coefficient of <math>f(n)</math>.
-Also note that since <math>r_1, r_2, \dots, r_{10000000010}</math> is the roots of <math>f(n)</math>, <math>f(n)=a(n-r_1)(n-r_2) \dots (n-r_{10000000010})</math>
-Now, notice that
-<math>m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})</math>
+[[2022 AMC 12B Problems/Problem 25]] Solution 5 (Now it's solution 6)
-<math>=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})</math>
+[[2023 AMC 12B Problems/Problem 20]] Solution 3
-<math>=\frac{f(-2)}{a}</math>
+[[2016 AIME I Problems/Problem 10]] Solution 3
-<math>=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}</math>
+[[2017 AIME I Problems/Problem 14]] Solution 2
-<math>=\frac{10000000010! \cdot a-2}{a}</math>
+[[2019 AIME I Problems/Problem 15]] Solution 6
-<math>=10000000010!-\frac{2}{a}</math>
+[[2022 AIME II Problems/Problem 3]] Solution 3
-Similarly, we have
+[[1978 USAMO Problems/Problem 1]] Solution 4
-<math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}</math>
+Restored diagram for [[1994 AIME Problems/Problem 7]]
-To minimize this, we minimize <math>m</math>. The minimum <math>m</math> can get is when <math>m=10000000011</math>, in which case
+[[Divergence Theorem]]
-<math>-\frac{2}{a}=10000000011!-10000000010!</math>
+[[Stokes' Theorem]]
-<math>=10000000011 \cdot 10000000010!-10000000010!</math>
+[[Principle of Insufficient Reasons]]
-<math>=10000000010 \cdot 10000000010!</math>
+==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).
-<math>\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>
+(ok :) :) this page is so cool!)
-<math>=-\frac{1}{a}</math>
+honestly i think your user page is very cool. :)
-<math>=\frac{10000000010 \cdot 10000000010}{2}</math>
+Hi Ddk001  [[User:zhenghua]] (Taking Oly Geo)
-<math>=5000000005 \cdot 10000000010!</math>
+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.
-, so there is <math>\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}</math> factors of <math>999999937</math>. <math>\square</math>
+==See also==
+* My [[User talk:Ddk001|talk page]]
+* [[Problems Collection|My problems collection]]
-====Problem 5====
+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]]
-<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>.
+<div style="clear:both;">
-Someone mind making a diagram for this?
+Can someone help me clear out [[Problems Collection|this page]]?
-====Solution 1====