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| − | ==Introduction==
| + | If you have a problem or solution to contribute, please go to [[Problems Collection|this page]]. |
| − | I am a 5th grader 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.
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| − | ==User Counts==
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| − | If this is you first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)
| + | 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 |
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| − | <math>\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{1}}}}}}}}}}}}}}}}</math> | + | <br> |
| | + | __NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"> |
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| − | (Please don't mess with the user count)
| + | ==<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"><math>\sqrt{1849}</math></font></center> |
| | + | <div style="margin-left: 10px; margin-bottom:10px"><font color="black">Please do not mess with the user count. I would know and I would find you.</font></div> |
| | + | </div> |
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| − | Doesn't that look like a number on a pyramid?
| + | Credits given to [[User:Firebolt360|Firebolt360]] for inventing the box above. |
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| | ==Cool asyptote graphs== | | ==Cool asyptote graphs== |
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| − | ==Problems Sharing Contest==
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| − | Here, you can post all the math problem 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:
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| − | 1. There is one and only one perfect square in the form
| + | ==Contributions== |
| − | | |
| − | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath>
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| − | | |
| − | where <math>p</math> and <math>q</math> are prime. Find that perfect square.
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| − | (DO NOT LOOK AT MY SOLUTIONS YET)
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| − | ==Contributions== | + | ===AMC=== |
| − | [[2005 AMC 8 Problems/Problem 21]] Solution 2
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| | [[2022 AMC 12B Problems/Problem 25]] Solution 5 (Now it's solution 6) | | [[2022 AMC 12B Problems/Problem 25]] Solution 5 (Now it's solution 6) |
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| | [[2023 AMC 12B Problems/Problem 20]] Solution 3 | | [[2023 AMC 12B Problems/Problem 20]] Solution 3 |
| | + | |
| | + | ===AIME=== |
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| | [[2016 AIME I Problems/Problem 10]] Solution 3 | | [[2016 AIME I Problems/Problem 10]] Solution 3 |
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| | Restored diagram for [[1994 AIME Problems/Problem 7]] | | Restored diagram for [[1994 AIME Problems/Problem 7]] |
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| − | [[Divergence Theorem]]
| + | ===USAMO/USAJMO=== |
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| − | [[Stokes' Theorem]] | + | [[1978 USAMO Problems/Problem 1]] Solution 4 |
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| − | [[Principle of Insufficient Reasons]] | + | [[2002 USAMO Problems/Problem 2]] Solution 2 |
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| − | ==Problems I made==
| + | 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 |
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| − | ===Introductory=== | + | ===IMO=== |
| − | 1. There is one and only one perfect square in the form
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| − | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath>
| + | [[1995 IMO Problems/Problem 6]] Solution 1 (I got the solution from a forum discussion) |
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| − | where <math>p</math> and <math>q</math> are prime. Find that perfect square.
| + | [[1986 IMO Problems/Problem 3]] Solution 2 (I may or may not had gotten the solution from another source) |
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| | + | [[1976 IMO Problems/Problem 6]] Solution 2 (I actually made this solution by myself! :) ) |
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| − | 2. <math>m</math> and <math>n</math> are positive integers. If <math>m^2=2^8+2^{11}+2^n</math>, find <math>m+n</math>.
| + | ===Theorems=== |
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| − | ===Intermediate===
| + | [[Divergence Theorem]] |
| − | 3.The fraction,
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| − | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath>
| + | [[Stokes' Theorem]] |
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| − | where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>.
| + | [[Principle of Insufficient Reasons]] |
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| | + | [[The Nested Sum Theorem]] |
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| − | 4. Suppose there is complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
| + | [[Hensel's Lemma]] |
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| − | <cmath>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</cmath>
| + | ===Definitions=== |
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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
| + | [[Laplace Transform]] |
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| | + | [[Polynomial Congruences]] |
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| − | 5. Suppose
| + | [[Primitive Roots]] |
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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>
| + | [[Order of a integer]] |
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| − | Find the remainder when <math>\min{x}</math> is divided by <math>1000</math>.
| + | [[Index]] |
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| | + | [[Diagonalizability]] |
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| − | 6. Suppose that there is <math>192</math> rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are <math>2</math> other pegs positioned sufficiently apart. A <math>move</math> is made if
| + | ===Others=== |
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| − | (i) <math>1</math> ring changed position (i.e., that ring is transferred from one peg to another)
| + | [[Proofs to Some Number Theory Facts]] |
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| − | (ii) No rings are on top of smaller rings.
| + | [[Chebyshev and the history of the Prime Number Theorem]] |
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| − | Then, let <math>x</math> be the minimum possible number <math>moves</math> that can transfer all <math>192</math> rings onto the second peg. Find the remainder when <math>x</math> is divided by <math>1000</math>.
| + | [[Proof of the Existence of Primitive Roots]] |
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| | + | [[Proof of Some Primitive Roots Facts]] |
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| − | 7. 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
| + | ==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: |
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| − | <cmath>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</cmath>
| + | 1. There is one and only one perfect square in the form |
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| − | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
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| − | <cmath>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</cmath>.
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| − | Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
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| − | ===Olympiad===
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| − | 8. (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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| − | Someone mind making a diagram for this?
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| − | 9. Suppose <cmath>\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]=\frac{p}{q}</cmath> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
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| − | I will leave a big gap below this sentence so you won't see the answers accidentally.
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| − | ==Answer key==
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| − | 1. 049 | |
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| − | 2. 019
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| − | 3. 092
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| − | 4. 170
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| − | 5. 736
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| − | 6. 895
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| − | 7. 011
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| − | 8. 054
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| − | 9. 077
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| − | ==Solutions==
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| − | *Note: All the solutions so far have been made by me :)
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| − | | |
| − | ===Problem 1===
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| − | There is one and only one perfect square in the form | |
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| | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> | | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> |
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| − | where <math>p</math> and <math>q</math> is prime. Find that perfect square. | + | where <math>p</math> and <math>q</math> are prime. Find that perfect square. ~[[Ddk001]] |
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| − | ===Solution 1===
| + | <math>\textbf{Solution by cxsmi}</math> |
| − | <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>.
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| − | Then,
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| − | <cmath>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)</cmath> | + | 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>. |
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| − | , 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>
| + | Note: I will be the first to admit that this solution is somewhat lucky. |
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| − | ===Problem 2===
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| − | <math>m</math> and <math>n</math> are positive integers. If <math>m^2=2^8+2^{11}+2^n</math>, find <math>m+n</math>.
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| − | ===Solution 1===
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| − | <cmath>m^2=2^8+2^{11}+2^n</cmath>
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| − | <cmath>\implies 2^n=m^2-2^8-2^{11}</cmath> | + | 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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| − | <cmath>\implies 2^n=(m+48)(m-48)</cmath> | + | <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> |
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| − | Let <math>m+48=2^t</math> and <math>m-48=2^s</math>. Then,
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| − | <cmath>2^t-2^s=96 \implies 2^s(2^{t-s}-1)=2^5 \cdot 3 \implies 2^{t-s}-1=3,2^s=2^5 \implies (t,s)=(7,5) \implies m+n=80+12=\boxed{092}</cmath> <math>\square</math> | + | <math>\textbf{Solution by Ddk001}</math> |
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| − | ===Problem 3===
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| − | The fraction,
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| − | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath>
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| − | where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>.
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| − | ===Solution 1(Probably official MAA, lots of proofs)===
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| − | '''Lemma 1: <math>\text{max} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{3}</math>'''
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| − | ''Proof:'' Since the sides of triangles have positive length, <math>a,b,c>0</math>. Hence,
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| − | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}>0 \implies \text{max} (\frac{ab+bc+ac}{(a+b+c)^2})= \frac{1}{\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})}</cmath>
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| − | , so now we just need to find <math>\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})</math>.
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| − | Since <math>(a-c)^2+(b-c)^2+(a-b)^2 \ge 0</math> by the [[Trivial Inequality]], we have
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| − |
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| − | <cmath>a^2-2ac+c^2+b^2-2bc+c^2+a^2-2ab+b^2 \ge 0</cmath>
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| − | <cmath>\implies a^2+b^2+c^2 \ge ac+bc+ab</cmath>
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| − | <cmath>\implies a^2+b^2+c^2+2(ac+bc+ab) \ge 3(ac+bc+ab)</cmath>
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| − | <cmath>\implies (a+b+c)^2 \ge 3(ac+bc+ab)</cmath>
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| − |
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| − | <cmath>\implies \frac{(a+b+c)^2}{ab+bc+ac} \ge 3</cmath>
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| − | <cmath>\implies \frac{ab+bc+ac}{(a+b+c)^2} \le \frac{1}{3}</cmath>
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| − | as desired. <math>\square</math>
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| − | To show that the minimum value is achievable, we see that if <math>a=b=c</math>, <math>\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{3}</math>, so the minimum is thus achievable.
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| − | Thus, <math>q=\frac{1}{3}</math>.
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| − | '''Lemma 2: <math>\frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}</math>'''
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| − | ''Proof:'' By the [[Triangle Inequality]], we have
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| − | <cmath>a+b>c</cmath>
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| − | <cmath>b+c>a</cmath>
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| − | <cmath>a+c>b</cmath>.
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| − | Since <math>a,b,c>0</math>, we have
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| − |
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| − | <cmath>c(a+b)>c^2</cmath>
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| − | <cmath>a(b+c)>a^2</cmath>
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| − | <cmath>b(a+c)>b^2</cmath>.
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| − | Add them together gives
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| − | <cmath>a^2+b^2+c^2<c(a+b)+a(b+c)+b(a+c)=2(ab+bc+ac)</cmath>
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| − | <cmath>\implies a^2+b^2+c^2+2(ab+bc+ac)<4(ab+bc+ac)</cmath>
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| − | <cmath>\implies (a+b+c)^2<4(ab+bc+ac)</cmath>
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| − | <cmath>\implies \frac{(a+b+c)^2}{ab+bc+ac}<4</cmath>
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| − | <cmath>\implies \frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}</cmath> <math>\square</math>
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| − | Even though unallowed, if <math>a=0,b=c</math>, then <math>\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{4}</math>, so
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| − | <cmath>\lim_{b=c,a \to 0} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{4}</cmath>.
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| − | Hence, <math>p=\frac{1}{4}</math>, since by taking <math>b=c</math> and <math>a</math> close <math>0</math>, we can get <math>\frac{ab+bc+ac}{(a+b+c)^2}</math> to be as close to <math>\frac{1}{4}</math> as we wish.
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| − |
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| − | <math>p+q=\frac{1}{3}+\frac{1}{4}=\frac{7}{12} \implies r+s=7+12=\boxed{019}</math> <math>\blacksquare</math>
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| − |
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| − | ===Solution 2 (Fast, risky, no proofs)===
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| − | By the [[Principle of Insufficient Reason]], taking <math>a=b=c</math> we get either the max or the min. Testing other values yields that we got the max, so <math>q=\frac{1}{3}</math>. Another extrema must occur when one of <math>a,b,c</math> (WLOG, <math>a</math>) is <math>0</math>. Again, using the logic of solution 1 we see <math>p=\frac{1}{4}</math> so <math>p+q=\frac{7}{12}</math> so our answer is <math>\boxed{019}</math>. <math>\square</math>
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| − |
| |
| − | ===Problem 4===
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| − | Suppose there are complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
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| − |
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| − | <cmath>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</cmath>
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| − |
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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
| |
| | ===Solution 1=== | | ===Solution 1=== |
| − | To make things easier, instead of saying <math>x_i</math>, we say <math>x</math>.
| + | The inradius of <math>\Delta ABC</math>, <math>r</math>, can be calculated as |
| | | | |
| − | Now, we have
| + | <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> |
| − | <cmath>(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}</cmath>. | |
| − | Expanding gives
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| | | | |
| − | <cmath>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</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 |
| | | | |
| − | To make things even simpler, let
| + | <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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| − | <cmath>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}</cmath> | + | The desired sum is <math>7072+2+4048=\boxed{11122}</math>. <math>\blacksquare</math> |
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| − | , so that <math>x^3-ax^2+bx-c=0</math>.
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| | | | |
| − | Then, if <math>P_n=x_{1}^n+x_{2}^n+x_{3}^n</math>, [[Newton's Sums]] gives
| + | [[Ddk001]] Presents |
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| − | <cmath>P_1+(-a)=0</cmath> <math>(1)</math>
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| − | <cmath>P_2+(-a) \cdot P_1+2 \cdot b=0</cmath> <math>(2)</math>
| + | THE FOLLOWING PROBLEM |
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| − | <cmath>P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0</cmath> <math>(3)</math>
| + | 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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| − | Therefore,
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| − |
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| − | <cmath>P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))</cmath>
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| − |
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| − | <cmath>=a \cdot P_2-b \cdot P_1+3 \cdot c</cmath>
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| − |
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| − | <cmath>=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c</cmath>
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| − |
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| − | <cmath>=a(a^2-2b)-ab+3c</cmath>
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| − |
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| − | <cmath>=a^3-3ab+3c</cmath>
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| − |
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| − | 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>
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| − |
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| − | <cmath>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})</cmath>.
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| − |
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| − | We substitute <math>x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}</math> to get
| |
| − |
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| − | <cmath>P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})</cmath>
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| − |
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| − | <cmath>=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</cmath>
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| − |
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| − | <cmath>=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</cmath>
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| − |
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| − | <cmath>=x^3+y^3+z^3+1</cmath>
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| − |
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| − | <cmath>=13+53+103+1</cmath>
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| − |
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| − | <cmath>=\boxed{170}</cmath>. <math>\square</math>
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| − |
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| − | Note: If you don't know [[Newton's Sums]], you can also use [[Vieta's Formulas]] to bash.
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| − |
| |
| − | ===Problem 5===
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| | Suppose | | Suppose |
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| Line 391: |
Line 167: |
| | 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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| − | ===Solution 1 (Euler's Totient Theorem)=== | + | ==Vandalism area== |
| − | We first simplify <math>2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:</math>
| + | Congratulations! You had reached the end of this boring user page. Time for vandalism! Write anything under 45000 bytes. |
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| − | <cmath>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)}</cmath>
| + | Hi Ddk001 ~ zhenghua |
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| − | so
| + | 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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| − | <cmath>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}</cmath>
| + | [[Vandalism Page]] |
| | | | |
| − | <cmath>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}</cmath>
| + | ==See also== |
| − | | + | * My [[User talk:Ddk001|talk page]] |
| − | <cmath>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}</cmath>.
| + | * [[Problems Collection|My problems collection]] |
| − | | |
| − | where the last step of all 3 congruences hold by the [[Euler's Totient Theorem]].
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| − | Hence,
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| − | | |
| − | <cmath>x \equiv 1 \pmod{5}</cmath>
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| − | | |
| − | <cmath>x \equiv 0 \pmod{6}</cmath>
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| − | <cmath>x \equiv 6 \pmod{7}</cmath>
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| − | | |
| − | 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>.
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| − | | |
| − | 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 :
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| − | | |
| − | <cmath>x \equiv 6 \pmod{210}</cmath>
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| − | <cmath>x \equiv 0 \pmod{144}</cmath>.
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| − | | |
| − | Solving yields <math>x \equiv 2\boxed{736} \pmod{7!}</math>, and we're done. <math>\square</math>
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| − | | |
| − | ===Problem 6===
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| − | Suppose that there is <math>192</math> rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are <math>2</math> other pegs positioned sufficiently apart. A <math>move</math> is made if
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| − | | |
| − | (i) <math>1</math> ring changed position (i.e., that ring is transferred from one peg to another)
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| − | | |
| − | (ii) No bigger rings are on top of smaller rings.
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| − | | |
| − | Then, let <math>x</math> be the minimum possible number <math>moves</math> that can transfer all <math>192</math> rings onto the second peg. Find the remainder when <math>x</math> is divided by <math>1000</math>.
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| − | ===Solution 1 (Recursion)=== | |
| − | Let <math>M_n</math> be the minimum possible number <math>moves</math> that can transfer <math>n</math> rings onto the second peg. To build the recursion, we consider what is the minimum possible number <math>moves</math> that can transfer <math>n+1</math> rings onto the second peg. If we use only legal <math>moves</math>, then <math>n+1</math> will be smaller on the top, bigger on the bottom. Hence, the largest ring have to be at the bottom of the second peg, or the largest peg will have nowhere to go. In order for the largest ring to be at the bottom, we must first move the top <math>n</math> rings to the third peg using <math>M_n</math> <math>moves</math>, then place the largest ring onto the bottom of the second peg using <math>1</math> <math>move</math>, and then get all the rings from the third peg on top of the largest ring using another <math>M_n</math> <math>moves</math>. This gives a total of <math>2M_n+1</math>, hence we have <math>M_{n+1}=2M_{n}+1</math>. Obviously, <math>M_1=1</math>. We claim that <math>M_n=2^n-1</math>. This is definitely the case for <math>n=1</math>. If this is true for <math>n</math>, then
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| − | <cmath>M_{n+1}=2M_{n}+1=2(2^n-1)+1=2^{n+1}-1</cmath>
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| − | | |
| − | so this is true for <math>n+1</math>. Therefore, by [[induction]], <math>M_n=2^n-1</math> is true for all <math>n</math>. Now, <math>x=M_{192}=2^{192}-1</math>. Therefore, we see that
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| − | <cmath>x+1 \equiv 0 \pmod{8}</cmath>.
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| − | | |
| − | But the <math>\text{mod 125}</math> part is trickier. Notice that by the [[Euler's Totient Theorem]],
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| − | <cmath>2^{\phi (125)}=2^{100} \equiv 1 \pmod{125} \implies 2^{200} \equiv 1 \pmod{125}</cmath> | + | 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;"> |
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| − | so <math>x+1=\frac{2^{200}}{256}</math> is equivalent to the inverse of <math>256</math> in <math>\text{mod 125}</math>, which is equivalent to the inverse of <math>6</math> in <math>\text{mod 125}</math>, which, by inspection, is simply <math>21</math>. Hence,
| + | Can someone help me clear out [[Problem Collection|this page]]? |
| − | | |
| − | <cmath>x+1 \equiv 0 \pmod{8}</cmath>
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| − | | |
| − | <cmath>x+1 \equiv 21 \pmod{125}</cmath>
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| − | | |
| − | , so by the [[Chinese Remainder Theorem]], <math>x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}</math>. <math>\square</math>
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| − | | |
| − | ===Problem 7===
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| − | 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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| − | <cmath>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</cmath>
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| − | | |
| − | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
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| − | | |
| − | <cmath>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</cmath>.
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| − | Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
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| − | | |
| − | ===Solution 1===
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| − | 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]],
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| − | <cmath>n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n</cmath>
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| − | | |
| − | <cmath>\implies (n+1)n \dots (n-10000000008)|f(n)-n</cmath>.
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| − | <cmath>\implies f(n)=a(n+1)n \dots (n-10000000008)+n</cmath>
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| − | 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>.
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| − | | |
| − | 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>
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| − | | |
| − | Now, notice that
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| − | | |
| − | <cmath>m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})</cmath>
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| − | <cmath>=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})</cmath>
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| − | <cmath>=\frac{f(-2)}{a}</cmath>
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| − | | |
| − | <cmath>=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}</cmath>
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| − | <cmath>=\frac{10000000010! \cdot a-2}{a}</cmath>
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| − | <cmath>=10000000010!-\frac{2}{a}</cmath>
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| − | | |
| − | Similarly, we have
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| − | <cmath>(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}</cmath>
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| − | To minimize this, we minimize <math>m</math>. The minimum <math>m</math> can get is when <math>m=10000000011</math>, in which case
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| − | <cmath>-\frac{2}{a}=10000000011!-10000000010!</cmath>
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| − | | |
| − | <cmath>=10000000011 \cdot 10000000010!-10000000010!</cmath>
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| − | <cmath>=10000000010 \cdot 10000000010!</cmath>
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| − | <cmath>\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})</cmath>
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| − | <cmath>=-\frac{1}{a}</cmath>
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| − | <cmath>=\frac{10000000010 \cdot 10000000010}{2}</cmath>
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| − | <cmath>=5000000005 \cdot 10000000010!</cmath>
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| − | | |
| − | , so there is <math>\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}</math> factors of <math>999999937</math>. <math>\square</math>
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| − | | |
| − | ===Problem 8===
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| − | <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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| − | Someone mind making a diagram for this?
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| − | ===Solution 1===
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| − | Line <math>IJ</math> is tangent to <math>\Omega</math> with point of tangency point <math>J</math> because <math>OJ=OA \implies \text{J is on } \Omega</math> and <math>IJ</math> is perpendicular to <math>OJ</math> so this is true by the definition of tangent lines. Both <math>G</math> and <math>K</math> are on <math>\Omega</math> and line <math>O’G</math>, so <math>O’G</math> intersects <math>\Omega</math> at both <math>G</math> and <math>K</math>, and since we’re given <math>O’G</math> intersects <math>\Omega</math> at one distinct point, <math>G</math> and <math>K</math> are not distinct, hence they are the same point.
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| − | | |
| − | Now, if the center of <math>2</math> tangent circles are connected, the line segment will pass through the point of tangency. In this case, if we connect the center of <math>2</math> tangent circles, <math>\Omega</math> and <math>\Omega_1</math> (<math>O</math> and <math>I</math> respectively), it is going to pass through the point of tangency, namely, <math>K</math>, which is the same point as <math>G</math>, so <math>O</math>, <math>I</math>, and <math>G</math> are collinear. Hence, <math>G</math> and <math>I</math> are on both lines <math>OI</math> and <math>CI</math>, so <math>CI</math> passes through point <math>O</math>, making <math>CG</math> a diameter of <math>\Omega</math>.
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| − | | |
| − | Now we state a few claims :
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| − | | |
| − | '''Claim 1: <math>\Delta O’IO</math> is equilateral. '''
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| − | | |
| − | ''Proof:''
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| − | | |
| − | <cmath>\frac{3}{4} (IK+O’L)^2</cmath>
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| − | | |
| − | <cmath>=\frac{3}{4} IK^2+\frac{3}{2} IK \cdot O’L+\frac{3}{4} O’L^2</cmath>
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| − | | |
| − | <cmath>=IG^2+IG \cdot GC</cmath>
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| − | | |
| − | <cmath>=IG \cdot (IG+GC)</cmath>
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| − | | |
| − | <cmath>=IG \cdot IC</cmath>
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| − | | |
| − | <cmath>=IJ^2</cmath>
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| − | | |
| − | where the last equality holds by the [[Power of a Point Theorem]].
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| − | | |
| − | Taking the square root of each side yields <math>IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2</math>.
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| − | | |
| − | Since, by the definition of point <math>L</math>, <math>L</math> is on <math>\Omega_1</math>. Hence, <math>IK=IL</math>, so
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| − | | |
| − | <math>IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2=\frac{\sqrt{3}}{2} (IL+O’L)^2=\frac{\sqrt{3}}{2} IO’^2</math>, and since <math>O’</math> is the reflection of point <math>O</math> over line <math>IJ</math>, <math>OJ=O’J=\frac{OO’}{2}</math>, and since <math>IJ=\frac{\sqrt{3}}{2} IO’^2</math>, by the [[Pythagorean Theorem]] we have
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| − | | |
| − | <math>JO’=\frac{IO’}{2} \implies \frac{OO’}{2}=\frac{IO’}{2} \implies OO’=IO’</math>
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| − | | |
| − | Since <math>IJ</math> is the perpendicular bisector of <math>OO’</math>, <math>IO’=IO</math> and we have <math>IO=IO’=OO’</math> hence <math>\Delta O’IO</math> is equilateral. <math>\square</math>
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| − | | |
| − | With this in mind, we see that
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| − | | |
| − | <cmath>2OJ=OO’=OI=OK+KI=OJ+GI=OJ+AC \implies OA=OJ=AC</cmath>
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| − | | |
| − | Here, we state another claim :
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| − | | |
| − | '''Claim 2 : <math>BH</math> is a diameter of <math>\Omega</math>'''
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| − | | |
| − | ''Proof:'' Since <math>OA=OC=AC</math>, we have
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| − | | |
| − | <cmath>\angle AOC =60^\circ \implies \angle ABC=\frac{1}{2} \angle AOC=30^\circ \implies AB=\sqrt{3} AC</cmath>
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| − | | |
| − | and the same reasoning with <math>\Delta CAH</math> gives <math>CH=\sqrt{3} AC</math> since <math>AH=IG=AC</math>.
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| − | | |
| − | Now, apply [[Ptolemy’s Theorem]] gives
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| − | | |
| − | <cmath>BH \cdot AC+BC \cdot AH=CH \cdot AB \implies BH \cdot AC+AC^2=3AC^2 \implies BH=2AC=2OA</cmath>
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| − | | |
| − | so <math>BH</math> is a diameter. <math>\square</math>
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| − | | |
| − | From that, we see that <math>\angle BEH=90^\circ</math>, so <math>\frac{EH}{BH}=\cos{BHE}</math>. Now,
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| − | | |
| − | <cmath>\angle BHE=\angle BAE=\frac{1}{2} \angle CAB=15^\circ</cmath>
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| − | | |
| − | , so
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| − | | |
| − | <cmath>\frac{EH}{BH}=\cos{15}=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{2}}{4} (\sqrt{3}+1)</cmath>
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| − | | |
| − | , so
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| − | | |
| − | <cmath>a=4, b=2, c=3, d=1 \implies a^2+b^2+c^2+d^2+abcd=1+4+9+16+24=\boxed{054}</cmath>
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| − | | |
| − | , and we’re done. <math>\blacksquare</math>
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| − | | |
| − | ===Problem 9===
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| − | Suppose <cmath>\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]=\frac{p}{q}</cmath> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
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| − | ===Solution 1(Wordless endless bash)===
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| − | <cmath>\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}</cmath>
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| − | | |
| − | <cmath>=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn(m+n+2)}</cmath>
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| − | | |
| − | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{m(m+n+2)}</cmath>
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| − | | |
| − | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{n+2} (\frac{1}{m}-\frac{1}{m+n+2})</cmath>
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| − | | |
| − | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \sum_{m=1}^{\infty} (\frac{1}{m}-\frac{1}{m+n+2})</cmath>
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| − | | |
| − | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot [(1-\frac{1}{n+3})+(\frac{1}{2}-\frac{1}{n+4})+ \dots]</cmath>
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| − | | |
| − | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})</cmath>
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| − | | |
| − | <cmath>=\sum_{n=1}^{\infty} (\frac{\frac{1}{2}}{n}-\frac{\frac{1}{2}}{n+2}) \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})</cmath>
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| − | | |
| − | <cmath>=\frac{1}{2} [(1-\frac{1}{3})(1+\frac{1}{2}+\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+ \dots</cmath>
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| − | | |
| − | <cmath>=\frac{1}{2} [[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\dots](1+\frac{1}{2}+\frac{1}{3})+[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{6})+\dots](1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+\dots](\frac{1}{4}+\frac{1}{5})+\dots]</cmath>
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| − | | |
| − | <cmath>=\frac{1}{2} [(1+\frac{1}{2}+\frac{1}{3})+\frac{1}{2} (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+\frac{1}{3} (\frac{1}{4}+\frac{1}{5})+\dots]</cmath>
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| − | | |
| − | <cmath>=\frac{1}{2} [\frac{11}{6}+\frac{1}{2} \cdot \frac{25}{12}+(\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\dots)+(\frac{1}{3 \cdot 5}+\frac{1}{4 \cdot 6}+\dots)]</cmath>
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| − | | |
| − | <cmath>=\frac{1}{2} [\frac{69}{24}+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\dots ]+\frac{1}{2} [(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+\dots ]</cmath>
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| − | | |
| − | <cmath>=\frac{1}{2} [\frac{69}{24}+\frac{1}{3}+\frac{1}{6}+\frac{1}{8}]</cmath>
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| − | | |
| − | <cmath>=\frac{1}{2} \cdot \frac{84}{24}</cmath>
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| − | | |
| − | <cmath>=\frac{7}{4}</cmath>
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| − | | |
| − | <cmath>(1+\frac{1}{x})^x=e^{x \cdot \ln (1+\frac{1}{x})}</cmath>
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| − | | |
| − | <cmath>=e^{x \cdot [(\frac{1}{x})-\frac{(\frac{1}{x})^2}{2}+\frac{(\frac{1}{x})^3}{3}+\dots]}</cmath>
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| − | | |
| − | <cmath>=e^{1-\frac{1}{2} (\frac{1}{x})+\frac{1}{3} (\frac{1}{x})^2+\dots}</cmath>
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| − | | |
| − | <cmath>=e \cdot e^{-\frac{1}{2} (\frac{1}{x})} \cdot e^{\frac{1}{3} (\frac{1}{x})^2} \dots</cmath>
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| − | | |
| − | <cmath>=e \cdot [1-\frac{1}{2x}+\frac{1}{2!} (\frac{1}{2x})^2- \dots] \cdot [1+\frac{1}{3x^2}+\frac{1}{2!} (\frac{1}{3x^2})^2+ \dots]</cmath>
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| − | | |
| − | <cmath>=e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]</cmath>
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| − | | |
| − | <cmath>\implies \lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]</cmath>
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| − | | |
| − | <cmath>=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]}{e}-1]]</cmath>
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| − | | |
| − | <cmath>=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots-1)]</cmath>
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| − | | |
| − | <cmath>=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots)]</cmath>
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| − | | |
| − | <cmath>=\lim_{x\rightarrow \infty} (\frac{x}{2}-\frac{x}{2}+\frac{11}{24}+\dots)</cmath>
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| − | | |
| − | <cmath>=\frac{11}{24}</cmath>
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| − | | |
| − | <cmath>\implies \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]</cmath>
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| − | | |
| − | <cmath>=\frac{7}{4}+\frac{11}{24}</cmath>
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| − | | |
| − | <cmath>=\frac{53}{24}</cmath>
| |
| − | | |
| − | <cmath>\implies p=53,q=24</cmath>
| |
| − | | |
| − | <cmath>\implies p+q=\boxed{077}</cmath> <math>\square</math>
| |
| − | | |
| − | ==See also==
| |
![[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]](http://latex.artofproblemsolving.com/d/6/1/d61604256e9de426afc94111a204811ff9d3cf0e.png)
![[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]](http://latex.artofproblemsolving.com/d/7/8/d78519cbbf9c22e2c4ea74ee089557db7a04e2ba.png)
![\[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]](http://latex.artofproblemsolving.com/b/1/1/b11bf3dce41593092737df0eea5f902de3989389.png)
and 
are prime. Find that perfect square. ~Ddk001
. 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 
in some order. The perfect square formed using these numbers is 
.
intersects 
. The perimeter of this diamond is expressible as 
, where 
, 
, and 
are integers, and 
is divided by 
?
![[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]](http://latex.artofproblemsolving.com/c/f/4/cf4b2b8da31fb32285d5b573ffbb35e870afea4c.png)
, 
, can be calculated as
![\[r=\frac{\textbf{Area}_{ABC}}{\textbf{Semiperimeter}} \implies r=\frac{2024^2/2}{(2024+2024+2024 \sqrt{2})/2}=\frac{2024}{2+\sqrt{2}}=2024-1012\sqrt{2}\]](http://latex.artofproblemsolving.com/d/0/2/d02fa3396a9eff50f834c674242087b858095775.png)
. 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
![\[4(4048-2024 \sqrt{2})-2 \cdot 2024 (3-2\sqrt{2})+2024 (3 \sqrt{2} -4)=2024 (8-4 \sqrt{2}-6+4\sqrt{2}+3\sqrt{2}-4)=2024(3\sqrt{2}-2)=7072\sqrt{2}-4048\]](http://latex.artofproblemsolving.com/b/c/8/bc8e18a5b46de191c8436a2d6d363517962ce15f.png)
. 
![\[x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}\]](http://latex.artofproblemsolving.com/2/a/4/2a4c33ddcd46bcf5c89a3e3115bc73c2c9c5d3d5.png)
is divided by 1000.
