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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">31</font></center> |
| | + | </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== | | ==Problems Sharing Contest== |
| − | 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: | + | 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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| | 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> | + | <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. | + | where <math>p</math> and <math>q</math> are prime. Find that perfect square. ~[[Ddk001]] |
| − | (DO NOT LOOK AT MY SOLUTIONS YET)
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| − | ==Contributions==
| + | <math>\textbf{Solution by cxsmi}</math> |
| − | [[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)
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| − | [[2023 AMC 12B Problems/Problem 20]] Solution 3
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| − | [[2016 AIME I Problems/Problem 10]] Solution 3
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| − | [[2017 AIME I Problems/Problem 14]] Solution 2
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| − | [[2019 AIME I Problems/Problem 15]] Solution 6
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| − | [[2022 AIME II Problems/Problem 3]] Solution 3
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| − | Restored diagram for [[1994 AIME Problems/Problem 7]]
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| − | [[Divergence Theorem]]
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| − | [[Stokes' Theorem]]
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| − | [[Principle of Insufficient Reasons]]
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| − | ==Problems I made==
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| − | ===Introductory===
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| − | 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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| − | where <math>p</math> and <math>q</math> are prime. Find that perfect square.
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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>.
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| − | ===Intermediate===
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| − | 3.The fraction,
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| − | <math>\frac{ab+bc+ac}{(a+b+c)^2}</math>
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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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| − | 4. 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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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
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| − | 5. 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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| − | Find the remainder when <math>\min{x}</math> is divided by <math>1000</math>.
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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
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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 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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| − | 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
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| − | <math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
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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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| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
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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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| − | dsf
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| − | fsd
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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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| − | <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>
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| − | where <math>p</math> and <math>q</math> is prime. Find that perfect square.
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| − | ===Solution 1===
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| − | <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>.
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| − | Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>.
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| − | 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>
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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>
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| − | <cmath>\implies 2^n=(m+48)(m-48)</cmath>
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| − | Let <math>m+48=2^t</math> and <math>m-48=2^s</math>. Then,
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| − | <math>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}</math> <math>\square</math>
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| − | ===Problem 3===
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| − | The fraction,
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| − | <math>\frac{ab+bc+ac}{(a+b+c)^2}</math> | |
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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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| − | <math>\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})}</math>
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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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| − | <math>a^2-2ac+c^2+b^2-2bc+c^2+a^2-2ab+b^2 \ge 0</math>
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| − | <math>\implies a^2+b^2+c^2 \ge ac+bc+ab</math>
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| − | <math>\implies a^2+b^2+c^2+2(ac+bc+ab) \ge 3(ac+bc+ab)</math>
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| − | <math>\implies (a+b+c)^2 \ge 3(ac+bc+ab)</math>
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| − | <math>\implies \frac{(a+b+c)^2}{ab+bc+ac} \ge 3</math>
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| − | <math>\implies \frac{ab+bc+ac}{(a+b+c)^2} \le \frac{1}{3}</math>
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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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| − | <math>a+b>c</math>
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| − | <math>b+c>a</math>
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| − | <math>a+c>b</math>.
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| − | Since <math>a,b,c>0</math>, we have
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| − | <math>c(a+b)>c^2</math>
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| − | <math>a(b+c)>a^2</math>
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| − | <math>b(a+c)>b^2</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>. |
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| − | Add them together gives
| + | Note: I will be the first to admit that this solution is somewhat lucky. |
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| − | <math>a^2+b^2+c^2<c(a+b)+a(b+c)+b(a+c)=2(ab+bc+ac)</math>
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| − | <math>\implies a^2+b^2+c^2+2(ab+bc+ac)<4(ab+bc+ac)</math> | + | 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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| − | <math>\implies (a+b+c)^2<4(ab+bc+ac)</math> | + | <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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| − | <math>\implies \frac{(a+b+c)^2}{ab+bc+ac}<4</math>
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| − | <math>\implies \frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}</math> <math>\square</math> | + | <math>\textbf{Solution by Ddk001}</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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| − | <math>\lim_{b=c,a \to 0} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{4}</math>.
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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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| − | <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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| − |
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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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| − | <math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</math>
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| − |
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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
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| | ===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 |
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| − | Now, we have
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| − | <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> | + | <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> |
| | | | |
| − | <math>P_2+(-a) \cdot P_1+2 \cdot b=0</math> <math>(2)</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 |
| | | | |
| − | <math>P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0</math> <math>(3)</math> | + | <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> |
| | | | |
| − | Therefore,
| + | The desired sum is <math>7072+2+4048=\boxed{11122}</math>. <math>\blacksquare</math> |
| | | | |
| − | <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>
| + | [[Ddk001]] Presents |
| | | | |
| − | <math>=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c</math>
| |
| | | | |
| − | <math>=a(a^2-2b)-ab+3c</math>
| + | THE FOLLOWING PROBLEM |
| | | | |
| − | <math>=a^3-3ab+3c</math>
| + | Note: This is one of my favorite problems. Very well designed and actually used two of my best tricks without looking weird. |
| | | | |
| − | 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>.
| |
| − |
| |
| − | 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 5===
| |
| | Suppose | | 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. | | Find the remainder when <math>\min{x}</math> is divided by 1000. |
| | | | |
| − | ===Solution 1 (Euler's Totient Theorem)=== | + | ==Contributions== |
| − | We first simplify <math>2^4 \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>
| + | [[2022 AMC 12B Problems/Problem 25]] Solution 5 (Now it's solution 6) |
| | | | |
| − | so
| + | [[2023 AMC 12B Problems/Problem 20]] Solution 3 |
| | | | |
| − | <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>
| + | [[2016 AIME I Problems/Problem 10]] Solution 3 |
| | | | |
| − | <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>
| + | [[2017 AIME I Problems/Problem 14]] Solution 2 |
| | | | |
| − | <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>.
| + | [[2019 AIME I Problems/Problem 15]] Solution 6 |
| | | | |
| − | where the last step of all 3 congruences hold by the [[Euler's Totient Theorem]].
| + | [[2022 AIME II Problems/Problem 3]] Solution 3 |
| − | Hence,
| |
| | | | |
| − | <math>x \equiv 1 \pmod{5}</math>
| + | [[1978 USAMO Problems/Problem 1]] Solution 4 |
| | | | |
| − | <math>x \equiv 0 \pmod{6}</math>
| + | Restored diagram for [[1994 AIME Problems/Problem 7]] |
| | | | |
| − | <math>x \equiv 6 \pmod{7}</math>
| + | [[Divergence Theorem]] |
| | | | |
| − | 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>.
| + | [[Stokes' Theorem]] |
| | | | |
| − | 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 :
| + | [[Principle of Insufficient Reasons]] |
| − | | |
| − | <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 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
| |
| − | | |
| − | (i) <math>1</math> ring changed position (i.e., that ring is transferred from one peg to another)
| |
| − | | |
| − | (ii) No bigger rings are on top of smaller rings.
| |
| − | | |
| − | 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>.
| |
| − | ===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
| |
| − | | |
| − | <math>M_{n+1}=2M_{n}+1=2(2^n-1)+1=2^{n+1}-1</math>
| |
| − | | |
| − | 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
| |
| − | | |
| − | <math>x+1 \equiv 0 \pmod{8}</math>.
| |
| − | | |
| − | But the <math>\text{mod 125}</math> part is trickier. Notice that by the [[Euler's Totient Theorem]],
| |
| − | | |
| − | <math>2^{\phi (125)}=2^{100} \equiv 1 \pmod{125} \implies 2^{200} \equiv 1 \pmod{125}</math>
| |
| − | | |
| − | 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,
| |
| − | | |
| − | <math>x+1 \equiv 0 \pmod{8}</math>
| |
| − | | |
| − | <math>x+1 \equiv 21 \pmod{125}</math>
| |
| − | | |
| − | , so by the [[Chinese Remainder Theorem]], <math>x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}</math>. <math>\square</math>
| |
| − | | |
| − | ===Problem 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
| |
| − | | |
| − | <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>
| |
| − | | |
| − | <math>=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})</math>
| |
| − | | |
| − | <math>=\frac{f(-2)}{a}</math>
| |
| − | | |
| − | <math>=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}</math>
| |
| − | | |
| − | <math>=\frac{10000000010! \cdot a-2}{a}</math>
| |
| − | | |
| − | <math>=10000000010!-\frac{2}{a}</math>
| |
| − | | |
| − | Similarly, we have
| |
| − | | |
| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}</math>
| |
| − | | |
| − | To minimize this, we minimize <math>m</math>. The minimum <math>m</math> can get is when <math>m=10000000011</math>, in which case
| |
| − | | |
| − | <math>-\frac{2}{a}=10000000011!-10000000010!</math>
| |
| − | | |
| − | <math>=10000000011 \cdot 10000000010!-10000000010!</math>
| |
| − | | |
| − | <math>=10000000010 \cdot 10000000010!</math>
| |
| − | | |
| − | <math>\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>
| |
| − | | |
| − | <math>=-\frac{1}{a}</math>
| |
| − | | |
| − | <math>=\frac{10000000010 \cdot 10000000010}{2}</math>
| |
| − | | |
| − | <math>=5000000005 \cdot 10000000010!</math>
| |
| − | | |
| − | , so there is <math>\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}</math> factors of <math>999999937</math>. <math>\square</math>
| |
| − | | |
| − | ===Problem 8===
| |
| − | <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>.
| |
| − | | |
| − | Someone mind making a diagram for this?
| |
| − | ===Solution 1===
| |
| − | 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.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | Now we state a few claims :
| |
| − | | |
| − | '''Claim 1: <math>\Delta O’IO</math> is equilateral. '''
| |
| − | | |
| − | ''Proof:'' <math>\frac{3}{4} (IK+O’L)^2</math>
| |
| − | | |
| − | <math>=\frac{3}{4} IK^2+\frac{3}{2} IK \cdot O’L+\frac{3}{4} O’L^2</math>
| |
| − | | |
| − | <math>=IG^2+IG \cdot GC</math>
| |
| − | | |
| − | <math>=IG \cdot (IG+GC)</math>
| |
| − | | |
| − | <math>=IG \cdot IC</math>
| |
| − | | |
| − | <math>=IJ^2</math>
| |
| − | | |
| − | where the last equality holds by the [[Power of a Point Theorem]].
| |
| − | | |
| − | Taking the square root of each side yields <math>IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2</math>.
| |
| − | | |
| − | Since, by the definition of point <math>L</math>, <math>L</math> is on <math>\Omega_1</math>. Hence, <math>IK=IL</math>, so
| |
| − | | |
| − | <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
| |
| − | | |
| − | <math>JO’=\frac{IO’}{2} \implies \frac{OO’}{2}=\frac{IO’}{2} \implies OO’=IO’</math>
| |
| − | | |
| − | 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>
| |
| − | | |
| − | With this in mind, we see that
| |
| − | | |
| − | <math>2OJ=OO’=OI=OK+KI=OJ+GI=OJ+AC \implies OA=OJ=AC</math>
| |
| − | | |
| − | Here, we state another claim :
| |
| − | | |
| − | '''Claim 2 : <math>BH</math> is a diameter of <math>\Omega</math>'''
| |
| − | | |
| − | ''Proof:'' Since <math>OA=OC=AC</math>, we have
| |
| − | | |
| − | <math>\angle AOC =60 \implies \angle ABC=\frac{1}{2} \angle AOC=30 \implies AB=\sqrt{3} AC</math>
| |
| − | | |
| − | and the same reasoning with <math>\Delta CAH</math> gives <math>CH=\sqrt{3} AC</math> since <math>AH=IG=AC</math>.
| |
| − | | |
| − | Now, apply [[Ptolemy’s Theorem]] gives
| |
| − | | |
| − | <math>BH \cdot AC+BC \cdot AH=CH \cdot AB \implies BH \cdot AC+AC^2=3AC^2 \implies BH=2AC=2OA</math>
| |
| − | | |
| − | so <math>BH</math> is a diameter. <math>\square</math>
| |
| − | | |
| − | From that, we see that <math>\angle BEH=90</math>, so <math>\frac{EH}{BH}=\cos{BHE}</math>. Now,
| |
| − | | |
| − | <math>\angle BHE=\angle BAE=\frac{1}{2} \angle CAB=15</math>
| |
| − | | |
| − | , so
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| − | | |
| − | <math>\frac{EH}{BH}=\cos{15}=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{2}}{4} (\sqrt{3}+1)</math>
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| − | | |
| − | , so
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| − | | |
| − | <math>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}</math>
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| − | | |
| − | , and we’re done. <math>\blacksquare</math>
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| − | | |
| − | Note: All angle measures are in degrees
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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===
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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>
| + | ==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). |
| | | | |
| − | <cmath>=\lim_{x\rightarrow \infty} \frac{x}{2}-\frac{x}{2}+\frac{11}{24}+\dots</cmath>
| + | (ok :) :) this page is so cool!) |
| | | | |
| − | <cmath>=\frac{11}{24}</cmath>
| + | honestly i think your user page is very cool. :) |
| | | | |
| − | <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>
| + | Hi Ddk001 [[User:zhenghua]] (Taking Oly Geo) |
| | | | |
| − | <cmath>=\frac{7}{4}+\frac{11}{24}</cmath>
| + | 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. |
| | | | |
| − | <cmath>=\frac{53}{24}</cmath>
| + | ==See also== |
| | + | * My [[User talk:Ddk001|talk page]] |
| | + | * [[Problems Collection|My problems collection]] |
| | | | |
| − | <cmath>\implies p=53,q=24</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;"> |
| | | | |
| − | <cmath>\implies p+q=\boxed{077}</cmath> <math>\square</math>
| + | Can someone help me clear out [[Problems Collection|this page]]? |
![[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.
