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| − | ==User Counts==
| + | If you have a problem or solution to contribute, please go to [[Problems Collection|this page]]. |
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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)
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| − | <math>\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{1}}}}}}</math>
| + | I am a aops user who likes making and doing problems, doing math, and redirecting pages (see [[Principle of Insufficient Reasons]]). I like geometry and don't like counting and probability. My number theory skill are also not bad |
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| − | (Please don't mess with the user count)
| + | <br> |
| | + | __NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"> |
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| − | Doesn't that look like a number on a pyramid?
| + | ==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>== |
| | + | <div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div> |
| | + | <center><font size="100px"><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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| | + | 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
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| − | <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>
| + | ==Contributions== |
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| − | where <math>p</math> and <math>q</math> are prime. Find that perfect square.
| + | ===AMC=== |
| − | (DO NOT LOOK AT MY SOLUTIONS)
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| − | ==Contibutions==
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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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| | [[2022 AIME II Problems/Problem 3]] Solution 3 | | [[2022 AIME II Problems/Problem 3]] Solution 3 |
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| − | ==Problems I made==
| + | Restored diagram for [[1994 AIME Problems/Problem 7]] |
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| | + | ===USAMO/USAJMO=== |
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| − | 1. (Much easier) There is one and only one perfect square in the form | + | [[1978 USAMO Problems/Problem 1]] Solution 4 |
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| − | <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>
| + | [[2002 USAMO Problems/Problem 2]] Solution 2 |
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| − | where <math>p</math> and <math>q</math> are prime. Find that perfect square.
| + | 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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| − | 2.The fraction,
| + | ===IMO=== |
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| − | <math>\frac{ab+bc+ac}{(a+b+c)^2}</math>
| + | [[1995 IMO Problems/Problem 6]] Solution 1 (I got the solution from a forum discussion) |
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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>.
| + | [[1986 IMO Problems/Problem 3]] Solution 2 (I may or may not had gotten the solution from another source) |
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| − | 3. Suppose there is complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy
| + | [[1976 IMO Problems/Problem 6]] Solution 2 (I actually made this solution by myself! :) ) |
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| − | <math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</math>
| + | ===Theorems=== |
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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
| + | [[Divergence Theorem]] |
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| − | 4. Suppose
| + | [[Stokes' Theorem]] |
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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>
| + | [[Principle of Insufficient Reasons]] |
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| − | Find the remainder when <math>\min{x}</math> is divided by 1000.
| + | [[The Nested Sum Theorem]] |
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| − | 5. 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
| + | [[Hensel's Lemma]] |
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| − | <math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
| + | ===Definitions=== |
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| − | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
| + | [[Laplace Transform]] |
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| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
| + | [[Polynomial Congruences]] |
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| − | Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
| + | [[Primitive Roots]] |
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| − | 6. (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>.
| + | [[Order of a integer]] |
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| − | Someone mind making a diagram for this?
| + | [[Index]] |
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| | + | [[Diagonalizability]] |
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| − | I will leave a big gap below this sentence so you won't see the answers accidentally.
| + | ===Others=== |
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| | + | [[Proofs to Some Number Theory Facts]] |
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| | + | [[Chebyshev and the history of the Prime Number Theorem]] |
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| | + | [[Proof of the Existence of Primitive Roots]] |
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| | + | [[Proof of Some Primitive Roots Facts]] |
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| | + | ==Problems Sharing Contest== |
| | + | Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start: |
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| | + | 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> |
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| | + | where <math>p</math> and <math>q</math> are prime. Find that perfect square. ~[[Ddk001]] |
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| | + | <math>\textbf{Solution by cxsmi}</math> |
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| | + | 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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| | + | Note: I will be the first to admit that this solution is somewhat lucky. |
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| | + | 2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle <math>\triangle ABC</math> intersects <math>\triangle ABC</math> itself. <math>\triangle ABC</math> has leg length <math>2024</math>. The perimeter of this diamond is expressible as <math>a\sqrt{b}-c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers, and <math>c</math> is not divisible by the square of any prime. What is the remainder when <math>a + b + c</math> is divided by <math>1000</math>? |
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| | + | <asy> |
| | + | 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>\textbf{Solution by Ddk001}</math> |
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| − | ==Answer key==
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| − | 1. 049
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| − | 2. 170
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| − | 3. 736
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| − | 4. 011
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| − | 5. 054
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| − | ==Solutions==
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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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| − |
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| − | ===Problem 2===
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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>.
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| − | Add them together gives
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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>
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| − | <math>\implies (a+b+c)^2<4(ab+bc+ac)</math>
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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>
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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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| − | ===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 3===
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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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| − | <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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| | ===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>.
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| − | Expanding gives
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| − | <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>.
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| − | To make things even simpler, let
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| − | <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>. | |
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| − | Then, if <math>P_n=x_{1}^n+x_{2}^n+x_{3}^n</math>, [[Newton's Sums]] gives
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| − | <math>P_1+(-a)=0</math> <math>(1)</math>
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| − | <math>P_2+(-a) \cdot P_1+2 \cdot b=0</math> <math>(2)</math>
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| − | <math>P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0</math> <math>(3)</math>
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| − | | |
| − | Therefore,
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| − | <math>P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))</math>
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| − | | |
| − | <math>=a \cdot P_2-b \cdot P_1+3 \cdot c</math>
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| − | <math>=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c</math>
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| − | | |
| − | <math>=a(a^2-2b)-ab+3c</math>
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| − | | |
| − | <math>=a^3-3ab+3c</math>
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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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| − | <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>. | + | <cmath>r=\frac{\textbf{Area}_{ABC}}{\textbf{Semiperimeter}} \implies r=\frac{2024^2/2}{(2024+2024+2024 \sqrt{2})/2}=\frac{2024}{2+\sqrt{2}}=2024-1012\sqrt{2}</cmath> |
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| − | As we have done many times before, we substitute <math>x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}</math> to get
| + | 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=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{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> |
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| − | <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> | + | The desired sum is <math>7072+2+4048=\boxed{11122}</math>. <math>\blacksquare</math> |
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| − | <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>
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| − | <math>=x^3+y^3+z^3+1</math>
| + | [[Ddk001]] Presents |
| | | | |
| − | <math>=13+53+103+1</math>
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| | | | |
| − | <math>=\boxed{170}</math>. <math>\square</math>
| + | THE FOLLOWING PROBLEM |
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| − | Note: If you don't know [[Newton's Sums]], you can also use [[Vieta's Formulas]] to bash. | + | Note: This is one of my favorite problems. Very well designed and actually used two of my best tricks without looking weird. |
| | | | |
| − | ===Problem 4===
| |
| | Suppose | | 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> | + | <cmath>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</cmath> |
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| | 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>\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. |
| − | | |
| − | <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>
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| − | | |
| − | so
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| − | | |
| − | <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>
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| − | | |
| − | <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>
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| − | | |
| − | <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>.
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| − | | |
| − | where the last step of all 3 congruences hold by the [[Euler's Totient Theorem]].
| |
| − | Hence,
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| − | | |
| − | <math>x \equiv 1 \pmod{5}</math>
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| − | | |
| − | <math>x \equiv 0 \pmod{6}</math>
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| − | | |
| − | <math>x \equiv 6 \pmod{7}</math>
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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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| − | | |
| − | <math>x \equiv 6 \pmod{210}</math>
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| − | <math>x \equiv 0 \pmod{144}</math>.
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| − | | |
| − | Solving yields <math>x \equiv 2\boxed{736} \pmod{7!}</math>, and we're done. <math>\square</math>
| |
| − | | |
| − | ===Problem 5===
| |
| − | 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 | |
| − | | |
| − | <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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| − | | |
| − | ===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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| − | | |
| − | <math>n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n</math>
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| − | | |
| − | <math>\implies (n+1)n \dots (n-10000000008)|f(n)-n</math>.
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| − | | |
| − | <math>\implies f(n)=a(n+1)n \dots (n-10000000008)+n</math>
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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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| − | | |
| − | <math>m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})</math>
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| − | | |
| − | <math>=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})</math>
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| − | | |
| − | <math>=\frac{f(-2)}{a}</math>
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| − | | |
| − | <math>=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}</math>
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| − | | |
| − | <math>=\frac{10000000010! \cdot a-2}{a}</math>
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| − | | |
| − | <math>=10000000010!-\frac{2}{a}</math>
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| − | | |
| − | Similarly, we have
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| − | | |
| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}</math>
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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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| − | <math>-\frac{2}{a}=10000000011!-10000000010!</math>
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| − | | |
| − | <math>=10000000011 \cdot 10000000010!-10000000010!</math>
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| − | <math>=10000000010 \cdot 10000000010!</math>
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| − | | |
| − | <math>\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>
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| − | | |
| − | <math>=-\frac{1}{a}</math>
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| − | <math>=\frac{10000000010 \cdot 10000000010}{2}</math>
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| − | | |
| − | <math>=5000000005 \cdot 10000000010!</math>
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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>
| |
| − | | |
| − | ===Problem 6===
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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 concurrent. 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: <math>\frac{3}{4} (IK+O’L)^2</math>
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| − | | |
| − | <math>=\frac{3}{4} IK^2+\frac{3}{2} IK \cdot O’L+\frac{3}{4} O’L^2</math>
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| − | | |
| − | <math>=IG^2+IG \cdot GC</math>
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| − | | |
| − | <math>=IG \cdot (IG+GC)</math>
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| − | | |
| − | <math>=IG \cdot IC</math>
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| − | | |
| − | <math>=IJ^2</math>
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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
| |
| − | | |
| − | <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>
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| − | | |
| − | With this in mind, we see that
| |
| − | | |
| − | <math>2OJ=OO’=OI=OK+KI=OJ+GI=OJ+AC \implies OA=OJ=AC</math>
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| − | | |
| − | 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
| + | Hi Ddk001 ~ zhenghua |
| | | | |
| − | <math>\frac{EH}{BH}=\cos{15}=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{2}}{4} (\sqrt{3}+1)</math>
| + | 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?) |
| | | | |
| − | , so
| + | [[Vandalism Page]] |
| | | | |
| − | <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>
| + | ==See also== |
| | + | * My [[User talk:Ddk001|talk page]] |
| | + | * [[Problems Collection|My problems collection]] |
| | | | |
| − | , and we’re done. <math>\blacksquare</math>
| + | 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;"> |
| | | | |
| − | Note: All angle measures are in degrees
| + | Can someone help me clear out [[Problem 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.
