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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{68}}}}}}</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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| − | Doesn't that look like a number on a pyramid?
| + | <br> |
| | + | __NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"> |
| | + | |
| | + | ==<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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| | <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> | | <asy>draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));</asy> |
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| − | ==Problems I made==
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| − | See if you can solve these:
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| − | 1. (Much easier) 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. 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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| − | 3. 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 1000.
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| − | 4. Suppose <math>f(x)</math> is a <math>10000000010</math>-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
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| − | <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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| − | 5. (Much harder) <math>\Delta ABC</math> is an isosceles triangle where <math>CB=CA</math>. Let the circumcircle of <math>\Delta ABC</math> be <math>\Omega</math>. Then, there is a point <math>E</math> and a point <math>D</math> on circle <math>\Omega</math> such that <math>AD</math> and <math>AB</math> trisects <math>\angle CAE</math> and <math>BE<AE</math>, and point <math>D</math> lies on minor arc <math>BC</math>. Point <math>F</math> is chosen on segment <math>AD</math> such that <math>CF</math> is one of the altitudes of <math>\Delta ACD</math>. Ray <math>CF</math> intersects <math>\Omega</math> at point <math>G</math> (not <math>C</math>) and is extended past <math>G</math> to point <math>I</math>, and <math>IG=AC</math>. Point <math>H</math> is also on <math>\Omega</math> and <math>AH=GI<HB</math>. Let the perpendicular bisector of <math>BC</math> and <math>AC</math> intersect at <math>O</math>. Let <math>J</math> be a point such that <math>OJ</math> is both equal to <math>OA</math> (in length) and is perpendicular to <math>IJ</math> and <math>J</math> is on the same side of <math>CI</math> as <math>A</math>. Let <math>O’</math> be the reflection of point <math>O</math> over line <math>IJ</math>. There exist a circle <math>\Omega_1</math> centered at <math>I</math> and tangent to <math>\Omega</math> at point <math>K</math>. <math>IO’</math> intersect <math>\Omega_1</math> at <math>L</math>. Now suppose <math>O’G</math> intersects <math>\Omega</math> at one distinct point, and <math>O’, G</math>, and <math>K</math> are collinear. If <math>IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2</math>, then <math>\frac{EH}{BH}</math> can be expressed in the form <math>\frac{\sqrt{b}}{a} (\sqrt{c} + d)</math>, where <math>b</math> and <math>c</math> are not divisible by the squares of any prime. Find <math>a^2+b^2+c^2+d^2+abcd</math>.
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| − | Someone mind making a diagram for this?
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| − | ==Answer key & solution to the problems==
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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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| | + | ==Contributions== |
| | | | |
| | + | ===AMC=== |
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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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| | + | ===AIME=== |
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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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| | + | ===USAMO/USAJMO=== |
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| | + | [[1978 USAMO Problems/Problem 1]] Solution 4 |
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| | + | [[2002 USAMO Problems/Problem 2]] Solution 2 |
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| | + | 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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| | + | ===IMO=== |
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| − | dsf
| + | [[1995 IMO Problems/Problem 6]] Solution 1 (I got the solution from a forum discussion) |
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| | + | [[1986 IMO Problems/Problem 3]] Solution 2 (I may or may not had gotten the solution from another source) |
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| | + | [[1976 IMO Problems/Problem 6]] Solution 2 (I actually made this solution by myself! :) ) |
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| | + | ===Theorems=== |
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| | + | [[Divergence Theorem]] |
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| | + | [[Stokes' Theorem]] |
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| | + | [[Principle of Insufficient Reasons]] |
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| | + | [[The Nested Sum Theorem]] |
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| | + | [[Hensel's Lemma]] |
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| | + | ===Definitions=== |
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| − | fsd
| + | [[Laplace Transform]] |
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| − | ===Answer key===
| + | [[Polynomial Congruences]] |
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| − | 1. 049
| + | [[Primitive Roots]] |
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| − | 2. 170
| + | [[Order of a integer]] |
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| − | 3. 736
| + | [[Index]] |
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| − | 4. 011
| + | [[Diagonalizability]] |
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| − | 5. 054
| + | ===Others=== |
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| − | ===Solutions===
| + | [[Proofs to Some Number Theory Facts]] |
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| − | ====Problem 1====
| + | [[Chebyshev and the history of the Prime Number Theorem]] |
| − | 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>
| + | [[Proof of the Existence of Primitive Roots]] |
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| − | where <math>p</math> and <math>q</math> is prime. Find that perfect square.
| + | [[Proof of Some Primitive Roots Facts]] |
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| − | ====Solution 1==== | + | ==Problems Sharing Contest== |
| − | <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>.
| + | 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: |
| − | 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====
| + | 1. There is one and only one perfect square in the form |
| − | 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> | + | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> |
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| − | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>.
| + | where <math>p</math> and <math>q</math> are prime. Find that perfect square. ~[[Ddk001]] |
| − | ====Solution 1====
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| − | To make things easier, instead of saying <math>x_i</math>, we say <math>x</math>.
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| − | Now, we have
| + | <math>\textbf{Solution by cxsmi}</math> |
| − | <math>(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}</math>. | |
| − | 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>. | + | 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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| − | To make things even simpler, let
| + | Note: I will be the first to admit that this solution is somewhat lucky. |
| − | <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> | + | 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>P_2+(-a) \cdot P_1+2 \cdot b=0</math> <math>(2)</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>P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0</math> <math>(3)</math>
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| − | Therefore,
| + | <math>\textbf{Solution by Ddk001}</math> |
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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> | + | ===Solution 1=== |
| | + | The inradius of <math>\Delta ABC</math>, <math>r</math>, can be calculated as |
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| − | <math>=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c</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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| − | <math>=a(a^2-2b)-ab+3c</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 |
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| − | <math>=a^3-3ab+3c</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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| − | 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>
| + | The desired sum is <math>7072+2+4048=\boxed{11122}</math>. <math>\blacksquare</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>.
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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
| + | [[Ddk001]] Presents |
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| − | <math>P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})</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-3(x^2y+y^2x+x^2z+z^2x+z^2y+y^2z+3xyz)+3xyz+1</math>
| + | THE FOLLOWING PROBLEM |
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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>
| + | Note: This is one of my favorite problems. Very well designed and actually used two of my best tricks without looking weird. |
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| − | <math>=x^3+y^3+z^3+1</math>
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| − | <math>=13+53+103+1</math>
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| − | <math>=\boxed{170}</math>. <math>\square</math>
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| − | Note: If you don't know [[Newton's Sums]], you can also use [[Vieta's Formulas]] to bash.
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| − | ====Problem 3====
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| | 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. |
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| − | <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]].
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| − | 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 :
| |
| − | | |
| − | <math>x \equiv 6 \pmod{210}</math>
| |
| − | | |
| − | <math>x \equiv 0 \pmod{144}</math>.
| |
| − | | |
| − | Solving yields <math>x \equiv 2\boxed{736} \pmod{7!}</math>, and we're done. <math>\square</math>
| |
| − | | |
| − | ====Problem 4====
| |
| − | Suppose <math>f(x)</math> is a <math>10000000010</math>-degrees polynomial. The [[Fundamental Theorem of Algebra]] tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
| |
| − | | |
| − | <math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
| |
| − | | |
| − | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of | |
| − | | |
| − | <math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
| |
| − | | |
| − | Find the number of factors of the prime <math>999999937</math> in <math>p</math>.
| |
| − | | |
| − | ====Solution 1====
| |
| − | Since all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>, we have that all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)-n=0</math>, so by the [[Factor Theorem]],
| |
| − | | |
| − | <math>n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n</math>
| |
| − | | |
| − | <math>\implies (n+1)n \dots (n-10000000008)|f(n)-n</math>.
| |
| − | | |
| − | <math>\implies f(n)=a(n+1)n \dots (n-10000000008)+n</math>
| |
| − | | |
| − | since <math>f(n)</math> is a <math>10000000010</math>-degrees polynomial, and we let <math>a</math> to be the leading coefficient of <math>f(n)</math>.
| |
| − | | |
| − | Also note that since <math>r_1, r_2, \dots, r_{10000000010}</math> is the roots of <math>f(n)</math>, <math>f(n)=a(n-r_1)(n-r_2) \dots (n-r_{10000000010})</math>
| |
| − | | |
| − | Now, notice that
| |
| − | | |
| − | <math>m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})</math>
| |
| − | | |
| − | <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>
| + | Hi Ddk001 ~ zhenghua |
| | | | |
| − | <math>=\frac{10000000010 \cdot 10000000010}{2}</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?) |
| | | | |
| − | <math>=5000000005 \cdot 10000000010!</math>
| + | [[Vandalism Page]] |
| | | | |
| − | , so there is <math>\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}</math> factors of <math>999999937</math>. <math>\square</math>
| + | ==See also== |
| | + | * My [[User talk:Ddk001|talk page]] |
| | + | * [[Problems Collection|My problems collection]] |
| | | | |
| − | ====Problem 5====
| + | The problems on this page are NOT copyrighted by the [http://www.maa.org Mathematical Association of America]'s [http://amc.maa.org American Mathematics Competitions]. [[File:AMC_logo.png|middle]] |
| − | <math>\Delta ABC</math> is an isosceles triangle where <math>CB=CA</math>. Let the circumcircle of <math>\Delta ABC</math> be <math>\Omega</math>. Then, there is a point <math>E</math> and a point <math>D</math> on circle <math>\Omega</math> such that <math>AD</math> and <math>AB</math> trisects <math>\angle CAE</math> and <math>BE<AE</math>, and point <math>D</math> lies on minor arc <math>BC</math>. Point <math>F</math> is chosen on segment <math>AD</math> such that <math>CF</math> is one of the altitudes of <math>\Delta ACD</math>. Ray <math>CF</math> intersects <math>\Omega</math> at point <math>G</math> (not <math>C</math>) and is extended past <math>G</math> to point <math>I</math>, and <math>IG=AC</math>. Point <math>H</math> is also on <math>\Omega</math> and <math>AH=GI<HB</math>. Let the perpendicular bisector of <math>BC</math> and <math>AC</math> intersect at <math>O</math>. Let <math>J</math> be a point such that <math>OJ</math> is both equal to <math>OA</math> (in length) and is perpendicular to <math>IJ</math> and <math>J</math> is on the same side of <math>CI</math> as <math>A</math>. Let <math>O’</math> be the reflection of point <math>O</math> over line <math>IJ</math>. There exist a circle <math>\Omega_1</math> centered at <math>I</math> and tangent to <math>\Omega</math> at point <math>K</math>. <math>IO’</math> intersect <math>\Omega_1</math> at <math>L</math>. Now suppose <math>O’G</math> intersects <math>\Omega</math> at one distinct point, and <math>O’, G</math>, and <math>K</math> are collinear. If <math>IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2</math>, then <math>\frac{EH}{BH}</math> can be expressed in the form <math>\frac{\sqrt{b}}{a} (\sqrt{c} + d)</math>, where <math>b</math> and <math>c</math> are not divisible by the squares of any prime. Find <math>a^2+b^2+c^2+d^2+abcd</math>.
| + | <div style="clear:both;"> |
| | | | |
| − | Someone mind making a diagram for this?
| + | Can someone help me clear out [[Problem Collection|this page]]? |
| − | ====Solution 1====
| |
If you have a problem or solution to contribute, please go to this page.
I am a aops user who likes making and doing problems, doing math, and redirecting pages (see Principle of Insufficient Reasons). I like geometry and don't like counting and probability. My number theory skill are also not bad
User Count
If this is your first time visiting this page, edit it by incrementing the user count below by one.
Please do not mess with the user count. I would know and I would find you.
Credits given to Firebolt360 for inventing the box above.
Cool asyptote graphs
Asymptote is fun!
![[asy]draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));[/asy]](//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]](//latex.artofproblemsolving.com/d/7/8/d78519cbbf9c22e2c4ea74ee089557db7a04e2ba.png)
Contributions
AMC
2022 AMC 12B Problems/Problem 25 Solution 5 (Now it's solution 6)
2023 AMC 12B Problems/Problem 20 Solution 3
AIME
2016 AIME I Problems/Problem 10 Solution 3
2017 AIME I Problems/Problem 14 Solution 2
2019 AIME I Problems/Problem 15 Solution 6
2022 AIME II Problems/Problem 3 Solution 3
Restored diagram for 1994 AIME Problems/Problem 7
USAMO/USAJMO
1978 USAMO Problems/Problem 1 Solution 4
2002 USAMO Problems/Problem 2 Solution 2
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
IMO
1995 IMO Problems/Problem 6 Solution 1 (I got the solution from a forum discussion)
1986 IMO Problems/Problem 3 Solution 2 (I may or may not had gotten the solution from another source)
1976 IMO Problems/Problem 6 Solution 2 (I actually made this solution by myself! :) )
Theorems
Divergence Theorem
Stokes' Theorem
Principle of Insufficient Reasons
The Nested Sum Theorem
Hensel's Lemma
Definitions
Laplace Transform
Polynomial Congruences
Primitive Roots
Order of a integer
Index
Diagonalizability
Others
Proofs to Some Number Theory Facts
Chebyshev and the history of the Prime Number Theorem
Proof of the Existence of Primitive Roots
Proof of Some Primitive Roots Facts
Problems Sharing Contest
Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start:
1. There is one and only one perfect square in the form
![\[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]](//latex.artofproblemsolving.com/b/1/1/b11bf3dce41593092737df0eea5f902de3989389.png)
where 
and 
are prime. Find that perfect square. ~Ddk001
1. We can expand the product in the expression. 
. 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 , we find that and must equal 
and 
in some order. The perfect square formed using these numbers is 
.
Note: I will be the first to admit that this solution is somewhat lucky.
2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle 
intersects itself. has leg length 
. The perimeter of this diamond is expressible as 
, where 
, 
, and 
are integers, and is not divisible by the square of any prime. What is the remainder when 
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]](//latex.artofproblemsolving.com/c/f/4/cf4b2b8da31fb32285d5b573ffbb35e870afea4c.png)
Solution 1
The inradius of 
, 
, can be calculated as
![\[r=\frac{\textbf{Area}_{ABC}}{\textbf{Semiperimeter}} \implies r=\frac{2024^2/2}{(2024+2024+2024 \sqrt{2})/2}=\frac{2024}{2+\sqrt{2}}=2024-1012\sqrt{2}\]](//latex.artofproblemsolving.com/d/0/2/d02fa3396a9eff50f834c674242087b858095775.png)
so the square have side length 
. 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
![\[4(4048-2024 \sqrt{2})-2 \cdot 2024 (3-2\sqrt{2})+2024 (3 \sqrt{2} -4)=2024 (8-4 \sqrt{2}-6+4\sqrt{2}+3\sqrt{2}-4)=2024(3\sqrt{2}-2)=7072\sqrt{2}-4048\]](//latex.artofproblemsolving.com/b/c/8/bc8e18a5b46de191c8436a2d6d363517962ce15f.png)
The desired sum is 
. 
Ddk001 Presents
THE FOLLOWING PROBLEM
Note: This is one of my favorite problems. Very well designed and actually used two of my best tricks without looking weird.
Suppose
![\[x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}\]](//latex.artofproblemsolving.com/2/a/4/2a4c33ddcd46bcf5c89a3e3115bc73c2c9c5d3d5.png)
Find the remainder when 
is divided by 1000.
Vandalism area
Congratulations! You had reached the end of this boring user page. Time for vandalism! Write anything under 45000 bytes.
Hi Ddk001 ~ zhenghua
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?)
Vandalism Page
See also
The problems on this page are NOT copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
![[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.
