problem
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|---|---|---|---|---|
Suppose $A,B\in M_3(\mathbb C)$ , $B\ne0$ , and $AB=0$ . Prove that there exists $D\in M_3(\mathbb C)$ with $D\ne0$ such that $AD=DA=0$ .
|
We can easily prove a much more general result, as follows.
<u>Proposition</u>. Let $k$ be a field, and let $R$ be an algebraic $k$ -algebra. If $a \in R,$ then either $a \in U(R),$ the group of units of $R,$ or $ab=ba=0$ for some $0 \ne b \in R.$ <u>Proof</u>. Suppose that $a \notin U(R).$ Since $R$ is algebraic over $k, \ a$ is algebraic over $k.$ Let $$ p(x):=x^n+c_1x^{n-1}+ \cdots + c_1x+c_0 \in k[x] $$
be the minimal polynomial of $a$ over $k.$ So $a^n+c_1a^{n-1}+ \cdots + c_1a+c_0=0.$ Since $a \notin U(R),$ we have $c_0=0.$ Let $b:=a^{n-1}+c_1a^{n-2} + \cdots + c_1.$ Then, by the minimality of $p(x),$ we have $b \ne 0$ and obviously $ab=ba=0.$
|
[] |
[
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] |
{
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577589.json"
}
|
Does there exist a twice differentiable function $f:\mathbb R\to\mathbb R$ such that $f'(x)=f(x+1)-f(x)$ for all $x$ and $f''(0)\ne0$ ? Justify your answer.
|
[] |
[
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577590.json"
}
|
|
A walker and a jogger travel along the same straight line in the same direction. The walker walks at one meter per second, while the jogger runs at two meters per second. The jogger starts one meter in front of the walker. A dog starts with the walker, and then runs back and forth between the walker and the jogger with constant speed of three meters per second. Let $f(n)$ meters denote the total distance travelled by the dog when it has returned to the walker for the nth time (so $f(0)=0$ ). Find a formula for $f(n)$ .
|
[] |
[
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577785.json"
}
|
|
Given that $40!=\overline{abcdef283247897734345611269596115894272pqrstuvwx}$ , find $a,b,c,d,e,f,p,q,r,s,t,u,v,w,x$ .
|
Indeed, the official solution is:
By examining the numbers $5,10,15,20,25,30,35,40$ (at most $40$ and divisible by $5$ ), we see that the power of $5$ dividing $40!$ is $9$ . Also it is easy to see that the power of $2$ dividing $40!$ is at least $9$ . Therefore $10^9$ divides $40!$ and we deduce that $p = q = r = s = t = u = v = w = x = 0$ .
Next note that $999999$ divides $40!$ . This is because $999999 = 9\cdot111111 = 9\cdot111\cdot1001 = 9\cdot3\cdot37\cdot11\cdot91 = 27\cdot7\cdot11\cdot13\cdot37$ . Since $999999 = 10^6-1$ , it follows that if we group the digits of $40!$ in sets of six starting from the units digit and working right to left, then the sum of these digits will be divisible by $999999$ . Therefore $abcdef + 283247 + 897734 + 345611 + 269596 + 115894 + 272000$ is divisible by $999999$ , that is $abcdef + 2184082 = 999999y$ for some integer $y$ . Clearly $y = 3$ and we conclude that $abcdef = 815915$ .
|
[
" $v_5(40!) = $ floor $(\\frac{40}{5}) + $ floor $(\\frac{40}{25}) + \\cdots$ $\\implies v_5(40!) = 9$ Observe there are exactly 9 variables at the end. \nSo $p = q = r = s = t = u = v = w = x = 0$ :):)\n<details><summary>please wait</summary>Still working on a,b,c,d,e,f.</details>",
"We already found the right Hand side letters which are all equal to zero then we have to work mod(9) and mod (11)\nAlso we have to set that all the left hand side letters are in [0;9]\n",
"just take a bunch mods and bash and you should get the answer",
"<blockquote>just take a bunch mods and bash and you should get the answer</blockquote>\n\nYeah \nwe get:\na+b+c+d+e+f congruent to 2(mod9)\na-b+c-d+e-f is congruent to -1(mod11)\n\n",
"<blockquote>just take a bunch mods and bash and you should get the answer</blockquote>\n\ntry $99, 101,999, 1001,9999,10001$ to get $815915$ ",
"Clearly from Legendre the last digits are all $0$ , but how do you find $a,b,c,d,e,f$ ?",
"By ~~wolfram~~ New Tab Theorem, we have\n40!=8.15915283e47, so $abcdef=815915$ ",
"<blockquote>By ~~wolfram~~ New Tab Theorem, we have\n40!=8.15915283e47, so $abcdef=815915$ </blockquote>\n\nWhat does it tell us",
"@above, if ur asking what the New Tab Theorem is, he just looked it up on wolfram alpha rip",
"<blockquote>@above, if ur asking what the New Tab Theorem is, he just looked it up on wolfram alpha rip</blockquote>\n\nI have looked for the theorem but nothing ",
"After we get that the last few letters are $0$ s, I'm guessing the intended solution is to use that $999999=1001 \\cdot 999 = 7 \\cdot 11 \\cdot 13 \\cdot 27 \\cdot 37$ divides $40!$ . This gives that $999999$ divides\n\\[abcdef+283247+897734+345611+269596+115894+272000\\]\nwhich gives $abcdef=815915$ ."
] |
[
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] |
{
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577786.json"
}
|
Two circles $\alpha,\beta$ touch externally at the point $X$ . Let $A,P$ be two distinct points on $\alpha$ different from $X$ , and let $AX$ and $PX$ meet $\beta$ again in the points $B$ and $Q$ respectively. Prove that $AP$ is parallel to $QB$ .
|
The homothety at $X$ transforming $\alpha$ to $\beta$ carries $AP$ to $BQ$ and hence the conclusion.
|
[] |
[
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] |
{
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577954.json"
}
|
Let $n$ be a nonzero integer. Prove that $n^4-7n^2+1$ can never be a perfect square.
|
Multiplying by four we want $(2n^2-7)^2-45$ to be a square. The only squares with difference $45$ are $23^2-22^2,9^2-6^2, 7^2-2^2$ . The equation $2n^2 - 7 = \pm 23, \pm 9, \pm 7$ has an integral solution only for $n=0$ , which is prohibited by the problem statement.
|
[
"Hint: just solve this: $(n^2-4)^2<n^4-7n^2+1<(n^2-3)^2$ "
] |
[
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] |
{
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577955.json"
}
|
Solve in $R$ the equation: $8x^3-4x^2-4x+1=0$
|
<blockquote><blockquote>Solve in $R$ the equation: $8x^3-4x^2-4x+1=0$ </blockquote> $\left\{\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}\right\}$ .</blockquote>
Because $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=$ $=\frac{\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}+\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{1}{2}$ ; $\cos\frac{\pi}{7}\cos\frac{3\pi}{7}+\cos\frac{\pi}{7}\cos\frac{5\pi}{7}+\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}=$ $=\frac{1}{2}\left(\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{2\pi}{7}\right)=$ $=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\left(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}\right)=-\frac{1}{2}$ and $\cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}=\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}=\frac{8\sin\frac{\pi}{7}\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}{8\sin\frac{\pi}{7}}=$ $=\frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}}=-\frac{1}{8}$ and we get the answer by the Viete's theorem.
|
[
"See here:\nhttps://www.wolframalpha.com/input/?i=8x%5E3-4x%5E2-4x%2B1%3D0",
"<blockquote>Solve in $R$ the equation: $8x^3-4x^2-4x+1=0$ </blockquote> $\\left\\{\\cos\\frac{\\pi}{7},\\cos\\frac{3\\pi}{7},\\cos\\frac{5\\pi}{7}\\right\\}$ .\n",
"For the collection.\nSolve the following equation. $$ x^5+x^4-12x^3-21x^2+x+5=0 $$ "
] |
[
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] |
{
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/1318258.json"
}
|
Let $d$ be a positive integer and let $A$ be a $d\times d$ matrix with integer entries. Suppose $I+A+A_2+\ldots+A_{100}=0$ (where $I$ denotes the identity $d\times d$ matrix, and $0$ denotes the zero matrix, which has all entries $0$ ). Determine the positive integers $n\le100$ for which $A_n+A_{n+1}+\ldots+A_{100}$ has determinant $\pm1$ .
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564178.json"
}
|
|
Define a sequence by $a_1=1,a_2=\frac12$ , and $a_{n+2}=a_{n+1}-\frac{a_na_{n+1}}2$ for $n$ a positive integer. Find $\lim_{n\to\infty}na_n$ .
|
It's clear $a_n>0$ and decreasing. It has limit so $\lim_{n\to \infty}\frac{a_n-a_{n+1}}{a_{n+1}-a_{n+2}}=1$ $\lim_{n\to\infty}na_n=\lim \frac{n}{\frac{1}{a_n}}=\lim \frac{n+1-n}{\frac{a_n-a_{n+1}}{a_n\cdot a_{n+1}}}=2$
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] |
{
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564179.json"
}
|
Let $\sum_{n=1}^\infty a_n$ be a convergent series of positive terms (so $a_i>0$ for all $i$ ) and set $b_n=\frac1{na_n^2}$ for $n\ge1$ . Prove that $\sum_{n=1}^\infty\frac n{b_1+b_2+\ldots+b_n}$ is convergent.
|
One idea that I get is comparing the convergence nature to that of harmonic series. So $a_n $ must be smaller than some $\frac{1}{n^{1+\epsilon}}$ for some $ \epsilon > 0$ . $\Rightarrow b_n \ge n^{1+2\epsilon}$ .
Now the sum $\sum_{i=1}^n b_i $ has as its $n$ th sum proportional to $n^{2+\epsilon_2}$ , for some $\epsilon_2 > 0$ . $\Rightarrow \sum_{n=1}^\infty \frac{n}{b_1+b_2+…+b_n}$ converges.
|
[
"Any ideas?",
"[https://artofproblemsolving.com/community/c7h562849p3285201](https://artofproblemsolving.com/community/c7h562849p3285201)"
] |
[
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] |
{
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564180.json"
}
|
For $n$ a positive integer, define $f_1(n)=n$ and then for $i$ a positive integer, define $f_{i+1}(n)=f_i(n)^{f_i(n)}$ . Determine $f_{100}(75)\pmod{17}$ . Justify your answer.
|
[
"we have f1(75)mod17=7^7mod17=12\n12^12mod17=4\n4^4mod17=1\nso f100(75)mod17=1"
] |
[
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564656.json"
}
|
|
Let $\triangle ABC$ be a triangle with sides $a,b,c$ and corresponding angles $A,B,C$ (so $a=BC$ and $A=\angle BAC$ etc.). Suppose that $4A+3C=540^\circ$ . Prove that $(a-b)^2(a+b)=bc^2$ .
|
$A=3B,C=180-4B, B<45$ $(a-b)^2(a+b)=bc^2 \to (\frac{ \sin 3B}{\sin B}-1)^2 (\frac{ \sin 3B}{\sin B}+1)=(\frac{\sin 4B}{\sin B})^2$ $(2-4 \sin^2 B)^2(4-4 \sin^2 B) = 16 cos^2 B \cos^2 2B$ $(1-2\sin^2 B)^2 (1- \sin^2B)=\cos^2B \cos^2 2B$ But $1-2\sin^2B=\cos 2B, 1-\sin^2B =\cos^2B$ or $ (\frac{ \sin 3B}{\sin B}-1)^2 (\frac{ \sin 3B}{\sin B}+1)=(\frac{\sin 4B}{\sin B})^2 \to (\sin 3B-\sin B)^2 (\sin 3B+\sin B)=\sin B\sin^2 4B$ $ (\sin 3B-\sin B)^2 (\sin 3B+\sin B) = 8 \sin^2 B \cos^2 2B \sin 2B \cos B=2 \sin B \cos 2B \sin 2B \sin 4B=\sin B \sin^2 4B$
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] |
{
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564658.json"
}
|
Let $A,B$ be two circles in the plane with $B$ inside $A$ . Assume that $A$ has radius $3$ , $B$ has radius $1$ , $P$ is a point on $A$ , $Q$ is a point on $B$ , and $A$ and $B$ touch so that $P$ and $Q$ are the same point. Suppose that $A$ is kept fixed and $B$ is rolled once round the inside of $A$ so that $Q$ traces out a curve starting and finishing at $P$ . What is the area enclosed by this curve?
|
[
"I thought something like this showed up on Mathical 2020."
] |
[
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564660.json"
}
|
|
Evaluate $\int^4_1\frac{x-2}{(x^2+4)\sqrt x}dx$
|
Let's bash this :blush:
set $u=\sqrt{x} \implies du=\frac{1}{2\sqrt{x}}$ so our integral changes to $I=\int_{1}^{2} \frac{2u^2-4}{u^{4}+4}du \implies I=\int_{1}^{2} \frac{2u^2-4}{(u^2-2u+2)(u^2+2u+2)}du$ $I=\int_{1}^{2} \frac{(u^2-2u+2)+(u^2+2u+2)-8}{(u^2-2u+2)(u^2+2u+2)}du \implies I=\int_{1}^{2} \frac{du}{u^2+2u+2}+\int_{1}^{2}\frac{du}{u^2-2u+2}+8\int_{1}^{2}\frac{du}{u^{4}+4}$ now this gives $\left[\arctan{(u+1)}\right]_{1}^{2}+\left[\arctan{(u-1)}\right]_{1}^{2}-8\left[\frac{\arctan{(u+1)}}{8}+\frac{\arctan{(u-1)}}{8}+\frac{\ln(u^2+2u+2)}{16}-\frac{\ln(u^2-2u+2)}{16}\right]_{1}^{2}=\frac{1}{2}\cdot (\ln(2)-\ln(2))=\boxed{0}$
|
[
"first substitute $x=t^2$ and divide the numerator and denominator by t^2 (to make it visually appealing) after which substituting $t+\\frac{2}{t}$ as $y$ should do the trick",
"Or, $\\frac{2u^2-4}{u^{4}+4}=\\frac{u-1}{u^2-2u+2}-\\frac{u+1}{u^2+2u+2}$ . This means $\\int\\frac{x-2}{(x^2+4)\\sqrt{x}}\\ dx=\\frac{1}{2}\\ln\\frac{x-2\\sqrt{x}+2}{x+2\\sqrt{x}+2}+C$ . $$ \\int_1^4\\frac{x-2}{(x^2+4)\\sqrt{x}}\\ dx=\\frac{1}{2}\\ln\\frac{2}{10}-\\frac{1}{2}\\ln\\frac{1}{5}=0 $$ "
] |
[
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] |
{
"answer_score": 1008,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561278.json"
}
|
Find $\sum_{k=1}^\infty\frac{k^2-2}{(k+2)!}$ .
|
$\frac{k^2 - 2}{(k+2)!}=\frac{k^2 - 4+2}{(k+2)!}=\frac{k- 2}{(k+1)!} + \frac{2}{(k+2)!}=\frac{k+1-3}{(k+1)!} + \frac{2}{(k+2)!}=\frac{1}{k!}-\frac{3}{(k+1)!} + \frac{2}{(k+2)!}$ .
Taking the sum over $k=1,2,...$ we have by the help of natural number expansion, $\sum_{k=1} \frac{1}{k!} - 3\sum_{k=1} \frac{1}{(k+1)!} + 2\sum_{k=1} \frac{1}{(k+2)!} = (e-1)-3(e-2)+2(e-1-1-\frac{1}{2}) = e-3e+2e-1+6-5=0$ .
|
[
"Isn't that $0$ ?",
"Obviously, and the proof?"
] |
[
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] |
{
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561279.json"
}
|
Find $\lim_{x\to\infty}\left((2x)^{1+\frac1{2x}}-x^{1+\frac1x}-x\right)$ .
|
This just approaches to $2x-x-x=\boxed{0}$ .
|
[
"It's actually <details><summary>answer</summary>$\\ln2$</details>, according to WA and the answer key.\nFor instance, you can't say that $\\lim_{x\\to\\infty}\\left(1+\\frac1x\\right)^x=\\lim_{x\\to\\infty}1^x=1$ .",
"Oh ok, I just learned calculus, havent done limits with exponents yet",
"This is a principled answer.\nBy changing the variable $y = \\frac1x$ we need to find the limit\n\\[\n\\lim_{y \\to 0^+} \\left( \\frac2y \\right)^{1 + \\frac{y}{2}} - \\left( \\frac1y \\right)^{1 + y} - \\frac1y\n\\]\nUsing Taylor's expansion we have\n\\begin{align*}\n\\left( \\frac2y \\right)^{1 + \\frac{y}{2}} &= \\frac2y \\cdot e^{\\frac{y}{2} \\log \\frac2y} = \\frac2y \\left(1 + \\frac{y}{2} \\log\\frac2y + \\mathcal{O}(y^2 \\log^2 y) \\right) \n&= \\frac2y + \\log 2 - \\log y + \\mathcal{O}(y \\log^2 y)\n\\end{align*}\nSimilarly\n\\begin{align*}\n\\left( \\frac1y \\right)^{1 + y} =& \\frac1y \\cdot e^{y \\log \\frac1y} = \\frac1y \\left(1 + y\\log\\frac1y + \\mathcal{O}(y^2 \\log^2 y) \\right) \n&= \\frac1y - \\log y + \\mathcal{O}(y \\log^2 y)\n\\end{align*}\nPutting everything together\n\\[\n\\left( \\frac2y \\right)^{1 + \\frac{y}{2}} - \\left( \\frac1y \\right)^{1 + y} - \\frac1y = \\log 2 + \\mathcal{O}(y \\log^2 y) \\to \\log 2 \\text{ as } y \\to 0^+\n\\]",
"Let be the function $f:(1,+\\infty)\\to\\mathbb{R}, f(x)=x^{1+\\frac1x}-x$ .\nThe requested limit is $\\lim_{x\\to\\infty}[f(2x)-f(x)]$ . $\\ln\\left(x^{^{\\frac1x}}\\right)=\\dfrac{\\ln x}{x}\\Longrightarrow x^{^{\\frac1x}}=e^{^{\\frac{\\ln x}{x}}}$ .\nUsing l'Hospital: $\\lim_{x\\to\\infty}\\dfrac{\\ln x}{x}=\\lim_{x\\to\\infty}\\dfrac{\\ln^2 x}{x}=0$ .\nUsing derivatives, results for $y\\in(0,1): 1+y<e^y<1+y+y^2\\quad(1)$ .\nFor $x>1: 0<\\dfrac{\\ln x}{x}<1$ .\n\nFor $y=\\dfrac{\\ln x}{x}$ , using $(1)$ , result the bounds of $f(x)$ : $x\\cdot\\dfrac{\\ln x}{x}<x\\cdot\\left(e^{^{\\frac{\\ln x}{x}}}-1\\right)<x\\cdot\\left(\\dfrac{\\ln x}{x}+\\dfrac{\\ln^2 x}{x^2}\\right)\\Longrightarrow$ $\\Longrightarrow \\ln x<f(x)<\\ln x+\\dfrac{\\ln^2 x}{x}\\Longrightarrow$ $\\Longrightarrow \\ln(2x)-\\ln x-\\dfrac{\\ln^2 x}{x}<f(2x)-f(x)<\\ln(2x)-\\ln x+\\dfrac{\\ln^2 (2x)}{2x}\\Longrightarrow$ $\\Longrightarrow \\lim_{x\\to\\infty}\\left(\\ln(2x)-\\ln x-\\dfrac{\\ln^2 x}{x}\\right)\\le\\lim_{x\\to\\infty}[f(2x)-f(x)]\\le\\lim_{x\\to\\infty}\\left(\\ln(2x)-\\ln x+\\dfrac{\\ln^2 (2x)}{2x}\\right)\\Longrightarrow$ $\\Longrightarrow \\ln2-\\lim_{x\\to\\infty}\\dfrac{\\ln^2 x}{x}\\le\\lim_{x\\to\\infty}[f(2x)-f(x)]\\le\\ln2+\\lim_{x\\to\\infty}\\dfrac{\\ln^2 (2x)}{2x}\\Longrightarrow$ $\\Longrightarrow\\lim_{x\\to\\infty}\\left((2x)^{1+\\frac1{2x}}-x^{1+\\frac1x}-x\\right)=\\lim_{x\\to\\infty}[f(2x)-f(x)]=\\ln2$ ."
] |
[
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] |
{
"answer_score": 1002,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561281.json"
}
|
A sequence $(a_n)$ is defined by $a_0=-1,a_1=0$ , and $a_{n+1}=a_n^2-(n+1)^2a_{n-1}-1$ for all positive integers $n$ . Find $a_{100}$ .
|
[
" $a_0=-1,a_1=1,a_2=3,a_3=8,a_4=15,\\hdots$ . We see $a_n=n^2-1$ .\n"
] |
[
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561538.json"
}
|
|
Let $m,n$ be positive integers and let $[a]$ denote the residue class $\pmod{mn}$ of the integer $a$ (thus $\{[r]|r\text{ is an integer}\}$ has exactly $mn$ elements). Suppose the set $\{[ar]|r\text{ is an integer}\}$ has exactly $m$ elements. Prove that there is a positive integer $q$ such that $q$ is coprime to $mn$ and $[nq]=[a]$ .
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561541.json"
}
|
|
Let $S$ be a set with an asymmetric relation $<$ ; this means that if $a,b\in S$ and $a<b$ , then we do not have $b<a$ . Prove that there exists a set $T$ containing $S$ with an asymmetric relation $\prec$ with the property that if $a,b\in S$ , then $a<b$ if and only if $a\prec b$ , and if $x,y\in T$ with $x\prec y$ , then there exists $t\in T$ such that $x\prec t\prec y$ .
|
[
"Is the relation transitive?\n"
] |
[
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561543.json"
}
|
|
Let $P(x)=x^{100}+20x^{99}+198x^{98}+a_{97}x^{97}+\ldots+a_1x+1$ be a polynomial where the $a_i~(1\le i\le97)$ are real numbers. Prove that the equation $P(x)=0$ has at least one nonreal root.
|
Let be $x_1,x_2,\dots,x_{100}$ the roots of $P(x)$ .
Using the Vieta's relations, results: $\sum_{i=1}^{100}x_i=-20\quad(1)$ ; $\sum_{1\le i<j\le100}x_ix_j=198\quad(2)$ ; $\prod_{i=1}^{100}x_i=1\quad(3)$ .
Assume $x_i\in\mathbb{R},\;\forall i\in\{1,2,\dots,100\}$ .
Then: $\sum_{1\le i<j\le100}(x_i-x_j)^2=99\cdot\sum_{i=1}^{100}x^2_i-2\cdot\sum_{1\le i<j\le100}x_ix_j=$ $=99\cdot\left(\sum_{i=1}^{100}x_i\right)^2-200\cdot\sum_{1\le i<j\le100}x_ix_j=99\cdot400-200\cdot198=0\Longrightarrow$ $\Longrightarrow x_i=x_j,\;\forall 1\le i<j\le100\Longrightarrow x_1=x_2=\dots=x_{100}=a,\forall i\in\{1,2,\dots,100\}\quad(4)$ .
From $(1)$ and $(4)$ results: $a=\dfrac{\sum_{i=1}^{100}x_i}{100}=-\dfrac{1}{5}$ .
But in this case: $\prod_{i=1}^{100}x_i=\dfrac{1}{5^{100}}$ , contradiction with $(3)$ .
Conclusion: the assumption $x_i\in\mathbb{R},\;\forall i\in\{1,2,\dots,100\}$ is false, hence $P(x)=0$ has at least a non-real root.
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] |
{
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561544.json"
}
|
Evaluate $\sum_{n=1}^\infty \frac{n}{(2^n-2^{-n})^2}+\frac{(-1)^nn}{(2^n-2^{-n})^2}$
|
For $|q|<1$ , we have $\sum_{k=1}^{\infty} q^k=\frac{q}{(q-1)}$ . Therefore for $|q|>1$ ,
\begin{align*}
\sum_{n=1}^{\infty} \frac{(-1)^n}{q^n-1}=\sum_{n=1}^{\infty} \frac{(-1)^nq^{-n}}{1-q^{-n}}
\end{align*}
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{q^n+1}&=\sum_{n=1}^{\infty} \frac{q^{-n}}{1+q^{-n}} &=\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} (-1)^{k+1}(q^{-n})^k &=-\sum_{k=1}^{\infty} \frac{(-1)^kq^{-k}}{1-q^{-k}}
\end{align*}
So, $\sum_{n=1}^{\infty}\left(\frac{(-1)^n}{q^n-1}+\frac{1}{q^n+1}\right)=0$ and since \[\frac{d}{dx}\left(\frac{(-1)^n}{x^n-1}+\frac{1}{x^n+1}\right)=\frac{n(-1)^{n+1}}{x(x^{\frac{n}{2}}-x^{-\frac{n}{2}})^2}-\frac{n}{x(x^{\frac{n}{2}}+x^{-\frac{n}{2}})^2}=0\] or \[\frac{n(-1)^{n}}{x(x^{\frac{n}{2}}-x^{-\frac{n}{2}})^2}+\frac{n}{x(x^{\frac{n}{2}}+x^{-\frac{n}{2}})^2}=0,\] we have \[\sum_{n=1}^{\infty} \left(\frac{n(-1)^{n}}{q(q^{\frac{n}{2}}-q^{-\frac{n}{2}})^2}+\frac{n}{q(q^{\frac{n}{2}}+q^{-\frac{n}{2}})^2}\right)=0\] and by setting $q=4$ , gives our desired result. $\quad \blacksquare$
|
[
" $S=\\sum\\limits_{n=1}^{+\\infty }{\\left( \\frac{n}{\\left( 2^{n}-2^{-n} \\right)^{2}}+\\frac{n\\left( -1 \\right)^{n}}{\\left( 2^{n}-2^{-n} \\right)^{2}} \\right)}=4\\sum\\limits_{n=1}^{+\\infty }{\\frac{n}{\\left( 2^{2n}-2^{-2n} \\right)^{2}}}=4\\sum\\limits_{n=1}^{+\\infty }{\\frac{n}{\\left( 4^{n}-4^{-n} \\right)^{2}}}$ $=4\\sum\\limits_{n=1}^{+\\infty }{\\frac{4^{2n}n}{\\left( 4^{2n}-1 \\right)^{2}}}=4\\sum\\limits_{n=1}^{+\\infty }{\\frac{16^{n}}{\\left( 16^{n}-1 \\right)^{2}}\\cdot n}$ perhaps it can serve the q-digamma function $\\psi _{q}\\left( z \\right)=-\\ln \\left( 1-q \\right)+\\ln q\\sum\\limits_{n=0}^{+\\infty }{\\frac{q^{n+z}}{1-q^{n+z}}}$ ",
"Ignore previous (edited) -_-",
"Sorry, the answer <details><summary>is</summary>0</details>.\n\nThe problem was from VTRMC 2013. See here: https://www.math.vt.edu/people/plinnell/Vtregional/E13/index.html",
"but you mistyped the problem.\nThe problem is $S=\\sum\\limits_{n=1}^{+\\infty }{\\left( \\frac{n}{\\left( 2^{n}+2^{-n} \\right)^{2}}+\\frac{\\left( -1 \\right)^{n}n}{\\left( 2^{n}-2^{-n} \\right)^{2}} \\right)}$ no $S=\\sum\\limits_{n=1}^{+\\infty }{\\left( \\frac{n}{\\left( 2^{n}-2^{-n} \\right)^{2}}+\\frac{\\left( -1 \\right)^{n}n}{\\left( 2^{n}-2^{-n} \\right)^{2}} \\right)}$ ",
"Sorry pprime, you are right. I even double checked my typing :rotfl: ",
"Find $\\sum_{n=1}^\\infty\\left(\\frac n{(2^n+2^{-n})^2}+\\frac{(-1)^nn}{(2^n-2^{-n})^2}\\right)$ ."
] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] |
{
"answer_score": 112,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/1279364.json"
}
|
Prove that $$ \frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}
\leq\frac{3\sqrt{3}}{2} $$ for any positive real numbers $x, y,z$ such that $x+y+z = xyz.$ [2008 VTRMC #1](https://artofproblemsolving.com/community/c7h236610p10925499)
[here](http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf)
|
<blockquote>Prove that $$ \frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}
\leq\frac{3\sqrt{3}}{2} $$ for any positive real numbers $x, y,z$ such that $x+y+z = xyz.$ [2008 VTRMC #1](https://artofproblemsolving.com/community/c7h236610p10925499)
[here](http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf)</blockquote> $$ LHS=\sum \frac{x}{\sqrt{x^2+\frac{xyz}{x+y+z}}}=\sum \sqrt{\frac{x(x+y+z)}{(x+y)(x+z)}} $$ $$ LHS=\frac{\sqrt{x+y+z}\left(\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)} \right)}{\sqrt{(x+y)(y+z)(z+x)}} $$ $$ LHS\le \frac{\sqrt{6(x+y+z)(xy+yz+zx)}}{\sqrt{\frac{8}{9}(x+y+z)(xy+yz+zx)}}=RHS $$
|
[
"<details><summary>Hint</summary>In a triangle we have $tanAtanBtanC=tanA+tanB+tanC$</details>\n<details><summary>If you still don't know</summary>So we let $x=tanA,y=tanB,z=tanC$ And $LHS=sinA+sinB+sinC$ Then we're left with a well-known ineq.\nJust use Jensen and we can get the result.</details>",
"<blockquote><blockquote>Prove that $$ \\frac{x}{\\sqrt{1+x^2}}+\\frac{y}{\\sqrt{1+y^2}}+\\frac{z}{\\sqrt{1+z^2}}\n\\leq\\frac{3\\sqrt{3}}{2} $$ for any positive real numbers $x, y,z$ such that $x+y+z = xyz.$ [2008 VTRMC #1](https://artofproblemsolving.com/community/c7h236610p10925499)\n[here](http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf)</blockquote> $$ LHS=\\sum \\frac{x}{\\sqrt{x^2+\\frac{xyz}{x+y+z}}}=\\sum \\sqrt{\\frac{x(x+y+z)}{(x+y)(x+z)}} $$ $$ LHS=\\frac{\\sqrt{x+y+z}\\left(\\sqrt{x(y+z)}+\\sqrt{y(z+x)}+\\sqrt{z(x+y)} \\right)}{\\sqrt{(x+y)(y+z)(z+x)}} $$ $$ LHS\\le \\frac{\\sqrt{6(x+y+z)(xy+yz+zx)}}{\\sqrt{\\frac{8}{9}(x+y+z)(xy+yz+zx)}}=RHS $$ </blockquote>\nNice.\n",
"<blockquote><blockquote><details><summary>Hint</summary>In a triangle we have $tanAtanBtanC=tanA+tanB+tanC$</details>\n<details><summary>If you still don't know</summary>So we let $x=tanA,y=tanB,z=tanC$ And $LHS=sinA+sinB+sinC$ Then we're left with a well-known ineq.\nJust use Jensen and we can get the result.</details></blockquote></blockquote>\n\nWell,it's impossible to open the document in China here...",
"https://artofproblemsolving.com/community/c6h1633381p10262290"
] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] |
{
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/1700826.json"
}
|
Let $I=3\sqrt2\int^x_0\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$ . If $0<x<\pi$ and $\tan I=\frac2{\sqrt3}$ , what is $x$ ?
|
Rewrite $I$ as \[I=6\int_{0}^{x} \frac{\cos \frac{t}{2}}{9+16\sin ^2 \frac{t}{2}}\, dt.\] Then by letting $u=\sin \frac{t}{2}, du=\frac{1}{2}\cos \frac{t}{2} \, dt$ , we get \[I=12 \int_{0}^{\sin \frac{x}{2}} \frac{du}{9+16u^2}.\] We then perform another substitution, $u=\frac{3}{4}v, du=\frac{3}{4} \, dv$ , \[I=\int_{0}^{\frac{4}{3}\sin \frac{x}{2}} \frac{dv}{1+v^2}=\tan ^{-1} \left(\frac{4}{3}\sin \frac{x}{2}\right).\] As $\tan I=\frac{2}{\sqrt 3}$ , we find that $\frac{4}{3}\sin \frac{x}{2}=\frac{2}{\sqrt 3} \implies \boxed{x=\frac{2\pi}{3}}.$
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] |
{
"answer_score": 1010,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559093.json"
}
|
Define a sequence $(a_n)$ for $n\ge1$ by $a_1=2$ and $a_{n+1}=a_n^{1+n^{-3/2}}$ . Is $(a_n)$ convergent (i.e. $\lim_{n\to\infty}a_n<\infty$ )?
|
Clearly $a_n$ is increasing so being convergent is equivalent to being bounded from above which is equivalent to $\log a_n$ being bounded from above. But
\[\log a_{n+1}=\log 2 \cdot \prod_{n=1}^N \left(1+n^{-3/2}\right) \le \log 2 \cdot \prod_{n=1}^N e^{n^{-3/2}} = \log 2 \cdot e^{\sum_{n=1}^N n^{-3/2}}\]
which is bounded since $\sum_{n=1}^{\infty} n^{-3/2}$ is convergent. Here we used that $1+x \le e^x$ .
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] |
{
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559094.json"
}
|
Let
\begin{align*}X&=\begin{pmatrix}7&8&98&-9&-7-7&-7&9\end{pmatrix}Y&=\begin{pmatrix}9&8&-98&-7&77&9&8\end{pmatrix}.\end{align*}Let $A=Y^{-1}X$ and let $B$ be the inverse of $X^{-1}+A^{-1}$ . Find a matrix $M$ such that $M^2=XY-BY$ (you may assume that $A$ and $X^{-1}+A^{-1}$ are invertible).
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559095.json"
}
|
|
Let $ABC$ be a right-angled triangle with $\angle ABC=90^\circ$ , and let $D$ be on $AB$ such that $AD=2DB$ . What is the maximum possible value of $\angle ACD$ ?
|
WLOG let $BD=1,AD=2$ . Consider a ray with vertex $B$ perpendicular to $AB$ . Let $O$ be a point on the perpendicular bisector of $AD$ , starting at its midpoint and moving in the direction of the ray. Considering the circle centered at $O$ through $A,D$ , all points on major arc $AD$ of that circle result in the same value of $\angle ACD$ , and this angle decreases as $O$ moves further away from $AD$ , so our maximum value occurs when the circle is first tangent to our ray. In this case, by PoP $BC^2=BD\cdot BA=3\implies BC=\sqrt{3}\implies \angle ACD=\boxed{30^{\circ}}$
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] |
{
"answer_score": 1024,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559448.json"
}
|
A positive integer $n$ is called special if it can be represented in the form $n=\frac{x^2+y^2}{u^2+v^2}$ , for some positive integers $x,y,u,v$ . Prove that
(a) $25$ is special;
(b) $2014$ is not special;
(c) $2015$ is not special.
|
(b) Notice that $19|2014$ and $19\equiv 3\text{(mod 4)}$ . We must have $19|x^2+y^2\implies 2|v_{19}(x^2+y^2)$ . This means that we also must have $19|u^2+v^2\implies 2|v_{19}(u^2+v^2)$ , but then $2|v_{19}(\frac{x^2+y^2}{u^2+v^2})$ , contradiction.
(c) Same thing as above, except for $31$ .
|
[
"(a) $\\frac{6^2+8^2}{2^2+2^2}=25$ ",
"<blockquote>(a) $\\frac{6^2+8^2}{2^2+2^2}=25$ </blockquote> $\\frac{6^2+8^2}{2^2+2^2}=\\frac{25}{2}$ Instead, consider $\\frac{5^2+5^2}{1^2+1^2}$ ",
"More universally, we can proof if for any p is 3(mod4),the power of such p in n is even,then n is special.\n(after revising the basic theory of $x^2+y^2=n$ ,I'm sure of this.)"
] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] |
{
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559450.json"
}
|
Find the least positive integer $n$ such that $2^{2014}$ divides $19^n-1$ .
|
LTE gives, if $n$ is even, $v_2(19^n-1)=2+v_2(n)\ge2014$ , so we must have $v_2(n)\ge2012$ , and the minimum for even $n$ is $2^{2012}$ . If $n$ is odd, then (again by LTE) we have $v_2(19^n-1)=1<2014$ , so the answer is $\boxed{2^{2012}}$ . (Can someone check this?)
|
[
"By LTE, $$ \\nu_2(19^n - 1) = \\nu_2(18) + \\nu_2(20) + \\nu_2(n) - 1 = 1 + 2 + \\nu_2(n) - 1 = 2 + \\nu_2(n). $$ We need this to be greater than $2014$ , so we must have $\\nu_2(n)\\geq 2012\\iff 2^{2012} | n$ . Since $n$ is positive, it follows that the smallest such $n$ is $\\boxed{2^{2012}}$ .\n\nedit: sniped",
"What case of LTE gives $v_2(19^n-1)=v_2(18)+v_2(20)+v_2(n)-1$ ? My book says it's true if $2\\mid x-y$ and $n$ is even. Is $v_2(x^n-y^n)=v_2(x-y)+v_2(x+y)+v_2(n)-1$ also true if $n$ is odd? (Sorry, I'm a beginner to LTE)",
"Bumping the question.",
"<blockquote>What case of LTE gives $v_2(19^n-1)=v_2(18)+v_2(20)+v_2(n)-1$ ? My book says it's true if $2\\mid x-y$ and $n$ is even. Is $v_2(x^n-y^n)=v_2(x-y)+v_2(x+y)+v_2(n)-1$ also true if $n$ is odd? (Sorry, I'm a beginner to LTE)</blockquote>\n\nI forgot about the odd case (oops), but that is easily ruled out by modulo $4$ .",
"With $x,y$ is odd\nIf $n$ is odd then by LTE, $v_2(x^n-y^n)=v_2(x-y)$ if $x-y$ is divisible by 4, n is even then $v_2(x^n-y^n)=v_2(x-y)+v_2(n)$ if $n$ is even, $v_2(x^n-y^n)=v_2(x-y)+v_2(x+y)+v_2(n)-1$ ",
"<blockquote>With $x,y$ is odd\nIf $n$ is odd then by LTE, $v_2(x^n-y^n)=v_2(x-y)$ if $x-y$ is divisible by 4, n is even then $v_2(x^n-y^n)=v_2(x-y)+v_2(n)$ if $n$ is even, $v_2(x^n-y^n)=v_2(x-y)+v_2(x+y)+v_2(n)-1$ </blockquote>\n\nAlso note that when $x,y$ odd and $x-y$ is divisible by $4$ , then $v_2(x+y)-1=0$ . Similarly if $x+y$ is divisible by $4$ then $v_2(x-y)-1=0$ .\nSo the second formula you gave is redundant when the third formula is known.",
"<blockquote>Also, $v_2(n)\\ge2012\\iff2^{2012}\\mid n$ isn't true.</blockquote>\n\nif $n$ is an integer, it is (and in fact, that's given)",
"<blockquote>Find the least positive integer $n$ such that $2^{2014}$ divides $19^n-1$ .</blockquote>\n\nLTE kills this problem. "
] |
[
"origin:aops",
"Undergraduate Contests",
"2014 VTRMC",
"VTRMC"
] |
{
"answer_score": 1016,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2014 VTRMC/2556216.json"
}
|
Find $\sum_{n=2}^\infty\frac{n^2-2n-4}{n^4+4n^2+16}$ .
|
$\frac{n^2-2n-4}{n^4+4n^2+16}=\frac{n^2-2n-4}{(n^2-2n+4)(n^2+2n+4)}=\frac 12\times (\frac{n-2}{n^2-2n+4}-\frac{n}{n^2+2n+4})$ .
Note that $\frac{n+2}{n^2+2n+4}=\frac{n+2}{(n+2)^2-2(n+2)+4}$ , thus $\sum_{n=2}^\infty\frac{n^2-2n-4}{n^4+4n^2+16}=\frac 12\sum_{n=2}^\infty(\frac{n-2}{n^2-2n+4}-\frac{n}{n^2+2n+4})=\frac 12\times (\frac{2-2}{2^2-2\times 2+4}+\frac{3-2}{3^2-2\times 3+4})=\boxed{\frac {1}{14}}.$ oops sorry @below
fixed.
|
[
"In the telescoping sum, it is not remain the first two members? ",
"<details><summary>solution</summary>The following telescoping works $$ \\sum_{n=2}^{\\infty} \\frac{n^2-2n-4}{(n^2-2n+4)(n^2+2n+4)} = \\frac{1}{2}\\sum_{n=2}^{\\infty}\\left(\\frac{n-2}{(n-1)^2+3}-\\frac{n}{(n+1)^2+3}\\right) = \\boxed{\\frac{1}{14}} $$</details>",
"I agree with the above answer, and Wolfram|Alpha confirms.",
"<blockquote><details><summary>solution</summary>I seem to have a different answer.\nThe following telescoping works $$ \\sum_{n=2}^{\\infty} \\frac{n^2-2n-4}{(n^2-2n+4)(n^2+2n+4)} = \\frac{1}{2}\\sum_{n=2}^{\\infty}\\left(\\frac{n-2}{(n-1)^2+3}-\\frac{n}{(n+1)^2+3}\\right) = \\boxed{\\frac{1}{14}} $$</details></blockquote>\n\nYeah you're right. My answer was wrong. Now fixed :blush: ",
"<blockquote><details><summary>solution</summary>The following telescoping works $$ \\sum_{n=2}^{\\infty} \\frac{n^2-2n-4}{(n^2-2n+4)(n^2+2n+4)} = \\frac{1}{2}\\sum_{n=2}^{\\infty}\\left(\\frac{n-2}{(n-1)^2+3}-\\frac{n}{(n+1)^2+3}\\right) = \\boxed{\\frac{1}{14}} $$</details></blockquote>\n\nCan u let me know how did you arrive at the last step?",
" $\\frac{n^2 - 2n -4}{n^4 +4n^2 +16}$ $= \\frac{n^2 -2n -4}{(n^2 -2n +4)(n^2 + 2n +4)}$ $= \\frac{An + B}{n^2-2n+4} + \\frac{Cn+D}{n^2 + 2n +4}$ $= \\frac{(An+B)(n^2+2n+4) +(Cn+D)(n^2 -2n+4)}{(n^2 -2n+4)(n^2 +2n +4)}$ So $(A+C)n^3 + (2A+B-2C-D)n^2 +(4A+2B+4C-2D)n +(4B+D)$ $\\equiv n^2 -2n -4$ Comparing coefficients, $C=-A$ $2A + B + 2A -D = 1$ $4A+2B -4A -2D = -2$ $B+D = -1$ Solving, $B= -1, D = 0$ $A= \\frac{1}{2}, C = \\frac{-1}{2}$ .\nTherefore given sum is $\\frac{1}{2} \\sum_{n=2}^{\\infty} \\left( \\frac{n-2}{n^2 - 2n +4} - \\frac{n}{n^2 +2n +4} \\right)$ $= \\frac{1}{2} \\left( 0 - \\frac{2}{12} + \\frac{1}{7} - \\frac{3}{19} + \\frac{2}{12} - \\frac{4}{28} + \\frac{3}{19} - … \\right)$ $= \\frac{1}{14}$ .\nHopefully, this makes W.R.O.N.G’s last step clear.\n( $(n-1)^2 + 3 = n^2 -2n +4, (n+1)^2 + 3 = n^2 +2n +4$ .)"
] |
[
"origin:aops",
"Undergraduate Contests",
"2014 VTRMC",
"VTRMC"
] |
{
"answer_score": 1006,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2014 VTRMC/2556461.json"
}
|
Evaluate $\int^2_0\frac{x(16-x^2)}{16-x^2+\sqrt{(4-x)(4+x)(12+x^2)}}dx$ .
|
<details><summary>solution</summary>Divide both the numerator and the denominator by $(16-x^2)$ and perform the substitution $t=x^2$ . The required integral then becomes $$ I = \frac{1}{2}\int_{0}^{4}\frac{dt}{1+\sqrt{\frac{12+t}{16-t}}} $$ Making the change of variables $t\mapsto 4-t$ , we get $$ I = \frac{1}{2}\int_{0}^{4}\frac{dt}{1+\sqrt{\frac{16-t}{12+t}}} $$ Adding the above two equations, we have $$ I = \frac{1}{4}\int_{0}^{4}dt = \boxed{1} $$</details>
|
[] |
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"origin:aops",
"Undergraduate Contests",
"2014 VTRMC",
"VTRMC"
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"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2014 VTRMC/2556462.json"
}
|
Suppose we are given a $19\times19$ chessboard (a table with $19^2$ squares) and remove the central square. Is it possible to tile the remaining $19^2-1=360$ squares with $4\times1$ and $1\times4$ rectangles? (So that each of the $360$ squares is covered by exactly one rectangle.) Justify your answer.
|
It can't be done.
Mod a $4\times 4$ binary array, initially containing all zeroes, each $1\times 4$ rectangle inverts the bits in a row and each $4\times 1$ rectangle inverts the bits in a column. In either case, for any row $(a,b,c,d)$ in the array, any other is always $(a,b,c,d)$ or $(1-a,1-b,1-c,1-d)$ . Ditto columns.
However, the $19\times 19$ chessboard with a hole removed is
\[ \begin{array}[t]{|c|c|c|c|}
\hline
1&1&1&0 \hline
1&0&1&0 \hline
1&1&1&0 \hline
0&0&0&0 \hline
\end{array} \]
modulo that binary array, so is unreachable.
|
[] |
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"origin:aops",
"Undergraduate Contests",
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"answer_score": 14,
"boxed": false,
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"path": "Contest Collections/Undergraduate Contests/VTRMC/2014 VTRMC/2556689.json"
}
|
Let $n\ge1$ and $r\ge2$ be positive integers. Prove that there is no integer $m$ such that $n(n+1)(n+2)=m^r$ .
|
<blockquote>Hint: because $(n,n+1,n+2)=1$ , we must have $n=a^r, n+1=b^r, n+2=c^r$ .</blockquote>
What! No, that is wrong on many levels. Take numbers $1, 7, 49$ . We have $\gcd(1, 7, 49) = 1$ but even though $1 \cdot 7 \cdot 49 = 7^3$ not all of $1, 7, 49$ are perfect cubes.
Here is a solution. We note that we have $(n+1)((n+1)^2 - 1) = m^r$ and so both $n+1$ and $(n+1)^2-1$ are perfect powers, say $t_1^r$ and $t_2^r$ . Thus $(n+1)^2 - 1 = t_2^r$ which by Mihailescu gives us $n = 2$ . However this clearly is not a solution as $24$ is NOT a perfect power.
|
[
"Hint: because $n,n+1,n+2$ are relative prime two by two, we must have $n=a^r, n+1=b^r, n+2=c^r$ .",
"I used the same idea, butit was wrong:\n<blockquote> both $n+1$ and $(n+1)^2-1$ are perfect powers</blockquote>\nIn your case why it is true?",
"<blockquote>I used the same idea, butit was wrong:\n<blockquote> both $n+1$ and $(n+1)^2-1$ are perfect powers</blockquote>\nIn your case why it is true?</blockquote>\n\nBecause that only works when there are $2$ terms. Your idea can work as well, but only when $n,n+1, n+2$ are **pairwise coprime**. Hope you got it :) You edited the post to reflect this, but it's false because if $n$ is even then $2$ divides both $n$ and $n+2$ .",
"Ok, I see! Thank you!"
] |
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|
Let $A,B$ be two points in the plane with integer coordinates $A=(x_1,y_1)$ and $B=(x_2,y_2)$ . (Thus $x_i,y_i\in\mathbb Z$ , for $i=1,2$ .) A path $\pi:A\to B$ is a sequence of **down** and **right** steps, where each step has an integer length, and the initial step starts from $A$ , the last step ending at $B$ . In the figure below, we indicated a path from $A_1=(4,9)$ to $B1=(10,3)$ . The distance $d(A,B)$ between $A$ and $B$ is the number of such paths. For example, the distance between $A=(0,2)$ and $B=(2,0)$ equals $6$ . Consider now two pairs of points in the plane $A_i=(x_i,y_i)$ and $B_i=(u_i,z_i)$ for $i=1,2$ , with integer coordinates, and in the configuration shown in the picture (but with arbitrary coordinates): $x_2<x_1$ and $y_1>y_2$ , which means that $A_1$ is North-East of $A_2$ ; $u_2<u_1$ and $z_1>z_2$ , which means that $B_1$ is North-East of $B_2$ .
Each of the points $A_i$ is North-West of the points $B_j$ , for $1\le i$ , $j\le2$ . In terms of inequalities, this means that $x_i<\min\{u_1,u_2\}$ and $y_i>\max\{z_1,z_2\}$ for $i=1,2$ .
(a) Find the distance between two points $A$ and $B$ as before, as a function of the coordinates of $A$ and $B$ . Assume that $A$ is North-West of $B$ .
(b) Consider the $2\times2$ matrix $M=\begin{pmatrix}d(A_1,B_1)&d(A_1,B_2)d(A_2,B_1)&d(A_2,B_2)\end{pmatrix}$ . Prove that for any configuration of points $A_1,A_2,B_1,B_2$ as described before, $\det M>0$ .
|
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|
Let $S$ denote the set of $2$ by $2$ matrices with integer entries and determinant $1$ , and let $T$ denote those matrices of $S$ which are congruent to the identity matrix $I\pmod3$ (so $\begin{pmatrix}a&bc&d\end{pmatrix}\in T$ means that $a,b,c,d\in\mathbb Z,ad-bc=1,$ and $3$ divides $b,c,a-1,d-1$ ).
(a) Let $f:T\to\mathbb R$ be a function such that for every $X,Y\in T$ with $Y\ne I$ , either $f(XY)>f(X)$ or $f(XY^{-1})>f(X)$ . Show that given two finite nonempty subsets $A,B$ of $T$ , there are matrices $a\in A$ and $b\in B$ such that if $a'\in A$ , $b'\in B$ and $a'b'=ab$ , then $a'=a$ and $b'=b$ .
(b) Show that there is no $f:S\to\mathbb R$ such that for every $X,Y\in S$ with $Y\ne\pm I$ , either $f(XY)>f(X)$ or $f(XY^{-1})>f(X)$ .
|
[] |
[
"origin:aops",
"Undergraduate Contests",
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"VTRMC"
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|
Find all n such that $n^{4}+6n^{3}+11n^{2}+3n+31$ is a perfect square.
|
So I think it's safe to assume that $n$ is an integer
(otherwise, the expression is equal to each perfect square $\geq 36$ for one positive and one negative $n$ value)
<details><summary>Solution</summary>Let that be $f(n)$ $(n^{2}+3n+1)^{2}-f(n)=3n+30$ , positive for $n>-10$ $f(n)-(n^{2}+3n)^{2}=2n^{2}+3n+31$ , always positive. $n$ cannot be greater than $-10$ , or else $f(n)$ lies between two perfect squares. $(n^{2}+3n+2)^{2}-f(n)=2n^{2}+9n-27$ is positive for all $n \leq-10$ , so for $n \leq-10$ , $f(n)$ is not a perfect square unless $f(n)=(n^{2}+3n+1)^{2}\implies (n^{2}+3n+1)^{2}-f(n) = 0$ $\implies 3n+30 = 0 \implies n=-10$ Thus, $\boxed{n=-10}$ is the only integer solution. Indeed, $f(-10)=71^{2}$</details>
|
[
"You made a mistake.\r $(n^{2}+3n+1)^{2}-f(n) = 3n-30$ \r\nThough it seems right other than that. Thanks.\r\n\r\nAnd n is integer."
] |
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"origin:aops",
"2015 VTRMC",
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"VTRMC"
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"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2015 VTRMC/139049.json"
}
|
The planar diagram below, with equilateral triangles and regular hexagons, sides length $2$ cm, is folded along the dashed edges of the polygons, to create a closed surface in three-dimensional Euclidean spaces. Edges on the periphery of the planar diagram are identified (or glued) with precisely one other edge on the periphery in a natural way. Thus, for example, $BA$ will be joined to $QP$ and $AC$ will be joined to $DC$ . Find the volume of the three-dimensional region enclosed by the resulting surface.
|
[
"Hint: It's a truncated tetrahedron."
] |
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"origin:aops",
"2015 VTRMC",
"Undergraduate Contests",
"VTRMC"
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"boxed": false,
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"path": "Contest Collections/Undergraduate Contests/VTRMC/2015 VTRMC/2551434.json"
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|
Let $(a_i)_{1\le i\le2015}$ be a sequence consisting of $2015$ integers, and let $(k_i)_{1\le i\le2015}$ be a sequence of $2015$ positive integers (positive integer excludes $0$ ). Let $$ A=\begin{pmatrix}a_1^{k_1}&a_1^{k_2}&\cdots&a_1^{k_{2015}}a_2^{k_1}&a_2^{k_2}&\cdots&a_2^{k_{2015}}\vdots&\vdots&\ddots&\vdotsa_{2015}^{k_1}&a_{2015}^{k_2}&\cdots&a_{2015}^{k_{2015}}\end{pmatrix}. $$ Prove that $2015!$ divides $\det A$ .
|
The first part itself is sufficient. For prime $p$ (we only care about $p\le2015$ ) and $k\in\mathbb N$ there are at least $\left\lceil\frac{2015}{p^k}\right\rceil$ of $a$ 's congruent $\bmod\,p^k$ by the pigeonhole principle, and so
\[v_p(\det A)
\ge v_p\left(\prod_{i<j}(a_i-a_j)\right)
\ge\sum_{k=1}^\infty\binom{\left\lceil\frac{2015}{p^k}\right\rceil}2
\ge\sum_{k=1}^\infty\left\lfloor\frac{2015}{p^k}\right\rfloor
=v_p(2015!),\]
where the inequality $\binom{\left\lceil\frac{2015}{p^k}\right\rceil}2\ge\left\lfloor\frac{2015}{p^k}\right\rfloor$ holds since:
1) If $p^k>2015$ then it's trivial;
2) $p^k=2015$ is impossible since $2015=5\cdot13\cdot31$ ;
3) If $p^k<2015$ and $p^k\nmid2015$ then $\left\lceil\frac{2015}{p^k}\right\rceil\ge2$ and so
\[\binom{\left\lceil\frac{2015}{p^k}\right\rceil}2
=\frac{\left\lceil\frac{2015}{p^k}\right\rceil\left(\left\lceil\frac{2015}{p^k}\right\rceil-1\right)}{2}
\ge\left\lceil\frac{2015}{p^k}\right\rceil-1
=\left\lfloor\frac{2015}{p^k}\right\rfloor;\]
4) If $p^k<2015$ and $p^k\mid2015$ then $p\in\{5,13,31\}$ and $k=1$ and the inequality is trivial to check.
|
[
"If some $a_i=0$ or some $a_i=a_j,i\\not= j$ or some $k_i=k_j,i\\not= j$ , then $\\det(A)=0$ ; then we assume that it is not the case.\nAs for the Vandermonde determinant, by recurrence and linear combinations of columns, we show that $\\Pi_{i<j}(a_i-a_j)$ divides $\\det(A)$ .\nUnlike the Vandermonde case, $\\Pi_i a_i$ divides $\\det(A)$ . \nIt remains to show that this implies the required result (I think that it is true).\n",
"<blockquote>If some $a_i=0$ or some $a_i=a_j,i\\not= j$ or some $k_i=k_j,i\\not= j$ , then $\\det(A)=0$ ; then we assume that it is not the case.\nAs for the Vandermonde determinant, by recurrence and linear combinations of columns, we show that $\\Pi_{i<j}(a_i-a_j)$ divides $\\det(A)$ .\nUnlike the Vandermonde case, $\\Pi_i a_i$ divides $\\det(A)$ . \nIt remains to show that this implies the required result (I think that it is true).</blockquote>\n\nCan we use the fact that $\\Pi \\frac{a_i - a_j}{i-j}$ is integer and the determinant of $A$ is $\\Pi (a_i - a_j)$ ?"
] |
[
"origin:aops",
"2015 VTRMC",
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"VTRMC"
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"answer_score": 34,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2015 VTRMC/2551438.json"
}
|
Consider the harmonic series $\sum_{n\ge1}\frac1n=1+\frac12+\frac13+\ldots$ . Prove that every positive rational number can be obtained as an unordered partial sum of this series. (An unordered partial sum may skip some of the terms $\frac1k$ .)
|
Well, for me a "partial sum" is always something finite. In any case, if we allow infinite sums, the problem becomes almost trivial (modulo the divergence of the harmonic series), while for finite sums one needs to be slightly more clever.
To put the problem in other words, we want to represent every positive rational number as a sum of distinct reciprocals. These things are called Egyptian Fractions and as you can imagine, they have been studied for hundreds (if not thousands) of years.
As such, if I understand correctly, the exact problem here was known to and solved by Fibonacci more than 800 years ago.
Indeed, there is a simple greedy algorithm to produce such a representation: Always subtract the largest reciprocal still available.
Now let's see why this process ends after finitely many steps.
Indeed, first of all, since the harmonic series diverges, your remainder becomes arbitrarily small and in particular from some point on is less than $1$ .
So suppose at some point you are at $0<\frac{p}{q}<1$ . We claim that the numerator strictly decreases from now on in each step. This would clearly show that we can only do finitely many steps.
Indeed, in the next step you subtract $\frac{1}{n}$ where $\frac{1}{n} \le \frac{p}{q}<\frac{1}{n-1}$ (note that $n>1$ since $\frac{p}{q}<1$ ).
This leaves you with $\frac{p}{q}-\frac{1}{n}=\frac{np-q}{nq}$ . We claimed that $np-q <p$ which is equivalent to $(n-1)p<q$ which is equivalent to $\frac{p}{q}<\frac{1}{n-1}$ which is true by assumption!
|
[
"If an infinite number of terms is allowed, this is true for any positive number, not just the rationals. Maybe you meant partial sum with finitely many terms?",
"All that I know is this. If $\\sum a_n$ is a series of real numbers that is converge but not absolutely, then for every $-\\infty \\le \\alpha \\le \\beta\\le +\\infty$ \nthere exist a rearrangement of $\\sum a_n$ , say $s_n'$ , such that $liminf s_n' = \\alpha, limsup s_n' = \\beta$ .\nYou may want to take $\\alpha = \\beta = r$ , the rational number that you want. ",
"As I mentioned, I think the problem is clearly not about (infinite) **subseries**but about (finite) **subsums**, so the rearrangement theorem does not help here!"
] |
[
"origin:aops",
"2015 VTRMC",
"Undergraduate Contests",
"VTRMC"
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{
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2015 VTRMC/2551440.json"
}
|
Evaluate $\int^\infty_0\frac{\operatorname{arctan}(\pi x)-\operatorname{arctan}(x)}xdx$ (where $0\le\operatorname{arctan}(x)<\frac\pi2$ for $0\le x<\infty$ ).
|
<details><summary>pretty standard approach if you've seen similar problems before</summary>This is equivalent to $$ \int_0^\infty \int_1^\pi \frac{da\, dx}{1+a^2x^2} $$ $$ \int_1^\pi\int_0^\infty \frac{dx\, da}{1+a^2x^2} $$ $$ \frac{\pi}{2}\int_1^\pi \frac{da}{a} $$ $$ \frac{\pi\ln\pi}{2} $$</details>
|
[
"Frullani integral will help"
] |
[
"origin:aops",
"2015 VTRMC",
"Undergraduate Contests",
"VTRMC"
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{
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2015 VTRMC/2551519.json"
}
|
Let $(a_1,b_1),\ldots,(a_n,b_n)$ be $n$ points in $\mathbb R^2$ (where $\mathbb R$ denotes the real numbers), and let $\epsilon>0$ be a positive number. Can we find a real-valued function $f(x,y)$ that satisfies the following three conditions?
1. $f(0,0)=1$ ;
2. $f(x,y)\ne0$ for only finitely many $(x,y)\in\mathbb R^2$ ;
3. $\sum_{r=1}^n\left|f(x+a_r,y+b_r)-f(x,y)\right|<\epsilon$ for every $(x,y)\in\mathbb R^2$ .
Justify your answer.
|
[] |
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"origin:aops",
"2015 VTRMC",
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"VTRMC"
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"path": "Contest Collections/Undergraduate Contests/VTRMC/2015 VTRMC/2551521.json"
}
|
|
Let $n$ be a positive integer and let $x_1,\ldots,x_n$ be $n$ nonzero points in $\mathbb R^2$ . Suppose $\langle x_i,x_j\rangle$ (scalar or dot product) is a rational number for all $i,j$ ( $1\le i,j\le n$ ). Let $S$ denote all points of $\mathbb R^2$ of the form $\sum_{i=1}^na_ix_i$ where the $a_i$ are integers. A closed disk of radius $R$ and center $P$ is the set of points at distance at most $R$ from $P$ (includes the points distance $R$ from $P$ ). Prove that there exists a positive number $R$ and closed disks $D_1,D_2,\ldots$ of radius $R$ such that
(a) Each disk contains exactly two points of $S$ ;
(b) Every point of $S$ lies in at least one disk;
(c) Two distinct disks intersect in at most one point.
|
[] |
[
"origin:aops",
"2015 VTRMC",
"Undergraduate Contests",
"VTRMC"
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"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2015 VTRMC/2551524.json"
}
|
|
Determine the number of real solutions to the equation $\sqrt{2 -x^2} = \sqrt[3]{3 -x^3}.$
|
[
"Just write $(2-x^2)^3=(3-x^3)^2$ i.e. $2x^6-6x^4-6x^3+12x^2+1=0$ "
] |
[
"origin:aops",
"Undergraduate Contests",
"2017 VTRMC",
"VTRMC"
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"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2017 VTRMC/1687621.json"
}
|
|
Evaluate $ \int _ { 0 } ^ { a } d x / ( 1 + \cos x + \sin x ) $ for $ - \pi / 2 < a < \pi $ . Use your answer to show that $ \int _ { 0 } ^ { \pi / 2 } d x / ( 1 + \cos x + \sin x ) = \ln 2 $ .
|
$-\dfrac{\pi}{4}<\dfrac{a}{2}<\dfrac{\pi}{2}\Longrightarrow -\dfrac{\pi}{4}<x\le 0$ or $0\le x<\dfrac{\pi}{2}$ .
Using the substitution $\tan \dfrac{x}{2}=t$ results: $dx=\dfrac{2dt}{1+t^2}, \cos x=\dfrac{1-t^2}{1+t^2}, \sin x=\dfrac{2t}{1+t^2}$ , $ \int _ { 0 } ^ { a } \dfrac{d x}{ 1 + \cos x + \sin x}=\int_0^{\tan \frac{a}{2}}\dfrac{2dt}{(1+t^2)\left(1+\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}\right)}=$ $=\int_0^{tan \frac{a}{2}}\dfrac{dt}{1+t}=\left[\ln(1+t)\right]_0^{\tan \frac{a}{2}}=\ln\left(1+\tan \dfrac{a}{2}\right)$ .
For $a=\dfrac{\pi}{2}$ results: $\int _ { 0 } ^ { \frac{\pi}{2} } \dfrac{d x}{ 1 + \cos x + \sin x}=\ln\left(1+\tan \dfrac{\pi}{4}\right)=\ln 2$ .
|
[
" $ \\int _ { 0 } ^ { \\pi / 2 } d x / ( 1 + \\cos x + \\sin x ) = \\ln 2 $ To integrate this just write $(1+cosx)=2cos^{2}\\frac{x}{2}$ and\n $sinx=2sin\\frac{x}{2}cos\\frac{x}{2}$ Then we get $I=\\int\\frac {1}{2cos\\frac{x}{2}\\bigg(cos\\frac{x}{2}+sin\\frac{x}{2}\\bigg)}$ Now just divide the numerator and denominator by $cos^2\\frac{x}{2}$ The integral becomes $I=\\int\\frac{\\frac{1}{2}sec^2\\frac{x}{2}}{\\bigg(1+tan\\frac{x}{2}\\bigg)}$ $I=ln\\bigg(1+tan\\frac{x}{2}\\bigg)+C$ Now put the limits $0 , \\frac{\\pi}{2}$ "
] |
[
"origin:aops",
"Undergraduate Contests",
"2017 VTRMC",
"VTRMC"
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{
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2017 VTRMC/1687825.json"
}
|
Let $ABC$ be a triangle and let $P$ be a point in its interior. Suppose $ \angle B A P = 10 ^ { \circ } , \angle A B P = 20 ^ { \circ } , \angle P C A = 30 ^ { \circ } $ and $ \angle P A C = 40 ^ { \circ } $ . Find $ \angle P B C $ .
|
Trigonometry does it. $\angle APB=150^{\circ}, \angle BPC=100^{\circ}$ . So, from sine law in triangle $ABP$ we have: $\frac{c}{\sin150^{\circ}}=\frac{AP}{\sin20^{\circ}}=\frac{BP}{\sin10^{\circ}}$ , so: $AP=2c\sin20^{\circ}$ and $BP=2c\sin10^{\circ}$ . From sine law in triangle $APC$ we have: $\frac{AP}{\sin30^{\circ}}=\frac{CP}{\sin40^{\circ}} \implies CP=2AP\sin40^{\circ}=4c\sin20^{\circ}\sin40^{\circ}=c(2\cos20^{\circ}-1)$ Let now $\angle PBC=x$ , so from sine law in triangle $BPC$ we have: $\frac{c(2\cos20^{\circ}-1)}{\sin x}=\frac{2c\sin10^{\circ}}{\sin(80^{\circ}-x)}=\frac{2c\sin10^{\circ}}{\sin(10^{\circ}+x)}$ .
After simple factorization you will get: $2\cos 20^{\circ}\cos(x+10^{\circ})=2\sin 10^{\circ}\sin x+\cos(x+10^{\circ})=\cos(x-10^{\circ})$ So, after factorization $\cos (30^{\circ}+x)=0 \implies 30^{\circ}+x=90^{\circ}$ , finally $x=60^{\circ}$ :)
|
[] |
[
"origin:aops",
"Undergraduate Contests",
"2017 VTRMC",
"VTRMC"
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"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2017 VTRMC/1687829.json"
}
|
Let $P$ be an interior point of a triangle of area $T$ . Through the point $P$ , draw lines parallel to the three sides, partitioning the triangle into three triangles and three parallelograms. Let $a$ , $b$ and $c$ be the areas of the three triangles. Prove that $ \sqrt { T } = \sqrt { a } + \sqrt { b } + \sqrt { c } $ .
|
Let the triangle be $ABC$ , and suppose the pints cut off from side $BC$ are $A_1, A_2$ and so on. Join $A$ to $P$ and extend it to meet $A_1A_2$ at $K$ ; then the triangles $PA_1A_2$ and $ABC$ being similar at ehomothetic w.r.t. point $K$ whence $\frac{a}{T} := \frac{[PA_1A_2]}{[ABC]} = (\frac{PL}{AL})^2$ whence $\sqrt{\frac{a}{T}} = \frac{PL}{AL}$ . Thus analogously constructing points $L:= BP \cap AC, M := CP \cap AB$ suffices to show that $\frac{PK}{AK}+ \frac{PL}{BL} + \frac{PM}{CM} =1$ , which is easy ( $\frac{PK}{AK} = \frac{[BPK]}{[BAK]} = \frac{[CPK]}{[CAK]} = \frac{[PBC]}{[ABC]}$ and sum cyclically).
|
[
"Let $AC \\parallel DE, BC \\parallel FG, AB \\parallel HI$ . Now, triangles $DFP, PIE, HPG, ABC$ are similar, so: $\\frac{DF}{PI}=\\frac{\\sqrt{a}}{\\sqrt{b}}$ and $\\frac{HP}{PI}=\\frac{\\sqrt{c}}{\\sqrt{b}}$ . Adding up these equations we get $\\frac{DF+HP}{PI}=\\frac{\\sqrt{a}+\\sqrt{c}}{\\sqrt{b}}$ Let's add $1$ on both sides. We have $\\frac{PI+DF+HP}{PI}=\\frac{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}{\\sqrt{b}}$ Now, $PI=FB, HP=AD$ , so $PI+DF+HP=FB+DF+AD=AB$ and $\\frac{AB}{PI}=\\frac{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}{\\sqrt{b}}$ But also $\\frac{AB}{PI}=\\frac{\\sqrt{T}}{\\sqrt{b}}$ $\\implies \\frac{\\sqrt{T}}{\\sqrt{b}}=\\frac{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}{\\sqrt{b}}$ . And finally, $\\sqrt{T}=\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$ "
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|
Let $ f ( x , y ) = ( x + y ) / 2 , g ( x , y ) = \sqrt { x y } , h ( x , y ) = 2 x y / ( x + y ) $ , and let $$ S = \{ ( a , b ) \in \mathrm { N } \times \mathrm { N } | a \neq b \text { and } f( a , b ) , g ( a , b ) , h ( a , b ) \in \mathrm { N } \} $$ where $\mathbb{N}$ denotes the positive integers. Find the minimum of $f$ over $S$ .
|
Given $(a,b) \in \mathbb{N}^2$ , let $d:=gcd(a,b)$ Then $g(a,b)=d\sqrt{xy} \in \mathbb{N} \implies \exists m, n \in \mathbb{N}$ s.t. $gcd(m, n)=1, a=m^2, b=n^2$ . So $h(x, y) \in \mathbb{Z} \implies (m^2+n^2)|d \iff d:= (m^2+n^2)\lambda$ for some $\lambda \in \mathbb{N}$ , whence $a= m^2(m^2+n^2)\lambda, b= n^2(m^2+n^2)\lambda$ , so we finally have to minimise $f(a,b)= \frac{\lambda (m^2+n^2)^2}{2} \in \mathbb{N}$ subject to $m, n \in \mathbb{N}; m \neq n; (m, n) =1$ . Thus, $m, n$ both can't be $1$ and should both be of same parity or $\lambda$ must be even. Hence for obtaining the minimal $f$ over $S$ the only two possible candidates are $(m, n, \lambda) = (1, 2, 2), (1, 3, 1)$ and the minimal $f$ ( $= 25$ ) is clearly obtained by the former (namely, for $a= 10, b= 40$ .
|
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|
Let $ f ( x ) \in \mathbb { Z } [ x ] $ be a polynomial with integer coefficients such that $ f ( 1 ) = - 1 , f ( 4 ) = 2 $ and $f ( 8 ) = 34 $ . Suppose $n\in\mathbb{Z}$ is an integer such that $ f ( n ) = n ^ { 2 } - 4 n - 18 $ . Determine all possible values for $n$ .
|
A slightly different solution:
Let $ g(x) = f(x) - x^2 + 4x + 18 = (x - n_i) h(x) $ , where $ n_i $ is a solution to the polynomial.
It's obvious that $h(x) \in \mathbb{Z}[x]$ , because $ x - n_i $ is monic.
And, $ \forall k \in \mathbb{Z}, g(k) \in \mathbb{Z} $ , because $ k^2 $ , $4k$ , and $18$ are all integers, and $ f(x) $ has integer coefficients.
So, $ (k - n_i) h(k) = g(k) $ . $ k - n_i $ and $ h(k) $ are integers, which means $ (k - n_i) \mid g(k) $ .
Plugging in 1, 4, and 8, we get
\begin{align*}
1 - n_i &\mid 20
4 - n_i &\mid 20
8 - n_i &\mid 20
\end{align*}
Thus, we are looking for the 3-tuple $(i, i +3, i + 4)$ , where $i \in \mathbb{Z}$ and each element divides 20.
There are two solutions: $(-5, -2, 2)$ and $(-2, 1, 5)$ . These two 3-tuples correspond to $ n = \fbox{6, 3}$
|
[
"From the formula $ f ( n ) = n ^ { 2 } - 4 n - 18 $ how we get $f(1)=-1$ ?",
"Let $g(x)=x^2-4x+2$ then $g(1)=f(1),g(4)=f(4),g(8)=f(8)$ $f(x)=(x-1)(x-4)(x-8)h(x)+g(x)$ $n^2-4n-18=f(n)=(n-1)(n-4)(n-8)h(n)+n^2-4n+2$ $-20=(n-1)(n-4)(n-8)h(n)$ $n-1$ is divisor of $20$ so $n=-19,-9,-4,-3,-1,0,2,3,5,6,11,21$ $n-4$ is divisor of $20$ so $n=-16,-6,-1,0,2,3,5,6,8,9,14,24$ So $n=-1,0,2,3,5,6$ $n-8$ is divisor of $20$ so $n=3,6$ Answer $n=3,6$ "
] |
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"answer_score": 36,
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"path": "Contest Collections/Undergraduate Contests/VTRMC/2017 VTRMC/1687834.json"
}
|
Find all pairs $(m, n)$ of nonnegative integers for which $ m ^ { 2 } + 2 \cdot 3 ^ { n } = m \left( 2 ^ { n + 1 } - 1 \right) $ .
|
[
"https://artofproblemsolving.com/community/c6h418640p2361999\nIMO Shortlist 2010 - Problem N2 "
] |
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|
It is known that $\int_1^2x^{-1}\arctan (1+x)\ dx = q\pi\ln(2)$ for some rational number $q.$ Determine $q.$ Here, $0\leq\arctan(x)<\frac{\pi}{2}$ for $0\leq x <\infty.$
|
You might be also interested in his "brother": https://artofproblemsolving.com/community/u380742h1735424p11265354 $$ I=\int_1^2 \frac{\arctan(1+x)}{x}dx $$ let us substitute $x=\frac{2} {t}\Rightarrow dx=-\frac{2} {t^2} dt $ $$ I=\int_1^2 \frac{\arctan\left(1+\frac{2 }{t}\right) } {\frac{2} {t}} \frac{2} {t^2} dt=\int_1^2 \frac{\arctan\left(1+\frac{2}{t}\right) } {t}dt $$ Adding with the intial integral gives: $$ 2 I=\int_1^2 \frac{\arctan(1+t)+\arctan\left(1+\frac{2 }{t}\right) } {t }dt $$ $$ \arctan(1+t)+\arctan\left(1+\frac{2 }{t}\right)=\pi - \arctan(1)=\frac{3 \pi} {4} $$ $$ I=\frac{3 \pi} {8}\int_1^2 \frac{dt} {t} =\frac38 \pi \ln 2 $$ Which gives $q=\frac38$
|
[] |
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|
Prove that there is no function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $f(f(n))=n+1.$ Here $\mathbb{N}$ is the positive integers $\{1,2,3,\dots\}.$
|
<blockquote>Prove that there is no function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $f(f(n))=n+1.$ Here $\mathbb{N}$ is the positive integers $\{1,2,3,\dots\}.$ </blockquote> $f(f(n))=n+1$ implies $f(n+1)=f(n)+1$ and so $f(n)=n+c$ And pluging this back in orginal equation, we get $c=\frac 12$ , impossible.
|
[] |
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|
Let $A, B \in M_6 (\mathbb{Z} )$ such that $A \equiv I \equiv B \text{ mod }3$ and $A^3 B^3 A^3 = B^3$ . Prove that $A = I$ . Here $M_6 (\mathbb{Z} )$ indicates the $6$ by $6$ matrices with integer entries, $I$ is the identity matrix, and $X \equiv Y \text{ mod }3$ means all entries of $X-Y$ are divisible by $3$ .
|
$\bullet$ Here $A,B\in M_n(\mathbb{Z})$ and the $U_i,V_i,W_i$ too. $A=I+3U_0,B=I+3V_0$ . Moreover $A^3=I+9U_1,B^3=I+9V_1$ . Then $A^3B^3A^3-B^3=0$ implies $2U_1+(9U_1^2+9U_1V_1+9V_1U_1+9^2U_1V_1U_1)=0$ . Thus $U_1=9U_2$ and therefore $2U_2+(9^2U_2^2+9U_2V_1+9V_1U_2+9^3U_2V_1U_2)=0$ . Thus $U_2=9U_3$ and so on...
-Note that, during step $i$ , each term in parentheses contains a $9U_i$ factor-
Finally, for every $k$ , $9U_1=9^2U_2=9^3U_3=\cdots=9^{k}U_k$ .
Let $\rho=\max_{i,j} |(A^3)_{i,j}|+1$ ; considering some $k$ s.t. $9^k>\rho$ , we deduce that $U_1=0$ and $A^3=I$ . $\bullet$ Thus $U_0+3U_0^2+3U_0^3=0$ . Therefore $U_0=3W_1$ , $W_1+9W_1^2+27W_1^3=0$ and $W_1=9W_2$ and so on...
We reason as above and obtain that $U_0=0$ and we are done.
|
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|
Let $m, n$ be integers such that $n \geq m \geq 1$ . Prove that $\frac{\text{gcd} (m,n)}{n} \binom{n}{m}$ is an integer. Here $\text{gcd}$ denotes greatest common divisor and $\binom{n}{m} = \frac{n!}{m!(n-m)!}$ denotes the binomial coefficient.
|
For all positive integers $m$ and $n$ , there exists some integers $x$ and $y$ such that $mx + ny = \mathrm{gcd}(m, n)$ . You can continue from here :D
|
[
"Consider separately the cases m=1, m=n, and n>m>1.\n\nIf m=1 or n then the given expression equals 1.\n\nIf n>m>1 then simplify the expression by breaking apart the factorials like n! = n*(n-1)!. Then using some reasoning involving divisibility and gcd, you can concluded the expression is an integer.",
"Isn't that Putnam 2000 B2?"
] |
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"Undergraduate Contests",
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"2018 VTRMC"
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|
For $n \in \mathbb{N}$ , let $a_n = \int _0 ^{1/\sqrt{n}} | 1 + e^{it} + e^{2it} + \dots + e^{nit} | \ dt$ . Determine whether the sequence $(a_n) = a_1, a_2, \dots$ is bounded.
|
<blockquote>For $n \in \mathbb{N}$ , let $a_n = \int _0 ^{1/\sqrt{n}} | 1 + e^{it} + e^{2it} + \dots + e^{nit} | \ dt$ . Determine whether the sequence $(a_n) = a_1, a_2, \dots$ is bounded.</blockquote>
Note that $\sin (x) \geqslant \frac{2}{\pi}x, \forall x \in [0, \frac{\pi}{2}]$ . Therefore, $a_n = \int _0 ^{1/\sqrt{n}} | 1 + e^{it} + e^{2it} + \dots + e^{nit} | \ dt = \int _0 ^{1/\sqrt{n}} | \frac{1-e^{(n+1)it}}{1-e^{it}} | dt = $ $\int _0 ^{1/\sqrt{n}} | \frac{\sin( \frac{(n+1)t}{2})}{\sin(\frac{t}{2} )} | dt \leqslant \pi \int _0 ^{1/\sqrt{n}} \frac{|\sin( \frac{(n+1)t}{2}) |}{t} dt = $ $\frac{2\pi}{(n+1)} \int _0 ^{(n+1)/2\sqrt{n}} \frac{|\sin u|}{u} du \leqslant \frac{2\pi}{(n+1)} \int _0 ^{n \pi} \frac{|\sin u|}{u} du = $ $\frac{2\pi}{(n+1)} \sum_{i=0}^{n-1} \int _{(i-1) \pi}^{i \pi} \frac{|\sin u|}{u} du \leqslant \frac{2n\pi}{(n+1)} \int _{0}^{\pi} \frac{|\sin u|}{u} du < \infty$ .
The answer is $\boxed{\textrm{bounded}}$ .
Footnote. I think if we define $b_n = \int _0 ^1 | 1 + e^{it} + e^{2it} + \dots + e^{nit} | \ dt$ , $(b_n)_{n \in \mathbb{N}}$ would also be bounded.
|
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|
For $n \in \mathbb{N}$ , define $a_n = \frac{1 + 1/3 + 1/5 + \dots + 1/(2n-1)}{n+1}$ and $b_n = \frac{1/2 + 1/4 + 1/6 + \dots + 1/(2n)}{n}$ . Find the maximum and minimum of $a_n - b_n$ for $1 \leq n \leq 999$ .
|
[
"Quick solution: See the below.\n<details><summary>Click to expand</summary>[https://personal.math.vt.edu/plinnell/Vtregional/S18/index.html](https://personal.math.vt.edu/plinnell/Vtregional/S18/index.html)\nIt was shocking that I didn’t have to find generalized term.</details>"
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|
A continuous function $f : [a,b] \to [a,b]$ is called piecewise monotone if $[a, b]$ can be subdivided into finitely many subintervals $$ I_1 = [c_0,c_1], I_2 = [c_1,c_2], \dots , I_\ell = [ c_{\ell - 1},c_\ell ] $$ such that $f$ restricted to each interval $I_j$ is strictly monotone, either increasing or decreasing. Here we are assuming that $a = c_0 < c_1 < \cdots < c_{\ell - 1} < c_\ell = b$ . We are also assuming that each $I_j$ is a maximal interval on which $f$ is strictly monotone. Such a maximal interval is called a lap of the function $f$ , and the number $\ell = \ell (f)$ of distinct laps is called the lap number of $f$ . If $f : [a,b] \to [a,b]$ is a continuous piecewise-monotone function, show that the sequence $( \sqrt[n]{\ell (f^n )})$ converges; here $f^n$ means $f$ composed with itself $n$ -times, so $f^2 (x) = f(f(x))$ etc.
|
**Lemma:** For all piecewise monotone $f, g \colon [a,b] \to [a,b]$ we have $\ell ( f \circ g ) \le \ell ( f ) \ell ( g )$ .**Proof:** Let $I_1 = [c_0,c_1], I_2 = [c_1,c_2], \dots , I_\ell = [ c_{\ell - 1},c_\ell ]$ with $\ell = \ell ( g )$ be such that on each interval $I_i$ we have that $g \colon I_i \to [a, b]$ is strictly monotone. Then $f \circ g$ can have at most $\ell ( f ) $ laps on each interval $I_i$ , and thus at most $\ell(f) \ell(g)$ overall.
Now let $a_n = \log \ell ( f^n )$ . The above lemma tells us that $$ a_{m+n} = \log \ell ( f^m \circ f^n ) \le \log ( \ell ( f^m ) \ell ( f^n ) ) = a_m + a_n. $$ So $(a_n)$ is subadditive. Furthermore, $a_n / n$ is bounded from below (by $0$ ). So by a well-known lemma/exercise, $a_n / n$ converges, and thus $e^{ a_n / n } = \sqrt[n]{\ell (f^n )}$ converges as well.
|
[] |
[
"origin:aops",
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|
Give all possible representations of $2022$ as a sum of at least two consecutive positive integers and prove that these are the only representations.
|
<blockquote>Give all possible representations of $2022$ as a sum of at least two consecutive positive integers and prove that these are the only representations.</blockquote>
If $2022$ is the sum of the integers between $m$ and $n$ (inclusive), we have $$ 2022=\sum \limits _{k=m}^nk=\frac{(n+m)(n+1-m)}{2}, $$ so $$ 4044=(n+m)(n+1-m). $$ Now $m\geq 1$ implies $2m>1$ , which implies $n+m>n+1-m$ . Now find the divisors of $4044$ and solve the $6$ systems of equations for $n+m$ and $n+1-m$ to find that the answers are $(m,n)=(673,675)$ , $(m,n)=(504,507)$ and $(m,n)=(163,174)$ .
|
[] |
[
"origin:aops",
"Undergraduate Contests",
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"2022 VTRMC"
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|
Let $A$ and $B$ be the two foci of an ellipse and let $P$ be a point on this ellipse. Prove that the focal radii of $P$ (that is, the segments $\overline{AP}$ and $\overline{BP}$ ) form equal angles with the tangent to the ellipse at $P$ .
|
*For the sake of completing the solution:*
Wlog work with the standard ellipse i.e $\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$ and assume that $a \geq b$ .
The foci are given by $S=(ae,0)$ and $S'=(-ae,0)$ where $e$ (eccentricity) satisfies $b^2=a^2\left(1-e^2\right)$ .
Parametrize the points on the ellipse as $P=(a \text{cos} \theta, b \text{sin} \theta)$ .
The slope of the tangent at point $P$ on the ellipse is given by $\frac{\text{d}y}{\text{d}x}=\frac{-b \text{cot}\theta}{a}=m$ .
The slopes of the lines $SP'$ and $SP$ are given by $\left(\frac{b \text{sin} \theta} {a \left(\text{cos}\theta+e\right)}, \frac{b \text{sin} \theta} {a \left(\text{cos}\theta-e\right)}\right)=(m_1,m_2)$ respectively.
A not-so-ugly check shows that $\left | \frac{m-m_1}{1+mm_1} \right| = \left |\frac{m-m_2}{1+mm_2} \right |$ thus showing that the two angles are equal.
|
[
"<blockquote>Let $A$ and $B$ be the two foci of an ellipse and let $P$ be a point on this ellipse. Prove that the focal radii of $P$ (that is, the segments $\\overline{AP}$ and $\\overline{BP}$ ) form equal angles with the tangent to the ellipse at $P$ .</blockquote>\n\nMore general [equality](https://math.stackexchange.com/questions/323574/equal-angles-formed-by-the-tangent-lines-to-an-ellipse-and-the-lines-through-the).",
"[url=https://davidaltizio.web.illinois.edu/geom-conics.pdf]Good handout (with that proof in it)."
] |
[
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"2022 VTRMC"
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|
Find all positive integers $a, b, c, d,$ and $n$ satisfying $n^a + n^b + n^c = n^d$ and prove that these are the only such solutions.
|
<blockquote>Find all positive integers $a, b, c, d,$ and $n$ satisfying $n^a + n^b + n^c = n^d$ and prove that these are the only such solutions.</blockquote>
Clearly $n>1$ . WLOG let $a\leq b\leq c$ and write $b=a+x$ and $c=b+y=a+x+y$ , with $x,y\in \mathbb{N}_0$ . Then $n^d=n^a(1+n^x+n^{x+y})$ . If $x\in \mathbb{N}$ , we have that $\gcd (1+n^x+n^{x+y},n)=\gcd (1,n)=1$ , contradiction because $1+n^x+n^{x+y}\mid n$ and is greater than $1$ . Hence $x=0$ , so $n^d=n^a(2+n^y)$ .
Assume first $y\in \mathbb{N}$ .
If $n$ is odd then $\gcd (2+n^y,n)=\gcd (2,n)=1$ , contradiction because $2+n^y\mid n^d$ and is greater than $1$ . Hence $n$ is even, say $n=2k$ with $k\in \mathbb{N}$ . Then $2^dk^d=2^ak^a(2+2^yk^y)=2^{a+1}k^a(1+2^{y-1}k^y)$ . If $y=1$ , we have $2^dk^d=2^{a+1}k^a(1+k)$ , but $\gcd (1+k,k)=\gcd (1,k)=1$ , so $1+k\mid 2$ , implying that $k=1$ , so $2^d=2^{a+2}$ , hence $d=a+2$ , giving $(a,b,c,d,n)=(a,a,a+1,a+2,2)$ . If $y>1$ , we have that $1+2^{y-1}k^y$ is relatively prime with $2$ and $k$ , which is a contradiction because it divides $2^dk^d$ .
Assume now $y=0$ .
Then $n^d=n^a+n^a+n^a=3n^a$ . Let $k$ be the exponent of $3$ in the prime factorization of $n$ , then $kd$ is the exponent of $3$ in the prime factorization of $n^d$ and $1+ka$ is the exponent of $3$ in the prime factorization of $3n^a$ , hence $kd=1+ka$ , so $k(d-a)=1$ , implying that $k\mid 1$ , so $k=1$ , hence $d-a=1$ , so $d=a+1$ , implying that $n^{a+1}=3n^a$ , so $n=3$ . Hence $(a,b,c,d,n)=(a,a,a,a+1,3)$ .
|
[
"Assume wlog that $a \\le b \\le c$ and write $b-a=q$ , $c-a=r$ , $d-a=s$ . Then $1 + n^q + n^r = n^s$ .\nSince $s > r$ we have $n^s \\le 3 n^r \\le 3 n ^{s-1}$ and so $n \\le 3$ . Also clearly $n>1$ .\n\nIf $n=3$ then we must have $q=r=0$ , since otherwise $1+3^q+3^r$ is not divisible by $3$ . So in this case we get the solutions $(a, a, a, a+1, 3)$ and permutations.\n\nIf $n=2$ we have $q=0$ and $r>0$ , since otherwise $1 + 2^q + 2^r$ is not divisible by $2$ . Then $2^s - 2^r = 2$ , which is only possible for $s=2$ , $r=1$ . Hence in this case we get the solutions $(a, a, a+1, a+2, 2)$ and permutations.\n(Thinking of numbers in base $n$ helps here.)"
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[
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|
Calculate the exact value of the series $\sum _{n=2} ^\infty \log (n^3 +1) - \log (n^3 - 1)$ and provide justification.
|
<blockquote><blockquote>Calculate the exact value of the series $\sum _{n=2} ^\infty \log (n^3 +1) - \log (n^3 - 1)$ and provide justification.</blockquote>
We have\begin{align*}\sum \limits _{n=2}^\infty \log (n^3+1)-\log (n^3-1) & =\lim \limits _{m\to \infty}\sum \limits _{n=2}^m \log (n^3+1)-\log (n^3-1)
& =\lim \limits _{m\to \infty}\sum \limits _{n=2}^m\log \frac{n^3+1}{n^3-1}
& =\lim \limits _{m\to \infty}\sum \limits _{n=2}^m\log \left (1+\frac{2}{n^3-1}\right )
& =\lim \limits _{m\to \infty}\log \left (\prod \limits _{n=2}^m\left (1+\frac{2}{n^3-1}\right )\right )
& =\lim \limits _{m\to \infty}\log \frac{3m(m+1)}{2(m^2+m+1)}
& =\log \left (\lim \limits _{m\to \infty}\frac{3m(m+1)}{2(m^2+m+1)}\right )
& =\log \frac{3}{2},
\end{align*}where we have used the formula $$ \prod \limits _{n=2}^m\left (1+\frac{2}{n^3-1}\right )=\frac{3m(m+1)}{2(m^2+m+1)}, $$ which is easily proved by induction.</blockquote>
Just simple with telescoping product: $L=\underset{m\to \infty }{\mathop{\lim }}\,\sum\limits_{n=2}^{m}{\log }({{n}^{3}}+1)-\log ({{n}^{3}}-1)=\underset{m\to \infty }{\mathop{\lim }}\,\sum\limits_{n=2}^{m}{\log }\frac{{{n}^{3}}+1}{{{n}^{3}}-1}=\underset{m\to \infty }{\mathop{\lim }}\,\log \prod\limits_{n=2}^{m}{\frac{{{n}^{3}}+1}{{{n}^{3}}-1}=}$ $=\underset{m\to \infty }{\mathop{\lim }}\,\log \prod\limits_{n=2}^{m}{\frac{n+1}{n-1}}\prod\limits_{n=2}^{m}{\frac{{{n}^{2}}-n+1}{{{(n+1)}^{2}}-(n+1)+1}}=\underset{m\to \infty }{\mathop{\lim }}\,\log \frac{m(m+1)}{2}\frac{3}{{{m}^{2}}+m+1}$ , etc
|
[
"<blockquote>Calculate the exact value of the series $\\sum _{n=2} ^\\infty \\log (n^3 +1) - \\log (n^3 - 1)$ and provide justification.</blockquote>\n\nWe have\\begin{align*}\\sum \\limits _{n=2}^\\infty \\log (n^3+1)-\\log (n^3-1) & =\\lim \\limits _{m\\to \\infty}\\sum \\limits _{n=2}^m \\log (n^3+1)-\\log (n^3-1) \n& =\\lim \\limits _{m\\to \\infty}\\sum \\limits _{n=2}^m\\log \\frac{n^3+1}{n^3-1} \n& =\\lim \\limits _{m\\to \\infty}\\sum \\limits _{n=2}^m\\log \\left (1+\\frac{2}{n^3-1}\\right ) \n& =\\lim \\limits _{m\\to \\infty}\\log \\left (\\prod \\limits _{n=2}^m\\left (1+\\frac{2}{n^3-1}\\right )\\right ) \n& =\\lim \\limits _{m\\to \\infty}\\log \\frac{3m(m+1)}{2(m^2+m+1)} \n& =\\log \\left (\\lim \\limits _{m\\to \\infty}\\frac{3m(m+1)}{2(m^2+m+1)}\\right ) \n& =\\log \\frac{3}{2},\n\\end{align*}where we have used the formula $$ \\prod \\limits _{n=2}^m\\left (1+\\frac{2}{n^3-1}\\right )=\\frac{3m(m+1)}{2(m^2+m+1)}, $$ which is easily proved by induction.",
"Heavily related with [this problem on Fmat](http://www.fmat.cl/index.php?showtopic=36536&st=0) (post #2 of the link is very shorter and telescopic)\n\nRegards\nClaudio",
"Also see [ 1977 Putnam B1](https://artofproblemsolving.com/community/c7h2818685p24900290).",
"Is there similar with exponent 2? 4? 5? 6? 7? …"
] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] |
{
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988736.json"
}
|
Let $A$ be an invertible $n \times n$ matrix with complex entries. Suppose that for each positive integer $m$ , there exists a positive integer $k_m$ and an $n \times n$ invertible matrix $B_m$ such that $A^{k_m m} = B_m A B_m ^{-1}$ . Show that all eigenvalues of $A$ are equal to $1$ .
|
<blockquote>Let $A$ be an invertible $n \times n$ matrix with complex entries. Suppose that for each positive integer $m$ , there exists a positive integer $k_m$ and an $n \times n$ invertible matrix $B_m$ such that $A^{k_m m} = B_m A B_m ^{-1}$ . Show that all eigenvalues of $A$ are equal to $1$ .</blockquote>
Let $\lambda$ be an arbitrary eigenvalue of $A$ and let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$ , then $$ A^{k_mm}B_mv=B_mAB_m^{-1}B_mv=B_mAv=B_m\lambda v=\lambda B_mv, $$ so $B_mv$ is an eigenvector of $A$ with eigenvalue $\lambda$ . Hence $\lambda$ is the $k_mm$ -th power of some eigenvalue of $A$ . This is true for all $m\in \mathbb{N}$ .
Now construct a sequence $(m_n)_{n\in \mathbb{N}}$ satisfying $m_{n+1}>k_{m_n}m_n$ for all $n\in \mathbb{N}$ (from now on, we ignore the original meaning of $n$ in the problem statement). Then this sequence is infinite and for each $n\in \mathbb{N}$ we have that $\lambda$ is a $k_{m_n}m_n$ -th power of some eigenvalue of $A$ . Since $A$ has only a finite number of eigenvalues we have that there exists $a,b\in \mathbb{N}$ , with $a<b$ , such that $\lambda$ is a $k_{m_a}m_a$ -th power and a $k_{m_b}m_b$ -th power of the same eigenvalue of $A$ , which we will call $\alpha$ , observe that $\alpha \neq 0$ because $A$ is invertible. Then $\alpha ^{k_{m_a}m_a}=\lambda =\alpha ^{k_{m_b}m_b}$ , so $1=\alpha ^{k_{m_b}m_b-k_{m_a}m_a}$ , hence $\alpha =1$ , because $k_{m_b}m_b-k_{m_a}m_a>0$ . Hence $\lambda =1$ .
Since $\lambda$ was arbitrary, we conclude that all eigenvalues of $A$ are equal to $1$ .
|
[
"Then $A=I+N$ , where $N$ is nilpotent. Yet, we cannot do better\nIndeed one has many $>0$ integers $k$ s.t. $A^k$ is similar to $A$ . But $(I+N)^k=I+kN+\\cdots$ is similar to $I+kN$ which is similar to $A$ .",
"Two things.\n<blockquote>\nLet $\\lambda$ be an arbitrary eigenvalue of $A$ and let $v$ be an eigenvector of $A$ with eigenvalue $\\lambda$ , then $$ A^{k_mm}B_mv=B_mAB_m^{-1}B_mv=B_mAv=B_m\\lambda v=\\lambda B_mv, $$ so $B_mv$ is an eigenvector of $A$ with eigenvalue $\\lambda$ .</blockquote>\nHere $B_mv$ is an eigenvector of $A^{k_mm}$ .\n<blockquote>\n Then $\\alpha ^{k_{m_a}m_a}=\\lambda =\\alpha ^{k_{m_b}m_b}$ , so $1=\\alpha ^{k_{m_b}m_b-k_{m_a}m_a}$ , hence $\\alpha =1$ , because $k_{m_b}m_b-k_{m_a}m_a>0$ .\n</blockquote>\nThis doesn't imply $\\alpha =1$ , just that $\\alpha$ is a root of unity.",
"<blockquote>Two things.\n<blockquote>\nLet $\\lambda$ be an arbitrary eigenvalue of $A$ and let $v$ be an eigenvector of $A$ with eigenvalue $\\lambda$ , then $$ A^{k_mm}B_mv=B_mAB_m^{-1}B_mv=B_mAv=B_m\\lambda v=\\lambda B_mv, $$ so $B_mv$ is an eigenvector of $A$ with eigenvalue $\\lambda$ .</blockquote>\nHere $B_mv$ is an eigenvector of $A^{k_mm}$ .\n<blockquote>\n Then $\\alpha ^{k_{m_a}m_a}=\\lambda =\\alpha ^{k_{m_b}m_b}$ , so $1=\\alpha ^{k_{m_b}m_b-k_{m_a}m_a}$ , hence $\\alpha =1$ , because $k_{m_b}m_b-k_{m_a}m_a>0$ .\n</blockquote>\nThis doesn't imply $\\alpha =1$ , just that $\\alpha$ is a root of unity.</blockquote>\n\nThe first one was a typo. About the second one, darn, I totally missed that.",
"Let $spectrum(A)=(\\lambda_i)$ . Since $\\lambda_i$ is a root of unity, let $r_i$ be its order. \nLet $lcm(r_1,\\cdots,r_n)=r$ . Note that ${\\lambda_i}^r=1$ .\nWe know that $A^{k_rr}$ and $A$ are similar; then, up to order, $(\\lambda_i)=({\\lambda_i}^{k_rr})=(1,\\cdots,1)$ and we are done."
] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] |
{
"answer_score": 98,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988738.json"
}
|
Let $f : \mathbb{R} \to \mathbb{R}$ be a function whose second derivative is continuous. Suppose that $f$ and $f''$ are bounded. Show that $f'$ is also bounded.
|
Suppose $|f(x)|,|f''(x)|\le M$ for all $x$ . Let $k$ be a positive number and suppose that $f'(x)\ge k$ for some $x$ .
For $a\ge x$ we have $ |f'(a)-f'(x)|=|\int_x^a f''(t) dt|\le M(a-x)$ , hence $f'(a)\ge f'(x)-M(a-x)\ge k-M(a-x)$ .
Hence $f'(a)\ge k/2$ on the interval $[x,x+{k\over {2M}}]$ .
Hence $2M\ge |f(x+{k\over {2M}})-f(x)|=|\int_x^{x+{k\over {2M}}}f'(t)dt|\ge {k\over {2M}}\cdot {k\over 2}={{k^2}\over {4M}}\implies k^2\le 8M^2$ . Choosing the $k$ in the begining with $k^2>8M^2$ we get a contradiction. Hence $f'$ is bounded from above. In fact we have $f'(x)\le 2M \sqrt{2}$ for all $x$ .
The same argument on $-f$ shows that $|f'(x)|\le 2M\sqrt{2}$ for all $x$ .
|
[
"By Taylor up to second order: $$ f(t+1)=f(t)+f'(t)+f''(t+u)/2 $$ some $u \\in ]0,1[$ Therefore $$ f'(t)=f(t+1)-f(t)-f''(t+u)/2 $$ Now if $|f(x)|\\leq a$ and $|f''(x)|\\leq b$ , $$ |f'(t)|=|f(t+1)-f(t)-f''(t+u)/2|\\leq 2a+b/2 $$ and hence is bounded\n\nRegards\nClaudio.",
"<blockquote>\n\nNow if $|f(x)|\\leq a$ and $|f''(x)|\\leq b$ , $$ |f'(t)|=|f(t+1)-f(t)-f''(t+u)/2|\\leq 2a+b/2 $$ and hence is bounded\n\nRegards\nClaudio.</blockquote>\n\nYou mean this?I see there is a typo\n",
"Where do you see any typo?",
"<blockquote>, $$ |f'(t)|=|f'(t+1)-f(t)-f''(t+u)/2|\\leq 2a+b/2 $$ and hence is bounded\n\nRegards\nClaudio.</blockquote>\n\n",
"<blockquote><blockquote>, $$ |f'(t)|=|f'(t+1)-f(t)-f''(t+u)/2|\\leq 2a+b/2 $$ and hence is bounded\n\nRegards\nClaudio.</blockquote></blockquote>\n\nyes, a silly typo. Thanks. "
] |
[
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] |
{
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988739.json"
}
|
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