Two Subsets of Squares of Reciprocals of Primes with Equal Sums












2












$begingroup$


Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago


















2












$begingroup$


Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago
















2












2








2





$begingroup$


Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers







share|cite|improve this question











$endgroup$




Let $$A={frac{1}{2^2},frac{1}{3^2},frac{1}{5^2},...}$$ be the set of squares of the reciprocals of prime numbers.



We have
$$sum_{xin A}x < infty$$
Do there exist $B subset A$, $C subset A$, $B cap C = emptyset$, such that
$$sum_{xin B}x = sum_{xin C}x ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions, as described in Wikipedia.






number-theory prime-numbers




number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago







co.sine

















asked 13 hours ago









co.sineco.sine

372112




372112








  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago
















  • 1




    $begingroup$
    The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
    $endgroup$
    – Servaes
    12 hours ago










1




1




$begingroup$
The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
$endgroup$
– Servaes
12 hours ago






$begingroup$
The answer is pathologically yes; take $B=C=varnothing$. These are the only finite sets with this property. There are infinitely many solutions with $B$ and $C$ infinite.
$endgroup$
– Servaes
12 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago



















3












$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3113041%2ftwo-subsets-of-squares-of-reciprocals-of-primes-with-equal-sums%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago
















4












$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago














4












4








4





$begingroup$

The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$






share|cite|improve this answer











$endgroup$



The answer is negative. Let $C=sum_{p}frac{1}{p^2}$. This constant is approximately $frac{1}{2}logfrac{5}{2}$ due to Euler's product, leading to:
$$ sum_{p}frac{1}{p^2}approx sum_pfrac{1}{2}logleft(frac{1+frac{1}{p^2}}{1-frac{1}{p^2}}right)=frac{1}{2}logfrac{zeta(2)^2}{zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $Ucup V$ with $Ucap V=emptyset$ and $U,Vneqemptyset$.
We may assume without loss of generality that $2in U$, hence
$$ sum_{pin U}frac{1}{p^2}geq frac{1}{4} > frac{1}{2}sum_{pin Ucup V}frac{1}{p^2}=frac{C}{2} $$
which contradicts
$$ sum_{pin U}frac{1}{p^2}=sum_{pin V}frac{1}{p^2}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 11 hours ago

























answered 12 hours ago









Jack D'AurizioJack D'Aurizio

290k33282662




290k33282662












  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago


















  • $begingroup$
    @co.sine: actually I was completely wrong before, I have fixed the answer.
    $endgroup$
    – Jack D'Aurizio
    11 hours ago










  • $begingroup$
    BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
    $endgroup$
    – co.sine
    11 hours ago










  • $begingroup$
    What about the case $U cup V subsetneq {text{primes}}$?
    $endgroup$
    – co.sine
    11 hours ago












  • $begingroup$
    I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
    $endgroup$
    – W-t-P
    11 hours ago










  • $begingroup$
    For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
    $endgroup$
    – qwr
    8 hours ago
















$begingroup$
@co.sine: actually I was completely wrong before, I have fixed the answer.
$endgroup$
– Jack D'Aurizio
11 hours ago




$begingroup$
@co.sine: actually I was completely wrong before, I have fixed the answer.
$endgroup$
– Jack D'Aurizio
11 hours ago












$begingroup$
BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
$endgroup$
– co.sine
11 hours ago




$begingroup$
BRILLIANT, Jack D'Aurizio. For the constant $C$: mathoverflow.net/a/53444/102549
$endgroup$
– co.sine
11 hours ago












$begingroup$
What about the case $U cup V subsetneq {text{primes}}$?
$endgroup$
– co.sine
11 hours ago






$begingroup$
What about the case $U cup V subsetneq {text{primes}}$?
$endgroup$
– co.sine
11 hours ago














$begingroup$
I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
$endgroup$
– W-t-P
11 hours ago




$begingroup$
I'd rather lean to believe your original answer: we do not have to partition the set of all primes, but only want to find two disjoint subsets thereof.
$endgroup$
– W-t-P
11 hours ago












$begingroup$
For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
$endgroup$
– qwr
8 hours ago




$begingroup$
For $U$, the sum must be at least $1/4$. $C<1/2$, so the maximum sum for $V$ must be less than $1/4$.
$endgroup$
– qwr
8 hours ago











3












$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago
















3












$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago














3












3








3





$begingroup$

Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.






share|cite|improve this answer











$endgroup$



Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2in B$, and by $Q$ the product of all primes $p$ with $1/p^2in C$. Then the LHS of $sum_{xin B}x=sum_{xin C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $prod_{pin P} p^2$, while the RHS has the denominator equal to $prod_{pin Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 12 hours ago

























answered 12 hours ago









W-t-PW-t-P

1,102610




1,102610












  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago


















  • $begingroup$
    What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
    $endgroup$
    – Servaes
    12 hours ago












  • $begingroup$
    @Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
    $endgroup$
    – W-t-P
    12 hours ago
















$begingroup$
What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
$endgroup$
– Servaes
12 hours ago






$begingroup$
What if $B$ and $C$ are empty or infinite? The latter is the interesting part of the question really, as your argument shows.
$endgroup$
– Servaes
12 hours ago














$begingroup$
@Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
$endgroup$
– W-t-P
12 hours ago




$begingroup$
@Servaes: I believe that in this case equality is possible, but this would be a little technical to explain in details.
$endgroup$
– W-t-P
12 hours ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3113041%2ftwo-subsets-of-squares-of-reciprocals-of-primes-with-equal-sums%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

"Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness.
Read more

Alcedinidae

Image
Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
Read more

RAC Tourist Trophy

Image
Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
Read more