Find number of triangles with integral sides and side lengths ≤ 2n
$begingroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
8241618
asked Dec 17 '18 at 12:49
tanyatanya
105
$endgroup$
add a comment |
$begingroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
8241618
asked Dec 17 '18 at 12:49
tanyatanya
105
$endgroup$
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
add a comment |
$begingroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
8241618
asked Dec 17 '18 at 12:49
tanyatanya
105
$endgroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
8241618
asked Dec 17 '18 at 12:49
tanyatanya
105
edited Dec 17 '18 at 14:43
Word Shallow
8241618
asked Dec 17 '18 at 12:49
tanyatanya
105
edited Dec 17 '18 at 14:43
Word Shallow
8241618
edited Dec 17 '18 at 14:43
Word Shallow
8241618
edited Dec 17 '18 at 14:43
Word Shallow
8241618
8241618
asked Dec 17 '18 at 12:49
tanyatanya
105
asked Dec 17 '18 at 12:49
tanyatanya
105
asked Dec 17 '18 at 12:49
tanyatanya
105
105
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
add a comment |
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043894%2ffind-number-of-triangles-with-integral-sides-and-side-lengths-%25e2%2589%25a4-2n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
$endgroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
13.2k1233
13.2k1233
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043894%2ffind-number-of-triangles-with-integral-sides-and-side-lengths-%25e2%2589%25a4-2n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55