Polygons defined by lengths of sides
$begingroup$
Can a polygon be defined by the lengths of its sides?
In other words, if given the lengths of the sides of a polygon, is there a way to figure out what polygon has those lengths and show that there is only one such polygon?
I know that this is true for triangles because of the SSS Congruence Theorem. If two triangles can be proven congruent by having the same sides, there must be only one way to make a triangle if given side lengths.
It seems like this should work for every type of polygon, but it would be great if anyone could provide proof of this.
geometry polygons
asked Dec 10 '18 at 5:33
6367763677
1
$endgroup$
add a comment |
$begingroup$
Can a polygon be defined by the lengths of its sides?
In other words, if given the lengths of the sides of a polygon, is there a way to figure out what polygon has those lengths and show that there is only one such polygon?
I know that this is true for triangles because of the SSS Congruence Theorem. If two triangles can be proven congruent by having the same sides, there must be only one way to make a triangle if given side lengths.
It seems like this should work for every type of polygon, but it would be great if anyone could provide proof of this.
geometry polygons
asked Dec 10 '18 at 5:33
6367763677
1
$endgroup$
2
$begingroup$
Since the case of triangles is already understood (triangles are "rigid"), I'd suggest considering the next simplest case, that of quadrilaterals.
$endgroup$
– hardmath
Dec 10 '18 at 5:36
1
$begingroup$
You can't do that even for a quadrilateral with all sides of length $1$. However, if we restrict ourself to cyclic polygons, there is at most one cyclic polygon with given side lengths in given order.
$endgroup$
– achille hui
Dec 10 '18 at 5:41
add a comment |
$begingroup$
Can a polygon be defined by the lengths of its sides?
In other words, if given the lengths of the sides of a polygon, is there a way to figure out what polygon has those lengths and show that there is only one such polygon?
I know that this is true for triangles because of the SSS Congruence Theorem. If two triangles can be proven congruent by having the same sides, there must be only one way to make a triangle if given side lengths.
It seems like this should work for every type of polygon, but it would be great if anyone could provide proof of this.
geometry polygons
asked Dec 10 '18 at 5:33
6367763677
1
$endgroup$
Can a polygon be defined by the lengths of its sides?
In other words, if given the lengths of the sides of a polygon, is there a way to figure out what polygon has those lengths and show that there is only one such polygon?
I know that this is true for triangles because of the SSS Congruence Theorem. If two triangles can be proven congruent by having the same sides, there must be only one way to make a triangle if given side lengths.
It seems like this should work for every type of polygon, but it would be great if anyone could provide proof of this.
geometry polygons
geometry polygons
asked Dec 10 '18 at 5:33
6367763677
1
asked Dec 10 '18 at 5:33
6367763677
1
asked Dec 10 '18 at 5:33
6367763677
1
asked Dec 10 '18 at 5:33
6367763677
1
asked Dec 10 '18 at 5:33
6367763677
1
1
2
$begingroup$
Since the case of triangles is already understood (triangles are "rigid"), I'd suggest considering the next simplest case, that of quadrilaterals.
$endgroup$
– hardmath
Dec 10 '18 at 5:36
1
$begingroup$
You can't do that even for a quadrilateral with all sides of length $1$. However, if we restrict ourself to cyclic polygons, there is at most one cyclic polygon with given side lengths in given order.
$endgroup$
– achille hui
Dec 10 '18 at 5:41
add a comment |
2
$begingroup$
Since the case of triangles is already understood (triangles are "rigid"), I'd suggest considering the next simplest case, that of quadrilaterals.
$endgroup$
– hardmath
Dec 10 '18 at 5:36
1
$begingroup$
You can't do that even for a quadrilateral with all sides of length $1$. However, if we restrict ourself to cyclic polygons, there is at most one cyclic polygon with given side lengths in given order.
$endgroup$
– achille hui
Dec 10 '18 at 5:41
2
2
$begingroup$
Since the case of triangles is already understood (triangles are "rigid"), I'd suggest considering the next simplest case, that of quadrilaterals.
$endgroup$
– hardmath
Dec 10 '18 at 5:36
$begingroup$
Since the case of triangles is already understood (triangles are "rigid"), I'd suggest considering the next simplest case, that of quadrilaterals.
$endgroup$
– hardmath
Dec 10 '18 at 5:36
1
1
$begingroup$
You can't do that even for a quadrilateral with all sides of length $1$. However, if we restrict ourself to cyclic polygons, there is at most one cyclic polygon with given side lengths in given order.
$endgroup$
– achille hui
Dec 10 '18 at 5:41
$begingroup$
You can't do that even for a quadrilateral with all sides of length $1$. However, if we restrict ourself to cyclic polygons, there is at most one cyclic polygon with given side lengths in given order.
$endgroup$
– achille hui
Dec 10 '18 at 5:41
add a comment |
0
active
oldest
votes
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%2f3033496%2fpolygons-defined-by-lengths-of-sides%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3033496%2fpolygons-defined-by-lengths-of-sides%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
2
$begingroup$
Since the case of triangles is already understood (triangles are "rigid"), I'd suggest considering the next simplest case, that of quadrilaterals.
$endgroup$
– hardmath
Dec 10 '18 at 5:36
1
$begingroup$
You can't do that even for a quadrilateral with all sides of length $1$. However, if we restrict ourself to cyclic polygons, there is at most one cyclic polygon with given side lengths in given order.
$endgroup$
– achille hui
Dec 10 '18 at 5:41