Single-Line Equation for Equilateral Triangle
$begingroup$
Is it possible to come up with a single-line equation in rectangular coordinates for an equilateral triangle with circumradius $R$, positioned symmetrical about the $y$-axis, as shown in the diagram below?
analytic-geometry absolute-value polygons
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
$endgroup$
add a comment |
$begingroup$
Is it possible to come up with a single-line equation in rectangular coordinates for an equilateral triangle with circumradius $R$, positioned symmetrical about the $y$-axis, as shown in the diagram below?
analytic-geometry absolute-value polygons
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
$endgroup$
$begingroup$
What do you mean by a “single-line equation?” This might be the sort of thing you’re looking for.
$endgroup$
– amd
Jan 8 at 19:36
$begingroup$
@amd single-line equation meaning not a piecewise equation. Thanks for the suggested link - I did come across it earlier but here I'm looking for the solution for a specific case, and in rectangular coordinates.
$endgroup$
– hypergeometric
Jan 9 at 14:20
$begingroup$
Can’t you adapt the accepted answer to that question to your specific case?
$endgroup$
– amd
Jan 9 at 19:34
$begingroup$
Will take a look when I have some time. In the meantime, have posted my solution.
$endgroup$
– hypergeometric
Jan 10 at 17:38
add a comment |
$begingroup$
Is it possible to come up with a single-line equation in rectangular coordinates for an equilateral triangle with circumradius $R$, positioned symmetrical about the $y$-axis, as shown in the diagram below?
analytic-geometry absolute-value polygons
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
$endgroup$
Is it possible to come up with a single-line equation in rectangular coordinates for an equilateral triangle with circumradius $R$, positioned symmetrical about the $y$-axis, as shown in the diagram below?
analytic-geometry absolute-value polygons
analytic-geometry absolute-value polygons
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
asked Jan 8 at 17:57
hypergeometrichypergeometric
17.8k1762
17.8k1762
$begingroup$
What do you mean by a “single-line equation?” This might be the sort of thing you’re looking for.
$endgroup$
– amd
Jan 8 at 19:36
$begingroup$
@amd single-line equation meaning not a piecewise equation. Thanks for the suggested link - I did come across it earlier but here I'm looking for the solution for a specific case, and in rectangular coordinates.
$endgroup$
– hypergeometric
Jan 9 at 14:20
$begingroup$
Can’t you adapt the accepted answer to that question to your specific case?
$endgroup$
– amd
Jan 9 at 19:34
$begingroup$
Will take a look when I have some time. In the meantime, have posted my solution.
$endgroup$
– hypergeometric
Jan 10 at 17:38
add a comment |
$begingroup$
What do you mean by a “single-line equation?” This might be the sort of thing you’re looking for.
$endgroup$
– amd
Jan 8 at 19:36
$begingroup$
@amd single-line equation meaning not a piecewise equation. Thanks for the suggested link - I did come across it earlier but here I'm looking for the solution for a specific case, and in rectangular coordinates.
$endgroup$
– hypergeometric
Jan 9 at 14:20
$begingroup$
Can’t you adapt the accepted answer to that question to your specific case?
$endgroup$
– amd
Jan 9 at 19:34
$begingroup$
Will take a look when I have some time. In the meantime, have posted my solution.
$endgroup$
– hypergeometric
Jan 10 at 17:38
$begingroup$
What do you mean by a “single-line equation?” This might be the sort of thing you’re looking for.
$endgroup$
– amd
Jan 8 at 19:36
$begingroup$
What do you mean by a “single-line equation?” This might be the sort of thing you’re looking for.
$endgroup$
– amd
Jan 8 at 19:36
$begingroup$
@amd single-line equation meaning not a piecewise equation. Thanks for the suggested link - I did come across it earlier but here I'm looking for the solution for a specific case, and in rectangular coordinates.
$endgroup$
– hypergeometric
Jan 9 at 14:20
$begingroup$
@amd single-line equation meaning not a piecewise equation. Thanks for the suggested link - I did come across it earlier but here I'm looking for the solution for a specific case, and in rectangular coordinates.
$endgroup$
– hypergeometric
Jan 9 at 14:20
$begingroup$
Can’t you adapt the accepted answer to that question to your specific case?
$endgroup$
– amd
Jan 9 at 19:34
$begingroup$
Can’t you adapt the accepted answer to that question to your specific case?
$endgroup$
– amd
Jan 9 at 19:34
$begingroup$
Will take a look when I have some time. In the meantime, have posted my solution.
$endgroup$
– hypergeometric
Jan 10 at 17:38
$begingroup$
Will take a look when I have some time. In the meantime, have posted my solution.
$endgroup$
– hypergeometric
Jan 10 at 17:38
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I’m not entirely satisfied with the following, but it works and generalizes to any cyclic polygon: $$left(sqrt3x+y-Rright)left(sqrt3x-y+Rright)left(2y+Rright)+sqrt{R^2-x^2-y^2}=sqrt{R^2-x^2-y^2}.$$ The idea is that you take the union of the extensions of the triangle’s sides (the term on the left) and add terms that restrict the domain (the two radicals).
answered Jan 11 at 0:25
amdamd
31.4k21052
$endgroup$
$begingroup$
Nice solution, esp the use of $sqrt{R^2-x^2-y^2}$ on both sides to restrict the domain. (+1)
$endgroup$
– hypergeometric
Jan 11 at 14:21
add a comment |
$begingroup$
A possible solution is
$$big|y+xsqrt3-Rbig|
; +;big|y-xsqrt3-Rbig|
; +;big|2y+Rbig|=3R+delta$$
where $delta$ is infinitesimal and $delta>0$.
See desmos implementation here.
Each of the three terms on LHS within absolute value signs when equated to zero form the equations of the three sides of the triangle.
Interestingly, when $delta=0$, the result is the entire region within the triangle.
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
$endgroup$
$begingroup$
You don’t find the “infinitesimal” fudge factor unsatisfactory? For any positive $delta$, this is a hexagon, and when $deltalt0$, there is no real solution to this equation.
$endgroup$
– amd
Jan 10 at 18:58
$begingroup$
I think it provides additional insight despite not being ideal. It's specified that $delta$ is infinitesimal. The comment that $delta>0$ has been added.
$endgroup$
– hypergeometric
Jan 11 at 14:19
$begingroup$
That’s exactly what I myself find unsatisfactory: what exactly does “infinitesimal” mean? There is no real number $delta$ for which this equation describes a triangle. The triangle is in some sense a limiting curve of these hexagons, but that’s a tricky thing to describe precisely.
$endgroup$
– amd
Jan 11 at 19:32
add a comment |
$begingroup$
Here's a neat solution by my friend YC:
$$frac 2{sqrt3}big| x big|
; +; Bigg|frac 43 y + frac 2{sqrt{3}} big|xbig| - frac R3 Bigg|
; = ; R$$
See Desmos implementation here.
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
$endgroup$
add a comment |
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%2f3066503%2fsingle-line-equation-for-equilateral-triangle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I’m not entirely satisfied with the following, but it works and generalizes to any cyclic polygon: $$left(sqrt3x+y-Rright)left(sqrt3x-y+Rright)left(2y+Rright)+sqrt{R^2-x^2-y^2}=sqrt{R^2-x^2-y^2}.$$ The idea is that you take the union of the extensions of the triangle’s sides (the term on the left) and add terms that restrict the domain (the two radicals).
answered Jan 11 at 0:25
amdamd
31.4k21052
$endgroup$
$begingroup$
Nice solution, esp the use of $sqrt{R^2-x^2-y^2}$ on both sides to restrict the domain. (+1)
$endgroup$
– hypergeometric
Jan 11 at 14:21
add a comment |
$begingroup$
I’m not entirely satisfied with the following, but it works and generalizes to any cyclic polygon: $$left(sqrt3x+y-Rright)left(sqrt3x-y+Rright)left(2y+Rright)+sqrt{R^2-x^2-y^2}=sqrt{R^2-x^2-y^2}.$$ The idea is that you take the union of the extensions of the triangle’s sides (the term on the left) and add terms that restrict the domain (the two radicals).
answered Jan 11 at 0:25
amdamd
31.4k21052
$endgroup$
$begingroup$
Nice solution, esp the use of $sqrt{R^2-x^2-y^2}$ on both sides to restrict the domain. (+1)
$endgroup$
– hypergeometric
Jan 11 at 14:21
add a comment |
$begingroup$
I’m not entirely satisfied with the following, but it works and generalizes to any cyclic polygon: $$left(sqrt3x+y-Rright)left(sqrt3x-y+Rright)left(2y+Rright)+sqrt{R^2-x^2-y^2}=sqrt{R^2-x^2-y^2}.$$ The idea is that you take the union of the extensions of the triangle’s sides (the term on the left) and add terms that restrict the domain (the two radicals).
answered Jan 11 at 0:25
amdamd
31.4k21052
$endgroup$
I’m not entirely satisfied with the following, but it works and generalizes to any cyclic polygon: $$left(sqrt3x+y-Rright)left(sqrt3x-y+Rright)left(2y+Rright)+sqrt{R^2-x^2-y^2}=sqrt{R^2-x^2-y^2}.$$ The idea is that you take the union of the extensions of the triangle’s sides (the term on the left) and add terms that restrict the domain (the two radicals).
answered Jan 11 at 0:25
amdamd
31.4k21052
edited Jan 11 at 0:33
answered Jan 11 at 0:25
amdamd
31.4k21052
answered Jan 11 at 0:25
amdamd
31.4k21052
answered Jan 11 at 0:25
amdamd
31.4k21052
31.4k21052
$begingroup$
Nice solution, esp the use of $sqrt{R^2-x^2-y^2}$ on both sides to restrict the domain. (+1)
$endgroup$
– hypergeometric
Jan 11 at 14:21
add a comment |
$begingroup$
Nice solution, esp the use of $sqrt{R^2-x^2-y^2}$ on both sides to restrict the domain. (+1)
$endgroup$
– hypergeometric
Jan 11 at 14:21
$begingroup$
Nice solution, esp the use of $sqrt{R^2-x^2-y^2}$ on both sides to restrict the domain. (+1)
$endgroup$
– hypergeometric
Jan 11 at 14:21
$begingroup$
Nice solution, esp the use of $sqrt{R^2-x^2-y^2}$ on both sides to restrict the domain. (+1)
$endgroup$
– hypergeometric
Jan 11 at 14:21
add a comment |
$begingroup$
A possible solution is
$$big|y+xsqrt3-Rbig|
; +;big|y-xsqrt3-Rbig|
; +;big|2y+Rbig|=3R+delta$$
where $delta$ is infinitesimal and $delta>0$.
See desmos implementation here.
Each of the three terms on LHS within absolute value signs when equated to zero form the equations of the three sides of the triangle.
Interestingly, when $delta=0$, the result is the entire region within the triangle.
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
$endgroup$
$begingroup$
You don’t find the “infinitesimal” fudge factor unsatisfactory? For any positive $delta$, this is a hexagon, and when $deltalt0$, there is no real solution to this equation.
$endgroup$
– amd
Jan 10 at 18:58
$begingroup$
I think it provides additional insight despite not being ideal. It's specified that $delta$ is infinitesimal. The comment that $delta>0$ has been added.
$endgroup$
– hypergeometric
Jan 11 at 14:19
$begingroup$
That’s exactly what I myself find unsatisfactory: what exactly does “infinitesimal” mean? There is no real number $delta$ for which this equation describes a triangle. The triangle is in some sense a limiting curve of these hexagons, but that’s a tricky thing to describe precisely.
$endgroup$
– amd
Jan 11 at 19:32
add a comment |
$begingroup$
A possible solution is
$$big|y+xsqrt3-Rbig|
; +;big|y-xsqrt3-Rbig|
; +;big|2y+Rbig|=3R+delta$$
where $delta$ is infinitesimal and $delta>0$.
See desmos implementation here.
Each of the three terms on LHS within absolute value signs when equated to zero form the equations of the three sides of the triangle.
Interestingly, when $delta=0$, the result is the entire region within the triangle.
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
$endgroup$
$begingroup$
You don’t find the “infinitesimal” fudge factor unsatisfactory? For any positive $delta$, this is a hexagon, and when $deltalt0$, there is no real solution to this equation.
$endgroup$
– amd
Jan 10 at 18:58
$begingroup$
I think it provides additional insight despite not being ideal. It's specified that $delta$ is infinitesimal. The comment that $delta>0$ has been added.
$endgroup$
– hypergeometric
Jan 11 at 14:19
$begingroup$
That’s exactly what I myself find unsatisfactory: what exactly does “infinitesimal” mean? There is no real number $delta$ for which this equation describes a triangle. The triangle is in some sense a limiting curve of these hexagons, but that’s a tricky thing to describe precisely.
$endgroup$
– amd
Jan 11 at 19:32
add a comment |
$begingroup$
A possible solution is
$$big|y+xsqrt3-Rbig|
; +;big|y-xsqrt3-Rbig|
; +;big|2y+Rbig|=3R+delta$$
where $delta$ is infinitesimal and $delta>0$.
See desmos implementation here.
Each of the three terms on LHS within absolute value signs when equated to zero form the equations of the three sides of the triangle.
Interestingly, when $delta=0$, the result is the entire region within the triangle.
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
$endgroup$
A possible solution is
$$big|y+xsqrt3-Rbig|
; +;big|y-xsqrt3-Rbig|
; +;big|2y+Rbig|=3R+delta$$
where $delta$ is infinitesimal and $delta>0$.
See desmos implementation here.
Each of the three terms on LHS within absolute value signs when equated to zero form the equations of the three sides of the triangle.
Interestingly, when $delta=0$, the result is the entire region within the triangle.
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
edited Jan 11 at 14:20
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
answered Jan 10 at 17:38
hypergeometrichypergeometric
17.8k1762
17.8k1762
$begingroup$
You don’t find the “infinitesimal” fudge factor unsatisfactory? For any positive $delta$, this is a hexagon, and when $deltalt0$, there is no real solution to this equation.
$endgroup$
– amd
Jan 10 at 18:58
$begingroup$
I think it provides additional insight despite not being ideal. It's specified that $delta$ is infinitesimal. The comment that $delta>0$ has been added.
$endgroup$
– hypergeometric
Jan 11 at 14:19
$begingroup$
That’s exactly what I myself find unsatisfactory: what exactly does “infinitesimal” mean? There is no real number $delta$ for which this equation describes a triangle. The triangle is in some sense a limiting curve of these hexagons, but that’s a tricky thing to describe precisely.
$endgroup$
– amd
Jan 11 at 19:32
add a comment |
$begingroup$
You don’t find the “infinitesimal” fudge factor unsatisfactory? For any positive $delta$, this is a hexagon, and when $deltalt0$, there is no real solution to this equation.
$endgroup$
– amd
Jan 10 at 18:58
$begingroup$
I think it provides additional insight despite not being ideal. It's specified that $delta$ is infinitesimal. The comment that $delta>0$ has been added.
$endgroup$
– hypergeometric
Jan 11 at 14:19
$begingroup$
That’s exactly what I myself find unsatisfactory: what exactly does “infinitesimal” mean? There is no real number $delta$ for which this equation describes a triangle. The triangle is in some sense a limiting curve of these hexagons, but that’s a tricky thing to describe precisely.
$endgroup$
– amd
Jan 11 at 19:32
$begingroup$
You don’t find the “infinitesimal” fudge factor unsatisfactory? For any positive $delta$, this is a hexagon, and when $deltalt0$, there is no real solution to this equation.
$endgroup$
– amd
Jan 10 at 18:58
$begingroup$
You don’t find the “infinitesimal” fudge factor unsatisfactory? For any positive $delta$, this is a hexagon, and when $deltalt0$, there is no real solution to this equation.
$endgroup$
– amd
Jan 10 at 18:58
$begingroup$
I think it provides additional insight despite not being ideal. It's specified that $delta$ is infinitesimal. The comment that $delta>0$ has been added.
$endgroup$
– hypergeometric
Jan 11 at 14:19
$begingroup$
I think it provides additional insight despite not being ideal. It's specified that $delta$ is infinitesimal. The comment that $delta>0$ has been added.
$endgroup$
– hypergeometric
Jan 11 at 14:19
$begingroup$
That’s exactly what I myself find unsatisfactory: what exactly does “infinitesimal” mean? There is no real number $delta$ for which this equation describes a triangle. The triangle is in some sense a limiting curve of these hexagons, but that’s a tricky thing to describe precisely.
$endgroup$
– amd
Jan 11 at 19:32
$begingroup$
That’s exactly what I myself find unsatisfactory: what exactly does “infinitesimal” mean? There is no real number $delta$ for which this equation describes a triangle. The triangle is in some sense a limiting curve of these hexagons, but that’s a tricky thing to describe precisely.
$endgroup$
– amd
Jan 11 at 19:32
add a comment |
$begingroup$
Here's a neat solution by my friend YC:
$$frac 2{sqrt3}big| x big|
; +; Bigg|frac 43 y + frac 2{sqrt{3}} big|xbig| - frac R3 Bigg|
; = ; R$$
See Desmos implementation here.
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
$endgroup$
add a comment |
$begingroup$
Here's a neat solution by my friend YC:
$$frac 2{sqrt3}big| x big|
; +; Bigg|frac 43 y + frac 2{sqrt{3}} big|xbig| - frac R3 Bigg|
; = ; R$$
See Desmos implementation here.
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
$endgroup$
add a comment |
$begingroup$
Here's a neat solution by my friend YC:
$$frac 2{sqrt3}big| x big|
; +; Bigg|frac 43 y + frac 2{sqrt{3}} big|xbig| - frac R3 Bigg|
; = ; R$$
See Desmos implementation here.
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
$endgroup$
Here's a neat solution by my friend YC:
$$frac 2{sqrt3}big| x big|
; +; Bigg|frac 43 y + frac 2{sqrt{3}} big|xbig| - frac R3 Bigg|
; = ; R$$
See Desmos implementation here.
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
answered Jan 11 at 14:56
hypergeometrichypergeometric
17.8k1762
17.8k1762
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%2f3066503%2fsingle-line-equation-for-equilateral-triangle%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
$begingroup$
What do you mean by a “single-line equation?” This might be the sort of thing you’re looking for.
$endgroup$
– amd
Jan 8 at 19:36
$begingroup$
@amd single-line equation meaning not a piecewise equation. Thanks for the suggested link - I did come across it earlier but here I'm looking for the solution for a specific case, and in rectangular coordinates.
$endgroup$
– hypergeometric
Jan 9 at 14:20
$begingroup$
Can’t you adapt the accepted answer to that question to your specific case?
$endgroup$
– amd
Jan 9 at 19:34
$begingroup$
Will take a look when I have some time. In the meantime, have posted my solution.
$endgroup$
– hypergeometric
Jan 10 at 17:38