Mixing
Prove without calculus: 2 tangent segments to convex curve longer than curve
$begingroup$
Consider a convex curve in the plane. Let B and C be any two points on it, and let A be the intersection of the tangent to the curve at B and C.
I would like to show, without calculus, that $AB + AC > BC$.
With calculus, it does not seem too bad. I assume we can rotate B and C around A to B' and C' so that, at the point on the curve $B'C'$ below A, the slope of is zero. Call it D.
In the diagram,
$AB' = int_y^z sqrt{1+slope(AB')^2} dx$
$B'D = int_z^w sqrt{1+slope(B'C')^2} dx$
$AC' = int_y^z sqrt{1+slope(AC')^2} dx$
$C'D = int_z^w sqrt{1+slope(B'C')^2} dx$
The absolute value of the slope of $B'C'$ is decreasing from the slope of $AB'$ from B' to D, and negative but increasing (up to the absolute value of the slope of $AC'$) from D to C'
This is a general case of this question, which helped me when trying to place an upper limit on Pi.
plane-geometry
asked Jan 31 at 23:26
aschultzaschultz
2691515
$endgroup$
add a comment |
$begingroup$
Consider a convex curve in the plane. Let B and C be any two points on it, and let A be the intersection of the tangent to the curve at B and C.
I would like to show, without calculus, that $AB + AC > BC$.
With calculus, it does not seem too bad. I assume we can rotate B and C around A to B' and C' so that, at the point on the curve $B'C'$ below A, the slope of is zero. Call it D.
In the diagram,
$AB' = int_y^z sqrt{1+slope(AB')^2} dx$
$B'D = int_z^w sqrt{1+slope(B'C')^2} dx$
$AC' = int_y^z sqrt{1+slope(AC')^2} dx$
$C'D = int_z^w sqrt{1+slope(B'C')^2} dx$
The absolute value of the slope of $B'C'$ is decreasing from the slope of $AB'$ from B' to D, and negative but increasing (up to the absolute value of the slope of $AC'$) from D to C'
This is a general case of this question, which helped me when trying to place an upper limit on Pi.
plane-geometry
asked Jan 31 at 23:26
aschultzaschultz
2691515
$endgroup$
$begingroup$
I don't see how you can avoid calculus when you talk about tangent segments and curve lengths.
$endgroup$
– TonyK
Jan 31 at 23:34
add a comment |
$begingroup$
Consider a convex curve in the plane. Let B and C be any two points on it, and let A be the intersection of the tangent to the curve at B and C.
I would like to show, without calculus, that $AB + AC > BC$.
With calculus, it does not seem too bad. I assume we can rotate B and C around A to B' and C' so that, at the point on the curve $B'C'$ below A, the slope of is zero. Call it D.
In the diagram,
$AB' = int_y^z sqrt{1+slope(AB')^2} dx$
$B'D = int_z^w sqrt{1+slope(B'C')^2} dx$
$AC' = int_y^z sqrt{1+slope(AC')^2} dx$
$C'D = int_z^w sqrt{1+slope(B'C')^2} dx$
The absolute value of the slope of $B'C'$ is decreasing from the slope of $AB'$ from B' to D, and negative but increasing (up to the absolute value of the slope of $AC'$) from D to C'
This is a general case of this question, which helped me when trying to place an upper limit on Pi.
plane-geometry
asked Jan 31 at 23:26
aschultzaschultz
2691515
$endgroup$
Consider a convex curve in the plane. Let B and C be any two points on it, and let A be the intersection of the tangent to the curve at B and C.
I would like to show, without calculus, that $AB + AC > BC$.
With calculus, it does not seem too bad. I assume we can rotate B and C around A to B' and C' so that, at the point on the curve $B'C'$ below A, the slope of is zero. Call it D.
In the diagram,
$AB' = int_y^z sqrt{1+slope(AB')^2} dx$
$B'D = int_z^w sqrt{1+slope(B'C')^2} dx$
$AC' = int_y^z sqrt{1+slope(AC')^2} dx$
$C'D = int_z^w sqrt{1+slope(B'C')^2} dx$
The absolute value of the slope of $B'C'$ is decreasing from the slope of $AB'$ from B' to D, and negative but increasing (up to the absolute value of the slope of $AC'$) from D to C'
This is a general case of this question, which helped me when trying to place an upper limit on Pi.
plane-geometry
plane-geometry
asked Jan 31 at 23:26
aschultzaschultz
2691515
asked Jan 31 at 23:26
aschultzaschultz
2691515
asked Jan 31 at 23:26
aschultzaschultz
2691515
asked Jan 31 at 23:26
aschultzaschultz
2691515
asked Jan 31 at 23:26
aschultzaschultz
2691515
2691515
$begingroup$
I don't see how you can avoid calculus when you talk about tangent segments and curve lengths.
$endgroup$
– TonyK
Jan 31 at 23:34
add a comment |
$begingroup$
I don't see how you can avoid calculus when you talk about tangent segments and curve lengths.
$endgroup$
– TonyK
Jan 31 at 23:34
$begingroup$
I don't see how you can avoid calculus when you talk about tangent segments and curve lengths.
$endgroup$
– TonyK
Jan 31 at 23:34
$begingroup$
I don't see how you can avoid calculus when you talk about tangent segments and curve lengths.
$endgroup$
– TonyK
Jan 31 at 23:34
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%2f3095623%2fprove-without-calculus-2-tangent-segments-to-convex-curve-longer-than-curve%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%2f3095623%2fprove-without-calculus-2-tangent-segments-to-convex-curve-longer-than-curve%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
$begingroup$
I don't see how you can avoid calculus when you talk about tangent segments and curve lengths.
$endgroup$
– TonyK
Jan 31 at 23:34