Expected area of triangle inscribed in a circle
$begingroup$
On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
Find the expected area of △ ABC .
geometric-probability
asked Jan 10 at 6:56
mrwongmrwong
5915
$endgroup$
add a comment |
$begingroup$
On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
Find the expected area of △ ABC .
geometric-probability
asked Jan 10 at 6:56
mrwongmrwong
5915
$endgroup$
add a comment |
$begingroup$
On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
Find the expected area of △ ABC .
geometric-probability
asked Jan 10 at 6:56
mrwongmrwong
5915
$endgroup$
On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
Find the expected area of △ ABC .
geometric-probability
geometric-probability
asked Jan 10 at 6:56
mrwongmrwong
5915
asked Jan 10 at 6:56
mrwongmrwong
5915
asked Jan 10 at 6:56
mrwongmrwong
5915
asked Jan 10 at 6:56
mrwongmrwong
5915
asked Jan 10 at 6:56
mrwongmrwong
5915
5915
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$
Let the line joining A and the centre makes an angle $theta$ with the x axis, then
For $theta$ between $0$ and $frac{pi}{2}$,
$$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$
For $theta$ between $0$ and $frac{pi}{2} - phi$
$$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$
For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$
$$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$
Hence the average area is
$$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$
$$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$
I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.
Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.
answered Jan 11 at 1:50
KY TangKY Tang
1245
$endgroup$
$begingroup$
Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
$endgroup$
– mrwong
Jan 12 at 4:04
$begingroup$
We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
$endgroup$
– KY Tang
Jan 13 at 1:37
$begingroup$
:Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
$endgroup$
– mrwong
Jan 13 at 5:08
$begingroup$
In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
$endgroup$
– KY Tang
Jan 13 at 17:16
$begingroup$
I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
$endgroup$
– mrwong
Jan 14 at 4:40
|
show 7 more comments
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%2f3068322%2fexpected-area-of-triangle-inscribed-in-a-circle%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
$begingroup$
For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$
Let the line joining A and the centre makes an angle $theta$ with the x axis, then
For $theta$ between $0$ and $frac{pi}{2}$,
$$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$
For $theta$ between $0$ and $frac{pi}{2} - phi$
$$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$
For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$
$$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$
Hence the average area is
$$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$
$$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$
I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.
Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.
answered Jan 11 at 1:50
KY TangKY Tang
1245
$endgroup$
$begingroup$
Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
$endgroup$
– mrwong
Jan 12 at 4:04
$begingroup$
We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
$endgroup$
– KY Tang
Jan 13 at 1:37
$begingroup$
:Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
$endgroup$
– mrwong
Jan 13 at 5:08
$begingroup$
In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
$endgroup$
– KY Tang
Jan 13 at 17:16
$begingroup$
I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
$endgroup$
– mrwong
Jan 14 at 4:40
|
show 7 more comments
$begingroup$
For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$
Let the line joining A and the centre makes an angle $theta$ with the x axis, then
For $theta$ between $0$ and $frac{pi}{2}$,
$$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$
For $theta$ between $0$ and $frac{pi}{2} - phi$
$$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$
For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$
$$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$
Hence the average area is
$$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$
$$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$
I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.
Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.
answered Jan 11 at 1:50
KY TangKY Tang
1245
$endgroup$
$begingroup$
Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
$endgroup$
– mrwong
Jan 12 at 4:04
$begingroup$
We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
$endgroup$
– KY Tang
Jan 13 at 1:37
$begingroup$
:Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
$endgroup$
– mrwong
Jan 13 at 5:08
$begingroup$
In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
$endgroup$
– KY Tang
Jan 13 at 17:16
$begingroup$
I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
$endgroup$
– mrwong
Jan 14 at 4:40
|
show 7 more comments
$begingroup$
For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$
Let the line joining A and the centre makes an angle $theta$ with the x axis, then
For $theta$ between $0$ and $frac{pi}{2}$,
$$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$
For $theta$ between $0$ and $frac{pi}{2} - phi$
$$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$
For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$
$$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$
Hence the average area is
$$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$
$$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$
I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.
Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.
answered Jan 11 at 1:50
KY TangKY Tang
1245
$endgroup$
For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$
Let the line joining A and the centre makes an angle $theta$ with the x axis, then
For $theta$ between $0$ and $frac{pi}{2}$,
$$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$
For $theta$ between $0$ and $frac{pi}{2} - phi$
$$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$
For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$
$$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$
Hence the average area is
$$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$
$$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$
I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.
Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.
answered Jan 11 at 1:50
KY TangKY Tang
1245
edited Jan 15 at 1:27
answered Jan 11 at 1:50
KY TangKY Tang
1245
answered Jan 11 at 1:50
KY TangKY Tang
1245
answered Jan 11 at 1:50
KY TangKY Tang
1245
1245
$begingroup$
Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
$endgroup$
– mrwong
Jan 12 at 4:04
$begingroup$
We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
$endgroup$
– KY Tang
Jan 13 at 1:37
$begingroup$
:Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
$endgroup$
– mrwong
Jan 13 at 5:08
$begingroup$
In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
$endgroup$
– KY Tang
Jan 13 at 17:16
$begingroup$
I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
$endgroup$
– mrwong
Jan 14 at 4:40
|
show 7 more comments
$begingroup$
Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
$endgroup$
– mrwong
Jan 12 at 4:04
$begingroup$
We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
$endgroup$
– KY Tang
Jan 13 at 1:37
$begingroup$
:Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
$endgroup$
– mrwong
Jan 13 at 5:08
$begingroup$
In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
$endgroup$
– KY Tang
Jan 13 at 17:16
$begingroup$
I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
$endgroup$
– mrwong
Jan 14 at 4:40
$begingroup$
Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
$endgroup$
– mrwong
Jan 12 at 4:04
$begingroup$
Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
$endgroup$
– mrwong
Jan 12 at 4:04
$begingroup$
We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
$endgroup$
– KY Tang
Jan 13 at 1:37
$begingroup$
We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
$endgroup$
– KY Tang
Jan 13 at 1:37
$begingroup$
:Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
$endgroup$
– mrwong
Jan 13 at 5:08
$begingroup$
:Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
$endgroup$
– mrwong
Jan 13 at 5:08
$begingroup$
In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
$endgroup$
– KY Tang
Jan 13 at 17:16
$begingroup$
In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
$endgroup$
– KY Tang
Jan 13 at 17:16
$begingroup$
I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
$endgroup$
– mrwong
Jan 14 at 4:40
$begingroup$
I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
$endgroup$
– mrwong
Jan 14 at 4:40
|
show 7 more comments
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%2f3068322%2fexpected-area-of-triangle-inscribed-in-a-circle%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
