All chords pass through the midpoint of each other
$begingroup$
Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.
Is it true that all $n$ chords are diameters?
I am not sure how to do these problems.
I tried drawing some circles and lines and it seems always true but I do not know why.
combinatorics circles
asked Dec 18 '18 at 12:32
user627514user627514
393
$endgroup$
add a comment |
$begingroup$
Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.
Is it true that all $n$ chords are diameters?
I am not sure how to do these problems.
I tried drawing some circles and lines and it seems always true but I do not know why.
combinatorics circles
asked Dec 18 '18 at 12:32
user627514user627514
393
$endgroup$
1
$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47
add a comment |
$begingroup$
Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.
Is it true that all $n$ chords are diameters?
I am not sure how to do these problems.
I tried drawing some circles and lines and it seems always true but I do not know why.
combinatorics circles
asked Dec 18 '18 at 12:32
user627514user627514
393
$endgroup$
Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.
Is it true that all $n$ chords are diameters?
I am not sure how to do these problems.
I tried drawing some circles and lines and it seems always true but I do not know why.
combinatorics circles
combinatorics circles
asked Dec 18 '18 at 12:32
user627514user627514
393
asked Dec 18 '18 at 12:32
user627514user627514
393
asked Dec 18 '18 at 12:32
user627514user627514
393
asked Dec 18 '18 at 12:32
user627514user627514
393
asked Dec 18 '18 at 12:32
user627514user627514
393
393
1
$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47
add a comment |
1
$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47
1
1
$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47
$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.
Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?
Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.
But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.
Here's a formal proof:
Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.
There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.
So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
$endgroup$
add a comment |
$begingroup$
Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.
answered Dec 18 '18 at 13:01
JoceJoce
834411
$endgroup$
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%2f3045114%2fall-chords-pass-through-the-midpoint-of-each-other%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.
Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?
Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.
But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.
Here's a formal proof:
Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.
There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.
So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
$endgroup$
add a comment |
$begingroup$
A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.
Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?
Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.
But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.
Here's a formal proof:
Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.
There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.
So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
$endgroup$
add a comment |
$begingroup$
A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.
Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?
Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.
But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.
Here's a formal proof:
Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.
There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.
So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
$endgroup$
A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.
Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?
Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.
But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.
Here's a formal proof:
Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.
There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.
So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
answered Dec 18 '18 at 13:03
Todor MarkovTodor Markov
2,420412
2,420412
add a comment |
add a comment |
$begingroup$
Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.
answered Dec 18 '18 at 13:01
JoceJoce
834411
$endgroup$
add a comment |
$begingroup$
Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.
answered Dec 18 '18 at 13:01
JoceJoce
834411
$endgroup$
add a comment |
$begingroup$
Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.
answered Dec 18 '18 at 13:01
JoceJoce
834411
$endgroup$
Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.
answered Dec 18 '18 at 13:01
JoceJoce
834411
answered Dec 18 '18 at 13:01
JoceJoce
834411
answered Dec 18 '18 at 13:01
JoceJoce
834411
answered Dec 18 '18 at 13:01
JoceJoce
834411
834411
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%2f3045114%2fall-chords-pass-through-the-midpoint-of-each-other%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
1
$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47