Mixing
Equilateral triangle that has its vertices on the centers of $3$ different chords of a circle
$begingroup$
$A$ is the center of the circle. The rest of the data are on the diagram. Using geogebra, it is easy to see that $triangle EZH$ is equilateral, but I can't prove it. Any idea?
geometry euclidean-geometry triangles circles
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
$endgroup$
add a comment |
$begingroup$
$A$ is the center of the circle. The rest of the data are on the diagram. Using geogebra, it is easy to see that $triangle EZH$ is equilateral, but I can't prove it. Any idea?
geometry euclidean-geometry triangles circles
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
$endgroup$
add a comment |
$begingroup$
$A$ is the center of the circle. The rest of the data are on the diagram. Using geogebra, it is easy to see that $triangle EZH$ is equilateral, but I can't prove it. Any idea?
geometry euclidean-geometry triangles circles
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
$endgroup$
$A$ is the center of the circle. The rest of the data are on the diagram. Using geogebra, it is easy to see that $triangle EZH$ is equilateral, but I can't prove it. Any idea?
geometry euclidean-geometry triangles circles
geometry euclidean-geometry triangles circles
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
edited Jan 27 at 12:10
Anubhab Ghosal
1,20619
1,20619
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
asked Jan 27 at 10:59
ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
583
583
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We will use complex numbers. Let $omega =e^{ifrac{pi}{3}}implies omega^3=-1implies omega^2-omega+1=0$. Let $O$ be the center of the circle and let the circle be the unit circle. Let the points be $a, aomega, b, bomega, c$ and $comega$ in counterclockwise order as shown in the figure. Therefore $D=frac{aomega+b}{2}$ and similarly $E$ and $F$.
We wish to show that $triangle DEF$ is equilateral. This is true as $$frac{D-E}{F-E}=frac{aomega+b-bomega-c}{comega+a-bomega-c}=frac{aomega-bomega^2+omega^3c}{a-bomega+omega^2c}=omega$$
$blacksquare$
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
$endgroup$
$begingroup$
Nice ........+1
$endgroup$
– Maria Mazur
Jan 27 at 12:29
$begingroup$
oh what an answer...thank you very much
$endgroup$
– ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
Jan 27 at 12:58
$begingroup$
Excellent solution (+1)
$endgroup$
– user376343
Jan 27 at 18:51
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%2f3089411%2fequilateral-triangle-that-has-its-vertices-on-the-centers-of-3-different-chord%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
$begingroup$
We will use complex numbers. Let $omega =e^{ifrac{pi}{3}}implies omega^3=-1implies omega^2-omega+1=0$. Let $O$ be the center of the circle and let the circle be the unit circle. Let the points be $a, aomega, b, bomega, c$ and $comega$ in counterclockwise order as shown in the figure. Therefore $D=frac{aomega+b}{2}$ and similarly $E$ and $F$.
We wish to show that $triangle DEF$ is equilateral. This is true as $$frac{D-E}{F-E}=frac{aomega+b-bomega-c}{comega+a-bomega-c}=frac{aomega-bomega^2+omega^3c}{a-bomega+omega^2c}=omega$$
$blacksquare$
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
$endgroup$
$begingroup$
Nice ........+1
$endgroup$
– Maria Mazur
Jan 27 at 12:29
$begingroup$
oh what an answer...thank you very much
$endgroup$
– ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
Jan 27 at 12:58
$begingroup$
Excellent solution (+1)
$endgroup$
– user376343
Jan 27 at 18:51
add a comment |
$begingroup$
We will use complex numbers. Let $omega =e^{ifrac{pi}{3}}implies omega^3=-1implies omega^2-omega+1=0$. Let $O$ be the center of the circle and let the circle be the unit circle. Let the points be $a, aomega, b, bomega, c$ and $comega$ in counterclockwise order as shown in the figure. Therefore $D=frac{aomega+b}{2}$ and similarly $E$ and $F$.
We wish to show that $triangle DEF$ is equilateral. This is true as $$frac{D-E}{F-E}=frac{aomega+b-bomega-c}{comega+a-bomega-c}=frac{aomega-bomega^2+omega^3c}{a-bomega+omega^2c}=omega$$
$blacksquare$
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
$endgroup$
$begingroup$
Nice ........+1
$endgroup$
– Maria Mazur
Jan 27 at 12:29
$begingroup$
oh what an answer...thank you very much
$endgroup$
– ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
Jan 27 at 12:58
$begingroup$
Excellent solution (+1)
$endgroup$
– user376343
Jan 27 at 18:51
add a comment |
$begingroup$
We will use complex numbers. Let $omega =e^{ifrac{pi}{3}}implies omega^3=-1implies omega^2-omega+1=0$. Let $O$ be the center of the circle and let the circle be the unit circle. Let the points be $a, aomega, b, bomega, c$ and $comega$ in counterclockwise order as shown in the figure. Therefore $D=frac{aomega+b}{2}$ and similarly $E$ and $F$.
We wish to show that $triangle DEF$ is equilateral. This is true as $$frac{D-E}{F-E}=frac{aomega+b-bomega-c}{comega+a-bomega-c}=frac{aomega-bomega^2+omega^3c}{a-bomega+omega^2c}=omega$$
$blacksquare$
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
$endgroup$
We will use complex numbers. Let $omega =e^{ifrac{pi}{3}}implies omega^3=-1implies omega^2-omega+1=0$. Let $O$ be the center of the circle and let the circle be the unit circle. Let the points be $a, aomega, b, bomega, c$ and $comega$ in counterclockwise order as shown in the figure. Therefore $D=frac{aomega+b}{2}$ and similarly $E$ and $F$.
We wish to show that $triangle DEF$ is equilateral. This is true as $$frac{D-E}{F-E}=frac{aomega+b-bomega-c}{comega+a-bomega-c}=frac{aomega-bomega^2+omega^3c}{a-bomega+omega^2c}=omega$$
$blacksquare$
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
edited Jan 27 at 13:16
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
answered Jan 27 at 11:51
Anubhab GhosalAnubhab Ghosal
1,20619
1,20619
$begingroup$
Nice ........+1
$endgroup$
– Maria Mazur
Jan 27 at 12:29
$begingroup$
oh what an answer...thank you very much
$endgroup$
– ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
Jan 27 at 12:58
$begingroup$
Excellent solution (+1)
$endgroup$
– user376343
Jan 27 at 18:51
add a comment |
$begingroup$
Nice ........+1
$endgroup$
– Maria Mazur
Jan 27 at 12:29
$begingroup$
oh what an answer...thank you very much
$endgroup$
– ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
Jan 27 at 12:58
$begingroup$
Excellent solution (+1)
$endgroup$
– user376343
Jan 27 at 18:51
$begingroup$
Nice ........+1
$endgroup$
– Maria Mazur
Jan 27 at 12:29
$begingroup$
Nice ........+1
$endgroup$
– Maria Mazur
Jan 27 at 12:29
$begingroup$
oh what an answer...thank you very much
$endgroup$
– ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
Jan 27 at 12:58
$begingroup$
oh what an answer...thank you very much
$endgroup$
– ΚΩΝΣΤΑΝΤΙΝΟΣ ΑΣΛΑΝΟΓΛΟΥ
Jan 27 at 12:58
$begingroup$
Excellent solution (+1)
$endgroup$
– user376343
Jan 27 at 18:51
$begingroup$
Excellent solution (+1)
$endgroup$
– user376343
Jan 27 at 18:51
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%2f3089411%2fequilateral-triangle-that-has-its-vertices-on-the-centers-of-3-different-chord%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
