Problem of three circles
$begingroup$
This geometrical problem was proposed in a Mathematics Contest for high school students of my country. It is truly hard to find its solution.
Let $ABC$ be an acute triangle inscribed in the circle with its center $O$. The line which is perpendicular to $AO$ at $O$ intersects $AB$ and $AC$ at $E$ and $F$ respectively.
Let $D$ be the intersection point of $BF$ and $CE$. The circumscribed circle of triangle $BDC$ intersects $AB$ and $AC$ at $M$ and $N$ respectively and the circumscribed circle of triangle $DEF$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively.
Let $S$ be the intersection point of $BC$ and $EF$, and $K$ be the intersection point of $PN$ and $MQ$.
Prove that $AKperp SD$.
I am happy if someone could propose some fresh ideas to attack this problem.
geometry contest-math euclidean-geometry triangle plane-geometry
edited Dec 30 '18 at 9:35
the_fox
2,57711432
asked Dec 30 '18 at 5:10
BlindBlind
287318
$endgroup$
add a comment |
$begingroup$
This geometrical problem was proposed in a Mathematics Contest for high school students of my country. It is truly hard to find its solution.
Let $ABC$ be an acute triangle inscribed in the circle with its center $O$. The line which is perpendicular to $AO$ at $O$ intersects $AB$ and $AC$ at $E$ and $F$ respectively.
Let $D$ be the intersection point of $BF$ and $CE$. The circumscribed circle of triangle $BDC$ intersects $AB$ and $AC$ at $M$ and $N$ respectively and the circumscribed circle of triangle $DEF$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively.
Let $S$ be the intersection point of $BC$ and $EF$, and $K$ be the intersection point of $PN$ and $MQ$.
Prove that $AKperp SD$.
I am happy if someone could propose some fresh ideas to attack this problem.
geometry contest-math euclidean-geometry triangle plane-geometry
edited Dec 30 '18 at 9:35
the_fox
2,57711432
asked Dec 30 '18 at 5:10
BlindBlind
287318
$endgroup$
add a comment |
$begingroup$
This geometrical problem was proposed in a Mathematics Contest for high school students of my country. It is truly hard to find its solution.
Let $ABC$ be an acute triangle inscribed in the circle with its center $O$. The line which is perpendicular to $AO$ at $O$ intersects $AB$ and $AC$ at $E$ and $F$ respectively.
Let $D$ be the intersection point of $BF$ and $CE$. The circumscribed circle of triangle $BDC$ intersects $AB$ and $AC$ at $M$ and $N$ respectively and the circumscribed circle of triangle $DEF$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively.
Let $S$ be the intersection point of $BC$ and $EF$, and $K$ be the intersection point of $PN$ and $MQ$.
Prove that $AKperp SD$.
I am happy if someone could propose some fresh ideas to attack this problem.
geometry contest-math euclidean-geometry triangle plane-geometry
edited Dec 30 '18 at 9:35
the_fox
2,57711432
asked Dec 30 '18 at 5:10
BlindBlind
287318
$endgroup$
This geometrical problem was proposed in a Mathematics Contest for high school students of my country. It is truly hard to find its solution.
Let $ABC$ be an acute triangle inscribed in the circle with its center $O$. The line which is perpendicular to $AO$ at $O$ intersects $AB$ and $AC$ at $E$ and $F$ respectively.
Let $D$ be the intersection point of $BF$ and $CE$. The circumscribed circle of triangle $BDC$ intersects $AB$ and $AC$ at $M$ and $N$ respectively and the circumscribed circle of triangle $DEF$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively.
Let $S$ be the intersection point of $BC$ and $EF$, and $K$ be the intersection point of $PN$ and $MQ$.
Prove that $AKperp SD$.
I am happy if someone could propose some fresh ideas to attack this problem.
geometry contest-math euclidean-geometry triangle plane-geometry
geometry contest-math euclidean-geometry triangle plane-geometry
edited Dec 30 '18 at 9:35
the_fox
2,57711432
asked Dec 30 '18 at 5:10
BlindBlind
287318
edited Dec 30 '18 at 9:35
the_fox
2,57711432
asked Dec 30 '18 at 5:10
BlindBlind
287318
edited Dec 30 '18 at 9:35
the_fox
2,57711432
edited Dec 30 '18 at 9:35
the_fox
2,57711432
edited Dec 30 '18 at 9:35
the_fox
2,57711432
2,57711432
asked Dec 30 '18 at 5:10
BlindBlind
287318
asked Dec 30 '18 at 5:10
BlindBlind
287318
asked Dec 30 '18 at 5:10
BlindBlind
287318
287318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $(WXYZ)$ denote the circumcircle of any cyclic quadrilateral $WXYZ$.
$EFCB$ is cyclic as $angle OEA=frac{pi}{2}-angle BAO=angle ACB.$ Therefore, $PQ|BC $ and $EF|MN$. Also, it follows that $MNPQ$ is cyclic.
Now, $angle DBM=angle DCN$ as $EFCB$ is cyclic. Therefore, $DM=DN$. Similarly, $angle DEP=angle DFQ$ as $EFCB$ is cyclic. Consequently, $DP=DQ$. Hence, $D$ is the centre of $(MNPQ)$.
Let $MN$ intersect $PQ$ at $T$. By construction, $S$ is the radical center of $(PQEF), (MNCB)$ and $(EFCB)$. So, $SD$ is the radical axis of $(PQEF)$ and $(MNCB)$. By construction, $T$ is the radical center of $(PQEF), (MNCB)$ and $(MNPQ)$. So, $TD$ is the radical axis of $(PQEF)$ and $(MNCB)$, whence, $S, T, D$ are collinear. Thus, $AKperp SD iff AK perp TD.$
Applying Brokard's theorem on $MNPQ$, we have, $AK perp TD$, as $D$ is the centre of $(MNPQ)$ and $A, K$ and $T$ are the three diagonal points of the complete quadrangle $MNPQ$.
$blacksquare$
Note that for any $E, F$ on $AB$ and $AC$ respectively such that $AEFsim ACB$, the problem statement is true. The circumcentre of $ABC$ is irrelevant.
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
$endgroup$
$begingroup$
Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem.
$endgroup$
– Blind
Dec 30 '18 at 12:37
$begingroup$
Which contest did this problem appear in?
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:53
1
$begingroup$
Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:56
1
$begingroup$
As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:57
1
$begingroup$
OK. I will wait one week to accept the solution if there are not another interesting solutions.
$endgroup$
– Blind
Dec 30 '18 at 13:59
|
show 4 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%2f3056521%2fproblem-of-three-circles%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
$begingroup$
Let $(WXYZ)$ denote the circumcircle of any cyclic quadrilateral $WXYZ$.
$EFCB$ is cyclic as $angle OEA=frac{pi}{2}-angle BAO=angle ACB.$ Therefore, $PQ|BC $ and $EF|MN$. Also, it follows that $MNPQ$ is cyclic.
Now, $angle DBM=angle DCN$ as $EFCB$ is cyclic. Therefore, $DM=DN$. Similarly, $angle DEP=angle DFQ$ as $EFCB$ is cyclic. Consequently, $DP=DQ$. Hence, $D$ is the centre of $(MNPQ)$.
Let $MN$ intersect $PQ$ at $T$. By construction, $S$ is the radical center of $(PQEF), (MNCB)$ and $(EFCB)$. So, $SD$ is the radical axis of $(PQEF)$ and $(MNCB)$. By construction, $T$ is the radical center of $(PQEF), (MNCB)$ and $(MNPQ)$. So, $TD$ is the radical axis of $(PQEF)$ and $(MNCB)$, whence, $S, T, D$ are collinear. Thus, $AKperp SD iff AK perp TD.$
Applying Brokard's theorem on $MNPQ$, we have, $AK perp TD$, as $D$ is the centre of $(MNPQ)$ and $A, K$ and $T$ are the three diagonal points of the complete quadrangle $MNPQ$.
$blacksquare$
Note that for any $E, F$ on $AB$ and $AC$ respectively such that $AEFsim ACB$, the problem statement is true. The circumcentre of $ABC$ is irrelevant.
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
$endgroup$
$begingroup$
Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem.
$endgroup$
– Blind
Dec 30 '18 at 12:37
$begingroup$
Which contest did this problem appear in?
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:53
1
$begingroup$
Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:56
1
$begingroup$
As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:57
1
$begingroup$
OK. I will wait one week to accept the solution if there are not another interesting solutions.
$endgroup$
– Blind
Dec 30 '18 at 13:59
|
show 4 more comments
$begingroup$
Let $(WXYZ)$ denote the circumcircle of any cyclic quadrilateral $WXYZ$.
$EFCB$ is cyclic as $angle OEA=frac{pi}{2}-angle BAO=angle ACB.$ Therefore, $PQ|BC $ and $EF|MN$. Also, it follows that $MNPQ$ is cyclic.
Now, $angle DBM=angle DCN$ as $EFCB$ is cyclic. Therefore, $DM=DN$. Similarly, $angle DEP=angle DFQ$ as $EFCB$ is cyclic. Consequently, $DP=DQ$. Hence, $D$ is the centre of $(MNPQ)$.
Let $MN$ intersect $PQ$ at $T$. By construction, $S$ is the radical center of $(PQEF), (MNCB)$ and $(EFCB)$. So, $SD$ is the radical axis of $(PQEF)$ and $(MNCB)$. By construction, $T$ is the radical center of $(PQEF), (MNCB)$ and $(MNPQ)$. So, $TD$ is the radical axis of $(PQEF)$ and $(MNCB)$, whence, $S, T, D$ are collinear. Thus, $AKperp SD iff AK perp TD.$
Applying Brokard's theorem on $MNPQ$, we have, $AK perp TD$, as $D$ is the centre of $(MNPQ)$ and $A, K$ and $T$ are the three diagonal points of the complete quadrangle $MNPQ$.
$blacksquare$
Note that for any $E, F$ on $AB$ and $AC$ respectively such that $AEFsim ACB$, the problem statement is true. The circumcentre of $ABC$ is irrelevant.
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
$endgroup$
$begingroup$
Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem.
$endgroup$
– Blind
Dec 30 '18 at 12:37
$begingroup$
Which contest did this problem appear in?
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:53
1
$begingroup$
Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:56
1
$begingroup$
As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:57
1
$begingroup$
OK. I will wait one week to accept the solution if there are not another interesting solutions.
$endgroup$
– Blind
Dec 30 '18 at 13:59
|
show 4 more comments
$begingroup$
Let $(WXYZ)$ denote the circumcircle of any cyclic quadrilateral $WXYZ$.
$EFCB$ is cyclic as $angle OEA=frac{pi}{2}-angle BAO=angle ACB.$ Therefore, $PQ|BC $ and $EF|MN$. Also, it follows that $MNPQ$ is cyclic.
Now, $angle DBM=angle DCN$ as $EFCB$ is cyclic. Therefore, $DM=DN$. Similarly, $angle DEP=angle DFQ$ as $EFCB$ is cyclic. Consequently, $DP=DQ$. Hence, $D$ is the centre of $(MNPQ)$.
Let $MN$ intersect $PQ$ at $T$. By construction, $S$ is the radical center of $(PQEF), (MNCB)$ and $(EFCB)$. So, $SD$ is the radical axis of $(PQEF)$ and $(MNCB)$. By construction, $T$ is the radical center of $(PQEF), (MNCB)$ and $(MNPQ)$. So, $TD$ is the radical axis of $(PQEF)$ and $(MNCB)$, whence, $S, T, D$ are collinear. Thus, $AKperp SD iff AK perp TD.$
Applying Brokard's theorem on $MNPQ$, we have, $AK perp TD$, as $D$ is the centre of $(MNPQ)$ and $A, K$ and $T$ are the three diagonal points of the complete quadrangle $MNPQ$.
$blacksquare$
Note that for any $E, F$ on $AB$ and $AC$ respectively such that $AEFsim ACB$, the problem statement is true. The circumcentre of $ABC$ is irrelevant.
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
$endgroup$
Let $(WXYZ)$ denote the circumcircle of any cyclic quadrilateral $WXYZ$.
$EFCB$ is cyclic as $angle OEA=frac{pi}{2}-angle BAO=angle ACB.$ Therefore, $PQ|BC $ and $EF|MN$. Also, it follows that $MNPQ$ is cyclic.
Now, $angle DBM=angle DCN$ as $EFCB$ is cyclic. Therefore, $DM=DN$. Similarly, $angle DEP=angle DFQ$ as $EFCB$ is cyclic. Consequently, $DP=DQ$. Hence, $D$ is the centre of $(MNPQ)$.
Let $MN$ intersect $PQ$ at $T$. By construction, $S$ is the radical center of $(PQEF), (MNCB)$ and $(EFCB)$. So, $SD$ is the radical axis of $(PQEF)$ and $(MNCB)$. By construction, $T$ is the radical center of $(PQEF), (MNCB)$ and $(MNPQ)$. So, $TD$ is the radical axis of $(PQEF)$ and $(MNCB)$, whence, $S, T, D$ are collinear. Thus, $AKperp SD iff AK perp TD.$
Applying Brokard's theorem on $MNPQ$, we have, $AK perp TD$, as $D$ is the centre of $(MNPQ)$ and $A, K$ and $T$ are the three diagonal points of the complete quadrangle $MNPQ$.
$blacksquare$
Note that for any $E, F$ on $AB$ and $AC$ respectively such that $AEFsim ACB$, the problem statement is true. The circumcentre of $ABC$ is irrelevant.
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
edited Dec 31 '18 at 18:21
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
answered Dec 30 '18 at 12:16
Anubhab GhosalAnubhab Ghosal
81618
81618
$begingroup$
Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem.
$endgroup$
– Blind
Dec 30 '18 at 12:37
$begingroup$
Which contest did this problem appear in?
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:53
1
$begingroup$
Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:56
1
$begingroup$
As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:57
1
$begingroup$
OK. I will wait one week to accept the solution if there are not another interesting solutions.
$endgroup$
– Blind
Dec 30 '18 at 13:59
|
show 4 more comments
$begingroup$
Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem.
$endgroup$
– Blind
Dec 30 '18 at 12:37
$begingroup$
Which contest did this problem appear in?
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:53
1
$begingroup$
Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:56
1
$begingroup$
As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:57
1
$begingroup$
OK. I will wait one week to accept the solution if there are not another interesting solutions.
$endgroup$
– Blind
Dec 30 '18 at 13:59
$begingroup$
Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem.
$endgroup$
– Blind
Dec 30 '18 at 12:37
$begingroup$
Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem.
$endgroup$
– Blind
Dec 30 '18 at 12:37
$begingroup$
Which contest did this problem appear in?
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:53
$begingroup$
Which contest did this problem appear in?
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:53
1
1
$begingroup$
Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:56
$begingroup$
Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:56
1
1
$begingroup$
As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:57
$begingroup$
As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem.
$endgroup$
– Anubhab Ghosal
Dec 30 '18 at 12:57
1
1
$begingroup$
OK. I will wait one week to accept the solution if there are not another interesting solutions.
$endgroup$
– Blind
Dec 30 '18 at 13:59
$begingroup$
OK. I will wait one week to accept the solution if there are not another interesting solutions.
$endgroup$
– Blind
Dec 30 '18 at 13:59
|
show 4 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%2f3056521%2fproblem-of-three-circles%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
