curiosity in circular analytical geometry
$begingroup$
Curiosity
Let $mathscr{C}$ be a circle of radius $r$, centered at the origin.
Assign point $O$ the cartesian coordinate $(0,0).$
Let $k$ be a fixed element in $(0,r).$
Set points A and B to have cartesian coordinates $(0,k)$ and $(0,-k)$ respectively.
Choose point C at random on $mathscr{C}.;$ Then $;left|;overline{OC};right|^2 = r^2.$
The end of this query gives an Algebraic Demonstration that
regardless of the point $C$ chosen,
$;left|;overline{AC};right|^2 + left|;overline{BC};right|^2 = 2(r^2 + k^2).$
Questions
Question 1 below added, per Blue's suggestion.
- see this query's Background section (the next section). My underlying desire is to find a non-algebraic way of predicting that
given $;z_1,z_2 in mathbb{C};$ the Locus of $;{z; :;
|z-z_1|^2 + |z-z_2|^2 ;=;$ a constant$};$ will be a circle. I have accepted Blue's answer, re Appollonious' (triangle-oriented) theorem, which does facilitate the prediction, though the logic is convoluted. I would welcome a separate circle-oriented geometry theorem that more directly facilitates the prediction.
The remaining questions in this section are now somewhat redundant, but are left in as a reference to how I originally posed my questions.
Can this result be demonstrated geometrically, without resorting to algebra?
Does this result have a (theorem) name?
Is there a free online geometry resource (e.g. pdf) that includes this result?
Background
In "An Introduction to Complex Function Theory", Bruce Palka, 1991, problem
4.6.vii, p.26 specifies:
geometrically describe $;{z;:;left|z-iright|^2 + left|z+iright|^2 = 4}.;$
After determining that the Locus was a circle of radius 1, centered at the origin,
I realized that
if the right hand side $; = R ;: ;R>2;$ then the Locus is a circle of radius
$;sqrt{(R-2)/2}.$
My (brief) subsequent online research found no mention of this curious result.
Algebraic Demonstration
In the diagram at the end of this query,
$;mathscr{C}, r, k,;$ point $A$ and point $B$ are as described
in the Curiousity section, at the start of this query.
Randomly choose $t$ in $;(k, r);$ and assign point $D$ the cartesian coordinates
$(0,t).$
Let point $C_1$ represent the right-hand-side intersection of $;mathscr{C};$
with the line $;y=t.$
Similarly, randomly choose $s$ in $;(0,k),;$ and assign point $E$ the cartesian
coordinate $;(0,s).$
Similarly, let point $C_2$ represent the left-hand-side intersection of $;mathscr{C};$
with the line $;y=s.$
$left|;overline{DC_1};right|^2 ;= ;r^2 - t^2.$
$left|;overline{AC_1};right|^2 ;= ;(t-k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{BC_1};right|^2 ;= ;(t+k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{AC_1};right|^2
+ left|;overline{BC_1};right|^2
;= ;2(t^2 + k^2) ;+ ;2(r^2 - t^2) ;= ;2(r^2 + k^2).$
$left|;overline{EC_2};right|^2 ;= ;r^2 - s^2.$
$left|;overline{AC_2};right|^2 ;= ;(k-s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{BC_2};right|^2 ;= ;(k+s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{AC_2};right|^2
+ left|;overline{BC_2};right|^2
;= ;2(k^2 + s^2) ;+ ;2(r^2 - s^2) ;= ;2(r^2 + k^2).$
geometry analytic-geometry
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
$endgroup$
add a comment |
$begingroup$
Curiosity
Let $mathscr{C}$ be a circle of radius $r$, centered at the origin.
Assign point $O$ the cartesian coordinate $(0,0).$
Let $k$ be a fixed element in $(0,r).$
Set points A and B to have cartesian coordinates $(0,k)$ and $(0,-k)$ respectively.
Choose point C at random on $mathscr{C}.;$ Then $;left|;overline{OC};right|^2 = r^2.$
The end of this query gives an Algebraic Demonstration that
regardless of the point $C$ chosen,
$;left|;overline{AC};right|^2 + left|;overline{BC};right|^2 = 2(r^2 + k^2).$
Questions
Question 1 below added, per Blue's suggestion.
- see this query's Background section (the next section). My underlying desire is to find a non-algebraic way of predicting that
given $;z_1,z_2 in mathbb{C};$ the Locus of $;{z; :;
|z-z_1|^2 + |z-z_2|^2 ;=;$ a constant$};$ will be a circle. I have accepted Blue's answer, re Appollonious' (triangle-oriented) theorem, which does facilitate the prediction, though the logic is convoluted. I would welcome a separate circle-oriented geometry theorem that more directly facilitates the prediction.
The remaining questions in this section are now somewhat redundant, but are left in as a reference to how I originally posed my questions.
Can this result be demonstrated geometrically, without resorting to algebra?
Does this result have a (theorem) name?
Is there a free online geometry resource (e.g. pdf) that includes this result?
Background
In "An Introduction to Complex Function Theory", Bruce Palka, 1991, problem
4.6.vii, p.26 specifies:
geometrically describe $;{z;:;left|z-iright|^2 + left|z+iright|^2 = 4}.;$
After determining that the Locus was a circle of radius 1, centered at the origin,
I realized that
if the right hand side $; = R ;: ;R>2;$ then the Locus is a circle of radius
$;sqrt{(R-2)/2}.$
My (brief) subsequent online research found no mention of this curious result.
Algebraic Demonstration
In the diagram at the end of this query,
$;mathscr{C}, r, k,;$ point $A$ and point $B$ are as described
in the Curiousity section, at the start of this query.
Randomly choose $t$ in $;(k, r);$ and assign point $D$ the cartesian coordinates
$(0,t).$
Let point $C_1$ represent the right-hand-side intersection of $;mathscr{C};$
with the line $;y=t.$
Similarly, randomly choose $s$ in $;(0,k),;$ and assign point $E$ the cartesian
coordinate $;(0,s).$
Similarly, let point $C_2$ represent the left-hand-side intersection of $;mathscr{C};$
with the line $;y=s.$
$left|;overline{DC_1};right|^2 ;= ;r^2 - t^2.$
$left|;overline{AC_1};right|^2 ;= ;(t-k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{BC_1};right|^2 ;= ;(t+k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{AC_1};right|^2
+ left|;overline{BC_1};right|^2
;= ;2(t^2 + k^2) ;+ ;2(r^2 - t^2) ;= ;2(r^2 + k^2).$
$left|;overline{EC_2};right|^2 ;= ;r^2 - s^2.$
$left|;overline{AC_2};right|^2 ;= ;(k-s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{BC_2};right|^2 ;= ;(k+s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{AC_2};right|^2
+ left|;overline{BC_2};right|^2
;= ;2(k^2 + s^2) ;+ ;2(r^2 - s^2) ;= ;2(r^2 + k^2).$
geometry analytic-geometry
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
$endgroup$
$begingroup$
This can be considered simple application of the Law of Cosines and the fact that supplementary angles have negative cosines. $$begin{align}0 &= 2kr (cosangle COA+cosangle COB) \ &= (|AC|^2-k^2-r^2)+(|BC|^2-k^2-r^2) \ &=|AC|^2+|BC|^2-2(k^2+r^2)end{align}$$ (I don't know if you'd consider this too "algebraic".) I'm not aware of any name for this result; it doesn't seem to need one.
$endgroup$
– Blue
Dec 19 '18 at 4:27
$begingroup$
By the way, this answer of mine gives a geometric derivation of the Law of Cosines.
$endgroup$
– Blue
Dec 19 '18 at 4:44
$begingroup$
@Blue slick...(but still algebraic). I am actually surprised that my result isn't less obscure.
$endgroup$
– user2661923
Dec 19 '18 at 5:18
$begingroup$
I'll note that the result has nothing to do with a circle; rather, it's simply a triangle relation. In particular, it relates the length of a median to other segments in a triangle. In that regard, it's a special case of Stewart's Theorem. (Ah! As the Wikipedia entry notes, the median case does have a name: Apollonius's Theorem.)
$endgroup$
– Blue
Dec 19 '18 at 5:34
add a comment |
$begingroup$
Curiosity
Let $mathscr{C}$ be a circle of radius $r$, centered at the origin.
Assign point $O$ the cartesian coordinate $(0,0).$
Let $k$ be a fixed element in $(0,r).$
Set points A and B to have cartesian coordinates $(0,k)$ and $(0,-k)$ respectively.
Choose point C at random on $mathscr{C}.;$ Then $;left|;overline{OC};right|^2 = r^2.$
The end of this query gives an Algebraic Demonstration that
regardless of the point $C$ chosen,
$;left|;overline{AC};right|^2 + left|;overline{BC};right|^2 = 2(r^2 + k^2).$
Questions
Question 1 below added, per Blue's suggestion.
- see this query's Background section (the next section). My underlying desire is to find a non-algebraic way of predicting that
given $;z_1,z_2 in mathbb{C};$ the Locus of $;{z; :;
|z-z_1|^2 + |z-z_2|^2 ;=;$ a constant$};$ will be a circle. I have accepted Blue's answer, re Appollonious' (triangle-oriented) theorem, which does facilitate the prediction, though the logic is convoluted. I would welcome a separate circle-oriented geometry theorem that more directly facilitates the prediction.
The remaining questions in this section are now somewhat redundant, but are left in as a reference to how I originally posed my questions.
Can this result be demonstrated geometrically, without resorting to algebra?
Does this result have a (theorem) name?
Is there a free online geometry resource (e.g. pdf) that includes this result?
Background
In "An Introduction to Complex Function Theory", Bruce Palka, 1991, problem
4.6.vii, p.26 specifies:
geometrically describe $;{z;:;left|z-iright|^2 + left|z+iright|^2 = 4}.;$
After determining that the Locus was a circle of radius 1, centered at the origin,
I realized that
if the right hand side $; = R ;: ;R>2;$ then the Locus is a circle of radius
$;sqrt{(R-2)/2}.$
My (brief) subsequent online research found no mention of this curious result.
Algebraic Demonstration
In the diagram at the end of this query,
$;mathscr{C}, r, k,;$ point $A$ and point $B$ are as described
in the Curiousity section, at the start of this query.
Randomly choose $t$ in $;(k, r);$ and assign point $D$ the cartesian coordinates
$(0,t).$
Let point $C_1$ represent the right-hand-side intersection of $;mathscr{C};$
with the line $;y=t.$
Similarly, randomly choose $s$ in $;(0,k),;$ and assign point $E$ the cartesian
coordinate $;(0,s).$
Similarly, let point $C_2$ represent the left-hand-side intersection of $;mathscr{C};$
with the line $;y=s.$
$left|;overline{DC_1};right|^2 ;= ;r^2 - t^2.$
$left|;overline{AC_1};right|^2 ;= ;(t-k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{BC_1};right|^2 ;= ;(t+k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{AC_1};right|^2
+ left|;overline{BC_1};right|^2
;= ;2(t^2 + k^2) ;+ ;2(r^2 - t^2) ;= ;2(r^2 + k^2).$
$left|;overline{EC_2};right|^2 ;= ;r^2 - s^2.$
$left|;overline{AC_2};right|^2 ;= ;(k-s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{BC_2};right|^2 ;= ;(k+s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{AC_2};right|^2
+ left|;overline{BC_2};right|^2
;= ;2(k^2 + s^2) ;+ ;2(r^2 - s^2) ;= ;2(r^2 + k^2).$
geometry analytic-geometry
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
$endgroup$
Curiosity
Let $mathscr{C}$ be a circle of radius $r$, centered at the origin.
Assign point $O$ the cartesian coordinate $(0,0).$
Let $k$ be a fixed element in $(0,r).$
Set points A and B to have cartesian coordinates $(0,k)$ and $(0,-k)$ respectively.
Choose point C at random on $mathscr{C}.;$ Then $;left|;overline{OC};right|^2 = r^2.$
The end of this query gives an Algebraic Demonstration that
regardless of the point $C$ chosen,
$;left|;overline{AC};right|^2 + left|;overline{BC};right|^2 = 2(r^2 + k^2).$
Questions
Question 1 below added, per Blue's suggestion.
- see this query's Background section (the next section). My underlying desire is to find a non-algebraic way of predicting that
given $;z_1,z_2 in mathbb{C};$ the Locus of $;{z; :;
|z-z_1|^2 + |z-z_2|^2 ;=;$ a constant$};$ will be a circle. I have accepted Blue's answer, re Appollonious' (triangle-oriented) theorem, which does facilitate the prediction, though the logic is convoluted. I would welcome a separate circle-oriented geometry theorem that more directly facilitates the prediction.
The remaining questions in this section are now somewhat redundant, but are left in as a reference to how I originally posed my questions.
Can this result be demonstrated geometrically, without resorting to algebra?
Does this result have a (theorem) name?
Is there a free online geometry resource (e.g. pdf) that includes this result?
Background
In "An Introduction to Complex Function Theory", Bruce Palka, 1991, problem
4.6.vii, p.26 specifies:
geometrically describe $;{z;:;left|z-iright|^2 + left|z+iright|^2 = 4}.;$
After determining that the Locus was a circle of radius 1, centered at the origin,
I realized that
if the right hand side $; = R ;: ;R>2;$ then the Locus is a circle of radius
$;sqrt{(R-2)/2}.$
My (brief) subsequent online research found no mention of this curious result.
Algebraic Demonstration
In the diagram at the end of this query,
$;mathscr{C}, r, k,;$ point $A$ and point $B$ are as described
in the Curiousity section, at the start of this query.
Randomly choose $t$ in $;(k, r);$ and assign point $D$ the cartesian coordinates
$(0,t).$
Let point $C_1$ represent the right-hand-side intersection of $;mathscr{C};$
with the line $;y=t.$
Similarly, randomly choose $s$ in $;(0,k),;$ and assign point $E$ the cartesian
coordinate $;(0,s).$
Similarly, let point $C_2$ represent the left-hand-side intersection of $;mathscr{C};$
with the line $;y=s.$
$left|;overline{DC_1};right|^2 ;= ;r^2 - t^2.$
$left|;overline{AC_1};right|^2 ;= ;(t-k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{BC_1};right|^2 ;= ;(t+k)^2
+ left|;overline{DC_1};right|^2.$
$left|;overline{AC_1};right|^2
+ left|;overline{BC_1};right|^2
;= ;2(t^2 + k^2) ;+ ;2(r^2 - t^2) ;= ;2(r^2 + k^2).$
$left|;overline{EC_2};right|^2 ;= ;r^2 - s^2.$
$left|;overline{AC_2};right|^2 ;= ;(k-s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{BC_2};right|^2 ;= ;(k+s)^2
+ left|;overline{EC_2};right|^2.$
$left|;overline{AC_2};right|^2
+ left|;overline{BC_2};right|^2
;= ;2(k^2 + s^2) ;+ ;2(r^2 - s^2) ;= ;2(r^2 + k^2).$
geometry analytic-geometry
geometry analytic-geometry
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
edited Dec 19 '18 at 7:41
user2661923
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
asked Dec 18 '18 at 21:56
user2661923user2661923
558112
558112
$begingroup$
This can be considered simple application of the Law of Cosines and the fact that supplementary angles have negative cosines. $$begin{align}0 &= 2kr (cosangle COA+cosangle COB) \ &= (|AC|^2-k^2-r^2)+(|BC|^2-k^2-r^2) \ &=|AC|^2+|BC|^2-2(k^2+r^2)end{align}$$ (I don't know if you'd consider this too "algebraic".) I'm not aware of any name for this result; it doesn't seem to need one.
$endgroup$
– Blue
Dec 19 '18 at 4:27
$begingroup$
By the way, this answer of mine gives a geometric derivation of the Law of Cosines.
$endgroup$
– Blue
Dec 19 '18 at 4:44
$begingroup$
@Blue slick...(but still algebraic). I am actually surprised that my result isn't less obscure.
$endgroup$
– user2661923
Dec 19 '18 at 5:18
$begingroup$
I'll note that the result has nothing to do with a circle; rather, it's simply a triangle relation. In particular, it relates the length of a median to other segments in a triangle. In that regard, it's a special case of Stewart's Theorem. (Ah! As the Wikipedia entry notes, the median case does have a name: Apollonius's Theorem.)
$endgroup$
– Blue
Dec 19 '18 at 5:34
add a comment |
$begingroup$
This can be considered simple application of the Law of Cosines and the fact that supplementary angles have negative cosines. $$begin{align}0 &= 2kr (cosangle COA+cosangle COB) \ &= (|AC|^2-k^2-r^2)+(|BC|^2-k^2-r^2) \ &=|AC|^2+|BC|^2-2(k^2+r^2)end{align}$$ (I don't know if you'd consider this too "algebraic".) I'm not aware of any name for this result; it doesn't seem to need one.
$endgroup$
– Blue
Dec 19 '18 at 4:27
$begingroup$
By the way, this answer of mine gives a geometric derivation of the Law of Cosines.
$endgroup$
– Blue
Dec 19 '18 at 4:44
$begingroup$
@Blue slick...(but still algebraic). I am actually surprised that my result isn't less obscure.
$endgroup$
– user2661923
Dec 19 '18 at 5:18
$begingroup$
I'll note that the result has nothing to do with a circle; rather, it's simply a triangle relation. In particular, it relates the length of a median to other segments in a triangle. In that regard, it's a special case of Stewart's Theorem. (Ah! As the Wikipedia entry notes, the median case does have a name: Apollonius's Theorem.)
$endgroup$
– Blue
Dec 19 '18 at 5:34
$begingroup$
This can be considered simple application of the Law of Cosines and the fact that supplementary angles have negative cosines. $$begin{align}0 &= 2kr (cosangle COA+cosangle COB) \ &= (|AC|^2-k^2-r^2)+(|BC|^2-k^2-r^2) \ &=|AC|^2+|BC|^2-2(k^2+r^2)end{align}$$ (I don't know if you'd consider this too "algebraic".) I'm not aware of any name for this result; it doesn't seem to need one.
$endgroup$
– Blue
Dec 19 '18 at 4:27
$begingroup$
This can be considered simple application of the Law of Cosines and the fact that supplementary angles have negative cosines. $$begin{align}0 &= 2kr (cosangle COA+cosangle COB) \ &= (|AC|^2-k^2-r^2)+(|BC|^2-k^2-r^2) \ &=|AC|^2+|BC|^2-2(k^2+r^2)end{align}$$ (I don't know if you'd consider this too "algebraic".) I'm not aware of any name for this result; it doesn't seem to need one.
$endgroup$
– Blue
Dec 19 '18 at 4:27
$begingroup$
By the way, this answer of mine gives a geometric derivation of the Law of Cosines.
$endgroup$
– Blue
Dec 19 '18 at 4:44
$begingroup$
By the way, this answer of mine gives a geometric derivation of the Law of Cosines.
$endgroup$
– Blue
Dec 19 '18 at 4:44
$begingroup$
@Blue slick...(but still algebraic). I am actually surprised that my result isn't less obscure.
$endgroup$
– user2661923
Dec 19 '18 at 5:18
$begingroup$
@Blue slick...(but still algebraic). I am actually surprised that my result isn't less obscure.
$endgroup$
– user2661923
Dec 19 '18 at 5:18
$begingroup$
I'll note that the result has nothing to do with a circle; rather, it's simply a triangle relation. In particular, it relates the length of a median to other segments in a triangle. In that regard, it's a special case of Stewart's Theorem. (Ah! As the Wikipedia entry notes, the median case does have a name: Apollonius's Theorem.)
$endgroup$
– Blue
Dec 19 '18 at 5:34
$begingroup$
I'll note that the result has nothing to do with a circle; rather, it's simply a triangle relation. In particular, it relates the length of a median to other segments in a triangle. In that regard, it's a special case of Stewart's Theorem. (Ah! As the Wikipedia entry notes, the median case does have a name: Apollonius's Theorem.)
$endgroup$
– Blue
Dec 19 '18 at 5:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's an alternative (but still very "algebraic") derivation that leverages Thales' theorem (the angle inscribed in a semicircle is right):
$$begin{align}
a^2+b^2
&= left(m^2+p^2right)+left(n^2+q^2right) \
&=left(m^2+n^2right)+left(p^2+q^2right) \
&=left(r+kright)^2+left(r-kright)^2 \
&=2left(r^2+k^2right)
end{align}$$
This result, which relates the length of a median to other segments in $triangle ABC$, is Apollonius's Theorem. It's a special case of Stewart's Theorem for the length of an arbitrary cevian.
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
$endgroup$
$begingroup$
unfortunately, I have to accept your answer as on-point. However, examine my question from the point of view of background. The idea is that given $;z_1, z_2 inmathbb{C};$, it is desired to predict in advance (without algebra) that the Locus of $;{z;:|z-z_1|^2 + |z-z_2|^2 = ;$a constant$};$ will be a circle. It takes some convoluted logic to use Appollonius' theorem to reach that conclusion.
$endgroup$
– user2661923
Dec 19 '18 at 6:43
$begingroup$
You don't "have to" accept any answer that fails to satisfy you, so feel free to un-accept and encourage others to answer. (One should usually allow a bit of time before accepting, anyway.) That said, you could/should make your desire clearer by asking up-front: "How might we anticipate that $|AC|^2+|BC|^2=text{constant}$ represents a circle?"
$endgroup$
– Blue
Dec 19 '18 at 7:00
$begingroup$
Good pt, re clarifying my desire; it never occurred to me that an on-point geometric theorem would exist that didn't easily fulfill my underlying desire. Re accept'g your answer, I feel compelled to for two reasons: (1) Your answer is directly on-point and can fulfill my underlying desire, though convoluted logic is involved. (2) Given the existence of Appollonius' (triangle-oriented) theorem, I strongly suspect that there will not be a separate circle-oriented geometry theorem that one might use. Assuming that my suspicion is correct, then your answer is the best I'll get.
$endgroup$
– user2661923
Dec 19 '18 at 7:20
$begingroup$
As to the logic being convoluted ... I don't think it is, necessarily. Symmetry requires that the center of the circle be the midpoint ($O$) of $AB$, so, "all we need" is a relation among $|AC|$, $|BC|$, $|OC|$, and, presumably, $|AB|$ (or half thereof). Stewart/Apollonius gives that relation, wherein the constancy of $|AC|^2+|BC|^2$ (and $|AB|$) immediately implies the constancy of $|OC|$. Done. That said, I'm actually inclined to believe that there's a more-satisfying diagrammatic demonstration of this fact.
$endgroup$
– Blue
Dec 19 '18 at 7:32
1
$begingroup$
I edited my query re your suggestion about clarifying my desire. I agree with the logic of your latest comment. My use of the adjective convoluted merely signifies that even if I had been aware of Appollonius' (triangle oriented) theorem, I could have easily overlooked that it can be used to show that the Locus is a circle. Further, since the Locus is a circle, my initial intuition is that some property of circles must be involved. This intuition is similar to the last sentence in your latest comment.
$endgroup$
– user2661923
Dec 19 '18 at 7:48
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%2f3045762%2fcuriosity-in-circular-analytical-geometry%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
$begingroup$
Here's an alternative (but still very "algebraic") derivation that leverages Thales' theorem (the angle inscribed in a semicircle is right):
$$begin{align}
a^2+b^2
&= left(m^2+p^2right)+left(n^2+q^2right) \
&=left(m^2+n^2right)+left(p^2+q^2right) \
&=left(r+kright)^2+left(r-kright)^2 \
&=2left(r^2+k^2right)
end{align}$$
This result, which relates the length of a median to other segments in $triangle ABC$, is Apollonius's Theorem. It's a special case of Stewart's Theorem for the length of an arbitrary cevian.
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
$endgroup$
$begingroup$
unfortunately, I have to accept your answer as on-point. However, examine my question from the point of view of background. The idea is that given $;z_1, z_2 inmathbb{C};$, it is desired to predict in advance (without algebra) that the Locus of $;{z;:|z-z_1|^2 + |z-z_2|^2 = ;$a constant$};$ will be a circle. It takes some convoluted logic to use Appollonius' theorem to reach that conclusion.
$endgroup$
– user2661923
Dec 19 '18 at 6:43
$begingroup$
You don't "have to" accept any answer that fails to satisfy you, so feel free to un-accept and encourage others to answer. (One should usually allow a bit of time before accepting, anyway.) That said, you could/should make your desire clearer by asking up-front: "How might we anticipate that $|AC|^2+|BC|^2=text{constant}$ represents a circle?"
$endgroup$
– Blue
Dec 19 '18 at 7:00
$begingroup$
Good pt, re clarifying my desire; it never occurred to me that an on-point geometric theorem would exist that didn't easily fulfill my underlying desire. Re accept'g your answer, I feel compelled to for two reasons: (1) Your answer is directly on-point and can fulfill my underlying desire, though convoluted logic is involved. (2) Given the existence of Appollonius' (triangle-oriented) theorem, I strongly suspect that there will not be a separate circle-oriented geometry theorem that one might use. Assuming that my suspicion is correct, then your answer is the best I'll get.
$endgroup$
– user2661923
Dec 19 '18 at 7:20
$begingroup$
As to the logic being convoluted ... I don't think it is, necessarily. Symmetry requires that the center of the circle be the midpoint ($O$) of $AB$, so, "all we need" is a relation among $|AC|$, $|BC|$, $|OC|$, and, presumably, $|AB|$ (or half thereof). Stewart/Apollonius gives that relation, wherein the constancy of $|AC|^2+|BC|^2$ (and $|AB|$) immediately implies the constancy of $|OC|$. Done. That said, I'm actually inclined to believe that there's a more-satisfying diagrammatic demonstration of this fact.
$endgroup$
– Blue
Dec 19 '18 at 7:32
1
$begingroup$
I edited my query re your suggestion about clarifying my desire. I agree with the logic of your latest comment. My use of the adjective convoluted merely signifies that even if I had been aware of Appollonius' (triangle oriented) theorem, I could have easily overlooked that it can be used to show that the Locus is a circle. Further, since the Locus is a circle, my initial intuition is that some property of circles must be involved. This intuition is similar to the last sentence in your latest comment.
$endgroup$
– user2661923
Dec 19 '18 at 7:48
add a comment |
$begingroup$
Here's an alternative (but still very "algebraic") derivation that leverages Thales' theorem (the angle inscribed in a semicircle is right):
$$begin{align}
a^2+b^2
&= left(m^2+p^2right)+left(n^2+q^2right) \
&=left(m^2+n^2right)+left(p^2+q^2right) \
&=left(r+kright)^2+left(r-kright)^2 \
&=2left(r^2+k^2right)
end{align}$$
This result, which relates the length of a median to other segments in $triangle ABC$, is Apollonius's Theorem. It's a special case of Stewart's Theorem for the length of an arbitrary cevian.
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
$endgroup$
$begingroup$
unfortunately, I have to accept your answer as on-point. However, examine my question from the point of view of background. The idea is that given $;z_1, z_2 inmathbb{C};$, it is desired to predict in advance (without algebra) that the Locus of $;{z;:|z-z_1|^2 + |z-z_2|^2 = ;$a constant$};$ will be a circle. It takes some convoluted logic to use Appollonius' theorem to reach that conclusion.
$endgroup$
– user2661923
Dec 19 '18 at 6:43
$begingroup$
You don't "have to" accept any answer that fails to satisfy you, so feel free to un-accept and encourage others to answer. (One should usually allow a bit of time before accepting, anyway.) That said, you could/should make your desire clearer by asking up-front: "How might we anticipate that $|AC|^2+|BC|^2=text{constant}$ represents a circle?"
$endgroup$
– Blue
Dec 19 '18 at 7:00
$begingroup$
Good pt, re clarifying my desire; it never occurred to me that an on-point geometric theorem would exist that didn't easily fulfill my underlying desire. Re accept'g your answer, I feel compelled to for two reasons: (1) Your answer is directly on-point and can fulfill my underlying desire, though convoluted logic is involved. (2) Given the existence of Appollonius' (triangle-oriented) theorem, I strongly suspect that there will not be a separate circle-oriented geometry theorem that one might use. Assuming that my suspicion is correct, then your answer is the best I'll get.
$endgroup$
– user2661923
Dec 19 '18 at 7:20
$begingroup$
As to the logic being convoluted ... I don't think it is, necessarily. Symmetry requires that the center of the circle be the midpoint ($O$) of $AB$, so, "all we need" is a relation among $|AC|$, $|BC|$, $|OC|$, and, presumably, $|AB|$ (or half thereof). Stewart/Apollonius gives that relation, wherein the constancy of $|AC|^2+|BC|^2$ (and $|AB|$) immediately implies the constancy of $|OC|$. Done. That said, I'm actually inclined to believe that there's a more-satisfying diagrammatic demonstration of this fact.
$endgroup$
– Blue
Dec 19 '18 at 7:32
1
$begingroup$
I edited my query re your suggestion about clarifying my desire. I agree with the logic of your latest comment. My use of the adjective convoluted merely signifies that even if I had been aware of Appollonius' (triangle oriented) theorem, I could have easily overlooked that it can be used to show that the Locus is a circle. Further, since the Locus is a circle, my initial intuition is that some property of circles must be involved. This intuition is similar to the last sentence in your latest comment.
$endgroup$
– user2661923
Dec 19 '18 at 7:48
add a comment |
$begingroup$
Here's an alternative (but still very "algebraic") derivation that leverages Thales' theorem (the angle inscribed in a semicircle is right):
$$begin{align}
a^2+b^2
&= left(m^2+p^2right)+left(n^2+q^2right) \
&=left(m^2+n^2right)+left(p^2+q^2right) \
&=left(r+kright)^2+left(r-kright)^2 \
&=2left(r^2+k^2right)
end{align}$$
This result, which relates the length of a median to other segments in $triangle ABC$, is Apollonius's Theorem. It's a special case of Stewart's Theorem for the length of an arbitrary cevian.
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
$endgroup$
Here's an alternative (but still very "algebraic") derivation that leverages Thales' theorem (the angle inscribed in a semicircle is right):
$$begin{align}
a^2+b^2
&= left(m^2+p^2right)+left(n^2+q^2right) \
&=left(m^2+n^2right)+left(p^2+q^2right) \
&=left(r+kright)^2+left(r-kright)^2 \
&=2left(r^2+k^2right)
end{align}$$
This result, which relates the length of a median to other segments in $triangle ABC$, is Apollonius's Theorem. It's a special case of Stewart's Theorem for the length of an arbitrary cevian.
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
edited Dec 19 '18 at 8:50
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
answered Dec 19 '18 at 6:36
BlueBlue
49.3k870157
49.3k870157
$begingroup$
unfortunately, I have to accept your answer as on-point. However, examine my question from the point of view of background. The idea is that given $;z_1, z_2 inmathbb{C};$, it is desired to predict in advance (without algebra) that the Locus of $;{z;:|z-z_1|^2 + |z-z_2|^2 = ;$a constant$};$ will be a circle. It takes some convoluted logic to use Appollonius' theorem to reach that conclusion.
$endgroup$
– user2661923
Dec 19 '18 at 6:43
$begingroup$
You don't "have to" accept any answer that fails to satisfy you, so feel free to un-accept and encourage others to answer. (One should usually allow a bit of time before accepting, anyway.) That said, you could/should make your desire clearer by asking up-front: "How might we anticipate that $|AC|^2+|BC|^2=text{constant}$ represents a circle?"
$endgroup$
– Blue
Dec 19 '18 at 7:00
$begingroup$
Good pt, re clarifying my desire; it never occurred to me that an on-point geometric theorem would exist that didn't easily fulfill my underlying desire. Re accept'g your answer, I feel compelled to for two reasons: (1) Your answer is directly on-point and can fulfill my underlying desire, though convoluted logic is involved. (2) Given the existence of Appollonius' (triangle-oriented) theorem, I strongly suspect that there will not be a separate circle-oriented geometry theorem that one might use. Assuming that my suspicion is correct, then your answer is the best I'll get.
$endgroup$
– user2661923
Dec 19 '18 at 7:20
$begingroup$
As to the logic being convoluted ... I don't think it is, necessarily. Symmetry requires that the center of the circle be the midpoint ($O$) of $AB$, so, "all we need" is a relation among $|AC|$, $|BC|$, $|OC|$, and, presumably, $|AB|$ (or half thereof). Stewart/Apollonius gives that relation, wherein the constancy of $|AC|^2+|BC|^2$ (and $|AB|$) immediately implies the constancy of $|OC|$. Done. That said, I'm actually inclined to believe that there's a more-satisfying diagrammatic demonstration of this fact.
$endgroup$
– Blue
Dec 19 '18 at 7:32
1
$begingroup$
I edited my query re your suggestion about clarifying my desire. I agree with the logic of your latest comment. My use of the adjective convoluted merely signifies that even if I had been aware of Appollonius' (triangle oriented) theorem, I could have easily overlooked that it can be used to show that the Locus is a circle. Further, since the Locus is a circle, my initial intuition is that some property of circles must be involved. This intuition is similar to the last sentence in your latest comment.
$endgroup$
– user2661923
Dec 19 '18 at 7:48
add a comment |
$begingroup$
unfortunately, I have to accept your answer as on-point. However, examine my question from the point of view of background. The idea is that given $;z_1, z_2 inmathbb{C};$, it is desired to predict in advance (without algebra) that the Locus of $;{z;:|z-z_1|^2 + |z-z_2|^2 = ;$a constant$};$ will be a circle. It takes some convoluted logic to use Appollonius' theorem to reach that conclusion.
$endgroup$
– user2661923
Dec 19 '18 at 6:43
$begingroup$
You don't "have to" accept any answer that fails to satisfy you, so feel free to un-accept and encourage others to answer. (One should usually allow a bit of time before accepting, anyway.) That said, you could/should make your desire clearer by asking up-front: "How might we anticipate that $|AC|^2+|BC|^2=text{constant}$ represents a circle?"
$endgroup$
– Blue
Dec 19 '18 at 7:00
$begingroup$
Good pt, re clarifying my desire; it never occurred to me that an on-point geometric theorem would exist that didn't easily fulfill my underlying desire. Re accept'g your answer, I feel compelled to for two reasons: (1) Your answer is directly on-point and can fulfill my underlying desire, though convoluted logic is involved. (2) Given the existence of Appollonius' (triangle-oriented) theorem, I strongly suspect that there will not be a separate circle-oriented geometry theorem that one might use. Assuming that my suspicion is correct, then your answer is the best I'll get.
$endgroup$
– user2661923
Dec 19 '18 at 7:20
$begingroup$
As to the logic being convoluted ... I don't think it is, necessarily. Symmetry requires that the center of the circle be the midpoint ($O$) of $AB$, so, "all we need" is a relation among $|AC|$, $|BC|$, $|OC|$, and, presumably, $|AB|$ (or half thereof). Stewart/Apollonius gives that relation, wherein the constancy of $|AC|^2+|BC|^2$ (and $|AB|$) immediately implies the constancy of $|OC|$. Done. That said, I'm actually inclined to believe that there's a more-satisfying diagrammatic demonstration of this fact.
$endgroup$
– Blue
Dec 19 '18 at 7:32
1
$begingroup$
I edited my query re your suggestion about clarifying my desire. I agree with the logic of your latest comment. My use of the adjective convoluted merely signifies that even if I had been aware of Appollonius' (triangle oriented) theorem, I could have easily overlooked that it can be used to show that the Locus is a circle. Further, since the Locus is a circle, my initial intuition is that some property of circles must be involved. This intuition is similar to the last sentence in your latest comment.
$endgroup$
– user2661923
Dec 19 '18 at 7:48
$begingroup$
unfortunately, I have to accept your answer as on-point. However, examine my question from the point of view of background. The idea is that given $;z_1, z_2 inmathbb{C};$, it is desired to predict in advance (without algebra) that the Locus of $;{z;:|z-z_1|^2 + |z-z_2|^2 = ;$a constant$};$ will be a circle. It takes some convoluted logic to use Appollonius' theorem to reach that conclusion.
$endgroup$
– user2661923
Dec 19 '18 at 6:43
$begingroup$
unfortunately, I have to accept your answer as on-point. However, examine my question from the point of view of background. The idea is that given $;z_1, z_2 inmathbb{C};$, it is desired to predict in advance (without algebra) that the Locus of $;{z;:|z-z_1|^2 + |z-z_2|^2 = ;$a constant$};$ will be a circle. It takes some convoluted logic to use Appollonius' theorem to reach that conclusion.
$endgroup$
– user2661923
Dec 19 '18 at 6:43
$begingroup$
You don't "have to" accept any answer that fails to satisfy you, so feel free to un-accept and encourage others to answer. (One should usually allow a bit of time before accepting, anyway.) That said, you could/should make your desire clearer by asking up-front: "How might we anticipate that $|AC|^2+|BC|^2=text{constant}$ represents a circle?"
$endgroup$
– Blue
Dec 19 '18 at 7:00
$begingroup$
You don't "have to" accept any answer that fails to satisfy you, so feel free to un-accept and encourage others to answer. (One should usually allow a bit of time before accepting, anyway.) That said, you could/should make your desire clearer by asking up-front: "How might we anticipate that $|AC|^2+|BC|^2=text{constant}$ represents a circle?"
$endgroup$
– Blue
Dec 19 '18 at 7:00
$begingroup$
Good pt, re clarifying my desire; it never occurred to me that an on-point geometric theorem would exist that didn't easily fulfill my underlying desire. Re accept'g your answer, I feel compelled to for two reasons: (1) Your answer is directly on-point and can fulfill my underlying desire, though convoluted logic is involved. (2) Given the existence of Appollonius' (triangle-oriented) theorem, I strongly suspect that there will not be a separate circle-oriented geometry theorem that one might use. Assuming that my suspicion is correct, then your answer is the best I'll get.
$endgroup$
– user2661923
Dec 19 '18 at 7:20
$begingroup$
Good pt, re clarifying my desire; it never occurred to me that an on-point geometric theorem would exist that didn't easily fulfill my underlying desire. Re accept'g your answer, I feel compelled to for two reasons: (1) Your answer is directly on-point and can fulfill my underlying desire, though convoluted logic is involved. (2) Given the existence of Appollonius' (triangle-oriented) theorem, I strongly suspect that there will not be a separate circle-oriented geometry theorem that one might use. Assuming that my suspicion is correct, then your answer is the best I'll get.
$endgroup$
– user2661923
Dec 19 '18 at 7:20
$begingroup$
As to the logic being convoluted ... I don't think it is, necessarily. Symmetry requires that the center of the circle be the midpoint ($O$) of $AB$, so, "all we need" is a relation among $|AC|$, $|BC|$, $|OC|$, and, presumably, $|AB|$ (or half thereof). Stewart/Apollonius gives that relation, wherein the constancy of $|AC|^2+|BC|^2$ (and $|AB|$) immediately implies the constancy of $|OC|$. Done. That said, I'm actually inclined to believe that there's a more-satisfying diagrammatic demonstration of this fact.
$endgroup$
– Blue
Dec 19 '18 at 7:32
$begingroup$
As to the logic being convoluted ... I don't think it is, necessarily. Symmetry requires that the center of the circle be the midpoint ($O$) of $AB$, so, "all we need" is a relation among $|AC|$, $|BC|$, $|OC|$, and, presumably, $|AB|$ (or half thereof). Stewart/Apollonius gives that relation, wherein the constancy of $|AC|^2+|BC|^2$ (and $|AB|$) immediately implies the constancy of $|OC|$. Done. That said, I'm actually inclined to believe that there's a more-satisfying diagrammatic demonstration of this fact.
$endgroup$
– Blue
Dec 19 '18 at 7:32
1
1
$begingroup$
I edited my query re your suggestion about clarifying my desire. I agree with the logic of your latest comment. My use of the adjective convoluted merely signifies that even if I had been aware of Appollonius' (triangle oriented) theorem, I could have easily overlooked that it can be used to show that the Locus is a circle. Further, since the Locus is a circle, my initial intuition is that some property of circles must be involved. This intuition is similar to the last sentence in your latest comment.
$endgroup$
– user2661923
Dec 19 '18 at 7:48
$begingroup$
I edited my query re your suggestion about clarifying my desire. I agree with the logic of your latest comment. My use of the adjective convoluted merely signifies that even if I had been aware of Appollonius' (triangle oriented) theorem, I could have easily overlooked that it can be used to show that the Locus is a circle. Further, since the Locus is a circle, my initial intuition is that some property of circles must be involved. This intuition is similar to the last sentence in your latest comment.
$endgroup$
– user2661923
Dec 19 '18 at 7:48
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%2f3045762%2fcuriosity-in-circular-analytical-geometry%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
$begingroup$
This can be considered simple application of the Law of Cosines and the fact that supplementary angles have negative cosines. $$begin{align}0 &= 2kr (cosangle COA+cosangle COB) \ &= (|AC|^2-k^2-r^2)+(|BC|^2-k^2-r^2) \ &=|AC|^2+|BC|^2-2(k^2+r^2)end{align}$$ (I don't know if you'd consider this too "algebraic".) I'm not aware of any name for this result; it doesn't seem to need one.
$endgroup$
– Blue
Dec 19 '18 at 4:27
$begingroup$
By the way, this answer of mine gives a geometric derivation of the Law of Cosines.
$endgroup$
– Blue
Dec 19 '18 at 4:44
$begingroup$
@Blue slick...(but still algebraic). I am actually surprised that my result isn't less obscure.
$endgroup$
– user2661923
Dec 19 '18 at 5:18
$begingroup$
I'll note that the result has nothing to do with a circle; rather, it's simply a triangle relation. In particular, it relates the length of a median to other segments in a triangle. In that regard, it's a special case of Stewart's Theorem. (Ah! As the Wikipedia entry notes, the median case does have a name: Apollonius's Theorem.)
$endgroup$
– Blue
Dec 19 '18 at 5:34