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
48.6k870156
$endgroup$
$begingroup$
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
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$begingroup$
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
48.6k870156
$endgroup$
$begingroup$
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
48.6k870156
$endgroup$
$begingroup$
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
48.6k870156
$endgroup$
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
48.6k870156
edited Dec 19 '18 at 8:50
answered Dec 19 '18 at 6:36
BlueBlue
48.6k870156
answered Dec 19 '18 at 6:36
BlueBlue
48.6k870156
answered Dec 19 '18 at 6:36
BlueBlue
48.6k870156
48.6k870156
$begingroup$
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 |
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$begingroup$
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