Computing a Geometrical ratio $fracab$.
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up vote
1
down vote
favorite
$XYZ$ is an equilateral triangle as shown on the image below.
The aim is to find the ratio $fracab$.
So far from the picture, it is easy to see that $b= fracYZ2$.
Does anyone have an idea on how to find $b$ in term of $a$?
It seems from discussion with my classmates that this might lead to another geometrical interpretation of the Golden ratio.
algebra-precalculus euclidean-geometry affine-geometry golden-ratio
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
add a comment |Â
up vote
1
down vote
favorite
$XYZ$ is an equilateral triangle as shown on the image below.
The aim is to find the ratio $fracab$.
So far from the picture, it is easy to see that $b= fracYZ2$.
Does anyone have an idea on how to find $b$ in term of $a$?
It seems from discussion with my classmates that this might lead to another geometrical interpretation of the Golden ratio.
algebra-precalculus euclidean-geometry affine-geometry golden-ratio
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
How do you see $b=(yz)/2$? And, what does this have to do with the golden ratio?
– Gerry Myerson
Sep 6 at 7:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$XYZ$ is an equilateral triangle as shown on the image below.
The aim is to find the ratio $fracab$.
So far from the picture, it is easy to see that $b= fracYZ2$.
Does anyone have an idea on how to find $b$ in term of $a$?
It seems from discussion with my classmates that this might lead to another geometrical interpretation of the Golden ratio.
algebra-precalculus euclidean-geometry affine-geometry golden-ratio
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
$XYZ$ is an equilateral triangle as shown on the image below.
The aim is to find the ratio $fracab$.
So far from the picture, it is easy to see that $b= fracYZ2$.
Does anyone have an idea on how to find $b$ in term of $a$?
It seems from discussion with my classmates that this might lead to another geometrical interpretation of the Golden ratio.
algebra-precalculus euclidean-geometry affine-geometry golden-ratio
algebra-precalculus euclidean-geometry affine-geometry golden-ratio
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
edited Sep 6 at 12:20
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
asked Sep 6 at 6:39
Guy Fsone
16.9k42672
16.9k42672
How do you see $b=(yz)/2$? And, what does this have to do with the golden ratio?
– Gerry Myerson
Sep 6 at 7:00
add a comment |Â
How do you see $b=(yz)/2$? And, what does this have to do with the golden ratio?
– Gerry Myerson
Sep 6 at 7:00
How do you see $b=(yz)/2$? And, what does this have to do with the golden ratio?
– Gerry Myerson
Sep 6 at 7:00
How do you see $b=(yz)/2$? And, what does this have to do with the golden ratio?
– Gerry Myerson
Sep 6 at 7:00
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Another way to find the ratio
is to solve
the $triangle ROS$,
for which we know two sides and an angle:
beginalign
triangle ROS:quad
|OR|&=r=tfrac13cdot2bcdottfracsqrt32
=tfracsqrt33,b
,\
|OS|&=2r=tfrac2sqrt33,b
,\
angle SRO&=angle ZRO+angle SRZ=90^circ+60^circ
=150^circ
,
endalign
so we can apply the cosine rule to get
beginalign
|OS|^2&=|OR|^2+|RS|^2-2cdot|OR|cdot|RS|cosangle SRO
,\
tfrac43,b^2&=
tfrac13,b^2+a^2-2cdottfracsqrt33,bcdot acdotcos150^circ
,\
tfrac43,b^2&=
tfrac13,b^2+a^2+2cdottfracsqrt33,bcdot acdotcos30^circ
,
endalign
beginalign
b^2
-bcdot a
-a^2
&=0
,\
(tfracba)^2
-
tfracba
-1
&=0
,
endalign
which gives the answer (positive root) as
beginalign
fracba
&=frac1+sqrt52
,
endalign
which is indeed the Golden ratio.
answered Sep 6 at 14:45
g.kov
5,7771718
can you elaborate on how you compute OR?
– Guy Fsone
Sep 7 at 8:42
@GuyFsone: $|OR|=r$ is the inradius of the equilateral triangle, and as such it is equal to one third of the height and the circumradius $R$ is twice as big. As far as the side $|XY|=|YZ|=|ZX|=2b$, the height is $h=2bsin60^circ=sqrt3, b$, then $|OR|=|OQ|=r=tfracsqrt33, b$.
– g.kov
Sep 7 at 10:01
add a comment |Â
up vote
3
down vote
Use the power of the point $R$ (or $Q$) with respect to the big circle.:
$$ (a+b)cdot a = s^2$$
where $s= YZ/2$.
Note that since $QRX$ is equlateral we have $s=b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q=1pm sqrt5over2$$
answered Sep 6 at 7:01
greedoid
28.7k93878
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Another way to find the ratio
is to solve
the $triangle ROS$,
for which we know two sides and an angle:
beginalign
triangle ROS:quad
|OR|&=r=tfrac13cdot2bcdottfracsqrt32
=tfracsqrt33,b
,\
|OS|&=2r=tfrac2sqrt33,b
,\
angle SRO&=angle ZRO+angle SRZ=90^circ+60^circ
=150^circ
,
endalign
so we can apply the cosine rule to get
beginalign
|OS|^2&=|OR|^2+|RS|^2-2cdot|OR|cdot|RS|cosangle SRO
,\
tfrac43,b^2&=
tfrac13,b^2+a^2-2cdottfracsqrt33,bcdot acdotcos150^circ
,\
tfrac43,b^2&=
tfrac13,b^2+a^2+2cdottfracsqrt33,bcdot acdotcos30^circ
,
endalign
beginalign
b^2
-bcdot a
-a^2
&=0
,\
(tfracba)^2
-
tfracba
-1
&=0
,
endalign
which gives the answer (positive root) as
beginalign
fracba
&=frac1+sqrt52
,
endalign
which is indeed the Golden ratio.
answered Sep 6 at 14:45
g.kov
5,7771718
can you elaborate on how you compute OR?
– Guy Fsone
Sep 7 at 8:42
@GuyFsone: $|OR|=r$ is the inradius of the equilateral triangle, and as such it is equal to one third of the height and the circumradius $R$ is twice as big. As far as the side $|XY|=|YZ|=|ZX|=2b$, the height is $h=2bsin60^circ=sqrt3, b$, then $|OR|=|OQ|=r=tfracsqrt33, b$.
– g.kov
Sep 7 at 10:01
add a comment |Â
up vote
1
down vote
accepted
Another way to find the ratio
is to solve
the $triangle ROS$,
for which we know two sides and an angle:
beginalign
triangle ROS:quad
|OR|&=r=tfrac13cdot2bcdottfracsqrt32
=tfracsqrt33,b
,\
|OS|&=2r=tfrac2sqrt33,b
,\
angle SRO&=angle ZRO+angle SRZ=90^circ+60^circ
=150^circ
,
endalign
so we can apply the cosine rule to get
beginalign
|OS|^2&=|OR|^2+|RS|^2-2cdot|OR|cdot|RS|cosangle SRO
,\
tfrac43,b^2&=
tfrac13,b^2+a^2-2cdottfracsqrt33,bcdot acdotcos150^circ
,\
tfrac43,b^2&=
tfrac13,b^2+a^2+2cdottfracsqrt33,bcdot acdotcos30^circ
,
endalign
beginalign
b^2
-bcdot a
-a^2
&=0
,\
(tfracba)^2
-
tfracba
-1
&=0
,
endalign
which gives the answer (positive root) as
beginalign
fracba
&=frac1+sqrt52
,
endalign
which is indeed the Golden ratio.
answered Sep 6 at 14:45
g.kov
5,7771718
can you elaborate on how you compute OR?
– Guy Fsone
Sep 7 at 8:42
@GuyFsone: $|OR|=r$ is the inradius of the equilateral triangle, and as such it is equal to one third of the height and the circumradius $R$ is twice as big. As far as the side $|XY|=|YZ|=|ZX|=2b$, the height is $h=2bsin60^circ=sqrt3, b$, then $|OR|=|OQ|=r=tfracsqrt33, b$.
– g.kov
Sep 7 at 10:01
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Another way to find the ratio
is to solve
the $triangle ROS$,
for which we know two sides and an angle:
beginalign
triangle ROS:quad
|OR|&=r=tfrac13cdot2bcdottfracsqrt32
=tfracsqrt33,b
,\
|OS|&=2r=tfrac2sqrt33,b
,\
angle SRO&=angle ZRO+angle SRZ=90^circ+60^circ
=150^circ
,
endalign
so we can apply the cosine rule to get
beginalign
|OS|^2&=|OR|^2+|RS|^2-2cdot|OR|cdot|RS|cosangle SRO
,\
tfrac43,b^2&=
tfrac13,b^2+a^2-2cdottfracsqrt33,bcdot acdotcos150^circ
,\
tfrac43,b^2&=
tfrac13,b^2+a^2+2cdottfracsqrt33,bcdot acdotcos30^circ
,
endalign
beginalign
b^2
-bcdot a
-a^2
&=0
,\
(tfracba)^2
-
tfracba
-1
&=0
,
endalign
which gives the answer (positive root) as
beginalign
fracba
&=frac1+sqrt52
,
endalign
which is indeed the Golden ratio.
answered Sep 6 at 14:45
g.kov
5,7771718
Another way to find the ratio
is to solve
the $triangle ROS$,
for which we know two sides and an angle:
beginalign
triangle ROS:quad
|OR|&=r=tfrac13cdot2bcdottfracsqrt32
=tfracsqrt33,b
,\
|OS|&=2r=tfrac2sqrt33,b
,\
angle SRO&=angle ZRO+angle SRZ=90^circ+60^circ
=150^circ
,
endalign
so we can apply the cosine rule to get
beginalign
|OS|^2&=|OR|^2+|RS|^2-2cdot|OR|cdot|RS|cosangle SRO
,\
tfrac43,b^2&=
tfrac13,b^2+a^2-2cdottfracsqrt33,bcdot acdotcos150^circ
,\
tfrac43,b^2&=
tfrac13,b^2+a^2+2cdottfracsqrt33,bcdot acdotcos30^circ
,
endalign
beginalign
b^2
-bcdot a
-a^2
&=0
,\
(tfracba)^2
-
tfracba
-1
&=0
,
endalign
which gives the answer (positive root) as
beginalign
fracba
&=frac1+sqrt52
,
endalign
which is indeed the Golden ratio.
answered Sep 6 at 14:45
g.kov
5,7771718
answered Sep 6 at 14:45
g.kov
5,7771718
answered Sep 6 at 14:45
g.kov
5,7771718
answered Sep 6 at 14:45
g.kov
5,7771718
5,7771718
can you elaborate on how you compute OR?
– Guy Fsone
Sep 7 at 8:42
@GuyFsone: $|OR|=r$ is the inradius of the equilateral triangle, and as such it is equal to one third of the height and the circumradius $R$ is twice as big. As far as the side $|XY|=|YZ|=|ZX|=2b$, the height is $h=2bsin60^circ=sqrt3, b$, then $|OR|=|OQ|=r=tfracsqrt33, b$.
– g.kov
Sep 7 at 10:01
add a comment |Â
can you elaborate on how you compute OR?
– Guy Fsone
Sep 7 at 8:42
@GuyFsone: $|OR|=r$ is the inradius of the equilateral triangle, and as such it is equal to one third of the height and the circumradius $R$ is twice as big. As far as the side $|XY|=|YZ|=|ZX|=2b$, the height is $h=2bsin60^circ=sqrt3, b$, then $|OR|=|OQ|=r=tfracsqrt33, b$.
– g.kov
Sep 7 at 10:01
can you elaborate on how you compute OR?
– Guy Fsone
Sep 7 at 8:42
can you elaborate on how you compute OR?
– Guy Fsone
Sep 7 at 8:42
@GuyFsone: $|OR|=r$ is the inradius of the equilateral triangle, and as such it is equal to one third of the height and the circumradius $R$ is twice as big. As far as the side $|XY|=|YZ|=|ZX|=2b$, the height is $h=2bsin60^circ=sqrt3, b$, then $|OR|=|OQ|=r=tfracsqrt33, b$.
– g.kov
Sep 7 at 10:01
@GuyFsone: $|OR|=r$ is the inradius of the equilateral triangle, and as such it is equal to one third of the height and the circumradius $R$ is twice as big. As far as the side $|XY|=|YZ|=|ZX|=2b$, the height is $h=2bsin60^circ=sqrt3, b$, then $|OR|=|OQ|=r=tfracsqrt33, b$.
– g.kov
Sep 7 at 10:01
add a comment |Â
up vote
3
down vote
Use the power of the point $R$ (or $Q$) with respect to the big circle.:
$$ (a+b)cdot a = s^2$$
where $s= YZ/2$.
Note that since $QRX$ is equlateral we have $s=b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q=1pm sqrt5over2$$
answered Sep 6 at 7:01
greedoid
28.7k93878
add a comment |Â
up vote
3
down vote
Use the power of the point $R$ (or $Q$) with respect to the big circle.:
$$ (a+b)cdot a = s^2$$
where $s= YZ/2$.
Note that since $QRX$ is equlateral we have $s=b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q=1pm sqrt5over2$$
answered Sep 6 at 7:01
greedoid
28.7k93878
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Use the power of the point $R$ (or $Q$) with respect to the big circle.:
$$ (a+b)cdot a = s^2$$
where $s= YZ/2$.
Note that since $QRX$ is equlateral we have $s=b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q=1pm sqrt5over2$$
answered Sep 6 at 7:01
greedoid
28.7k93878
Use the power of the point $R$ (or $Q$) with respect to the big circle.:
$$ (a+b)cdot a = s^2$$
where $s= YZ/2$.
Note that since $QRX$ is equlateral we have $s=b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q=1pm sqrt5over2$$
answered Sep 6 at 7:01
greedoid
28.7k93878
edited Sep 6 at 7:07
answered Sep 6 at 7:01
greedoid
28.7k93878
answered Sep 6 at 7:01
greedoid
28.7k93878
answered Sep 6 at 7:01
greedoid
28.7k93878
28.7k93878
add a comment |Â
add a comment |Â
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How do you see $b=(yz)/2$? And, what does this have to do with the golden ratio?
– Gerry Myerson
Sep 6 at 7:00