Sangaku: Find the Radii of the Inner Circles
up vote
2
down vote
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Sangaku (算額) are Japanese geometric puzzles written on wooden tablets over 150 years ago. There have been several previous puzzles, but I didn't see this one.
Find the radii of the two inner circles in terms of $x$:
geometry puzzle sangaku
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
asked Jan 31 '14 at 6:44
jmac
1618
add a comment |
up vote
2
down vote
favorite
Sangaku (算額) are Japanese geometric puzzles written on wooden tablets over 150 years ago. There have been several previous puzzles, but I didn't see this one.
Find the radii of the two inner circles in terms of $x$:
geometry puzzle sangaku
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
asked Jan 31 '14 at 6:44
jmac
1618
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Sangaku (算額) are Japanese geometric puzzles written on wooden tablets over 150 years ago. There have been several previous puzzles, but I didn't see this one.
Find the radii of the two inner circles in terms of $x$:
geometry puzzle sangaku
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
asked Jan 31 '14 at 6:44
jmac
1618
Sangaku (算額) are Japanese geometric puzzles written on wooden tablets over 150 years ago. There have been several previous puzzles, but I didn't see this one.
Find the radii of the two inner circles in terms of $x$:
geometry puzzle sangaku
geometry puzzle sangaku
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
asked Jan 31 '14 at 6:44
jmac
1618
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
asked Jan 31 '14 at 6:44
jmac
1618
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
edited Nov 22 at 20:59
Jean-Claude Arbaut
14.8k63362
14.8k63362
asked Jan 31 '14 at 6:44
jmac
1618
asked Jan 31 '14 at 6:44
jmac
1618
asked Jan 31 '14 at 6:44
jmac
1618
1618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Let the radius of the smaller circle be $r$, and the larger be $R$. It is immediately obvious that $r$ satisfies $$(x-r)^2 + (x/2)^2 = (r+x)^2,$$ and $R$ satisfies $$(x/2+R)^2 + (x/2)^2 = (x-R)^2.$$ The solutions of these are a straightforward algebraic exercise.
Slightly less trivial would be finding the radius of the circle inscribed in either of the two "ear-shaped" regions (inside the semicircle and one of the two quarter circles, but outside the other quarter circle). However, it too is amenable to the same solving technique.
Addendum. Since an image was requested, I have attached it below: 
The horizontal red line has length $x/2$, and the diagonal line has length $r+x$, since it joins the centers of two tangent circles with radius $r$ (blue) and $x$ (green/yellow). The third side is simply $x - r$, because the red horizontal line is parallel to the top edge of the square.
An analogous procedure for the larger circle results in the second equation. It really does not get any easier than that.
answered Jan 31 '14 at 6:51
heropup
62k65998
What is $x$ in your solution?
– nbubis
Jan 31 '14 at 7:00
As shown in the diagram, $x$ is the side length of the square.
– heropup
Jan 31 '14 at 7:00
Sorry, I think I'm a bit dim here and/or I need a bit more explanation (which you are of course not obligated to give). Where are you getting these formulas from? Any chance you could attach some words to the concept here, because I have absolutely no idea where these came from.
– jmac
Jan 31 '14 at 7:03
@jmac, elementary analytic geometry after choosing a system of coordinates.
– Martín-Blas Pérez Pinilla
Jan 31 '14 at 7:07
Use the Pythagorean theorem for a suitable right triangle. For the small circle, draw a line from the circle's center to the lower right vertex, and then drop a perpendicular from the circle's center to the right edge of the square. For the large circle, I leave it to you to find the appropriate triangle. Look at the equations.
– heropup
Jan 31 '14 at 7:07
|
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let the radius of the smaller circle be $r$, and the larger be $R$. It is immediately obvious that $r$ satisfies $$(x-r)^2 + (x/2)^2 = (r+x)^2,$$ and $R$ satisfies $$(x/2+R)^2 + (x/2)^2 = (x-R)^2.$$ The solutions of these are a straightforward algebraic exercise.
Slightly less trivial would be finding the radius of the circle inscribed in either of the two "ear-shaped" regions (inside the semicircle and one of the two quarter circles, but outside the other quarter circle). However, it too is amenable to the same solving technique.
Addendum. Since an image was requested, I have attached it below:
The horizontal red line has length $x/2$, and the diagonal line has length $r+x$, since it joins the centers of two tangent circles with radius $r$ (blue) and $x$ (green/yellow). The third side is simply $x - r$, because the red horizontal line is parallel to the top edge of the square.
An analogous procedure for the larger circle results in the second equation. It really does not get any easier than that.
answered Jan 31 '14 at 6:51
heropup
62k65998
What is $x$ in your solution?
– nbubis
Jan 31 '14 at 7:00
As shown in the diagram, $x$ is the side length of the square.
– heropup
Jan 31 '14 at 7:00
Sorry, I think I'm a bit dim here and/or I need a bit more explanation (which you are of course not obligated to give). Where are you getting these formulas from? Any chance you could attach some words to the concept here, because I have absolutely no idea where these came from.
– jmac
Jan 31 '14 at 7:03
@jmac, elementary analytic geometry after choosing a system of coordinates.
– Martín-Blas Pérez Pinilla
Jan 31 '14 at 7:07
Use the Pythagorean theorem for a suitable right triangle. For the small circle, draw a line from the circle's center to the lower right vertex, and then drop a perpendicular from the circle's center to the right edge of the square. For the large circle, I leave it to you to find the appropriate triangle. Look at the equations.
– heropup
Jan 31 '14 at 7:07
|
show 6 more comments
up vote
4
down vote
accepted
Let the radius of the smaller circle be $r$, and the larger be $R$. It is immediately obvious that $r$ satisfies $$(x-r)^2 + (x/2)^2 = (r+x)^2,$$ and $R$ satisfies $$(x/2+R)^2 + (x/2)^2 = (x-R)^2.$$ The solutions of these are a straightforward algebraic exercise.
Slightly less trivial would be finding the radius of the circle inscribed in either of the two "ear-shaped" regions (inside the semicircle and one of the two quarter circles, but outside the other quarter circle). However, it too is amenable to the same solving technique.
Addendum. Since an image was requested, I have attached it below:
The horizontal red line has length $x/2$, and the diagonal line has length $r+x$, since it joins the centers of two tangent circles with radius $r$ (blue) and $x$ (green/yellow). The third side is simply $x - r$, because the red horizontal line is parallel to the top edge of the square.
An analogous procedure for the larger circle results in the second equation. It really does not get any easier than that.
answered Jan 31 '14 at 6:51
heropup
62k65998
What is $x$ in your solution?
– nbubis
Jan 31 '14 at 7:00
As shown in the diagram, $x$ is the side length of the square.
– heropup
Jan 31 '14 at 7:00
Sorry, I think I'm a bit dim here and/or I need a bit more explanation (which you are of course not obligated to give). Where are you getting these formulas from? Any chance you could attach some words to the concept here, because I have absolutely no idea where these came from.
– jmac
Jan 31 '14 at 7:03
@jmac, elementary analytic geometry after choosing a system of coordinates.
– Martín-Blas Pérez Pinilla
Jan 31 '14 at 7:07
Use the Pythagorean theorem for a suitable right triangle. For the small circle, draw a line from the circle's center to the lower right vertex, and then drop a perpendicular from the circle's center to the right edge of the square. For the large circle, I leave it to you to find the appropriate triangle. Look at the equations.
– heropup
Jan 31 '14 at 7:07
|
show 6 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let the radius of the smaller circle be $r$, and the larger be $R$. It is immediately obvious that $r$ satisfies $$(x-r)^2 + (x/2)^2 = (r+x)^2,$$ and $R$ satisfies $$(x/2+R)^2 + (x/2)^2 = (x-R)^2.$$ The solutions of these are a straightforward algebraic exercise.
Slightly less trivial would be finding the radius of the circle inscribed in either of the two "ear-shaped" regions (inside the semicircle and one of the two quarter circles, but outside the other quarter circle). However, it too is amenable to the same solving technique.
Addendum. Since an image was requested, I have attached it below:
The horizontal red line has length $x/2$, and the diagonal line has length $r+x$, since it joins the centers of two tangent circles with radius $r$ (blue) and $x$ (green/yellow). The third side is simply $x - r$, because the red horizontal line is parallel to the top edge of the square.
An analogous procedure for the larger circle results in the second equation. It really does not get any easier than that.
answered Jan 31 '14 at 6:51
heropup
62k65998
Let the radius of the smaller circle be $r$, and the larger be $R$. It is immediately obvious that $r$ satisfies $$(x-r)^2 + (x/2)^2 = (r+x)^2,$$ and $R$ satisfies $$(x/2+R)^2 + (x/2)^2 = (x-R)^2.$$ The solutions of these are a straightforward algebraic exercise.
Slightly less trivial would be finding the radius of the circle inscribed in either of the two "ear-shaped" regions (inside the semicircle and one of the two quarter circles, but outside the other quarter circle). However, it too is amenable to the same solving technique.
Addendum. Since an image was requested, I have attached it below:
The horizontal red line has length $x/2$, and the diagonal line has length $r+x$, since it joins the centers of two tangent circles with radius $r$ (blue) and $x$ (green/yellow). The third side is simply $x - r$, because the red horizontal line is parallel to the top edge of the square.
An analogous procedure for the larger circle results in the second equation. It really does not get any easier than that.
answered Jan 31 '14 at 6:51
heropup
62k65998
edited Jan 31 '14 at 7:48
answered Jan 31 '14 at 6:51
heropup
62k65998
answered Jan 31 '14 at 6:51
heropup
62k65998
answered Jan 31 '14 at 6:51
heropup
62k65998
62k65998
What is $x$ in your solution?
– nbubis
Jan 31 '14 at 7:00
As shown in the diagram, $x$ is the side length of the square.
– heropup
Jan 31 '14 at 7:00
Sorry, I think I'm a bit dim here and/or I need a bit more explanation (which you are of course not obligated to give). Where are you getting these formulas from? Any chance you could attach some words to the concept here, because I have absolutely no idea where these came from.
– jmac
Jan 31 '14 at 7:03
@jmac, elementary analytic geometry after choosing a system of coordinates.
– Martín-Blas Pérez Pinilla
Jan 31 '14 at 7:07
Use the Pythagorean theorem for a suitable right triangle. For the small circle, draw a line from the circle's center to the lower right vertex, and then drop a perpendicular from the circle's center to the right edge of the square. For the large circle, I leave it to you to find the appropriate triangle. Look at the equations.
– heropup
Jan 31 '14 at 7:07
|
show 6 more comments
What is $x$ in your solution?
– nbubis
Jan 31 '14 at 7:00
As shown in the diagram, $x$ is the side length of the square.
– heropup
Jan 31 '14 at 7:00
Sorry, I think I'm a bit dim here and/or I need a bit more explanation (which you are of course not obligated to give). Where are you getting these formulas from? Any chance you could attach some words to the concept here, because I have absolutely no idea where these came from.
– jmac
Jan 31 '14 at 7:03
@jmac, elementary analytic geometry after choosing a system of coordinates.
– Martín-Blas Pérez Pinilla
Jan 31 '14 at 7:07
Use the Pythagorean theorem for a suitable right triangle. For the small circle, draw a line from the circle's center to the lower right vertex, and then drop a perpendicular from the circle's center to the right edge of the square. For the large circle, I leave it to you to find the appropriate triangle. Look at the equations.
– heropup
Jan 31 '14 at 7:07
What is $x$ in your solution?
– nbubis
Jan 31 '14 at 7:00
What is $x$ in your solution?
– nbubis
Jan 31 '14 at 7:00
As shown in the diagram, $x$ is the side length of the square.
– heropup
Jan 31 '14 at 7:00
As shown in the diagram, $x$ is the side length of the square.
– heropup
Jan 31 '14 at 7:00
Sorry, I think I'm a bit dim here and/or I need a bit more explanation (which you are of course not obligated to give). Where are you getting these formulas from? Any chance you could attach some words to the concept here, because I have absolutely no idea where these came from.
– jmac
Jan 31 '14 at 7:03
Sorry, I think I'm a bit dim here and/or I need a bit more explanation (which you are of course not obligated to give). Where are you getting these formulas from? Any chance you could attach some words to the concept here, because I have absolutely no idea where these came from.
– jmac
Jan 31 '14 at 7:03
@jmac, elementary analytic geometry after choosing a system of coordinates.
– Martín-Blas Pérez Pinilla
Jan 31 '14 at 7:07
@jmac, elementary analytic geometry after choosing a system of coordinates.
– Martín-Blas Pérez Pinilla
Jan 31 '14 at 7:07
Use the Pythagorean theorem for a suitable right triangle. For the small circle, draw a line from the circle's center to the lower right vertex, and then drop a perpendicular from the circle's center to the right edge of the square. For the large circle, I leave it to you to find the appropriate triangle. Look at the equations.
– heropup
Jan 31 '14 at 7:07
Use the Pythagorean theorem for a suitable right triangle. For the small circle, draw a line from the circle's center to the lower right vertex, and then drop a perpendicular from the circle's center to the right edge of the square. For the large circle, I leave it to you to find the appropriate triangle. Look at the equations.
– heropup
Jan 31 '14 at 7:07
|
show 6 more comments
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