Japanese Temple Geometry
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Hello, I was trying to solve this problem using descarte circle theorem for my maths report. I looked through the solution but I don't understand the part in the answer, where it says the two solutions are pn+1, pn-1. Can someone explain it for me. Thanks!
geometry euclidean-geometry circle quadratics sangaku
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
asked Jun 12 at 13:08
Rosa Kang
161
add a comment |
up vote
3
down vote
favorite
Hello, I was trying to solve this problem using descarte circle theorem for my maths report. I looked through the solution but I don't understand the part in the answer, where it says the two solutions are pn+1, pn-1. Can someone explain it for me. Thanks!
geometry euclidean-geometry circle quadratics sangaku
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
asked Jun 12 at 13:08
Rosa Kang
161
Instead of pictures, please type out the problem definition and solution, and please use MathJax when typing the equations. This way, your question is much more readible and will more likely get answers.
– Matti P.
Jun 12 at 13:16
What are those "remarks above"?
– Aretino
Jun 12 at 13:59
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Hello, I was trying to solve this problem using descarte circle theorem for my maths report. I looked through the solution but I don't understand the part in the answer, where it says the two solutions are pn+1, pn-1. Can someone explain it for me. Thanks!
geometry euclidean-geometry circle quadratics sangaku
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
asked Jun 12 at 13:08
Rosa Kang
161
Hello, I was trying to solve this problem using descarte circle theorem for my maths report. I looked through the solution but I don't understand the part in the answer, where it says the two solutions are pn+1, pn-1. Can someone explain it for me. Thanks!
geometry euclidean-geometry circle quadratics sangaku
geometry euclidean-geometry circle quadratics sangaku
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
asked Jun 12 at 13:08
Rosa Kang
161
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
asked Jun 12 at 13:08
Rosa Kang
161
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
edited Nov 23 at 6:34
Jean-Claude Arbaut
14.7k63363
14.7k63363
asked Jun 12 at 13:08
Rosa Kang
161
asked Jun 12 at 13:08
Rosa Kang
161
asked Jun 12 at 13:08
Rosa Kang
161
161
Instead of pictures, please type out the problem definition and solution, and please use MathJax when typing the equations. This way, your question is much more readible and will more likely get answers.
– Matti P.
Jun 12 at 13:16
What are those "remarks above"?
– Aretino
Jun 12 at 13:59
add a comment |
Instead of pictures, please type out the problem definition and solution, and please use MathJax when typing the equations. This way, your question is much more readible and will more likely get answers.
– Matti P.
Jun 12 at 13:16
What are those "remarks above"?
– Aretino
Jun 12 at 13:59
Instead of pictures, please type out the problem definition and solution, and please use MathJax when typing the equations. This way, your question is much more readible and will more likely get answers.
– Matti P.
Jun 12 at 13:16
Instead of pictures, please type out the problem definition and solution, and please use MathJax when typing the equations. This way, your question is much more readible and will more likely get answers.
– Matti P.
Jun 12 at 13:16
What are those "remarks above"?
– Aretino
Jun 12 at 13:59
What are those "remarks above"?
– Aretino
Jun 12 at 13:59
add a comment |
1 Answer
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Consider this form of the equation
$$2(p_1^2 + p_n^2 + p_{n+1}^2 + a^2 ) = ( p_1 + p_n + p_{n+1} - a )^2 tag{1}$$
and take it back a step by replacing $nto n-1$:
$$2(p_1^2 + p_{n-1}^2 + p_{n}^2 + a^2 ) = ( p_1 + p_{n-1} + p_{n} - a )^2 tag{2}$$
We see that $(2)$ is identical to $(1)$, except that $p_{n-1}$ replaces $p_{n+1}$. Therefore, $p_{n+1}$ and $p_{n-1}$ are the roots of the monic quadratic equation
$$2(p_1^2 + p_{n}^2 + x^2 + a^2 ) = ( p_1 + p_{n} + x - a )^2 tag{3}$$
(Perhaps this trick is discussed in the "remarks above" or Yoshida's solution.) From there, Vieta's formulas allow us to write the coefficient of $x$ in terms of those roots, which gives this recursion:
$$p_{n-1} + p_{n+1} = -2(a-p_1-p_n)$$
Since $p_1$ has evidently been assigned the value of $2a$ in the quoted solution, we get the recursion in the stated form. $square$
answered Jun 12 at 15:20
Blue
47.3k870149
add a comment |
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1 Answer
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Consider this form of the equation
$$2(p_1^2 + p_n^2 + p_{n+1}^2 + a^2 ) = ( p_1 + p_n + p_{n+1} - a )^2 tag{1}$$
and take it back a step by replacing $nto n-1$:
$$2(p_1^2 + p_{n-1}^2 + p_{n}^2 + a^2 ) = ( p_1 + p_{n-1} + p_{n} - a )^2 tag{2}$$
We see that $(2)$ is identical to $(1)$, except that $p_{n-1}$ replaces $p_{n+1}$. Therefore, $p_{n+1}$ and $p_{n-1}$ are the roots of the monic quadratic equation
$$2(p_1^2 + p_{n}^2 + x^2 + a^2 ) = ( p_1 + p_{n} + x - a )^2 tag{3}$$
(Perhaps this trick is discussed in the "remarks above" or Yoshida's solution.) From there, Vieta's formulas allow us to write the coefficient of $x$ in terms of those roots, which gives this recursion:
$$p_{n-1} + p_{n+1} = -2(a-p_1-p_n)$$
Since $p_1$ has evidently been assigned the value of $2a$ in the quoted solution, we get the recursion in the stated form. $square$
answered Jun 12 at 15:20
Blue
47.3k870149
add a comment |
up vote
3
down vote
Consider this form of the equation
$$2(p_1^2 + p_n^2 + p_{n+1}^2 + a^2 ) = ( p_1 + p_n + p_{n+1} - a )^2 tag{1}$$
and take it back a step by replacing $nto n-1$:
$$2(p_1^2 + p_{n-1}^2 + p_{n}^2 + a^2 ) = ( p_1 + p_{n-1} + p_{n} - a )^2 tag{2}$$
We see that $(2)$ is identical to $(1)$, except that $p_{n-1}$ replaces $p_{n+1}$. Therefore, $p_{n+1}$ and $p_{n-1}$ are the roots of the monic quadratic equation
$$2(p_1^2 + p_{n}^2 + x^2 + a^2 ) = ( p_1 + p_{n} + x - a )^2 tag{3}$$
(Perhaps this trick is discussed in the "remarks above" or Yoshida's solution.) From there, Vieta's formulas allow us to write the coefficient of $x$ in terms of those roots, which gives this recursion:
$$p_{n-1} + p_{n+1} = -2(a-p_1-p_n)$$
Since $p_1$ has evidently been assigned the value of $2a$ in the quoted solution, we get the recursion in the stated form. $square$
answered Jun 12 at 15:20
Blue
47.3k870149
add a comment |
up vote
3
down vote
up vote
3
down vote
Consider this form of the equation
$$2(p_1^2 + p_n^2 + p_{n+1}^2 + a^2 ) = ( p_1 + p_n + p_{n+1} - a )^2 tag{1}$$
and take it back a step by replacing $nto n-1$:
$$2(p_1^2 + p_{n-1}^2 + p_{n}^2 + a^2 ) = ( p_1 + p_{n-1} + p_{n} - a )^2 tag{2}$$
We see that $(2)$ is identical to $(1)$, except that $p_{n-1}$ replaces $p_{n+1}$. Therefore, $p_{n+1}$ and $p_{n-1}$ are the roots of the monic quadratic equation
$$2(p_1^2 + p_{n}^2 + x^2 + a^2 ) = ( p_1 + p_{n} + x - a )^2 tag{3}$$
(Perhaps this trick is discussed in the "remarks above" or Yoshida's solution.) From there, Vieta's formulas allow us to write the coefficient of $x$ in terms of those roots, which gives this recursion:
$$p_{n-1} + p_{n+1} = -2(a-p_1-p_n)$$
Since $p_1$ has evidently been assigned the value of $2a$ in the quoted solution, we get the recursion in the stated form. $square$
answered Jun 12 at 15:20
Blue
47.3k870149
Consider this form of the equation
$$2(p_1^2 + p_n^2 + p_{n+1}^2 + a^2 ) = ( p_1 + p_n + p_{n+1} - a )^2 tag{1}$$
and take it back a step by replacing $nto n-1$:
$$2(p_1^2 + p_{n-1}^2 + p_{n}^2 + a^2 ) = ( p_1 + p_{n-1} + p_{n} - a )^2 tag{2}$$
We see that $(2)$ is identical to $(1)$, except that $p_{n-1}$ replaces $p_{n+1}$. Therefore, $p_{n+1}$ and $p_{n-1}$ are the roots of the monic quadratic equation
$$2(p_1^2 + p_{n}^2 + x^2 + a^2 ) = ( p_1 + p_{n} + x - a )^2 tag{3}$$
(Perhaps this trick is discussed in the "remarks above" or Yoshida's solution.) From there, Vieta's formulas allow us to write the coefficient of $x$ in terms of those roots, which gives this recursion:
$$p_{n-1} + p_{n+1} = -2(a-p_1-p_n)$$
Since $p_1$ has evidently been assigned the value of $2a$ in the quoted solution, we get the recursion in the stated form. $square$
answered Jun 12 at 15:20
Blue
47.3k870149
edited Nov 23 at 7:27
answered Jun 12 at 15:20
Blue
47.3k870149
answered Jun 12 at 15:20
Blue
47.3k870149
answered Jun 12 at 15:20
Blue
47.3k870149
47.3k870149
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Instead of pictures, please type out the problem definition and solution, and please use MathJax when typing the equations. This way, your question is much more readible and will more likely get answers.
– Matti P.
Jun 12 at 13:16
What are those "remarks above"?
– Aretino
Jun 12 at 13:59