# Japanese Temple Geometry

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 $$p_{n+1}, p_{n-1}.$$ Can someone explain it for me. Thanks!

• 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. Commented Jun 12, 2018 at 13:16
• What are those "remarks above"? Commented Jun 12, 2018 at 13:59

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 $$n\to n-1$$: $$2(p_1^2 + p_{n-1}^2 + p_{n}^2 + a^2 ) = ( p_1 + p_{n-1} + p_{n} - a )^2 \tag{2}$$ 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. $$\square$$