MHB An application of the law of cosines?

[Kobzar]

Hello, everybody:

I am a philologist who is fond of mathematics, but who unfortunately has just an elementary high school knowledge of them. I am translating La leçon de Platon, by Dom Néroman (La Bégude de Mazenc, Arma Artis, 2002), which deals with music theory and mathematics in the works of Plato. The problem which brings me here is not about translation, but about mathematics. I am attaching a file with my problem and what I have tried so far to solve it.

Thank you very much in advance and best greetings.

 

[Opalg]

In fact, your answer is essentially the same as Néroman's. You have shown that $$\begin{aligned}\frac{PM}{PH} &= \frac{6\pm\sqrt{36 + 64(1 + \tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{100 + 64\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4(25 + 16\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4}\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm2\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}}. \end{aligned}$$ Now divide top and bottom of the fraction by $2$ to get $$\frac{PM}{PH} = \frac{3\pm\sqrt{25 + 16\tan^2i}}{4\sqrt{1+\tan^2i}}.$$ That is exactly Néroman's formula apart from two things. One of these is the ambiguous $\pm$ sign. That is resolved by the fact that the $+$ sign gives the ratio $\frac{PM}{PH}$, and the $-$ sign gives the ratio $\frac{Pm}{PH}$. The other anomaly is the $PH$ (or $h$) that occurs in Néroman's formula. That must be a mistake on his part. The formula should be either $$\tfrac{PM}{PH} = \tfrac1{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr)$$ or $$PM = \tfrac{PH}{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr).$$ But he appears to have mistakenly put the $PH$ on both sides of the formula.

 

[Kobzar]

In fact, your answer is essentially the same as Néroman's. You have shown that $$\begin{aligned}\frac{PM}{PH} &= \frac{6\pm\sqrt{36 + 64(1 + \tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{100 + 64\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4(25 + 16\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4}\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm2\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}}. \end{aligned}$$ Now divide top and bottom of the fraction by $2$ to get $$\frac{PM}{PH} = \frac{3\pm\sqrt{25 + 16\tan^2i}}{4\sqrt{1+\tan^2i}}.$$ That is exactly Néroman's formula apart from two things. One of these is the ambiguous $\pm$ sign. That is resolved by the fact that the $+$ sign gives the ratio $\frac{PM}{PH}$, and the $-$ sign gives the ratio $\frac{Pm}{PH}$. The other anomaly is the $PH$ (or $h$) that occurs in Néroman's formula. That must be a mistake on his part. The formula should be either $$\tfrac{PM}{PH} = \tfrac1{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr)$$ or $$PM = \tfrac{PH}{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr).$$ But he appears to have mistakenly put the $PH$ on both sides of the formula.

Thank you very much!