# How to prove this algebraic version of the sine law?

How to solve the following problem from Hall and Knight's Higher Algebra?

Suppose that \begin{align} a&=zb+yc,\tag{1}\\ b&=xc+za,\tag{2}\\ c&=ya+xb.\tag{3} \end{align} Prove that $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$

(I suppose that $$x,y,z$$ are real numbers whose moduli are not equal to $$1$$.)

I discovered this problem from chapter 3 of Prelude to Mathematics by W. W. Sawyer. Sawyer thought that this problem arose from the sine law: let $$a,b,c$$ be respectively the lengths of the edges opposite to three vertices $$A,B,C$$ of a triangle. Define $$x=\cos A$$ and define $$y,z$$ analogously. Now equalities $$(1)-(3)$$ simply relate $$a,b$$ and $$c$$ to each other by the cosines of the angles and $$(4)$$ is just a rewrite of the sine law $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$$ However, the algebraic version $$(4)$$ looks more general. For example, it does not state that $$a,b,c$$ must be positive or that they must satisfy the triangle inequality.

Sawyer wrote that this isn't a hard problem, but he didn't provide any solution. I can prove $$(4)$$ using linear algebra. Suppose that $$(a,b,c)\ne(0,0,0)$$ (otherwise $$(4)$$ is obvious). Rewrite $$(1)-(3)$$ in the form of $$M\mathbf a=0$$: $$\begin{bmatrix}-1&z&y\\ z&-1&x\\ y&x&-1\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix}=0.$$ Since $$x^2,y^2,z^2\ne1$$, $$M$$ has rank $$2$$ and $$D=\operatorname{adj}(M)$$ has rank $$1$$. Hence all columns of $$D$$ are parallel to $$(a,b,c)^T$$ and $$\frac{d_{11}}{d_{21}}=\frac{d_{12}}{d_{22}}=\frac{a}{b}$$. Since $$M$$ is symmetric, $$D$$ is symmetric too. Therefore $$\frac{1-x^2}{1-y^2}=\frac{d_{11}}{d_{22}}=\frac{d_{11}d_{12}}{d_{21}d_{22}}=\frac{a^2}{b^2}$$, i.e. $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$$.

As this problem comes from Hall and Knight's book, I think there should be a more elementary solution. Any ideas?

• In Hall and Knight, 1957, At the end of Chapter I, example 18 at the top of page 12 is "If $x=cy+bz, y=az+cx, z=bx+ay$, shew that $\frac{x^2}{1-a^2}=\frac{y^2}{1-b^2}=\frac{z^2}{1-c^2}.$" May 14 '20 at 11:57

Let $$a=0$$.

Thus, $$xc=b$$ and $$xb=c,$$ which gives $$x^2bc=bc$$ or $$(x^2-1)bc=0$$ and since $$x^2\neq1,$$ we obtain $$bc=0$$ and from here $$a=b=c=0,$$ which gives that our statement is true.

Let $$abc\neq0$$.

Thus, $$\frac{zb}{a}+\frac{yc}{a}=1$$ and $$\frac{xc}{b}+\frac{za}{b}=1,$$ which gives $$z^2+\frac{xyc^2}{ab}+\frac{xzc}{a}+\frac{yzc}{b}=1$$ or $$\frac{1-z^2}{c^2}=\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca},$$ which gives $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}=\frac{1}{\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}}.$$

• This is nice. The common value of $\frac{a^2}{1-x^2}$ and the like is beautiful! May 14 '20 at 9:53

It turns out that I solved the equations for the wrong variables. If I rewrite $$(1)-(3)$$ as $$\begin{bmatrix}0&c&b\\ c&0&a\\ b&a&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}a\\ b\\ c\end{bmatrix}$$ and solve for $$x,y,z$$ instead, I will get the law of cosines, i.e. $$x=\frac{b^2+c^2-a^2}{2bc}$$ etc.. Therefore $$\frac{a^2}{1-x^2}=\frac{4a^2b^2c^2}{2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)}.$$ As Roman Odaisky has pointed out, this expression can be rewritten as $$\frac{a^2b^2c^2}{4s(s-a)(s-b)(s-c)}$$, where $$s=\frac12(a+b+c)$$. By symmetry, $$\frac{b^2}{1-y^2}$$ and $$\frac{c^2}{1-z^2}$$ are also equal to the same expression. Geometrically (and according to Heron's formula), this means the common ratio in the law of sines is equal to $$\frac{abc}{2T}$$ where $$T$$ is the area of the triangle.

• What happens if $abc=0$? May 14 '20 at 9:52
• @MichaelRozenberg This corner case can be easily dealt with, as you do in your answer. May 14 '20 at 9:54
• I agree, but I think, without this case the solution is not full. May 14 '20 at 9:56
• If you rewrite the denominator as $16s(s-a)(s-b)(s-c)$ where $s=(a+b+c)/2$, the relationship to geometry is even more clear (as $R = abc/4A$ and the area comes from Heron’s formula). May 14 '20 at 18:05

We can write

I) $$z=\frac{a-yc}{b}$$ from (1)

Now, multiplying $$y$$ on both sides of (3) we get $$yc=y^2a+xyb$$.

So, from I) we get $$\frac{a-ay^2}{b}-xy=z$$.....(1')

Similarly from equation (2) we get

II) $$z=\frac{b-xc}{a}$$ from (2)

Now, multiplying $$x$$ on both sides of (3) we get $$xc=xya+x^2b$$.

And we get from (2) $$\frac{b-bx^2}{a}-xy=z$$.....(2')

From (1') and (2') we get,

$$\frac{a-ay^2}{b}=\frac{b-bx^2}{a} \rightarrow \frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$$.

Similarly, $$\frac{a^2}{1-x^2}=\frac{c^2}{1-z^2}$$

So, ultimately we have proved $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}$$

Suppose, $$b=0$$, then $$\frac{a}{c}=y=\frac{c}{a}$$ or $$y=1$$. So, $$\frac{b^2}{1-y^2}$$ will be undefined.

If also $$c=0$$ then $$a=0$$.

• What happens if $b=0$? May 14 '20 at 9:54
• Then $\frac{a}{c}=y=\frac{c}{a}$ . Means $y=1$. That will be undefined. May 14 '20 at 9:56
• Now, there is a problem with $c=0$. May 14 '20 at 9:57
• If $b,c=0$ then $a=0$. If only $c=0$ then $1-z^2=0$. May 14 '20 at 9:58

By (1) and (3), $$a=ay^2 + bxy +bz.$$ Thus, $$a(1-y^2)=b(xy+z)$$ so that $$a^2(1-y^2)=ab(xy+z).$$ In a similar way, we derive from (2) and (3) that $$b^2(1-x^2)=ab(xy+z).$$ Thus, the left sides of the two displayed equations are equal, yielding the first equality in (4). By symmetry, we're done.

IOW replace $$(a,c)$$ by $$(c,a)$$ and $$(x,z)$$ by $$(z,x)$$ above.