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?

  • 2
    $\begingroup$ 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 $\frac{x^2}{1-a^2}=\frac{y^2}{1-b^2}=\frac{z^2}{1-c^2}.$" $\endgroup$
    – Somos
    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 $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}=\frac{1}{\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}}.$$

  • 3
    $\begingroup$ This is nice. The common value of $\frac{a^2}{1-x^2}$ and the like is beautiful! $\endgroup$ 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.

  • $\begingroup$ What happens if $abc=0$? $\endgroup$ May 14 '20 at 9:52
  • $\begingroup$ @MichaelRozenberg This corner case can be easily dealt with, as you do in your answer. $\endgroup$ May 14 '20 at 9:54
  • $\begingroup$ I agree, but I think, without this case the solution is not full. $\endgroup$ May 14 '20 at 9:56
  • 1
    $\begingroup$ 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). $\endgroup$ 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 $a=0$.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.