On the impossibility of constructing a triangle given its internal bisectors

TL;DR

This paper proves that it is impossible to construct a triangle with given internal bisectors using only straightedge and compass, highlighting a fundamental geometric limitation.

Contribution

The paper provides a simple proof demonstrating the impossibility of triangle construction from internal bisectors with classical geometric tools.

Findings

Construction from internal bisectors is impossible with straightedge and compass.

The proof applies even to isosceles triangles.

This result clarifies a fundamental geometric constraint.

Abstract

We give a simple proof to the fact that it is impossible to use straightedge and compass to construct a triangle given the lengths of its internal bisectors, even if the triangle is isosceles.

Equations

$$(2l+b)(2l-b)=4q^2.$$

$$\begin{split}p^{2}&\,=b^{2}+x^{2}-2bx\cos\theta\\ &\,=b^{2}+\Big{(}\frac{bl}{b+l}\Big{)}^{2}-2b\Big{(}\frac{bl}{b+l}\Big{)}\cdot\frac{b}{2l}\\ &\,=\frac{b^{2}l(b+2l)}{(b+l)^{2}}.\end{split}$$

$$\frac{b^{2}l}{2l-b}=\frac{p^2(b+l{)}^2}{4q^2}.$$

$$2p^{2}\Big{(}\frac{l}{b}\Big{)}^{3}+3p^{2}\Big{(}\frac{l}{b}\Big{)}^{2}-4q^{2}\Big{(}\frac{l}{b}\Big{)}-p^{2}=0.$$

$$f(X)=2X^3+3X^2-4q^{2}X-1.$$

$$[\mathbb{Q}(q)(l/b):\mathbb{Q}(q)]=3,$$

$$2+3h(q)-4q^{2}h(q{)}^2=h(q{)}^3.$$

$$h(X{)}^2(4X^2+h(X))=3h(X)+2,$$

$$c^2(4X^2+c)=3c+2,$$

Full text

We consider a triangle $ABC$, isosceles of basis $BC$, and let $AM$ and $AP$ be internal bisectors of it (cf. Figure 1). We assume the lengths $p$ of $AP$ and $q$ of $AM$ to be known.

We let $AB=AC=l$ and $BC=b$. Since $AM$ is also height and median of $ABC$, we have $\cos\theta=\frac{b}{2l}$; also, Pythagoras’ theorem applied to triangle $ACM$ gives $l^{2}-\frac{b^{2}}{4}=q^{2}$ or, which is the same,

[TABLE]

With respect to the internal bisector AP, the interior angle bisector theorem furnishes $\frac{AP}{CP}=\frac{AB}{BC}$ or, letting $x=CP$, $\frac{l-x}{x}=\frac{l}{b}$. Solving for $x$, we obtain $x=\frac{bl}{b+l}$, and applying the cosine law to triangle $BPC$, we get

[TABLE]

From (1) and (2), it comes that

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By cross-multiplying, expanding and performing some elementary algebra, we arrive at the equality $2p^{2}l^{3}+3p^{2}bl^{2}-4q^{2}b^{2}l-p^{2}b^{3}=0$ and, upon division by $b^{3}$,

[TABLE]

Assuming, without loss of generality, that $p=1$, we conclude that $\frac{l}{b}$ is a root of the third degree polynomial

[TABLE]

Now, for the sake of contradiction, suppose that one can use straightedge and compass to construct $ABC$, knowing the lengths $p=1$ and $q$. By recalling the usual analysis of the classical Greek construction problems (cf. [1], for instance), this means that there exists a finite sequence of elementary constructions that allows us to obtain the length $\frac{lp}{b}=\frac{l}{b}$ (since we are assuming $p=1$). Yet in another way, $\frac{l}{b}$ is constructible from $\mathbb{Q}(p,q)=\mathbb{Q}(q)$, so that (by arguing again as in the analysis of the Greek problems) the degree $[\mathbb{Q}(q)(l/b):\mathbb{Q}(q)]$ must be a power of $2$. However, if we show that $f$ is irreducible in $\mathbb{Q}(q)[X]$, then

[TABLE]

which will be a contradiction.

We are left to establishing the irreducibility of $f$ in $\mathbb{Q}(q)[X]$, at least for some $q>0$. To this end, from now on we take $q>0$ to be transcendental. Then, $\mathbb{Q}[q]\simeq\mathbb{Q}[X]$ is a UFD, and Gauss’ theorem assures that it suffices to show that $f$ is irreducible in $\mathbb{Q}[q][X]$. If this is not so, then $f$ has a root $\alpha\in\mathbb{Q}(q)$, say $\alpha=\frac{g(q)}{h(q)}$, for some $g,h\in\mathbb{Q}[X]\setminus\{0\}$.

Applying the searching criterion for roots of $f$ belonging to the field of fractions of the UFD $\mathbb{Q}[q]$ assures that $g(q)\mid 1$ in $\mathbb{Q}[q]$. Therefore, we can assume $g(q)=1$, so that $\alpha=\frac{1}{h(q)}$ and $f(\alpha)=0$ gives

[TABLE]

This is the same as

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so that $\partial h=0$, say $h(X)=c$ for some $c\in\mathbb{Q}$. But this gives

[TABLE]

which is clearly impossible.