Prove that length of three bisectors determine triangle.

Question

All of us know that length of 3 bisectors determine triangle. But actually all proofs that I heard are sufficient large. So I'm interested in short and smart proof.

Any ideas?

Answer 1

One can always look for a smarter proof, but unfortunately a geometric proof is unlikely to be available because with the help of more Galois theory than I want to go into one can show that no geometric construction (with ruler and compasses) will allow you to construct the triangle given only the lengths of the angle bisectors. That strongly suggests that any proof will necessarily use algebra/analysis rather than geometry.

Let the sides of the triangle be $a,b,c$ and let the lengths of the three angle bisectors be $l,m,n$. We show first that $$l=\frac{2}{b+c}\sqrt{bcs(s-a)}\ \ (*)$$ where $2s=a+b+c$.

Let the bisector $AD$ have length $l$ (and sides $BC,CA,AB$ have lengths $a,b,c$ respectively). Then area $ABD+\text{area}\ ADC=\text{area}\ ABC$. So we have $bc\sin A=bl\sin\frac{A}{2}+cl\sin\frac{A}{2}$ and hence $l=\frac{2bc}{b+c}\cos\frac{A}{2}$. The cosine formula gives $\cos A=\frac{b^2+c^2-a^2}{2bc}$, so $2\cos^2\frac{A}{2}=1+\cos A=\frac{(b+c)^2-a^2}{2bc}$ and hence $\cos\frac{A}{2}=\frac{\sqrt{s(s-a)}}{bc}$ which gives $(*)$.

Taking $4(*)^2$ we have $4l^2=\frac{4}{(b+c)^2}bc(b+c+a)(b+c-a)=4bc-\frac{4a^2bc}{(b+c)^2}$. Adding $a^2+(b-c)^2$ to both sides we get $4l^2+a^2+(b-c)^2=a^2+(b+c)^2-\frac{4a^2bc}{(b+c)^2}=(b+c)^2+\frac{a^2(b-c)^2}{(b+c)^2}$. We now add $\pm2a(b-c)$ to both sides to get $$4l^2+\left(a\pm(b-c)\right)^2=\left(b+c\pm\frac{a(b-c)}{b+c}\right)^2$$ Taking the square root for each choice of sign and adding we get $$b+c=\sqrt{l^2+(s-b)^2}+\sqrt{l^2+(s-c)^2}$$ [note that $\frac{1}{2}(a+(b-c))=s-c$ and $\frac{1}{2}(a-(b-c))=s-b$]. Putting $x=s-a,y=s-b,z=s-c$ and defining $f(u,v)=\frac{1}{2}\left(\sqrt{u^2+v^2}-u\right)$, we can write this as $$x=f(y,l)+f(z,l)$$ We obviously get also the corresponding equations $$y=f(z,m)+f(x,m)\text{ and }z=f(x,n)+f(y,n)$$ Now fix $l,m,n$ and regard $x,y,z$ simply as real variables. Let $K=[0,l]\times[0,m]\times[0,n]\subset\mathbb{R}^3$ and define $F:K\to\mathbb{R}^3$ by $$F(x,y,z)=(f(y,l)+f(z,l),f(z,m)+f(x,m),f(x,n)+f(y,n))$$ The preceding work shows that $(x,y,z)$ is a fixed point of $F$ iff the triangle with side lengths $a=y+z,b=z+x,c=x+y$ has angle bisector lengths $l,m,n$.

Note that $0\le f(u,v)\le \frac{1}{2}v$, so $F(x,y,z)\in K$ for $(x,y,z)\in K$. So by the Brouwer fixed point theorem, $F$ must have a fixed point.

We have now established that some triangle has the given bisector lengths. It remains to show that it is unique. Fortunately only a little more work is required. We show that if $(x,y,z)\ne(x',y',z')$ then the distance between $(x,y,z)$ and $(x',y',z')$ is strictly less than the distance between $F(x,y,z)$ and $F(x',y',z')$. It follows that they cannot both be fixed points.

Note that $\left(\sqrt{y^2+l^2}-\sqrt{y'^2+l^2}\right)\left(\sqrt{y^2+l^2}+\sqrt{y'^2+l^2}\right)=y^2-y'^2=(y-y')(y+y')$, so we have $2|f(y,l)-f(y',l)|=|\sqrt{y^2+l^2}-y-\sqrt{y'^2+l^2}+y'|=|y-y'|\left|1-\frac{y+y'}{\sqrt{y^2+l^2}+\sqrt{y'^2+l^2}}\right|\le|y-y'|$ with the inequality strict for $y\ne y'$.

So we have $|F(x,y,z)-F(x',y',z')|^2\le |f(y,l)-f(y',l)+f(z,l)-f(z',l)|^2+|f(z,m)-f(z',m)+f(x,m)-f(x',m)|^2+|f(x,n)-f(x',n)+f(y,n)-f(y',n)|^2<\left(\frac{1}{2}|y-y'|+\frac{1}{2}|z-z'|\right)^2+\left(\frac{1}{2}|z-z'|+\frac{1}{2}|x-x'|\right)^2+\left(\frac{1}{2}|x-x'|+\frac{1}{2}|y-y'|\right)^2$.

Note that the inequality is strict because at least one of $|x-x'|,|y-y'|,|z-z'|$ is non-zero.

Adding the non-negative quantity $\left(\frac{1}{2}|y-y'|-\frac{1}{2}|z-z'|\right)^2+\left(\frac{1}{2}|z-z'|-\frac{1}{2}|x-x'|\right)^2+\left(\frac{1}{2}|x-x'|-\frac{1}{2}|y-y'|\right)^2$ the rhs becomes $|x-x'|^2+|y-y'|^2+|z-z'|^2$ and we have established $|F(x,y,z)-F(x',y',z')|<|(x,y,z)-(x',y',z')|$ as promised.

Answer 2

This is really a comment, but it needs a picture.

As per my comments, I'd love for someone to show how the following "bisecting" segments create a triangle. Because they clearly don't, the proof needs to include when segments do and do not create a triangle to be complete. You need to figure out under what conditions do segments create a triangle and then you need to figure out whether or not bisectors determine a triangle. It's not an easy proof and there is not going to be an easy proof of it.

It's one thing to assume a triangle and prove that its bisectors uniquely determine that triangle. It's quite a different thing to start with segments and say 1) do the segments form bisectors of a triangle and 2) if so, then do they uniquely determine the triangle.