shortest distance in triangle

Question

Given is a triangle ABC with $a<b<c$ and midpoint of incircle D. Starting in D, I am looking for the shortest way to A, B and C and back to D. I draw a picture with geogebra and it seems to be the shortest way to go not over the longest side (DACBD), but I cannot prove it.

This is MO GER 1967 10th grade 4. Round. Any ideas?

Thanks!

Answer 1

EXE : Consider a triangle of side lengths : $b+c,\ \sqrt{b^2+\varepsilon^2},\ \sqrt{ c^2+\varepsilon^2}$ s.t. the area is $\frac{1}{2}\varepsilon (b+c)$

Prove that $b+ \sqrt{c^2+\varepsilon^2} -\{ c+ \sqrt{ b^2+\varepsilon^2} \} >0$ where $b>c$

Proof : Consider \begin{align*}&(b+ \sqrt{c^2+\varepsilon^2} )^2- (c+ \sqrt{ b^2+\varepsilon^2})^2\\&= 2b\sqrt{c^2+\varepsilon^2} -2c\sqrt{ b^2+\varepsilon^2}\\&>0 \end{align*} where $b>c$

EXE : Consider a triangle $\Delta\ xyz$ whose incircle has a center $o$

Further, $o$ has a foot $x'$ on $[yz]$ i.e. $[ox']\perp [yz]$

When $|y-x'|=b,\ |y-o|=B,\ |z-x'|=c,\ |z-o|=C,\ |x-z'|=a,\ |x-o|=A$, then consider two paths $oyzxo,\ oyxzo$, then by cancelling $${\rm length}\ oyzxo -{\rm length}\ oyxzo =c+A-\{a+C\}$$

By previous EXE, we can determine the sign of the above.

Answer 2

If you consider an equal detour point (X(176) in https://faculty.evansville.edu/ck6/encyclopedia/ETC.html) then any one of the three possible tours are of equal length. Let's denote this point by $P$ and let $BC$ be the longest side. Then the assertion in question is equivalent to $I$ lying inside the triangle $BPC$. Since (by the Encyclopedia of Triangle Centers) the trilinear coordinates of $P$ are: $$1+\frac{\cos (\frac{B}{2}) \cdot \cos (\frac{C}{2})}{\cos (\frac{A}{2})}:1+\frac{\cos (\frac{C}{2}) \cdot \cos (\frac{A}{2})}{\cos (\frac{B}{2})}:1+\frac{\cos (\frac{A}{2}) \cdot \cos (\frac{B}{2})}{\cos (\frac{C}{2})}$$ And it suffices to show that the first coordinate is the largest of them. To see that substract $1$ from all of them and divide by $\cos (\frac{A}{2}) \cdot \cos (\frac{B}{2}) \cdot \cos (\frac{C}{2})$ and use the fact that since $BC$ is the largest side, $A$ is the largest angle (and so is $\frac{A}{2}$) and therefore $\frac{1}{\cos (\frac{A}{2})^2}$ is the largest value because $f(x)=\frac{1}{(\cos x)^2}$ is increasing.