How to prove that triangle inscribed in another triangle (were both have one shared side) have lower perimeter?

Question

This question looks really simple, but to my (and my co-workers) frustration we were not able to prove this in any way. I tried all triangle formulas known to me but I feel I'm missing the point, and proof will be or much simpler or much more complicated than what I tried.

So, the question:

Given a triangle ABC and point P inside that triangle, prove that for triangle APB the following inequality holds: |AB| + |BC| > |AP| + |PC|

(Actually it doesn't matter for me if it's > or >=).

Answer 1

One way to see that this must be true is to look at the ellipse through $B$ with foci $A$ and $C$. The interior of this ellipse is precisely all those point $P$ such that $|AP| + |PC| < |AB| + |BC|$. And if $P$ is inside the triangle, then it is inside the ellipse too.

Answer 2

This is a special case of "Archimedes' axiom", that if one convex curve $\gamma_1$ is inside another ($\gamma_2$), then $\gamma_1$ is shorter than $\gamma_2.$ Archimedes needed this to justify his computing the perimeter of a circle by inscribed/circumscribed polygons, and could not prove it, so made it an axiom. In general, this is a nontrivial fact, in this case it is a tedious computation (if you don't want to use the general machine).

EDIT the most elegant proof is via Crofton's formula, which says that the length of a convex curve is equal (up to normalizing constant) to the measure of the lines which intersect the interior -- that measure is obviously monotonic under containment...

Answer 3

Since $P$ is inside $\triangle ABC$, the convex hull span by the three vertices $A$, $B$, $C$, there exists 3 numbers $\alpha, \beta, \gamma \ge 0$ such that

$$\alpha+\beta+\gamma = 1\quad\text{ and }\quad \vec{P} = \alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}$$

This implies $$ |AP| = |(1-\alpha)\vec{A} - (1-\alpha-\gamma)\vec{B} - \gamma\vec{C}| = |(1-\alpha)(\vec{A}-\vec{B}) + \gamma (\vec{B}-\vec{C})|\\ \le (1-\alpha) |AB| + \gamma |BC| $$ and $$ |CP| = |-\alpha \vec{A} - (1-\alpha-\gamma)\vec{B} + (1-\gamma)\vec{C}| = |\alpha (\vec{B}-\vec{A}) + (1-\gamma)(\vec{C}-\vec{B})|\\ \le \alpha |AB| + (1-\gamma)|BC| $$ Summing these two inequalities immediately gives us $|AP| + |CP| \le |AB| + |BC|$.

Since numbers like $\alpha$ are ratios of distance of $P$ to $BC$ versus that of $A$ to $BC$, it is easy to translate above vector based inequalities to a pure geometric proof.

Construct a line through $P$ parallel to $BC$ and let it intersect $AB$ at $D$. Construct another line through $P$ parallel to $AB$ and let it intersect $BC$ at $E$. It is easy to see $|BE| = |PD|$ and $|BD| = |EP|$. Apply triangle inequalities to $\triangle APD$ and $\triangle CPE$, we have

$$|AP| + |CP| \le \Big(|AD| + |PD|\Big) + \Big(|CE| + |EP|\Big) = \Big(|AD| + |BE|\Big) + \Big(|CE| + |BD|\Big)\\ = \Big(|AD| + |DB|\Big) + \Big(|CE| + |BE|\Big) = |AB| + |BC|$$