A triangle with sides $a$, $b$, and $c$ is given. Denote by $s$ the semiperimeter, that is $s = \frac{a + b + c}{2}$. Construct a triangle with sides $s - a$, $s - b$, and $s - c$. This process is repeated until a triangle can no longer be constructed with the side lengths given. For which original triangles can this process be repeated indefinitely?
Problem
Source: APMO 1992
Tags: geometry, inequalities, geometric inequality, APMO
Davron
12.03.2006 07:57
Note that (s-a)+(s-b)+(a-c)=s so the perimeters are halving each time. Since (s-a)-(s-b)=b-a it is clear that the side length differences stay the same. If b-a differs from zero, then the process must eventually terminate because a side length difference would otherwise become greater than the perimeter. If a=b=c at each step the side dimensions are halved. Thus the procedure can only be continued indefinitely for equilateral triangles.
myceliumful
28.02.2012 04:06
(TYPO) To avoid confusion: (s-a)+(s-b)+(a-c)=s should be (s-a)+(s-b)+(s-c)=s. Thank you, Davron for the effort in providing the solution! Thank you Shobber as the OP!
BOGTRO
17.03.2013 09:00
Define by $T_n$ the triangle obtained by performing the process $n$ times. By routine calculations (and everything is taken cyclically over a,b,c to save effort), $T_1$ has sides $s-a$, $T_2$ has sides $a-\frac{1}{2}s$. We claim that for even $n$, $T_n$ has sides $a-s(\frac{1}{2}+\frac{1}{8}+\hdots+\frac{1}{2^{n-1}})$ and for odd $n$, $T_n$ has sides $(1-\frac{1}{4}-\frac{1}{16}-\hdots-\frac{1}{4^{n-1}})s-a$. These are both easily provable by induction.
Therefore, $T_n$ approaches having sides of $a-s(\frac{1}{2}+\frac{1}{8}+\hdots)=a-s(\frac{\frac{1}{2}}{1-\frac{1}{4}})=a-\frac{2}{3}s$ for even $n$ and sides of $(1-\frac{1}{4}-\frac{1}{16}-\hdots)s-a=(1-\frac{\frac{1}{4}}{1-\frac{1}{4}})s-a=\frac{2}{3}s-a$ for odd $n$.
WLOG let $a \leq b \leq c$. Then,
$(a-\frac{2}{3}s)+(b-\frac{2}{3}s) \geq c-\frac{2}{3}s$. Note that we may have equality as we are taking a limit. Similarly,
$(\frac{2}{3}s-c)+(\frac{2}{3}s-b) \geq \frac{2}{3}s-a$.
Adding these inequalities gives $a-c \geq c-a \iff 2a \geq 2c \iff a \geq c$. But $a \leq c$, so $a=c$. Therefore, $a=b=c$, and the only possible solution is an equilateral triangle.
Naysh
11.02.2015 22:45
The solution of the above poster can further be sped up my keeping track of $u,v,w$ instead of $a,b,c$, where the first three reals satisfy \[\begin{cases} u+v=a \\ v+w=b \\ w+u=c \end{cases}\]
djb86
27.06.2015 23:02
Since the triangles with sides $(s-a,s-b,s-c)$ and $(b+c-a,c+a-b,a+b-c)$ are similar, we use the transformation $$(a,b,c)\mapsto (b+c-a,c+a-b,a+b-c)$$ instead, noting that this keeps the perimeter invariant. Note that if $a\le b\le c$, then $a+b-c\le c+a-b\le b+c-a$ and that the difference between the largest and the smallest changes from $c-a$ to $2(c-a)$. If $c\ne a$, then repeating the process will eventually force the difference to become larger than the perimeter, which obviously cannot happen in a triangle.
joshualiu315
06.05.2024 09:36
The only triangles are equilateral triangles. Notice that the differences between the sidelengths remain constant throughout the transformations. Moreover, the perimeter of the triangle halves each iteration. Consider $|a-b|$; it must never be greater than the perimeter for any of the triangles in the process, but the perimeter tends to $0$. Thus, $|a-b|=0$ and similarly for the other differences.