2011 AMC 10B Problem 25

Step A: find the new sides.

1. From a corner of a triangle, the two tangent pieces to the inscribed circle are equal. So each side is split into two pieces, and each piece matches a piece on a neighboring side.
2. If the sides are $a = BC$, $b = CA$, $c = AB$, and $p$ is half the perimeter, these facts give
$$AD = p - a, \qquad BE = p - b, \qquad CF = p - c.$$
(Check: $AD + DB = (p - a) + (p - b) = 2p - a - b = c = AB$.)

Step B: try the first step with real numbers.

1. $T_1$ has sides $2011, 2012, 2013$. Half the perimeter is $6036 \div 2 = 3018$.
2. The new sides are $3018 - 2013 = 1005$, $3018 - 2012 = 1006$, $3018 - 2011 = 1007$.
3. So $T_2$ has sides $1005, 1006, 1007$: again three numbers in a row, with the middle one halved from $2012$ to $1006$.

Step C: the pattern keeps going.

1. If the sides are $x - 1, x, x + 1$, the perimeter is $3x$ and half of it is $\frac{3x}{2}$. The new sides are $\frac{3x}{2} - (x + 1) = \frac{x}{2} - 1$, then $\frac{x}{2}$, then $\frac{x}{2} + 1$.
2. So the middle side halves each time, and the sides always differ by $1$. The middle sides go $2012, 1006, 503, 251.5, \ldots$, and $T_n$ has middle side $\frac{2012}{2^{n-1}}$.

Step D: when does it stop being a triangle?

1. Sides $x - 1, x, x + 1$ make a triangle only if the two smaller ones add to more than the biggest: $(x - 1) + x > x + 1$, which means $x > 2$.
2. Keep halving $2012$ while it stays above $2$: dividing by $2^9 = 512$ gives $\frac{2012}{512} \approx 3.93$, which is fine. Dividing by $2^{10} = 1024$ gives about $1.96$, which is too small.
3. So the last triangle is $T_{10}$, with middle side $\frac{2012}{512}$.

Step E: its perimeter.

11. The perimeter is $3$ times the middle side:
$$3 \times \frac{2012}{512} = \frac{6036}{512} = \frac{1509}{128}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1509}{128}}$.

"Tangent piece from a corner equals half the perimeter minus the opposite side" turns a geometry setup into simple arithmetic. Because the sides are three numbers in a row, the new sides are again three numbers in a row, and each perimeter is half the previous one. The only real work is finding the moment the triangle rule breaks.

Each new perimeter equals $p$, half the old perimeter: $6036$, $3018$, $1509$, $\frac{1509}{2}$, and so on. So $T_n$ has perimeter $\frac{6036}{2^{n-1}}$, and $T_{10}$ has perimeter $\frac{6036}{512} = \frac{1509}{128}$. Every choice has the form $\frac{1509}{2^k}$, so the only question is where to stop, which Step D answers.

Tangent lengths. The two tangent segments from a vertex to the incircle are equal, so with $a = BC$, $b = CA$, $c = AB$ and semiperimeter $s$,
$$AD = s - a,\qquad BE = s - b,\qquad CF = s - c.$$

The pattern. Suppose $T_n$ has sides $x - 1, x, x + 1$. Then $s = \frac{3x}{2}$ and the new sides are
$$\frac{3x}{2} - (x + 1) = \frac{x}{2} - 1,\qquad \frac{3x}{2} - x = \frac{x}{2},\qquad \frac{3x}{2} - (x - 1) = \frac{x}{2} + 1.$$
So $T_{n+1}$ again has three sides spaced by $1$, now centered at $\frac{x}{2}$. Starting from $x = 2012$, the middle side of $T_n$ is $\frac{2012}{2^{\,n-1}}$, and the perimeter of $T_n$ is $3 \cdot \frac{2012}{2^{\,n-1}}$.

When does it stop? Sides $x - 1, x, x + 1$ form a triangle exactly when $(x - 1) + x > x + 1$, i.e. $x > 2$. We need the last $n$ with $\frac{2012}{2^{\,n-1}} > 2$, i.e. $2^{\,n-1} < 1006$. Since $2^9 = 512 < 1006 < 1024 = 2^{10}$, the last valid triangle is $T_{10}$, with middle side $\frac{2012}{512} = \frac{503}{128} \approx 3.93$. (For $T_{11}$ the middle side would be about $1.96 < 2$, so its sides would fail the triangle inequality.)

Perimeter.
$$3 \cdot \frac{2012}{512} = \frac{6036}{512} = \frac{1509}{128}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1509}{128}}$.