2011 AMC 10B Problem 25

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

The identity "tangent length $= s - \text{opposite side}$" turns a geometric construction into pure arithmetic on the side lengths, and the special shape of the sides (consecutive, spaced by $1$) is preserved because subtracting each side from the same $s$ just reverses and shifts them. The perimeter of $T_{n+1}$ is $3s - 2s = s$, half the previous perimeter, so the perimeters form a geometric sequence and the only real work is finding where the triangle inequality first breaks.

Each perimeter is half the one before: $6036, 3018, 1509, \frac{1509}{2}, \frac{1509}{4}, \ldots$, so the answer is $\frac{1509}{2^k}$ for some $k$, matching the form of every choice. The stopping condition needs the smallest side positive and the two smallest to beat the largest: sides $\frac{2012}{2^{n-1}} \pm 1$ and $\frac{2012}{2^{n-1}}$ work while $\frac{2012}{2^{n-1}} > 2$. Counting halvings from $2012$ down past $2$ gives nine, so $k = 7$ after the first three perimeters, i.e. $\frac{1509}{128}$.

• Halving one step too many (or too few): the triangle with middle side $\frac{2012}{512}$ is valid but the next, with middle side about $1.96$, violates the triangle inequality.
• Using the condition "smallest side positive" ($x > 1$) instead of the triangle inequality ($x > 2$), which incorrectly allows one more triangle and gives choice (E).
• Mixing up which tangent length goes with which vertex; since only the multiset of new side lengths matters, this does not change the answer, but it can derail the derivation of the pattern.