2011 AMC 10B Problem 25

P21-25~8 minPrint

Circles Sequences & Series Triangles: Area & Pythagorean Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Bound the quantity above/below or estimate to pin it down Compute small cases, spot the pattern, generalize

Let $T_1$ be a triangle with sides $2011, 2012,$ and $2013$ . For $n \ge 1$ , if $T_n = \triangle ABC$ and $D, E,$ and $F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB, BC$ and $AC,$ respectively, then $T_{n+1}$ is a triangle with side lengths $AD, BE,$ and $CF,$ if it exists. What is the perimeter of the last triangle in the sequence $( T_n )$ ?

Show solution without answering

Key insight

Each new side is half the perimeter minus an old side. Three sides in a row stay three sides in a row, and the middle side halves each time.

Solution

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.

$T_1$ has sides $2011, 2012, 2013$ . Half the perimeter is $6036 \div 2 = 3018$ .
The new sides are $3018 - 2013 = 1005$ , $3018 - 2012 = 1006$ , $3018 - 2011 = 1007$ .
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.

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 $\frac{x}{2} + 1$ .
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?

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 $x > 2$ .
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.
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 $\boxed{\textbf{(D)}\ \frac{1509}{128}}$ .

Why this works

"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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 11: Special Factorizations