AMC 10 · 2011 · #25
Grade 8 geometry-2dLet T1 be a triangle with side lengths 2011,2012, and 2013. For n≥1, if Tn=△ABC and D,E, and F are the points of tangency of the incircle of △ABC to the sides AB,BC, and AC, respectively, then Tn+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 (Tn)?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Start with triangle $T_1$ whose side lengths are $2011$, $2012$, $2013$. For any triangle $T_n = \triangle ABC$, let its incircle touch sides $AB$, $BC$, $AC$ at $D$, $E$, $F$; the next triangle $T_{n+1}$ has side lengths $AD$, $BE$, $CF$, but only when those three lengths actually form a triangle. The sequence stops at the last triangle that exists. Find that last triangle's perimeter.
Givens: $T_1$ has side lengths $2011$, $2012$, $2013$; $D$, $E$, $F$ are the incircle tangency points on sides $AB$, $BC$, $AC$; $T_{n+1}$ has side lengths $AD$, $BE$, $CF$ whenever they form a triangle; Answer choices: $\dfrac{1509}{8},\ \dfrac{1509}{32},\ \dfrac{1509}{64},\ \dfrac{1509}{128},\ \dfrac{1509}{256}$
Unknowns: The perimeter of the last triangle in the sequence $(T_n)$
Understand
Restated: Start with triangle $T_1$ whose side lengths are $2011$, $2012$, $2013$. For any triangle $T_n = \triangle ABC$, let its incircle touch sides $AB$, $BC$, $AC$ at $D$, $E$, $F$; the next triangle $T_{n+1}$ has side lengths $AD$, $BE$, $CF$, but only when those three lengths actually form a triangle. The sequence stops at the last triangle that exists. Find that last triangle's perimeter.
Givens: $T_1$ has side lengths $2011$, $2012$, $2013$; $D$, $E$, $F$ are the incircle tangency points on sides $AB$, $BC$, $AC$; $T_{n+1}$ has side lengths $AD$, $BE$, $CF$ whenever they form a triangle; Answer choices: $\dfrac{1509}{8},\ \dfrac{1509}{32},\ \dfrac{1509}{64},\ \dfrac{1509}{128},\ \dfrac{1509}{256}$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #4 Introduce a Variable, #14 Extreme Principle
Listing all the triangles by hand would be slow, so Tool #5 (Look for a Pattern) is the anchor: work out how one triangle turns into the next and find the rule that repeats. Tool #4 (Introduce a Variable) makes that rule clean by naming the middle side length, which exposes that the perimeter simply halves each step and the sides keep the same shape. Then the only question is when the shrinking sides get too lopsided to form a triangle, and Tool #14 (Extreme Principle) pins down that last surviving step by pushing the triangle inequality to its breaking point.
Execute — Answer: D
8.G.A.5 Step 1 Turn tangency points into new side lengths
- In any triangle the two tangent segments from a vertex to the incircle are equal, so each vertex has one tangent length.
- Write the sides as $a = BC$, $b = CA$, $c = AB$ with semiperimeter $s = \tfrac{a+b+c}{2}$.
- Then the tangent length from $A$ is $s-a$, from $B$ is $s-b$, and from $C$ is $s-c$.
- Since $D$, $E$, $F$ sit on $AB$, $BC$, $AC$, the pieces cut off are $AD = s-a$, $BE = s-b$, $CF = s-c$.
- So $T_{n+1}$ has side lengths $s-a$, $s-b$, $s-c$.
💡 Each vertex owns a single tangent length, so the new sides are just the old semiperimeter minus each old side.
7.EE.A.1 Step 2 See the shape stays the same and halves
- $T_1$ has sides $2011$, $2012$, $2013$ — three numbers spaced $1$ apart.
- Name the middle one $m$, so the sides are $m-1$, $m$, $m+1$ and $s = \tfrac{3m}{2}$.
- The new sides become $s-(m+1) = \tfrac{m}{2}-1$, $s-m = \tfrac{m}{2}$, and $s-(m-1) = \tfrac{m}{2}+1$.
- That is again three numbers spaced $1$ apart, with the new middle equal to $\tfrac{m}{2}$.
- So the perimeter goes from $3m$ to $\tfrac{3m}{2}$: exactly half, every step.
💡 The sides always differ by $1$, and the middle value keeps getting cut in half.
8.EE.A.1 Step 3 Write the middle side after n steps
- The middle side starts at $m_1 = 2012$ and halves each time, so after $n-1$ halvings $m_n = \dfrac{2012}{2^{\,n-1}}$.
- Because the sides are always $m_n-1$, $m_n$, $m_n+1$, the perimeter is $P_n = 3m_n = \dfrac{6036}{2^{\,n-1}}$.
- Every triangle in the sequence is captured by this one formula.
💡 Repeated halving is the same as dividing by a power of two.
7.G.A.2 Step 4 Find when the triangle stops existing
- Sides $m-1$, $m$, $m+1$ form a triangle only if the two shorter ones beat the longest: $(m-1)+m > m+1$, which simplifies to $m > 2$.
- (The largest side is $m+1$, and this is the tightest of the three inequalities.) So $T_n$ exists exactly when its middle side satisfies $m_n > 2$; once $m_n$ drops to $2$ or below, the three lengths can no longer close into a triangle.
💡 When the two shorter sides can just barely reach across the longest one, the triangle is at its breaking point.
7.EE.B.4 Step 5 Solve for the last surviving triangle
- Require $m_n = \dfrac{2012}{2^{\,n-1}} > 2$, i.e.
- $2^{\,n-1} < 1006$.
- Since $2^{9} = 512 < 1006 < 1024 = 2^{10}$, the largest allowed exponent is $n-1 = 9$, so $n = 10$ is the last triangle.
- Checking the next one, $m_{11} = \dfrac{2012}{1024} \approx 1.965 \le 2$, so $T_{11}$ indeed fails.
- The last triangle is $T_{10}$, with perimeter $P_{10} = \dfrac{6036}{2^{9}} = \dfrac{6036}{512} = \dfrac{1509}{128}$, which is choice $\textbf{(D)}$.
💡 Find the last power of two below $1006$, and that step is where the shrinking triangle just survives.
8.G.A.5 In any triangle the two tangent segments from a vertex to the incircle are equal 7.EE.A.1 $T_1$ has sides $2011$, $2012$, $2013$ — three numbers spaced $1$ apart. Name th 8.EE.A.1 The middle side starts at $m_1 = 2012$ and halves each time, so after $n-1$ halv 7.G.A.2 Sides $m-1$, $m$, $m+1$ form a triangle only if the two shorter ones beat the lo 7.EE.B.4 Require $m_n = \dfrac{2012}{2^{\,n-1}} > 2$, i.e. $2^{\,n-1} < 1006$. Since $2^{ Review
Reasonableness: The perimeters halve cleanly: $6036, 3018, 1509, \dots$ down to $P_{10} = \tfrac{1509}{128} \approx 11.79$. The very next value $P_{11} \approx 5.89$ would give sides near $0.96, 1.96, 2.96$, where $0.96 + 1.96 = 2.92 < 2.96$ — the two short sides can't reach across, so no triangle, confirming the stop at $T_{10}$. Every answer choice has numerator $1509 = 3 \times 503$ and differs only by the power of two, so the whole problem reduces to finding the right exponent; $128 = 2^{7}$ matches $\textbf{(D)}$.
Alternative: Track only the perimeter. Since the new perimeter equals the old semiperimeter, $P_{n+1} = \tfrac{1}{2}P_n$, giving $P_n = \dfrac{6036}{2^{\,n-1}}$. A triangle whose sides differ by $1$ has middle side $\tfrac{P}{3}$, and it satisfies the triangle inequality exactly when $\tfrac{P}{3} > 2$, i.e. $P > 6$. The smallest $P_n$ still above $6$ is $P_{10} = \dfrac{6036}{512} = \dfrac{1509}{128} \approx 11.79$ (the next, $\approx 5.89$, drops below $6$), giving $\textbf{(D)}$ without listing any triangle individually.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using the equal-tangent-segments fact of the incircle to write each new side as the semiperimeter minus an old side.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Simplifying $s-(m\pm1)$ to show the new sides are $\tfrac{m}{2}-1, \tfrac{m}{2}, \tfrac{m}{2}+1$ and the perimeter halves.)8.EE.A.1Know and apply the properties of integer exponents (Expressing the repeated halving of the middle side and perimeter as division by $2^{\,n-1}$.)7.G.A.2Draw geometric shapes with given conditions including triangles (Applying the triangle inequality to three side lengths differing by $1$ to get the existence condition $m > 2$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Solving $2^{\,n-1} < 1006$ to locate the last surviving triangle and compute its perimeter $\tfrac{1509}{128}$.)
⭐ Each new triangle's perimeter is exactly half the last one's, so keep halving until the three sides get too lopsided to close up — the last one that still closes is the answer.
⭐ Each new triangle's perimeter is exactly half the last one's, so keep halving until the three sides get too lopsided to close up — the last one that still closes is the answer.
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