2019 AMC 12A Problem 25

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Concepts:altituderecursionangle chasing

Difficulty rating: 2520

25.

Let A0B0C0\triangle A_0 B_0 C_0 be a triangle whose angle measures are exactly 59.999,59.999^\circ, 60,60^\circ, and 60.001.60.001^\circ. For each positive integer nn define AnA_n to be the foot of the altitude from An1A_{n-1} to line Bn1Cn1.B_{n-1}C_{n-1}. Likewise, define BnB_n to be the foot of the altitude from Bn1B_{n-1} to line An1Cn1,A_{n-1}C_{n-1}, and CnC_n to be the foot of the altitude from Cn1C_{n-1} to line An1Bn1.A_{n-1}B_{n-1}. What is the least positive integer nn for which AnBnCn\triangle A_n B_n C_n is obtuse?

1010

1111

1313

1414

1515

Solution:

For an acute triangle, the orthic triangle (feet of the altitudes) has angles 1802α180^\circ - 2\alpha for each original angle α.\alpha.

Writing an angle as 60+x,60^\circ + x, the new angle is 602x,60^\circ - 2x, so each deviation from 6060^\circ is multiplied by 2.-2. The initial deviations are ±0.001.\pm 0.001^\circ.

After nn steps a deviation has magnitude 0.0012n0.001 \cdot 2^n degrees. The triangle first becomes obtuse when this exceeds 30,30^\circ, i.e. 2n>30000.2^n \gt 30000. Since 214=163842^{14} = 16384 and 215=32768,2^{15} = 32768, the least such nn is 15.15.

Thus, the correct answer is E.

Problem 25 in Other Years