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2019 AMC 12A Problem 25

Problem 25 of 25HarderAlgebraGeometry

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 An−1A_{n-1} to line Bn−1Cn−1.B_{n-1}C_{n-1}. Likewise, define BnB_n to be the foot of the altitude from Bn−1B_{n-1} to line An−1Cn−1,A_{n-1}C_{n-1}, and CnC_n to be the foot of the altitude from Cn−1C_{n-1} to line An−1Bn−1.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?

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Solution

For an acute triangle, the orthic triangle (feet of the altitudes) has angles 180∘−2α180^\circ - 2\alpha for each original angle α.\alpha. Writing an angle as 60∘+x,60^\circ + x, the new angle is 60∘−2x,60^\circ - 2x, so each deviation from 60∘60^\circ is multiplied by −2.-2. The initial deviations are ±0.001∘.\pm 0.001^\circ. After nn steps a deviation has magnitude 0.001⋅2n0.001 \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.
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Tagged: altitude · recursion · angle chasing

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