Don't Be Intimidated (Fall 2021 AMC 12B #5, 8, 13)

Fall 2021 AMC 12B #5

Problem: Call a fraction \frac{a}{b}, not necessarily in the simplest form, special if a and b are positive integers whose sum is 15. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

Solution: We can list out all the special fractions: \{\frac{1}{14},\frac{2}{13}, \frac{3}{12}, \frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3}, \frac{13}{2}, \frac{14}{1}\}. Simplifying, we have the fractions \{\frac{1}{14},\frac{2}{13}, \frac{1}{4}, \frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},\frac{8}{7},\frac{3}{2},2,\frac{11}{4},4, \frac{13}{2}, 14\}. We see that we can only add fractions with the same denominator, so we only need to focus on \{\frac{1}{4},\frac{1}{2},\frac{2}{3},\frac{3}{2},2,\frac{11}{4},4, \frac{13}{2}, 14\}. From the whole numbers \{2,4,14\}, we can achieve 4, 6, 16, 8, 18, 28. From the fractions with denominator 2 including \{\frac{1}{2},\frac{3}{2}, \frac{13}{2}\}, we obtain the additional integers 1,2,3,7,13. From the fractions with denominator 4 including \{\frac{1}{4}, \frac{11}{4}\}, we get 3, which has already been found. Thus, we have 6 + 5 = 11 distinct integers that can be written as the sum of two special fractions.

Fall 2021 AMC 12B #8

Problem: The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?

Solution: We can draw a diagram and label the congruent sides x, the base b, and the height to the base h.

From here, we can find that x^2=2bh as stated in the problem. We can find the area of the triangle in two ways since Area = \frac{1}{2}bh = \frac{1}{2}ab\sin{\theta}, which gives us bh = x^2\sin{\theta}. Substituting 2bh = x^2, we get \sin{\theta} = \frac{1}{2}. The only solutions \theta here are \theta = 30^\circ, 150^\circ. Since the triangle is obtuse, we know that the vertex angle \theta = 150^\circ.

Fall 2021 AMC 12B #13

Problem: Let c = \frac{2\pi}{11}. What is the value of \frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?

Solution: While this problem may look intimidating and prompt you to find the sine triple angle formula, there is actually no need for that. Since we are given a value of c, we should just plug in c into the expression. Doing so, we get \frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}} We can note that \frac{22\pi}{11} = 2\pi and \sin\theta = -\sin(2\pi-\theta), and simplify this to \frac{\sin \frac{6\pi}{11} \cdot - \sin \frac{10\pi}{11} \cdot -\sin \frac{4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=1 

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