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Tiebreaker Round Solutions 2014 CHMMC SolutionSolution 2. We have y = b(a + d) and z = c(a + d), so b = y/(a + d) and c = z/(a + d). Thus b and c are entirely
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Document Date: 2014-12-15 00:22:49
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Tiebreaker Round Solutions 2014 CHMMC SolutionSolution 2. We have y = b(a + d) and z = c(a + d), so b = y/(a + d) and c = z/(a + d). Thus b and c are entirely