Problem 23 · AMC 8 Stretch
Stretch
Fractions, Decimals & Percents
Number Theory
logical-reasoningpattern-recognition
The 'mediant' of two fractions adds the tops and adds the bottoms: \(\frac{a}{b} \oplus \frac{c}{d} = \frac{a+c}{b+d}\). (This is NOT how you add fractions, but the result always lands strictly between them.) What is the mediant of \(\frac{1}{3}\) and \(\frac{1}{2}\)? Give it as a fraction in lowest terms.
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Answer: 2/5
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Hint 1 of 4
Compute the mediant of \(\frac{1}{3}\) and \(\frac{1}{2}\) by adding tops and adding bottoms.
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Hint 2 of 4
Turn all three fractions into decimals (or a common denominator) and put them in order. Is the mediant in the middle?
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Hint 3 of 4
To see why it always works, compare \(\frac{a}{b}\) with \(\frac{a+c}{b+d}\) using cross-multiplication (the bigger fraction has the bigger cross-product).
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Approach: Compute the mediant, verify it lies between via cross-multiplication
- Add tops and bottoms: \(\frac{1 + 1}{3 + 2} = \frac{2}{5}\), already in lowest terms.
- Check the order as decimals: \(\frac{1}{3} \approx 0.333\), \(\frac{2}{5} = 0.4\), \(\frac{1}{2} = 0.5\). The mediant sits right between them.
- Why it always works: if \(\frac{a}{b} < \frac{c}{d}\) then \(ad < bc\). Comparing \(\frac{a}{b}\) with \(\frac{a+c}{b+d}\) by cross-multiplying gives \(a(b+d) < b(a+c)\), i.e. \(ab + ad < ab + bc\), i.e. \(ad < bc\) — exactly what we know.
- The same check shows the mediant is below \(\frac{c}{d}\), so the mediant of \(\frac{1}{3}\) and \(\frac{1}{2}\) is \(\frac{2}{5}\) and always lands strictly between.
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