Problem 27 · 2009 Math Kangaroo
Stretch
Algebra & Patterns
caseworksubstitution
If \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=k\), how many possible real values exist for k?
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Answer: B — 2
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Hint 1 of 2
Either the three numbers add to something nonzero, or they add to zero—handle the two cases separately.
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Hint 2 of 2
Add all three given fractions’ numerators and denominators to find k when a + b + c ≠ 0.
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Approach: split into a + b + c ≠ 0 and a + b + c = 0
- If a + b + c ≠ 0, adding the equal fractions gives k = (a+b+c)/[2(a+b+c)] = 1/2.
- If a + b + c = 0, each denominator equals minus its numerator, so k = −1.
- Thus k takes 2 possible values.
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