# FMM20

### From charlesreid1

# Friday Morning Math Problem

## Arrange in Order

Given the positive integers a, b, c, d, with (a/b) < (c/d) < 1, arrange the following quantities in order of increasing magnitude:

(b/a)

(d/c)

(bd)/(ac)

(b+d)/(a+c)

1

Solution |
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Since (c/d) < 1, we know (d/c) > 1.
Since (a/b) < (c/d), we know (b/a) > (d/c). Hence, 1 < (d/c) < (b/a). If we know (d/c) > 1 and (b/a) > 1, then their product will be greater than both. Hence, 1 < (d/c) < (b/a) < (bd)/(ac). Next, we can show that the fraction (b+d)/(a+c) is greater than (d/c) and less than (b/a). Since (d/c) < (b/a), we know (ad < bc), and (cd + ad) < (cd + bc). Rearranging gives (d/c) < (b+d)/(a+c). Likewise, since (bc > ad), we know (ab + bc) > (ab + ad), which can be rearranged to (b+d)/(a+c) < (b/a). The order is therefore 1 < (d/c) < (b+d)/(a+c) < (b/a) < (bd)/(ac) |

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Friday Morning Math ProblemsSpider Socks and Shoes FMM1 Sums of Powers of 2 FMM2 Fifty Coins FMM3 The Zeta Monogram FMM4 The Cthulhus Monogram FMM4B Multiplication Logic FMM5 The Termite and the Cube FMM6 Sharing Dump Trucks FMM7 The Flippant Juror FMM8 Bus Routes FMM9 A Robust Bus System FMM9B Square-Free Sequence FMM10 Inferring Rule from Sequence FMM11 Checkerboard Color Schemes FMM12 One-Handed Chords FMM13 First Ace FMM14 Which Color Cab FMM15 Petersburg Paradox Revisited FMM16 A Binomial Challenge FMM17 A Radical Sum FMM18 Memorable Phone Numbers FMM19 Arrange in Order FMM20 A Pair of Dice Games: FMM21
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Source: Problem 2-5, Challenging Problems in Algebra

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