Maths Olympiad Prep

Track / Stage 7 / 72 of 300 #1472 of 1964

Problem 1472

National Olympiad second round; IMO P1/P4
Algebra Difficulty 7.1 Find the answer

a,b,ca, b, c are positive real numbers such that (ab−1)(bc−1)(ca−1)=1(\sqrt {ab}-1)(\sqrt {bc}-1)(\sqrt {ca}-1)=1
At most, how many of the numbers: a−bc,a−cb,b−ac,b−ca,c−ab,c−baa-\frac {b}{c}, a-\frac {c}{b}, b-\frac {a}{c}, b-\frac {c}{a}, c-\frac {a}{b}, c-\frac {b}{a} can be bigger than 11?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

1. Given Condition:
We start with the given condition:
(ab−1)(bc−1)(ca−1)=1 (\sqrt{ab} - 1)(\sqrt{bc} - 1)(\sqrt{ca} - 1) = 1
where a,b,ca, b, c are positive real numbers.

2. Claim:
We need to determine how many of the numbers:
a−bc,a−cb,b−ac,b−ca,c−ab,c−ba a - \frac{b}{c}, \quad a - \frac{c}{b}, \quad b - \frac{a}{c}, \quad b - \frac{c}{a}, \quad c - \frac{a}{b}, \quad c - \frac{b}{a}
can be greater than 1.

3. Example to Achieve 4:
Consider the specific values b=c=32b = c = \frac{3}{2} and a=2+423a = 2 + \frac{4\sqrt{2}}{3}. We need to check if these values satisfy the given condition and if they allow four of the expressions to be greater than 1.

4. Assumption and Simplification:
Assume without loss of generality that a≥b≥ca \geq b \geq c. We analyze the case when c≥2c \geq 2:
- For a=b=c=2a = b = c = 2, all expressions are equal to 1.
- For a>2a > 2, the product (ab−1)(ac−1)(bc−1)(\sqrt{ab} - 1)(\sqrt{ac} - 1)(\sqrt{bc} - 1) becomes greater than 1, which contradicts the given condition. Thus, c<2c < 2.

5. Case Analysis:
- If c≤1c \leq 1, then c−bac - \frac{b}{a} and c−abc - \frac{a}{b} are both less than 1. Additionally, b−ac≤b−a<0b - \frac{a}{c} \leq b - a < 0, so at most three expressions can be greater than 1.
- If 1<c<21 < c < 2, we analyze further:
- c−ab<1c - \frac{a}{b} < 1, so at most five expressions can be greater than 1.

6. Checking for 5 Expressions:
Suppose five expressions can be greater than 1. This implies:
- c−ba>1c - \frac{b}{a} > 1
- b−ac>1b - \frac{a}{c} > 1
- a−bc>1a - \frac{b}{c} > 1

These yield:
bc>a+candac>a+b bc > a + c \quad \text{and} \quad ac > a + b

7. Further Implications:
- Since b>2b > 2 (otherwise b−ac>1b - \frac{a}{c} > 1 would yield a contradiction), we have a>b>2>ca > b > 2 > c.
- From c−ba>1c - \frac{b}{a} > 1, we get ac>a+bac > a + b.
- From b−ac>1b - \frac{a}{c} > 1, we get bc>a+cbc > a + c.

8. Contradiction:
Combining these, we get:
ac>a+b>2abandbc>a+c>2ac ac > a + b > 2\sqrt{ab} \quad \text{and} \quad bc > a + c > 2\sqrt{ac}
This implies:
abc2>4abcorbc3>16 abc^2 > 4a\sqrt{bc} \quad \text{or} \quad bc^3 > 16
Since c2<4c^2 < 4, we get bc>4bc > 4. Finally, ab>ac>bc>4ab > ac > bc > 4, which contradicts the given condition (ab−1)(bc−1)(ca−1)=1(\sqrt{ab} - 1)(\sqrt{bc} - 1)(\sqrt{ca} - 1) = 1.

9. Conclusion:
Therefore, at most four of the given expressions can be greater than 1.

The final answer is 4 \boxed{ 4 } .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.