Maths Olympiad Prep

Track / Stage 7 / 64 of 300 #1464 of 1964

Problem 1464

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 (ab1)(bc1)(ca1)=1(\sqrt {ab}-1)(\sqrt {bc}-1)(\sqrt {ca}-1)=1
At most, how many of the numbers: abc,acb,bac,bca,cab,cbaa-\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.

Official solution

1. Given Condition:
We start with the given condition:
(ab1)(bc1)(ca1)=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:
abc,acb,bac,bca,cab,cba 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 abca \geq b \geq c. We analyze the case when c2c \geq 2:
- For a=b=c=2a = b = c = 2, all expressions are equal to 1.
- For a>2a > 2, the product (ab1)(ac1)(bc1)(\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 c1c \leq 1, then cbac - \frac{b}{a} and cabc - \frac{a}{b} are both less than 1. Additionally, bacba<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:
- cab<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:
- cba>1c - \frac{b}{a} > 1
- bac>1b - \frac{a}{c} > 1
- abc>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 bac>1b - \frac{a}{c} > 1 would yield a contradiction), we have a>b>2>ca > b > 2 > c.
- From cba>1c - \frac{b}{a} > 1, we get ac>a+bac > a + b.
- From bac>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 (ab1)(bc1)(ca1)=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 and solution reproduced as published; topic, difficulty and ordering added by this site.