Difference between revisions of "1975 Canadian MO Problems/Problem 4"

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The number we're searching for it's the golden ratio.
The number we're searching for is the golden ratio.
{{Old CanadaMO box|num-b = 3|num-a=5|year=1975}}
{{Old CanadaMO box|num-b = 3|num-a=5|year=1975}}

Latest revision as of 11:15, 24 January 2018

Problem 4

For a positive number such as $3.27$, $3$ is referred to as the integral part of the number and $.27$ as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression.


Let $a$ be the integer part and $b$ be the decimal part, thus, we have the G.P.





There we have a quadratic equation. We must isolate $a$ or $b$.


$b=\frac{-a\pm a\sqrt{5}}{2}$

If the number must be positive, thus we'll consider only the solution


As $b$ is the decimal part, then it must be lower than 1





This is the golden ratio and it's approximately $1.618$.

As $a$ must be an integer, thus $a=1$. Therefore



To find our number we must sum $a+b$, so




The number we're searching for is the golden ratio.

1975 Canadian MO (Problems)
Preceded by
Problem 3
1 2 3 4 5 6 7 8 Followed by
Problem 5

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