1975 Canadian MO Problems/Problem 4

Revision as of 19:03, 23 January 2018 by Alevini98 (talk | contribs) (Added solution)

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.

Solution

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


$(b,a,a+b)$


So,


$b(a+b)=a^2$

$ab+b^2=a^2$


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

$a^2-ab-b^2=0$

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


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

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


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


$b<1$

$\frac{-a+a\sqrt{5}}{2}<1$

$a<\frac{2}{\sqrt{5}-1}$

$a<\frac{1+\sqrt{5}}{2}$


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


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


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

$b=\frac{-1+\sqrt{5}}{2}$


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


$a+b$

$1+\frac{-1+\sqrt5}{2}$

$\boxed{\frac{1+\sqrt{5}}{2}}$


The number we're searching for it's the golden ratio.


1975 Canadian MO (Problems)
Preceded by
Problem 3
1 2 3 4 5 6 7 8 9 10 Followed by
Problem 5


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