1975 Canadian MO Problems/Problem 4

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 •	Followed by Problem 5

Page

Toolbox

Search

1975 Canadian MO Problems/Problem 4

Problem 4

Solution