Difference between revisions of "1975 Canadian MO Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | + | Let <math>a</math> be the integer part and <math>b</math> be the decimal part, thus, we have the G.P. | |
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| + | <math>(b,a,a+b)</math> | ||
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| + | So, | ||
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| + | <math>b(a+b)=a^2</math> | ||
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| + | <math>ab+b^2=a^2</math> | ||
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| + | There we have a quadratic equation. We must isolate <math>a</math> or <math>b</math>. | ||
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| + | <math>a^2-ab-b^2=0</math> | ||
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| + | <math>b=\frac{-a\pm a\sqrt{5}}{2}</math> | ||
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| + | If the number must be positive, thus we'll consider only the solution | ||
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| + | <math>b=\frac{-a+a\sqrt{5}}{2}</math> | ||
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| + | As <math>b</math> is the decimal part, then it must be lower than 1 | ||
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| + | <math>b<1</math> | ||
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| + | <math>\frac{-a+a\sqrt{5}}{2}<1</math> | ||
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| + | <math>a<\frac{2}{\sqrt{5}-1}</math> | ||
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| + | <math>a<\frac{1+\sqrt{5}}{2}</math> | ||
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| + | This is the golden ratio and it's approximately <math>1.618</math>. | ||
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| + | As <math>a</math> must be an integer, thus <math>a=1</math>. Therefore | ||
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| + | <math>b=\frac{-a+a\sqrt{5}}{2}</math> | ||
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| + | <math>b=\frac{-1+\sqrt{5}}{2}</math> | ||
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| + | To find our number we must sum <math>a+b</math>, so | ||
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| + | <math>a+b</math> | ||
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| + | <math>1+\frac{-1+\sqrt5}{2}</math> | ||
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| + | <math>\boxed{\frac{1+\sqrt{5}}{2}}</math> | ||
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| + | The number we're searching for is the golden ratio. | ||
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{{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 10:15, 24 January 2018
Problem 4
For a positive number such as 
, 
is referred to as the integral part of the number and 
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 
be the integer part and 
be the decimal part, thus, we have the G.P.
So,
There we have a quadratic equation. We must isolate or .
If the number must be positive, thus we'll consider only the solution
As is the decimal part, then it must be lower than 1
This is the golden ratio and it's approximately 
.
As must be an integer, thus 
. Therefore
To find our number we must sum 
, 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 |
