1973 Canadian MO Problems/Problem 1
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Problem
Solve the simultaneous inequalities,
and
; i.e. find a single inequality equivalent to the two simultaneous inequalities.
What is the greatest integer that satisfies both inequalities
and
.
Give a rational number between
and
.
Express
as a product of two integers neither of which is an integral multiple of
.
Without the use of logarithm tables evaluate
.
Solution
![$x<\frac{1}{4x} \Rightarrow 4x^2 < 1 \Rightarrow x^2< \frac{1}{4} \Rightarrow -\frac{1}{2} < x < \frac{1}{2}$](http://latex.artofproblemsolving.com/6/0/3/603a2592ecd822f17a7367213edb24958f8eb582.png)
Since from the second inequality , our solution is thus:
.
From
, we get that
. From
, we get:
![$x^{2}+3x > 16 \Rightarrow x^{2}+3x - 16 > 0 \Rightarrow x > \frac{-3 + \sqrt{71}}{2} \approx 2 \text{ or } x < \frac{-3-\sqrt{71}}{2}\approx -5$](http://latex.artofproblemsolving.com/1/9/7/19788ea1133f5623218c1328f38b304d29be09d9.png)
With these two inequalities, we see that the greatest integer satisfying the requirements is .
. Thus, a rational number in between
and
is
Thus,
See also
1973 Canadian MO (Problems) | ||
Preceded by 1973 Canadian MO Problems |
1 • 2 • 3 • 4 • 5 | Followed by Problem 2 |