Difference between revisions of "1959 IMO Problems/Problem 2"
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<cmath>A^2 = 2(x+|x-1|)</cmath> | <cmath>A^2 = 2(x+|x-1|)</cmath> | ||
− | Case I: If <math>x \le 1</math>, then <math>|x-1| = 1 - x</math>, and the equation reduces to <math>A^2 = 2</math>. This is precisely part (a) of the question, for which the valid interval is now <math>x \in \left[ \frac{1}{2}, 1 \right]</math> | + | '''Case I:''' If <math>x \le 1</math>, then <math>|x-1| = 1 - x</math>, and the equation reduces to <math>A^2 = 2</math>. This is precisely part (a) of the question, for which the valid interval is now <math>x \in \left[ \frac{1}{2}, 1 \right]</math> |
− | Case II: If <math>x > 1</math>, then <math>|x-1| = x - 1</math> and we have | + | '''Case II:''' If <math>x > 1</math>, then <math>|x-1| = x - 1</math> and we have |
<cmath>x = \frac{A^2 + 2}{4} > 1</cmath> | <cmath>x = \frac{A^2 + 2}{4} > 1</cmath> | ||
which simplifies to | which simplifies to | ||
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i.e., <math>\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}</math>. | i.e., <math>\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}</math>. | ||
− | Case I: when <math>2x-1\ge 1</math> (i.e., <math>x\ge 1</math>), the equation becomes <math>2\sqrt{2x-1}=\sqrt{2}A</math>. For (a), we have <math>x=1</math>; for (b) we have <math>x=3/4</math> but <math>x\ge 1</math>, thus no solution; for (c) we have <math>x=\frac{3}{2}</math>. | + | '''Case I:''' when <math>2x-1\ge 1</math> (i.e., <math>x\ge 1</math>), the equation becomes <math>2\sqrt{2x-1}=\sqrt{2}A</math>. For (a), we have <math>x=1</math>; for (b) we have <math>x=3/4</math> but <math>x\ge 1</math>, thus no solution; for (c) we have <math>x=\frac{3}{2}</math>. |
− | Case II: when <math>0\le 2x-1 <1</math> (i.e., <math>1/2\le x <1</math>), the equation becomes <math>2=\sqrt{2}A</math>, which only works for (a) <math>A=\sqrt{2}</math>. | + | '''Case II:''' when <math>0\le 2x-1 <1</math> (i.e., <math>1/2\le x <1</math>), the equation becomes <math>2=\sqrt{2}A</math>, which only works for (a) <math>A=\sqrt{2}</math>. |
In summary, any <math>x \in \left[1/2, 1\right]</math> is a solution for (a); there is no solution for (b); there is one solution for (c), which is <math>x=\frac{3}{2}</math>. | In summary, any <math>x \in \left[1/2, 1\right]</math> is a solution for (a); there is no solution for (b); there is one solution for (c), which is <math>x=\frac{3}{2}</math>. |
Revision as of 01:36, 22 November 2023
Contents
Problem
For what real values of is
given (a) , (b)
, (c)
, where only non-negative real numbers are admitted for square roots?
Solution 1
The square roots imply that .
Square both sides of the given equation:
Add the first and the last terms to get:
Multiply the middle terms, and use to get:
Since the term inside the square root is a perfect square, and by factoring 2 out, we get
Use the property that
to get
Case I: If , then
, and the equation reduces to
. This is precisely part (a) of the question, for which the valid interval is now
Case II: If , then
and we have
which simplifies to
This tells there that there is no solution for (b), since we must have
For (c), we have , which means that
, so the only solution is
.
~flamewavelight (Expanded)
~phoenixfire (edited)
Solution 2
Note that the equation can be rewritten to
i.e.,
.
Case I: when (i.e.,
), the equation becomes
. For (a), we have
; for (b) we have
but
, thus no solution; for (c) we have
.
Case II: when (i.e.,
), the equation becomes
, which only works for (a)
.
In summary, any is a solution for (a); there is no solution for (b); there is one solution for (c), which is
.
~zhaoxiayu
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |