Difference between revisions of "1959 IMO Problems/Problem 2"
(→Solution) |
(→Solution) |
||
| Line 13: | Line 13: | ||
Firstly, the square roots imply that a valid domain for x is <math>x\ge \frac{1}{2}</math>. | Firstly, the square roots imply that a valid domain for x is <math>x\ge \frac{1}{2}</math>. | ||
| − | Square both sides of the given equation: <cmath> \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) = A^2 | + | Square both sides of the given equation: <cmath> \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) = A^2</cmath> |
Add the first and the last terms to get <cmath>2x + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} = A^2</cmath> | Add the first and the last terms to get <cmath>2x + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} = A^2</cmath> | ||
| − | Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get:<cmath>2x + 2 \sqrt{x^2 + 2x - 1} = A^2</cmath> | + | Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get:<cmath>2x + 2 \sqrt{x^2 + 2x - 1} = A^2</cmath> |
| + | |||
| + | Simplify to obtain <cmath>A^2 = 2(x+|x-1|)</cmath> | ||
If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>. Otherwise, we have | If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>. Otherwise, we have | ||
| Line 23: | Line 25: | ||
<cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | <cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | ||
<cmath>A^2 > 2 </cmath> | <cmath>A^2 > 2 </cmath> | ||
| − | |||
Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math>A^2 \ge 2</math>, and for (c), the only solution is <math> x=\frac{3}{2}</math>. Q.E.D. | Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math>A^2 \ge 2</math>, and for (c), the only solution is <math> x=\frac{3}{2}</math>. Q.E.D. | ||
Revision as of 13:33, 15 December 2019
Problem
For what real values of 
is
given (a) 
, (b) 
, (c) 
, where only non-negative real numbers are admitted for square roots?
Solution
Firstly, the square roots imply that a valid domain for x is 
.
Square both sides of the given equation: ![\[\Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) = A^2\]](http://latex.artofproblemsolving.com/8/f/d/8fd224adfc76efccbea2530a204e6cb1ab1981ee.png)
Add the first and the last terms to get ![\[2x + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} = A^2\]](http://latex.artofproblemsolving.com/c/2/4/c2448cf30f2865c8e296e5351c36d067804afbc0.png)
Multiply the middle terms, and use 
to get:![\[2x + 2 \sqrt{x^2 + 2x - 1} = A^2\]](http://latex.artofproblemsolving.com/8/b/4/8b430696cc7fcd3af6ee1f4f2e5e02d8195e724f.png)
Simplify to obtain ![\[A^2 = 2(x+|x-1|)\]](http://latex.artofproblemsolving.com/8/0/a/80aa38fd1f8fd60f0b24ee1d2cf5cb4486b5ab49.png)
If 
, then we must clearly have 
. Otherwise, we have
![\[x = \frac{A^2 + 2}{4} > 1,\]](http://latex.artofproblemsolving.com/6/3/e/63e3fa17646a88fd17dacf722e355f4f04391608.png)
![\[A^2 > 2\]](http://latex.artofproblemsolving.com/2/2/a/22ab9dfef568f64933f1d1b717b53e0b597fb233.png)
Hence for (a) the solution is ![$x \in \left[ \frac{1}{2}, 1 \right]$](http://latex.artofproblemsolving.com/c/c/b/ccb86723dfdd4a38b0ae3a8ec3ec4815cf6c5ade.png)
, for (b) there is no solution, since we must have 
, and for (c), the only solution is 
. Q.E.D.
~flamewavelight (Expanded)
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 | ||
