Difference between revisions of "1985 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | {{ | + | The equation can be re-written as |
+ | <cmath>\begin{align}\label{eqn1} | ||
+ | (x+10^5)^2(x-10^5)^2 -(x+10^5)(x-10^5) -x-1=0. | ||
+ | \end{align}</cmath> | ||
+ | |||
+ | We first prove that the equation has no negative roots. | ||
+ | Let <math>x\le 0.</math> The equation above can be further re-arranged as | ||
+ | <cmath>\begin{align*}[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]=x-1.\end{align*}</cmath> | ||
+ | The right hand side of the equation is negative. Therefore <cmath>[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2)]<0,</cmath> and we have | ||
+ | <math>-1<(x+10^5)(x-10^5) <2.</math> Then the left hand side of the equation is bounded by | ||
+ | <cmath>|[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]|\le 3\times 3.</cmath> | ||
+ | However, since <math>|(x+10^5)(x-10^5)|\le 2</math> and <math>x<0,</math> it follows that <math>|x+10^5| <\frac{2}{|x-10^5|}<2\times 10^{-5}</math> for negative <math>x.</math> Then <math>x<2\times 10^{-5}-10^5.</math> The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9. | ||
+ | |||
+ | Now let <math>x>0.</math> When <math>x=10^5,</math> the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of <math>10^5</math>, as its leading coefficient is positive. We will prove that <math>x=10^5</math> is a good approximation of the roots (within <math>10^{-2}</math>). In fact, we can solve the "quadratic" equation (1) for <math>(x+10^5)(x-10^5)</math>: | ||
+ | <cmath>(x+10^5)(x-10^5)=\frac{1\pm\sqrt{1+4(x+1)}}{2}.</cmath> | ||
+ | Then <cmath>x-10^5=\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}.</cmath> | ||
+ | Easy to see that | ||
+ | <math>|x-10^5| <1</math> for positve <math>x.</math> Therefore, <math>10^5-1<x<10^5+1.</math> Then | ||
+ | <cmath>\begin{align*} | ||
+ | |x-10^5|&=\left|\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ | ||
+ | &\le \left |\frac{1}{2(x+10^5)}\right |+\left |\frac{\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ | ||
+ | &\le \frac{1}{2(10^5-1+10^5)} +\frac{\sqrt{1+4(10^5+1+1)}}{2(10^5-1+10^5)} \\ | ||
+ | &<10^{-2}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Let <math>x_1</math> be a root of the equation with <math>x_1<10^5.</math> Then <math>0<10^5-x_1<10^{-2}</math> and | ||
+ | <cmath>x_1-10^5=\frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}.</cmath> | ||
+ | An aproximation of <math>x_1</math> is defined as follows: | ||
+ | <cmath>\tilde{x}_1=10^5+\frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.</cmath> | ||
+ | We check the error of the estimate: | ||
+ | <cmath>\begin{align*} | ||
+ | |\tilde{x}_1-x_1|&=\left | \frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right | \\ | ||
+ | &\le \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right |+\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | The first absolute value | ||
+ | <cmath> \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right | =\frac{|x_1- 10^5|}{2(10^5+10^5)(x_1+10^5)}<10^{-12}.</cmath> | ||
+ | |||
+ | The second absolute value | ||
+ | <cmath>\begin{align*} | ||
+ | &\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)} - \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} | ||
+ | \right |\\ | ||
+ | &\le \left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}\right |+\left |\frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |\\ | ||
+ | &\le 10^{-7}+10^{-9}, | ||
+ | \end{align*}</cmath> | ||
+ | through a rationalized numerator.Therefore <math>|\tilde{x}_1-x_1|\le 10^{-6}.</math> | ||
+ | |||
+ | For a real root <math>x_2</math> with <math>x_2>10^5,</math> we choose | ||
+ | <cmath>\tilde{x}_2=10^5+\frac{1+\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.</cmath> | ||
+ | We can similarly prove it has the desired approximation. | ||
== See Also == | == See Also == |
Revision as of 06:38, 18 May 2018
Problem
Determine each real root of
correct to four decimal places.
Solution
The equation can be re-written as
We first prove that the equation has no negative roots. Let The equation above can be further re-arranged as The right hand side of the equation is negative. Therefore and we have Then the left hand side of the equation is bounded by However, since and it follows that for negative Then The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.
Now let When the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of , as its leading coefficient is positive. We will prove that is a good approximation of the roots (within ). In fact, we can solve the "quadratic" equation (1) for : Then Easy to see that for positve Therefore, Then
Let be a root of the equation with Then and An aproximation of is defined as follows: We check the error of the estimate:
The first absolute value
The second absolute value through a rationalized numerator.Therefore
For a real root with we choose We can similarly prove it has the desired approximation.
See Also
1985 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.