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Difference between revisions of "2019 CIME I Problems/Problem 13"

(Created page with "Suppose <math>\text{P}</math> is a monic polynomial whose roots <math>a</math>, <math>b</math>, and <math>c</math> are real numbers, at least two of which are positive, that s...")
 
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Suppose <math>\text{P}</math> is a monic polynomial whose roots <math>a</math>, <math>b</math>, and <math>c</math> are real numbers, at least two of which are positive, that satisfy the relation <cmath>a(a-b)=b(b-c)=c(c-a)=1.</cmath> Find the greatest integer less than or equal to <cmath>100|P(\sqrt{3})|</cmath>.
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Suppose <math>\text{P}</math> is a monic polynomial whose roots <math>a</math>, <math>b</math>, and <math>c</math> are real numbers, at least two of which are positive, that satisfy the relation <cmath>a(a-b)=b(b-c)=c(c-a)=1.</cmath> Find the greatest integer less than or equal to <math>100|P(\sqrt{3})|</math>.
  
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==Solution==
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We can rearrange the equations to yield <math>b=a-\frac{1}{a}</math> and equivalent. Adding all three of these equations and simplifying results in <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0</math>, so <math>\frac{ab+bc+ca}{abc}=0</math> and thus <math>ab+bc+ca=0</math>.
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Next, take the three given equations and add them, resulting in <math>a^2+b^2+c^2-ab-bc-ca=3</math>. This is equivalent to <math>(a+b+c)^2-3(ab-bc-ca)=3</math> by direct expansion, so by substituting what we found earlier, we know that <math>a+b+c=\pm\sqrt{3}</math>.
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Finally, return to the three equations <math>b=a-\frac{1}{a}</math> and equivalent. Squaring all three results in <math>b^2=a^2-2+\frac{1}{a^2}</math> and equivalent. Adding the equations and simplifying results in <math>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=6</math>, so <math>a^2b^2+b^2c^2+c^2a^2=6(abc)^2</math>. But notice that
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<cmath>a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=\mp2\sqrt{3}abc</cmath>
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Thus, <math>\mp2\sqrt{3}abc=6(abc)^2</math>. Notice that if any one of <math>a,b,c</math> is <math>0</math>, then the given equations cannot be true; thus we can divide <math>abc</math> from both sides, resulting in <math>\mp2\sqrt{3}=6abc</math>, so <math>abc=\mp\frac{\sqrt{3}}{3}</math>.
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This finally means that by Vieta’s Formulas:
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<cmath>P(x)=x^3\pm\sqrt{3}x^2\mp\frac{\sqrt{3}}{3}</cmath>
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We can easily show using derivatives that the sign of the <math>x^2</math>-coefficient must be negative and vice versa for the constant (due to the existence of two or more positive roots). Thus,
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<cmath>P(x)=x^3-\sqrt{3}x^2+\frac{\sqrt{3}}{3}</cmath>
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Then, <math>P(\sqrt{3})=\frac{\sqrt{3}}{3}</math>, so the answer is <math>\frac{100\sqrt{3}}{3}>\boxed{057}</math>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
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==See also==
 
{{CIME box|year=2019|n=I|num-b=12|num-a=14}}
 
{{CIME box|year=2019|n=I|num-b=12|num-a=14}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
{{MAC Notice}}
 
{{MAC Notice}}

Latest revision as of 03:55, 21 March 2025

Suppose $\text{P}$ is a monic polynomial whose roots $a$, $b$, and $c$ are real numbers, at least two of which are positive, that satisfy the relation \[a(a-b)=b(b-c)=c(c-a)=1.\] Find the greatest integer less than or equal to $100|P(\sqrt{3})|$.

Solution

We can rearrange the equations to yield $b=a-\frac{1}{a}$ and equivalent. Adding all three of these equations and simplifying results in $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, so $\frac{ab+bc+ca}{abc}=0$ and thus $ab+bc+ca=0$.

Next, take the three given equations and add them, resulting in $a^2+b^2+c^2-ab-bc-ca=3$. This is equivalent to $(a+b+c)^2-3(ab-bc-ca)=3$ by direct expansion, so by substituting what we found earlier, we know that $a+b+c=\pm\sqrt{3}$.

Finally, return to the three equations $b=a-\frac{1}{a}$ and equivalent. Squaring all three results in $b^2=a^2-2+\frac{1}{a^2}$ and equivalent. Adding the equations and simplifying results in $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=6$, so $a^2b^2+b^2c^2+c^2a^2=6(abc)^2$. But notice that \[a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=\mp2\sqrt{3}abc\] Thus, $\mp2\sqrt{3}abc=6(abc)^2$. Notice that if any one of $a,b,c$ is $0$, then the given equations cannot be true; thus we can divide $abc$ from both sides, resulting in $\mp2\sqrt{3}=6abc$, so $abc=\mp\frac{\sqrt{3}}{3}$.

This finally means that by Vieta’s Formulas: \[P(x)=x^3\pm\sqrt{3}x^2\mp\frac{\sqrt{3}}{3}\] We can easily show using derivatives that the sign of the $x^2$-coefficient must be negative and vice versa for the constant (due to the existence of two or more positive roots). Thus, \[P(x)=x^3-\sqrt{3}x^2+\frac{\sqrt{3}}{3}\] Then, $P(\sqrt{3})=\frac{\sqrt{3}}{3}$, so the answer is $\frac{100\sqrt{3}}{3}>\boxed{057}$.

~ eevee9406

See also

2019 CIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions

Template:MAC Notice

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