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2023 WSMO Speed Round Problems/Problem 9: Difference between revisions

Pinkpig (talk | contribs)
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Created page with "==Problem== Suppose that <math>b</math> and <math>c</math> are the roots of the equation <math>x^2-\log(16)x+\log(64).</math> If <math>\sqrt{a+b}+\sqrt{a+c} = \sqrt{b+c},</ma..."
 
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==Solution==
==Solution==
We have
<cmath>\begin{align*}
\sqrt{a+b}+\sqrt{a+c} &= \sqrt{b+c}\implies\\
\left(\sqrt{a+b}+\sqrt{a+c}\right)^2 &= \left(\sqrt{b+c}\right)^2\implies\\
(a+b)+(a+c)+2\sqrt{(a+b)(a+c)} &= b+c\implies\\
2a+2\sqrt{(a+b)(a+c)} &= 0\implies\\
2\sqrt{(a+b)(a+c)} &= -2a\implies\\
\left(\sqrt{(a+b)(a+c)}\right)^2 &= (-a)^2\implies\\
(a+b)(a+c) &= a^2\implies\\
a^2+ab+ac+bc &= a^2\implies\\
ab+ac+bc &= 0\implies\\
a &= -\frac{bc}{b+c}.
\end{align*}</cmath>
From Vieta's Formulas, we have
<cmath>\begin{align*}
a &= -\frac{\log(64)}{\log(16)}=-\log_{16}64 = -\log_{4^2}4^3=-\frac{3}{2}\implies\\
2^a &= 2^{-\frac{3}{2}} = \frac{1}{\sqrt{8}} = \frac{\sqrt{2}}{4}\implies 2+4 = \boxed{6}.
\end{align*}</cmath>
~pinkpig

Latest revision as of 10:12, 13 September 2025

Problem

Suppose that $b$ and $c$ are the roots of the equation $x^2-\log(16)x+\log(64).$ If $\sqrt{a+b}+\sqrt{a+c} = \sqrt{b+c},$ then $2^a = \frac{\sqrt{m}}{n},$ find $m+n.$

Solution

We have \begin{align*} \sqrt{a+b}+\sqrt{a+c} &= \sqrt{b+c}\implies\\ \left(\sqrt{a+b}+\sqrt{a+c}\right)^2 &= \left(\sqrt{b+c}\right)^2\implies\\ (a+b)+(a+c)+2\sqrt{(a+b)(a+c)} &= b+c\implies\\ 2a+2\sqrt{(a+b)(a+c)} &= 0\implies\\ 2\sqrt{(a+b)(a+c)} &= -2a\implies\\ \left(\sqrt{(a+b)(a+c)}\right)^2 &= (-a)^2\implies\\ (a+b)(a+c) &= a^2\implies\\ a^2+ab+ac+bc &= a^2\implies\\ ab+ac+bc &= 0\implies\\ a &= -\frac{bc}{b+c}. \end{align*} From Vieta's Formulas, we have \begin{align*} a &= -\frac{\log(64)}{\log(16)}=-\log_{16}64 = -\log_{4^2}4^3=-\frac{3}{2}\implies\\ 2^a &= 2^{-\frac{3}{2}} = \frac{1}{\sqrt{8}} = \frac{\sqrt{2}}{4}\implies 2+4 = \boxed{6}. \end{align*}

~pinkpig

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