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LMNAS $25^{th}$ UGM, Indonesian

Suppose that $x,y\in\mathbb{R}$, so that :

$x^2+y^2+Ax+By+C=0$

with $A,B,C>2014$. Find the minimum value of $7x-24y$

$x^2+y^2+Ax+By+C=0$

can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$

Stuck,:>

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    $\begingroup$ To other mathSE reviewers: please respect the query tags : no Calculus please. Personally, I do not know how to attack such a problem without Calculus. $\endgroup$ Dec 21, 2020 at 4:55
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    $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Dec 21, 2020 at 4:55
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    $\begingroup$ @BookOfFlames normally, I would agree with your comment. Given that the problem is from a contest, and that Calculus is (presumably) outlawed, I feel that the OP deserves a serious hint, if anyone has one. $\endgroup$ Dec 21, 2020 at 4:57
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    $\begingroup$ It should be $(x+\frac A2)^2$ and $(y+\frac B2)^2$ $\endgroup$ Dec 21, 2020 at 5:19

3 Answers 3

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Since $$ x^2+Ax+\left( \frac{A}{2} \right) ^2+y^2+By+\left( \frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C\\ \left( x+\frac{A}{2} \right) ^2+\left( y+\frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C $$ Let $\displaystyle r=\sqrt{\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C}$, we can set $x=r\cos \theta -\dfrac{A}{2},y=r\sin \theta -\dfrac{B}{2}$, then $$ \begin{aligned} 7x-24y&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} (7 \cos \theta-24 \sin \theta)-7 A+24 B\right)\\ &=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} \sqrt{7^2+24^2}\sin(\theta+\phi)-7 A+24 B\right)\\ &\geq \frac{1}{2} \left(25\sqrt{A^2+B^2-4 C} -7 A+24 B\right) \end{aligned} $$ The second step uses The auxiliary Angle formula.

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Using Cauchy-Schwarz inequality, $$\left(x+\frac{A}{2}\right)^2+\left(y+\frac{B}{2}\right)^2+C-\left(\frac{A}{2}\right)^2-\left(\frac{B}{2}\right)^2=0\\ \implies 7x-24y=7\left(x+\frac A2\right)-24\left(y+\frac B2\right)+\frac{-7A+24B}{2}\\ \ge -\sqrt{7^2+(-24)^2} \cdot \sqrt{\left(x+\frac A2\right)^2+\left(y+\frac B2\right)^2}+ \frac{-7A+24B}{2}\\ =-25 \frac{1}{2} \sqrt{A^2+B^2-4C} + \frac{-7A+24B}{2}\\ =\frac 12 \left(-25\sqrt{A^2+B^2-4C}-7A+24B\right).\blacksquare$$

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Hint:

Let $7x-24y=z\iff x=?$

Replace the value of $x$ in $$x^2+y^2+Ax+By+C=0$$ to form a Quadratic Equation in $y$

As $y$ is real, the discriminant must be $\ge0$

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