Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 21, 2011"
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=== First Solution === | === First Solution === | ||
<math>(2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26</math>. Hence <math>(2x+3y)^2 \le 26</math> or <math>2x+3y = \sqrt{26}</math>. | <math>(2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26</math>. Hence <math>(2x+3y)^2 \le 26</math> or <math>2x+3y = \sqrt{26}</math>. | ||
− | If <math>3x-2y=0</math> and <math>2x+3y=\sqrt{26}</math>, then <math>2x+3y</math> attains this maximum value on the circle <math>x^2+y^2=2</math>. | + | If <math>3x-2y=0</math> and <math>2x+3y=\boxed{\sqrt{26}}</math>, then <math>2x+3y</math> attains this maximum value on the circle <math>x^2+y^2=2</math>. |
=== Second Solution === | === Second Solution === | ||
Let <math>x</math> and <math>y</math> be real numbers such that <math>x^2+y^2=2</math>. Note that | Let <math>x</math> and <math>y</math> be real numbers such that <math>x^2+y^2=2</math>. Note that | ||
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<cmath>\sqrt{26}\ge 2x+3y</cmath> | <cmath>\sqrt{26}\ge 2x+3y</cmath> | ||
so the maximum is <math>\boxed{\sqrt{26}}</math> and it is reached when <math>\frac{4}{x^2}=\frac{9}{y^2}\implies 2y=3x</math>. | so the maximum is <math>\boxed{\sqrt{26}}</math> and it is reached when <math>\frac{4}{x^2}=\frac{9}{y^2}\implies 2y=3x</math>. | ||
+ | ===Third Solution=== | ||
+ | Imagine the equations graphed in the coordinate plane. <math> x^2+y^2=2 </math> is a circle centered at the origin with | ||
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+ | radius <math> \sqrt{2} </math>. <math> 2x+3y=k </math> is a line. We want to find the largest value of <math> k </math> such that the line | ||
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+ | intersects the circle, giving real number solutions for <math> x </math> and <math> y </math>. This occurs when <math> 2x+3y=k </math> is tangent | ||
+ | |||
+ | to the circle, and thus when the distance from the line to the origin is <math> \sqrt{2} </math>. The distance from a point <math> | ||
+ | |||
+ | (x_0, y_0) </math> to a line <math> Ax+By+C=0 </math> is | ||
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+ | |||
+ | |||
+ | <math> \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} </math>. | ||
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+ | Plugging in <math> A=2, B=3, C=-k, x_0=0 </math>, and <math> y_0=0 </math> and setting the expression equal to <math> \sqrt{2} </math> yields | ||
+ | |||
+ | <math> \pm\frac{k}{\sqrt{13}}=\sqrt{2} </math>, or <math> k=\pm\sqrt{26} </math>. We want the largest value of <math> k </math>, so | ||
+ | |||
+ | <math> k=\boxed{\sqrt{26}} </math> is the highest possible value. |
Revision as of 19:43, 21 June 2011
Problem
AoPSWiki:Problem of the Day/June 21, 2011
Solutions
First Solution
. Hence or . If and , then attains this maximum value on the circle .
Second Solution
Let and be real numbers such that . Note that thus, we may assume that and are positive. Furthermore, by the Cauchy-Schwarz Inequality, we have but since , the inequality is equivalent with or so the maximum is and it is reached when .
Third Solution
Imagine the equations graphed in the coordinate plane. is a circle centered at the origin with
radius . is a line. We want to find the largest value of such that the line
intersects the circle, giving real number solutions for and . This occurs when is tangent
to the circle, and thus when the distance from the line to the origin is . The distance from a point to a line is
.
Plugging in , and and setting the expression equal to yields
, or . We want the largest value of , so
is the highest possible value.