Minimum Value

Another way of writing $\sqrt{x^{2}+y^{2}}$ would be the distance from the origin to the point $(x,y)$ on a function $f(x)$ .

The shortest distance from a point $(p,q)$ to the line $ax+by+c=0$ is given by the formula $\frac{\left|ap+bq+c\right|}{\sqrt{a^{2}+b^{2}}}$

In this case, out point is $(0,0)$ and our line is $5x+12y-60=0$ , so after plugging those values in we get our distance is $\frac{60}{13}$ .

$13$ is a prime number, and $60$ is not a multiple of $13$ , so we cannot reduce our fraction. Our answer is then $60+13=\boxed{73}$

I liked the solution of Brandon Monsen sir but here is an algebraic way

By Cauchy-Scwarz Inequality

$\large{\left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { 5 }^{ 2 }+{ 12 }^{ 2 } \right) \ge { \left( 5x+12y \right) }^{ 2 }\\ \left( { x }^{ 2 }+{ y }^{ 2 } \right) \ge \frac { { \left( 5x+12y \right) }^{ 2 } }{ \left( { 13 }^{ 2 } \right) } \\ \sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) } \ge \frac { 60 }{ 13 } }$

This gives $A+B=73$

The minimum value of $\sqrt{x^2+y^2}$ is the distance from the origin to the line $5x+12y=60$ along the line perpendicular to this line that crosses through $(0,0)$ . This line is $12x-5y=0$ . Plugging in $y=\frac{12}{5}x$ into the original line and solving yields $x=\frac{300}{169}$ and $y=\frac{720}{169}$ , which yields a minimum value of $\frac{60}{13}$ , so the answer is $73$ .

don't ask simple questions please,even a child can do this one ...

This is the first time i have successfully applied cauchy-schwarz

The minimum of sqrt(x^2 + y^2) is the same as minimizing x^2 + y^2, Since 12y = 60 - 5x, y = 5 - (5/12)x, and x^2 + y^2 = 169x^2/144 - 25x/6 + 25, Defining the latter as f(x), f'(x) =0 implies 169x/72 = 25/6, or x = 300/169. then y = 720/169, x^2 + y^2 = 608400/(169^2),and sqrt(x^2 + y^2) = 780/169 = 60/13, whence a + b = 73. Ed Gray