Triangle With Reciprocal Sides

Relevant wiki: Solving Triangles - Problem Solving - Medium

Let's denote $a$ as $\frac { 1 }{ 2015 }$ , $b$ as $\frac { 1 }{ x }$ , and $c$ as $\frac { 1 }{ y }$ , where $x$ and $y$ are distinct positive integers.

By the Triangle Inequality,

$\frac { 1 }{ x } +\frac { 1 }{ y } >\frac { 1 }{ 2015 }$ ..... $(*)$

$\frac { 1 }{ x } +\frac { 1 }{ 2015 } >\frac { 1 }{ y }$ ..... $(**)$

$\frac { 1 }{ y } +\frac { 1 }{ 2015 } >\frac { 1 }{ x }$ ..... $(***)$

WLOG, let's say that $\frac { 1 }{ x } >\frac { 1 }{ y } \Rightarrow y>x$ , so $(**)$ is always true.

Working with $(***)$ ,

$\frac { 1 }{ 2015 } >\frac { 1 }{ x } -\frac { 1 }{ y }$

$\frac { 1 }{ 2015 } >\frac { y-x }{ xy }$

$\frac { xy }{ 2015 } >y-x$

$xy>2015y-2015x$

$xy-2015y+2015x>0$

$(x-2015)(y+2015)>-4060225$

Let's experiment and say that $y=x+1$ .

$(x-2015)(x+2016)>-4060225$

${ x }^{ 2 }+x-2015>0$

$x>\frac { -1+\sqrt { { (1) }^{ 2 }-(4\cdot 1\cdot -2015) } }{ 2 } \Rightarrow x>\frac { -1+\sqrt { 8061 } }{ 2 } \approx 44.392$

We are trying to minimize $x+y$ , so we observe that if the amount that $y$ is greater than $x$ by is denoted by $z$ , then the solution of the quadratic is $x>\frac { -z+\sqrt { { z }^{ 2 }+8060z } }{ 2 }$ , where the minimum value of $x$ is minimized at $z=1$ ( $z$ cannot be less than one because $y$ is an integer greater than $x$ ).

Therefore, $x=45$ , and $y=46$ . Plugging these values back into $(*)$ , $\frac { 1 }{ 45 } +\frac { 1 }{ 46 } >\frac { 1 }{ 2015 }$ , so $x+y=\boxed {91}$

Note that $\frac{1}{x}-\frac{1}{y} \leq \frac{1}{2015}$ Now, we need two successive integers. Let, $x=z$ and $y=z+1$ Then, $\frac{1}{z}-\frac{1}{z+1} \leq \frac{1}{2015}$ Solving the above equation we get, $z \geq 44.392$ Hence, we have $x=45$ and $y=46$