A Pre-RMO question! -25

• We draw the diagram and complete the sides

• The triangle indicated are similar $\triangle AEB \sim \triangle AFG$ . Hence by similarity,

$\begin{aligned} \dfrac{x}{2} &= \dfrac{x+3}{5} \\ 5x &= 2x + 6 \\ x&= 2 \end{aligned}$

• Area of $\triangle AEB$ is

$\begin{aligned} &= \dfrac{1}{2} \times (x+3)(5) \\ &= \dfrac{25}{2} \end{aligned}$

• Answer to be entered is

$25 + 2 = \boxed{27}$

Let $A$ be the origin. Since the slope of a line is equal to $\frac{\text{rise}}{\text{run}}$ , line $AF$ has the equation $y = -3x$ , and line $GB$ has the equation $y = \frac{3}{2} (x - 5) = \frac{3}{2}x - \frac{15}{2}$ . Therefore $-3x = \frac{3}{2}x - \frac{15}{2}$ , so $-\frac{9}{2}x = -\frac{15}{2}$ and $x = \frac{5}{3}$ . As $y = -3x$ , $y = -5$ .

This means that the height of triangle $AEB$ is $5$ , and the base is $5$ . Thus its area is $\frac{1}{2} (5)(5) = \frac{25}{2}$ , and $m + n = 25 + 2 = \boxed{27}$ .

$|\overline {FG}|=5-(2+1)=2$ cm.

So, the ratio of heights of $\triangle {EFG}$ and $\triangle {EAB}$ is $2:5$ . Hence the height of $\triangle {EAB}$ is $5$ cm.

Therefore it's area is $\dfrac 12 \times 5\times 5=\dfrac {25}{2}$

Hence the required answer is $25+2=\boxed {27}$ .