# 2006 UNCO Math Contest II Problems/Problem 1

## Problem

If a dart is thrown at the $6\times 6$ target, what is the probability that it will hit the shaded area? $[asy] filldraw((2,2)--(4,6)--(6,6)--(6,4)--cycle,blue); filldraw((2,2)--(6,2)--(6,1)--cycle,blue); filldraw((2,2)--(0,0)--(0,1)--cycle,blue); filldraw((2,2)--(0,4)--(0,6)--cycle,blue); for(int i=0;i<7;++i){ draw((0,i)--(6,i),black); draw((i,0)--(i,6),black); } dot((2,2));dot((0,4));dot((0,6));dot((4,6));dot((6,6)); dot((6,4));dot((6,2));dot((6,1));dot((0,0));dot((0,1)); [/asy]$

## Solution

To solve this we start by breaking it up into it's four shaded areas.

Starting with the bottom right one, we see it splits four squares. It hits the vertex on the left end, and it hits both vertexes on the right end so we know that the total area of that region in two squares. The reason for this is because we can take the triangles and rearrange them into two squares. Moving to the bottom left region, we see that the triangles can be put together to form a square, so we get one square for that region. For the top left region, we see that we can get two squares by rearranging. For the top right region, we see there are five squares already shaded in. We also see that the triangles can be moved to give us three more squares.

Thus, the total shaded area is $2+1+2+5+3=13$. Because there are $36$ squares, the probability is $\frac {13}{36}$.