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 $\frac {13}{36}$.

Solution 2 (Simple Geometry)

Knowing that this is a probability question, try to think of it as the ratio of the area of the shaded region to the area of the total region. In order to find the shaded region, we need to find the area of the four shapes.

When it comes to obtuse triangles, the altitude (height) is out of the shape and is always perpendicular to the base of the triangle. Notice how the two obtuse triangles each share an altitude with the length of 2, that means the height of the obtuse triangle is 2. We can now find the area of the two obtuse triangles and their areas are 1 and 2.

There is one more triangle that is shaded, but luckily, it is a right triangle so the height is basically given to us, the area of the right triangle is 2.

There is one last shape in this diagram, and notice it is shaped like a kite. In order to find the area of the kite, we need to find their diameters ($d_{1}$ as the shorter diameter and $d_{2}$ as the longer diameter). Notice how the diagonal of the square is $\sqrt 2$, so $d_{1}=2\sqrt 2$ and $d_{2}=4\sqrt 2$. Now we can find the area of the kite, which is $\frac{d_{1}d_{2}}{2}=8$.

Now we find the total area of the shaded region $1+2+2+8=13$ and the entire area of the grid is 36 which means our answer is $\frac {13}{36}$.

~ghfhgvghj10

See Also

2006 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
Invalid username
Login to AoPS