Six points $A, B, C, D, E$, and $F$ are placed on a square grid, as shown. How many triangles that are not right-angled can be drawn by using 3 of these 6 points as vertices?
(A) 2
(B) 1
(C) 6
(D) 0
(E) 4

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-086.jpg?height=228&width=304&top_left_y=469&top_left_x=1474)

## Part C: Each correct answer is worth 8.
We first make a list of all of the possible triangles with an edge lying on $D E F$ (that is, with an edge that is one of $D E, D F$ or $E F$ ):

$D A E, D B E, D C E, D A F, D B F, D C F, E A F, E B F, E C F$

Of these, triangles $D A E, D B E, D A F, D C F, E B F, E C F$ are obviously right-angled, while triangles $D C E$ and $E A F$ are not right-angled because they each contain an angle of $135^{\circ}$. What about triangle $D B F$ ? In fact, triangle $D B F$ is right-angled since $\angle D B E=\angle E B F=45^{\circ}$, and so $\angle D B F=90^{\circ}$.

Therefore, there are 2 triangles among this group which are not right-angled.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-367.jpg?height=238&width=298&top_left_y=1865&top_left_x=1445)

If we now look at the possible triangles with an edge lying on $A B C$, we will again find 2 non right-angled triangles among these triangles. (We can see this by reflecting all of the earlier triangles to move their base from $D E F$ to $A B C$.)

We can also see that it is impossible to make a triangle without an edge lying on $A B C$ or $D E F$, since the only other possible edges are $A D, A E, A F, B D, B E, B F, C D, C E$, and $C F$. No three of these nine edges can be joined to form a triangle since each edge has one end at $A, B$ or $C$ and the other end at $D$, $E$ or $F$, and there are no edges joining two points among either set of three points.

Therefore, there are only 4 triangles that can be formed that are not right-angled.

ANSWER: (E)

## Part C