In the diagram, the points are evenly spaced vertically and horizontally. A segment $A B$ is drawn using two of the points, as shown. Point $C$ is chosen to be one of the remaining 18 points. For how many of these 18 possible points is triangle $A B C$ isosceles?
(A) 6
(B) 4
(C) 5
(D) 2
(E) 3
In how many ways can we place $C$ so that $A B=A C$ ?

To get from $A$ to $B$ we go 1 unit in one direction and 2 units in the perpendicular direction from $A$.

To choose $C$ so that $A B=A C$, we must also go 1 unit in one direction and 2 units in the perpendicular direction.

This gives the 3 possible points:

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-329.jpg?height=225&width=263&top_left_y=1113&top_left_x=991)

In how many ways can we place $C$ so that $B A=B C$ ?

To choose $C$ so that $B A=B C$, we again also go 1 unit in one direction and 2 units in the other direction from $B$.

This gives the 3 possible points:

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-329.jpg?height=220&width=241&top_left_y=1641&top_left_x=996)

So in total we have 6 possible points so far, which is the largest of the possible answers.

So 6 must be the answer.

(There is one more possibility: can $C$ be placed so that $C A=C B$ ? Since 6 must be the answer, then the answer to this question is no. We can also check this by trying to find such points in the diagram.)

ANSWER: (A)