+27 In the diagram, the points are evenly spaced vertically and horizontally. A segment $AB$ 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 $ABC$ isosceles?
+1911 There are two cases to consider:
Case 1: C is on the horizontal line through A or the vertical line through B.
Without loss of generality, suppose C is on the horizontal line through A. If AB is a horizontal segment, then △ABC cannot be isosceles.
Otherwise, AC=AB if and only if C is one of the two points on this horizontal line that are the same distance from A as B is.
The same argument applies if C is on the vertical line through B. Thus, there are at most 2+2=4 such points C.
Case 2: C is not on the horizontal line through A or the vertical line through B.
Then △ABC is isosceles if and only if C lies on the perpendicular bisector of AB. There are 4 such points.
Hence, there are 4+4=8 such points C.
