Compare the product of the two triangle numbers with the product of the square and circle numbers in each group. One is always exactly double the other.
Step 1: Identify the shapes involved in each group.
Each group has two triangles, one square, and one circle. Call the two triangle numbers \(t_1\) and \(t_2\), the square number \(s\), and the circle number \(c\).
Group 1: \(t_1=6, t_2=8, s=8, c=12\). Group 2: \(t_1=5, t_2=6, s=10, c=6\). Group 3: \(t_1=3, t_2=12, s=4, c=\;?\).
Step 2: Test a relationship between the four numbers using Group 1.
Multiply the two triangle numbers: \(6 \times 8 = 48\).
Multiply the square and circle numbers: \(8 \times 12 = 96\).
Notice that \(96 = 2 \times 48\), so the product of the square and the circle is exactly twice the product of the two triangles.
Step 3: Confirm the same relationship with Group 2.
Triangle product: \(5 \times 6 = 30\). Square times circle: \(10 \times 6 = 60\).
Again, \(60 = 2 \times 30\), so the pattern holds:
\[ s \times c = 2 \times t_1 \times t_2 \]
Step 4: Apply the rule to Group 3.
Triangle product: \(3 \times 12 = 36\), so \(2 \times 36 = 72\).
The square number is 4, so:
\[ 4 \times c = 72 \]
\[ c = \frac{72}{4} = 18 \]
Final Answer:
The missing circle number is 18.
\[ \boxed{18} \]
