Sample Question 6:

In a given plane, points \(A\) and \(B\) are \(10\) units apart. How many points \(C\) are there in the plane such that the perimeter of \(\triangle ABC\) is \(50\) units and the area of \(\triangle ABC\) is \(100\) square units?

\(\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}\)

---

Answer Keys

Question 6: A

---

Solutions

Question 6
Step 1: Observe that regardless of the point \(C\) we choose, \(AB\) will be the base of the triangle. Without loss of generality, set points \(A\) and \(B\) as (0,0) and (10,0), respectively. This is because, for any other combination of points, we can simply rotate the plane to fit them to (0,0) and (10,0) under a new coordinate system.

Step 2: When we select point \(C\), it must have a y-coordinate of \(\pm20\) in order to secure the triangle's area of 100 units.

Step 3: For the perimeter to be at its minimum, place \(C\) in the middle at (5, 20) by symmetry. In this case, \(AC\) and \(BC\) both become \(\sqrt{20^2+5^2} = \sqrt{425}\).

Step 4: Calculate the perimeter of this minimal triangle: \(2\sqrt{425} + 10\), which is greater than 50.

Step 5: Observe that if the minimum possible perimeter is more than 50, there is no possible triangle that would satisfy the conditions.

So, the result is \(\boxed{\textbf{(A) }0}\).