2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 10 Easy

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? (2019 AMC 10B Problem, Question#10)

  • A.

    0

  • B.

    2

  • C.

    4

  • D.

    8

  • E.

    infnitely many

Answer:A

Notice that whatever point we pick for C, AB will be the base of the triangle. Without loss of generality, let points A and B be (0,0) and (0, 10), since for any other combination of points, we can just rotate the plane to make them (0,0) and (0, 10) under a new coordinate system. When we pick point C, we have to make sure that its y-coordinate is \pm 20, because that's the only way the area of the triangle can be 100.

Now when the perimeter is minimized, by symmetry, we put C in the middle, at (5,20).  We can easily see that AC and BC will both be \sqrt {{20}^2+5^2}=\sqrt {425}. The perimeter of this minimal tiangle is 2 \sqrt{425}+10. which is larger than 50. Since the minimum perimeter is greater than 50, there is no triangle that satisfies the condition, giving us \text {(A)}0.

Without loss of generality, let AB be a horizontal segment of length 10. Now realize that C has to lie on one of the lines parallel to AB and vertically 20 units away from it. But 10 + 20 + 20 is already 50, and this doesn't form a triangle. Otherwise, without loss of generality,  AC< 20. Dropping altitude CD, we have a right triangle ACD with hypotenuse AC< 20 and leg CD=20, which is clearly impossible, again giving the answer as \text {(A)}0.