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2019 AMC 12B Problem 6

Problem 6 of 25EasierAlgebraGeometry

In a given plane, points AA and BB are 1010 units apart. How many points CC are there in the plane such that the perimeter of ABC\triangle ABC is 5050 units and the area of ABC\triangle ABC is 100100 square units?

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Solution

The perimeter condition gives CA+CB=5010=40,CA+CB=50-10=40, so CC lies on an ellipse with foci A,BA,B and major axis 2a=40.2a=40. Thus a=20a=20 and c=5,c=5, so the semi-minor axis is b=a2c2=37519.36. b=\sqrt{a^2-c^2}=\sqrt{375}\approx19.36. For area 100100 with base AB=10,AB=10, the height from CC must be 210010=20.\dfrac{2\cdot100}{10}=20. But the greatest possible height on the ellipse is b19.36<20,b\approx19.36\lt20, so no such CC exists. Thus, A is the correct answer.

More practice

Concepts: ellipse · bounding to limit cases

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.