2019 AMC 12B Problem 6

Attempt Problem 6 of the 2019 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

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?
00
22
44
88
infinitely many
Answer: A
Concepts:ellipsebounding to limit cases
Small hint:
With ABAB fixed, the perimeter condition puts CC on an ellipse with foci AA and BB
Big hint:
Area 100100 with base 1010 needs height 20;20; compare that to the ellipse’s greatest height
Solution:
The perimeter condition gives CA+CB=50−10=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=a2−c2=375≈19.36. 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 2⋅10010=20.\dfrac{2\cdot100}{10}=20. But the greatest possible height on the ellipse is b≈19.36<20,b\approx19.36\lt20, so no such CC exists.
Thus, A is the correct answer.
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