2018 AMC 10A Problem 21

Below is the professionally curated solution for Problem 21 of the 2018 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:parabola circle substitution

Difficulty rating: 1820

21.

Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?

$a = \dfrac14$

$\dfrac14 \lt a \lt \dfrac12$

$a \gt \dfrac14$

$a = \dfrac12$

$a \gt \dfrac12$

Solution:

Substitute $y=x^2-a$ into $x^2+y^2=a^2$ . This gives $x^2+(x^2-a)^2=a^2$ , so $x^2(x^2-(2a-1))=0$ .

The factor $x^2=0$ always gives the single point $(0,-a)$ . The other factor gives two additional real points exactly when $2a-1>0$ .

There are exactly three intersection points when $a>\dfrac12$ . Thus, E is the correct answer.

← Problem 20 Full Exam Problem 22 →

Problem 21 in Other Years

2000 AMC 10 · 2001 AMC 10 · 2002 AMC 10A · 2002 AMC 10B · 2003 AMC 10A · 2003 AMC 10B · 2004 AMC 10A · 2004 AMC 10B · 2005 AMC 10A · 2005 AMC 10B · 2006 AMC 10A · 2006 AMC 10B · 2007 AMC 10A · 2007 AMC 10B · 2008 AMC 10A · 2008 AMC 10B · 2009 AMC 10A · 2009 AMC 10B · 2010 AMC 10A · 2010 AMC 10B · 2011 AMC 10A · 2011 AMC 10B · 2012 AMC 10A · 2012 AMC 10B · 2013 AMC 10A · 2013 AMC 10B · 2014 AMC 10A · 2014 AMC 10B · 2015 AMC 10A · 2015 AMC 10B · 2016 AMC 10A · 2016 AMC 10B · 2017 AMC 10A · 2017 AMC 10B · 2018 AMC 10B · 2019 AMC 10A · 2019 AMC 10B · 2020 AMC 10A · 2020 AMC 10B · 2021 AMC 10A Spring · 2021 AMC 10B Spring · 2021 AMC 10A Fall · 2021 AMC 10B Fall · 2022 AMC 10A · 2022 AMC 10B · 2023 AMC 10A · 2023 AMC 10B · 2024 AMC 10A · 2024 AMC 10B · 2025 AMC 10A · 2025 AMC 10B