Difference between revisions of "2021 AIME II Problems"

(Problem 3)
(Problem 4)
Line 21: Line 21:
  
 
==Problem 4==
 
==Problem 4==
These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.
+
There are real numbers <math>a, b, c, </math> and <math>d</math> such that <math>-20</math> is a root of <math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \cdot i, </math> where <math>m</math> and <math>n</math> are positive integers and <math>i = \sqrt{-1}.</math> Find <math>m+n.</math>
  
 
[[2021 AIME II Problems/Problem 4|Solution]]
 
[[2021 AIME II Problems/Problem 4|Solution]]

Revision as of 13:51, 22 March 2021

These are not the problems you are looking for.

2021 AIME II (Answer Key)
Printable version | AoPS Contest CollectionsPDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Problem 1

Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$.)

Solution

Problem 2

Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$. The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$, $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$. Find $AF$.

Someone please help with the diagram

Solution

Problem 3

Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\]is divisible by $3$.

Solution

Problem 4

There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$

Solution

Problem 5

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 6

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 7

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 8

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 9

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 10

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 11

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 12

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 13

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 14

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

Problem 15

These problems will not be available until the 2021 AIME II is released on Thursday, March 18, 2021.

Solution

See also

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
2021 AIME I
Followed by
2022 AIME I
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png