AMC 12

Conducting Body: Mathematical Association of America (MAA)
For: Students in grade 12 or below
Subjects: Algebra, geometry, number theory, combinatorics
Question Type: 25 multiple-choice questions
Duration: 75 minutes
Scoring: Points for correct answers, fewer for blank questions
Frequency: Twice annually (A and B competitions)
Qualifies For: American Invitational Mathematics Examination (AIME)

The American Mathematics Competition 12 (AMC 12) is a high school mathematics exam designed to promote problem-solving skills. It is intended for students in grade 12 or below who have a strong command of the high school mathematics curriculum. The AMC 12 serves as the initial qualifying test for a series of competitions that ultimately select the United States team for the International Mathematical Olympiad. A high score is a significant accomplishment that students often include in their college applications.

The exam consists of 25 multiple-choice questions to be answered in 75 minutes. The questions are arranged in order of increasing difficulty, with the first ten being relatively straightforward and the last five being particularly demanding. The test covers a wide range of topics, including algebra, geometry, number theory, and combinatorics, often requiring insightful and creative solutions rather than just procedural knowledge.

In practice, the exam's difficulty comes from its steep learning curve and the intense time pressure. Students must solve complex problems quickly and decide which questions to attempt and which to leave blank. The later problems often combine concepts from different areas of mathematics and require a level of ingenuity that goes beyond typical classroom exercises. Success depends on both deep conceptual understanding and strategic test-taking.

How a question gets worked here

From a mathematics paper rather than AMC 12, worked the way every question on this site is. AMC 12 questions are being added.

JEE Advanced 2022

Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let \ $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ , \ $\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$ , $b_{2}, b_{3} \in \mathbb{R}$ \ $\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}$ , $c_{1}, c_{2}, c_{3} \in \mathbb{R}$ be three vectors such that $b_{2}b_{3} > 0$ , $\vec{a} \cdot \vec{b} = 0$ and $\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}.$ Then, which of the following is/are TRUE ?

Vector Algebra Let

\hat{i}, \hat{j}

and

\hat{k}

be the unit vectors along the three positive coordinate axes. Let

\vec{a} = 3\hat{i} + \hat{j} - \hat{k}

\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}

b_{2}, b_{3} \in \mathbb{R}

\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}

c_{1}, c_{2}, c_{3} \in \mathbb{R}

be three vectors such that

b_{2}b_{3} > 0

\vec{a} \cdot \vec{b} = 0

and

\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}.

Then, which of the following is/are TRUE ?

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

$\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ $\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$ $\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}$ $\text{Given: } b_{2}b_{3} > 0,\quad \vec{a} \cdot \vec{b} = 0$ $\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}$ $\text{To find: which of (1) - (4) is/are TRUE}$ $\text{Plan: } (1)\ \text{Matrix} \Rightarrow \vec{c}\times\vec{b} = \vec{a}-\vec{c}$ $(2)\ \text{dot } \vec{b} \Rightarrow \vec{b}\cdot\vec{c}=0\,;\quad (3)\ \text{dot } \vec{c} \Rightarrow \vec{a}\cdot\vec{c}=|\vec{c}|^2$ $(4)\ |\vec{b}|>\sqrt{10}\ \text{via } b_2b_3>0\,;\quad (5)\ |\vec{c}|\le|\vec{a}|=\sqrt{11}$ $\vec{a}\cdot\vec{b}=0 \Rightarrow \vec{a}\perp\vec{b}$ $[\vec{c}]_\times = \begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}$ $[\vec{c}]_\times\,\vec{b} = \vec{c}\times\vec{b}$ $(3-c_1,\ 1-c_2,\ -1-c_3) = \vec{a}-\vec{c}$ $\vec{c}\times\vec{b} = \vec{a}-\vec{c}$ $\vec{b}\cdot(\vec{c}\times\vec{b}) = \vec{b}\cdot(\vec{a}-\vec{c})$ $0 = \vec{b}\cdot\vec{a} - \vec{b}\cdot\vec{c}$ $\vec{b}\cdot\vec{a} = 0 \Rightarrow \vec{b}\cdot\vec{c} = 0$ $\vec{b}\cdot\vec{c}=0 \quad \checkmark\ \text{(option 2 is TRUE)}$ $\vec{c}\cdot(\vec{c}\times\vec{b}) = \vec{c}\cdot(\vec{a}-\vec{c})$ $0 = \vec{c}\cdot\vec{a} - \vec{c}\cdot\vec{c} = \vec{a}\cdot\vec{c} - |\vec{c}|^2$ $\vec{a}\cdot\vec{c} = |\vec{c}|^2$ $\vec{c}=0 \Rightarrow \vec{c}\times\vec{b}=0=\vec{a}-\vec{c}=\vec{a}\ \text{(false)}$ $\vec{a}\cdot\vec{c}=|\vec{c}|^2>0 \Rightarrow \text{option (1) is FALSE}$ $\vec{a}\cdot\vec{b} = 3(1) + 1(b_2) + (-1)(b_3) = 0$ $3 + b_2 - b_3 = 0$ $b_3 = b_2 + 3$ $b_2 b_3>0 \Rightarrow b_2(b_2+3)>0$ $b_2<-3 \ \text{ or } \ b_2>0$ $|\vec{b}|^2 = 1+b_2^2+b_3^2 = 1+b_2^2+(b_2+3)^2$ $|\vec{b}|^2 = 2b_2^2+6b_2+10$ $|\vec{b}|^2-10 = 2b_2(b_2+3) = 2b_2b_3$ $|\vec{b}|^2-10 = 2b_2b_3 > 0 \Rightarrow |\vec{b}|^2 > 10$ $|\vec{b}| > \sqrt{10} \quad \checkmark\ \text{(option 3 is TRUE)}$ $\vec{a}\cdot\vec{c} = |\vec{c}|^2 \le |\vec{a}|\,|\vec{c}|$ $|\vec{c}| \le |\vec{a}| = \sqrt{9+1+1} = \sqrt{11}$ $|\vec{c}| \le \sqrt{11} \quad \checkmark\ \text{(option 4 is TRUE)}$ $\vec{c}\ \text{lies inside the sphere of radius}\ |\vec{a}|=\sqrt{11}$ $\text{Answer: } (2),\ (3),\ (4)$ $\vec{b}\cdot\vec{c}=0\,;\quad |\vec{b}|>\sqrt{10}\,;\quad |\vec{c}|\le\sqrt{11}$

Get a question like this solved →