August 6 - AMC 12 Topics

August 6 - AMC 12 Topics

Roman Hall

AMC 12 Topics

There are 4 topics tested on the AMC 12 but not the AMC 10: they are logs, trig, complex numbers, and vectors. You won't see a lot of them in middle school, but it's good to know some things about these topics before you go to high school.

That being said, I designed this lesson with the assumption that you already have a basic understanding of some of these topics. I'm not going to explain what they are, but rather how they're applied in competition. If you know nothing about one of the topics, feel free to do your own research or just skip that part. If you are familiar with these topics, then I hope this lesson gives you a sense of some the more advanced topics beyond the AMC 10.

Log Properties

•  $\log_b1=0$  

•  $\log_bb=1$  
  

•  $\log_bm^n=n\log_bm$  

•  $\log_bmn=\log_bm+\log_bn$  

•  $\log_b\frac{m}{n}=\log_bm-\log_bn$  

•  $\log_mn=\frac{\log_bn}{\log_bm}$  (base change)

Open Ended

Log Warmup
Compute  $\log_{\frac{1}{4}}\sqrt[3]{1024}$ .

Solution to Log Warmup

Use log properties to write  $\log_{\frac{1}{4}}\sqrt[3]{1024}=\log_{\frac{1}{4}}2^{\frac{10}{3}}=\frac{10}{3}\log_{\frac{1}{4}}2$ . Exponent properties tell us that  $\left(\frac{1}{4}\right)^{-\frac{1}{2}}=2$ . It follows that  $\log_{\frac{1}{4}}2=-\frac{1}{2}$ , and the requested answer is  $\frac{10}{3}\left(-\frac{1}{2}\right)=-\frac{5}{3}$ .

Open Ended

2015 AMC 12A Problem 14
Compute the value of  $a$  for which  $\frac{1}{\log_2a}+\frac{1}{\log_3a}+\frac{1}{\log_4a}=1$ .

Solution to 2015 AMC 12A Problem 14

Do a base change to get
  $1=\frac{1}{\log_2a}+\frac{1}{\log_3a}+\frac{1}{\log_4a}=\log_a2+\log_a3+\log_a4=\log_a24$ 

It follows that  $\log_a24=1\Longrightarrow a=24$ .

Open Ended

Log Bonus (SEND YOUR SOLUTION TO WIN A PRIZE)
Compute the real number  $x>1$  which satisfies the equation  $\log_2x\log_4x\log_6x=\log_2x\log_4x+\log_2x\log_6x+\log_4x\log_6x$ .

Trig Identities

•  $\sin-x=-\sin x$  

•  $\cos-x=\cos x$  

•  $\sin^2x+\cos^2x=1$  

•  $\sin2x=2\sin x\cos x$  
  

•  $\cos2x=1-2\sin^2x=2\cos^2x-1$  

•  $\sin\left(\frac{\pi}{2}-x\right)=\cos x$  

•  $\cos\left(\frac{\pi}{2}-x\right)=\sin x$  

Trig Identities

•  $\sin\left(\pi-x\right)=\sin x$  

•  $\cos\left(\pi-x\right)=-\cos x$  

•  $\sec x=\frac{1}{\cos x}$  

•  $\tan x=\frac{\sin x}{\cos x}$  

•  $1+\tan^2x=\sec^2x$  
  

•  $\csc x=\frac{1}{\sin x}$  

•  $\cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}$  

Trig Identities

•  $\sin\left(a+b\right)=\sin a\cos b+\cos a\sin b$  

•  $\sin\left(a-b\right)=\sin a\cos b-\cos a\sin b$  

•  $\cos\left(a+b\right)=\cos a\cos b-\sin a\sin b$  

•  $\cos\left(a-b\right)=\cos a\cos b+\sin a\sin b$  

•  $\tan\left(a+b\right)=\frac{\tan a+\tan b}{1-\tan a\tan b}$  

•  $\tan\left(a-b\right)=\frac{\tan a-\tan b}{1+\tan a\tan b}$  

•  $\tan2x=\frac{2\tan x}{1-\tan^2x}$  

Open Ended

1999 AHSME Problem 27
In triangle  $ABC$ ,  $3\sin A+4\cos B=6$  and  $4\sin B+3\cos A=1$ . Compute the degree measure of  $<C$ .

Solution to 1999 AHSME Problem 27

Square the given equations and add, simplifying with the Pythagorean Identity  $\sin^2x+\cos^2x=1$ .

 $9\sin^2A+16\cos^2B+24\sin A\cos B=36$ 
 $9\cos^2A+16\sin^2B+24\cos A\sin B=1$  
 $25+24\left(\sin A\cos B+\cos A\sin B\right)=37$  
 $\frac{1}{2}=\sin A\cos B+\cos A\sin B$  

Replace the right side with the sine addition identity to get  $\frac{1}{2}=\sin\left(A+B\right)=\sin\left(180°-C\right)=\sin C$  . It follows that  $<C=30°,150°$ . The latter is not possible, so the answer is  $30$ .

Law of Sines

In triangle  $ABC$ -- where side  $a$  is opposite angle  $A$ ,  $b$  is opposite  $B$ , and  $c$  is opposite  $C$ -- and where  $R$  is the circumradius

 $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$   

Open Ended

2019 AMC 12A Problem 19
In triangle  $ABC$  with integer side lengths,  $\cos A=\frac{11}{16}$ ,  $\cos B=\frac{7}{8}$ , and  $\cos C=-\frac{1}{4}$ . Compute the least possible perimeter for triangle  $ABC$ .

Solution to 2019 AMC 12A Problem 19

Use the Pythagorean Identity  $\sin^2x+\cos^2x=1$  on each angle to get

 $\sin A=\frac{3\sqrt{15}}{16},\ \sin B=\frac{\sqrt{15}}{8},\ \sin C=\frac{\sqrt{15}}{4}$  

By the Law of Sines,  $a:b:c=\sin A:\sin B:\sin C=\frac{3\sqrt{15}}{16}:\frac{\sqrt{15}}{8}:\frac{\sqrt{15}}{4}=3:2:4$  

The least possible perimeter is  $3+2+4=9$ 

Law of Cosines

For a triangle with sides of lengths  $a,b,c$  opposite angles of measures  $A,B,C$ 

 $c^2=a^2+b^2-2ab\cos C$  
 $b^2=a^2+c^2-2ac\cos B$  
 $a^2=b^2+c^2-2bc\cos A$  

Open Ended

ARML 2009 Team Round Problem 2
In triangle  $ABC$ ,  $AB=4$ ,  $BC=6$ , and  $AC=8$ . Squares  $ABQR$  and  $BCST$  are drawn external to and lie in the same plane as triangle  $ABC$ . Compute  $QT$ .

Solution to ARML 2009 Team Round Problem 2

Set  $m<ABC=x$  and  $m<TBQ=y$ . Then  $x+y=180°$  and so  $\cos x+\cos y=0$ . Applying the Law of Cosines to triangles  $ABC$  and  $TBQ$  gives  $AC^2=AB^2+BC^2-2AB\cdot BC\cos x$  and  $QT^2=BT^2+BQ^2-2BT\cdot BQ\cos y$ , which, after substituting values, become  $8^2=4^2+6^2-48\cos x$  and  $QT^2=4^2+6^2-48\cos y$ . Adding the last two equations yields  $QT^2+8^2=2\left(4^2+6^2\right)$  or  $QT=2\sqrt{10}$ .

De Moive's Theorem

 $\left(\cos\theta+i\sin\theta\right)^n=\cos n\theta+i\sin n\theta$  

Open Ended

Compute  $\left(\frac{1+i\sqrt{3}}{2}\right)^{2021}$  

Solution

Note that  $\frac{1+i\sqrt{3}}{2}=\frac{1}{2}+\frac{\sqrt{3}}{2}i=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$  

Applying De Moive's Theorem gives

  $\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)^{2021}=\cos\frac{2021\pi}{3}+i\sin\frac{2021\pi}{3}=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}$  

Simplifying that gives the final answer of  $\frac{1}{2}-\frac{\sqrt{3}}{2}i$  

Vectors

We'll close with a vectors problem

Open Ended

ARML 2009 Team Round Problem 5
Let  $A_1A_2A_3A_4A_5A_6A_7A_8$  be a regular octagon. Let  $u$  be the vector from  $A_1$  to  $A_2$  and let  $v$  be the vector from  $A_1$  to  $A_8$ . The vector from  $A_1$  to  $A_4$  can be written as  $au+bv$  for a unique ordered pair of real numbers  $\left(a,b\right)$ . Compute  $\left(a,b\right)$ .

Solution to ARML 2009 Team Round Problem 5

Solution is too much to type but the answer is  $\left(2+\sqrt{2},1+\sqrt{2}\right)$