2014 AIME II Problems/Problem 12

Problem

Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$

Solution 1

Note that $\cos{3C}=-\cos{(3A+3B)}$ . Thus, our expression is of the form $\cos{3A}+\cos{3B}-\cos{(3A+3B)}=1$ . Let $\cos{3A}=x$ and $\cos{3B}=y$ .

Using the fact that $\cos(3A+3B)=\cos 3A\cos 3B-\sin 3A\sin 3B=xy-\sqrt{1-x^2}\sqrt{1-y^2}$ , we get $x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}=1$ , or $\sqrt{1-x^2}\sqrt{1-y^2}=xy-x-y-1=(x-1)(y-1)$ .

Squaring both sides, we get $(1-x^2)(1-y^2) = [(x-1)(y-1)]^2$ . Cancelling factors, $(1+x)(1+y) = (1-x)(1-y)$ .

Expanding, $1+x+y+xy=1-x-y+xy\rightarrow x+y=-x-y\rightarrow 2x=-2y$ .

Simplification leads to $x=-y$ and $x+y=0$ .

Therefore, $\cos(3C)=1$ . So $\angle C$ could be $0^\circ$ or $120^\circ$ . We eliminate $0^\circ$ and use law of cosines to get our answer:

$m=10^2+13^2-2\cdot 10\cdot 13\cos\angle C$ $\rightarrow m=269-260\cos 120^\circ=269-260\left(\frac{1}{2}\right)$ $\rightarrow m=269+130=399$

$\framebox{399}$

Solution 2

As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$

Expanding, we get

$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$

$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$

$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$

$\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1$

$\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$

Note that $\tan{x}=\frac{1}{\tan(90-x)}$ , or $\tan{x}\tan(90-x)=1$

Thus $\frac{3A}{2}+\frac{3B}{2}=90$ , or $A+B=60$ .

Now we know that $C=120$ , so we can just use Law or Cosines to get $\boxed{399}$

2014 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2014 AIME II Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See also