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Difference between revisions of "2014 AIME II Problems/Problem 12"

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Suppose that the angles of <math>\triangle ABC</math> satisfy <math>\cos(3A)+\cos(3B)+\cos(3C)=1.</math> Two sides of the triangle have lengths 10 and 13. There is a positive integer <math>m</math> so that the maximum possible length for the remaining side of <math>\triangle ABC</math> is <math>\sqrt{m}.</math> Find <math>m.</math>
 
Suppose that the angles of <math>\triangle ABC</math> satisfy <math>\cos(3A)+\cos(3B)+\cos(3C)=1.</math> Two sides of the triangle have lengths 10 and 13. There is a positive integer <math>m</math> so that the maximum possible length for the remaining side of <math>\triangle ABC</math> is <math>\sqrt{m}.</math> Find <math>m.</math>
  
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== Solution ==
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== Solution 1 ==
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Note that <math>\cos{3C}=-\cos{(3A+3B)}</math>. Thus, our expression is of the form <math>\cos{3A}+\cos{3B}-\cos{(3A+3B)}=1</math>. Let <math>\cos{3A}=x</math> and <math>\cos{3B}=y</math>. Expanding, we get <math>x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}=1</math>, or <math>-\sqrt{1-x^2}\sqrt{1-y^2}=(x-1)(y-1)</math>.  (why the minus sign at the left side ???) Since the right side of our equation must be non-positive, <math>x</math> or <math>y</math> is <math>\leq{1}</math>, and the other variable must take on a value of 1. WLOG, we can set <math>x=\cos{3A}=1</math>, or <math>A=0,120</math>; however, only 120 is valid, as we want a nondegenerate triangle. Since <math>B+C=60</math>, the maximum angle in the triangle is 120 degrees. Using Law of Cosines, we get <math>S^{2}=169+100+2(10)(13)(\frac{1}{2})=\framebox{399}</math>, and we're done.
+
Note that <math>\cos{3C}=-\cos{(3A+3B)}</math>. Thus, our expression is of the form <math>\cos{3A}+\cos{3B}-\cos{(3A+3B)}=1</math>. Let <math>\cos{3A}=x</math> and <math>\cos{3B}=y</math>.
  
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I believe there's a mistake in the solution. There shouldn't be a negative at the left hand side of the equation. We should get: 
+
Using the fact that <math>\cos(3A+3B)=\cos 3A\cos 3B-\sin 3A\sin 3B=xy-\sqrt{1-x^2}\sqrt{1-y^2}</math>, we get <math>x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}=1</math>, or <math>\sqrt{1-x^2}\sqrt{1-y^2}=xy-x-y+1=(x-1)(y-1)</math>.
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<math>\sqrt{1-x^2}\sqrt{1-y^2}=(x-1)(y-1)</math>
 
  
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squaring both sides we get:
+
Squaring both sides, we get <math>(1-x^2)(1-y^2) = [(x-1)(y-1)]^2</math>. Cancelling factors, <math>(1+x)(1+y) = (1-x)(1-y)</math>.
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<math>(1-x^2)(1-y^2) = [(x-1)(y-1)]^2</math>
 
  
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factoring:
+
*Notice here that we cancelled out one factor of (x-1) and (y-1), which implies that (x-1) and (y-1) were not 0. If indeed they were 0 though, we would have
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<math>(1+x)(1+y) = (1-x)(1-y).</math>
+
 
−
Then:
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<math>cos(3A)-1=0, cos(3A)=1</math>  
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<math>xy+x+y+1 = 1-x-y+xy.</math>
+
 
−
Then:
+
For this we could say that A must be 120 degrees for this to work. This is one case. The B case follows in the same way, where B must be equal to 120 degrees. This doesn't change the overall solution though, as then the other angles are irrelevant (this is the largest angle, implying that this will have the longest side and so we would want to have the 120 degreee angle opposite of the unknown side).
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<math>2x+2y=0.</math>
+
 
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Then:
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Expanding, <math>1+x+y+xy=1-x-y+xy\rightarrow x+y=-x-y</math>.
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<math>x=-y</math>
+
 
−
Therefore, cos(3C) must equal 1. So C could be 0 or 120 degrees. We eliminate 0 and use law of cosines to get our answer: <math>\framebox{399}</math>
+
Simplification leads to <math>x+y=0</math>.
 +
 
 +
Therefore, <math>\cos(3C)=1</math>. So <math>\angle C</math> could be <math>0^\circ</math> or <math>120^\circ</math>. We eliminate <math>0^\circ</math> and use law of cosines to get our answer:  
 +
 
 +
<cmath>m=10^2+13^2-2\cdot 10\cdot 13\cos\angle C</cmath>
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<cmath>\rightarrow m=269-260\cos 120^\circ=269-260\left(\text{-}\frac{1}{2}\right)</cmath>
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<cmath>\rightarrow m=269+130=399</cmath>
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 +
<math>\framebox{399}</math>
 +
 
 +
NOTE: This solution forgot the case of dividing by 0
 +
 
 +
NOTE FROM DIFFERENT PERSON: Pretty sure they addressed that in the notice
 +
 
 +
==Solution 2==
 +
As above, we can see that <math>\cos3A+\cos3B-\cos(3A+3B)=1</math>
 +
 
 +
Expanding, we get
 +
 
 +
<math>\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1</math>
 +
 
 +
<math>\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B</math>
 +
 
 +
<math>(\cos3A-1)(\cos3B-1)=\sin3A\sin3B</math>
 +
 
 +
<math>\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1</math>
 +
 
 +
<math>\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1</math>
 +
 
 +
Note that <math>\tan{x}=\frac{1}{\tan(90-x)}</math>, or <math>\tan{x}\tan(90-x)=1</math>
 +
 
 +
Thus <math>\frac{3A}{2}+\frac{3B}{2}=90</math>, or <math>A+B=60</math>.
 +
 
 +
Now we know that <math>C=120</math>, so we can just use the Law of Cosines to get <math>\boxed{399}</math>
 +
 
 +
Note: This solution also forgets that <math>A</math> or <math>B</math> might be 120 when dividing by <math>\sin 3A</math> and <math>\sin 3B</math>
 +
==Solution 3==
 +
<cmath>\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)</cmath>
 +
<cmath>2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)</cmath>
 +
If <math>\cos\frac{3}{2}(A+B) = 0</math>, then <math>\frac{3}{2}(A+B)=90</math>, <math>A+B=60</math>, so <math>C=120</math>; otherwise,
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<cmath>2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)</cmath>
 +
<cmath>\sin\frac{3}{2}A\sin\frac{3}{2}B=0</cmath>
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so either <math>\sin\frac{3}{2}A=0</math> or <math>\sin\frac{3}{2}B=0</math>, i.e., either <math>A=120</math> or <math>B=120</math>. In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. <math>\boxed{399}</math>
 +
 
 +
-Mathdummy
  
 
== See also ==
 
== See also ==

Revision as of 00:36, 20 July 2020

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, $(1+x)(1+y) = (1-x)(1-y)$.

  • Notice here that we cancelled out one factor of (x-1) and (y-1), which implies that (x-1) and (y-1) were not 0. If indeed they were 0 though, we would have

$cos(3A)-1=0, cos(3A)=1$

For this we could say that A must be 120 degrees for this to work. This is one case. The B case follows in the same way, where B must be equal to 120 degrees. This doesn't change the overall solution though, as then the other angles are irrelevant (this is the largest angle, implying that this will have the longest side and so we would want to have the 120 degreee angle opposite of the unknown side).

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

Simplification leads to $x+y=0$.

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:

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

$\framebox{399}$

NOTE: This solution forgot the case of dividing by 0

NOTE FROM DIFFERENT PERSON: Pretty sure they addressed that in the notice

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 the Law of Cosines to get $\boxed{399}$

Note: This solution also forgets that $A$ or $B$ might be 120 when dividing by $\sin 3A$ and $\sin 3B$

Solution 3

\[\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)\] \[2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)\] If $\cos\frac{3}{2}(A+B) = 0$, then $\frac{3}{2}(A+B)=90$, $A+B=60$, so $C=120$; otherwise, \[2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)\] \[\sin\frac{3}{2}A\sin\frac{3}{2}B=0\] so either $\sin\frac{3}{2}A=0$ or $\sin\frac{3}{2}B=0$, i.e., either $A=120$ or $B=120$. In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. $\boxed{399}$

-Mathdummy

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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