Difference between revisions of "2013 AIME II Problems/Problem 15"

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(Problem 15)
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
\cos^2A + \cos^2B + 2\sinA\sinB\cosC &= 1-\sin^2A + 1-\sin^2B + 2\sinA\sinB\cosC \\
+
\cos^2A + \cos^2B + 2\sin A\sin B\cos C &= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C \\
&= 2-(\sin^2A + \sin^2B - 2\sinA\sinB\cosC)
+
&= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
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Therefore: <cmath>\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.</cmath> Similarly, <cmath>\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.</cmath> Note that the desired value is equivalent to <math>2-\sin^2B</math>, which is <math>2-\sin^2(A+C)</math>. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of <math>\dfrac{111-4\sqrt{35}}{72}</math>. Thus, the answer is <math>111+4+35+72 = \boxed{222}</math>.
 
Therefore: <cmath>\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.</cmath> Similarly, <cmath>\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.</cmath> Note that the desired value is equivalent to <math>2-\sin^2B</math>, which is <math>2-\sin^2(A+C)</math>. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of <math>\dfrac{111-4\sqrt{35}}{72}</math>. Thus, the answer is <math>111+4+35+72 = \boxed{222}</math>.
  
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Note that the problem has a flaw because <math>\cos B < 0</math> which contradicts with the statement that it's an acute triangle. Would be more accurate to state that <math>A</math> and <math>C</math> are smaller than 90. -Mathdummy
  
 
===Solution 2===
 
===Solution 2===
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Thus, as <cmath>\begin{align*} \cos A\cos B-\sin A\sin B=\cos (A+B)=-\cos C ,\end{align*}</cmath>  
 
Thus, as <cmath>\begin{align*} \cos A\cos B-\sin A\sin B=\cos (A+B)=-\cos C ,\end{align*}</cmath>  
we have <cmath>\begin{align*} \dfrac{15}{8}-2\cos^2 C +\cos^2 C=1  
+
we have <cmath>\begin{align*} \dfrac{15}{8}-2\cos^2 C +\cos^2 C=1,
\end{align*}</cmath>, so <math>\cos C</math> is <math>\sqrt{\dfrac{7}{8}}</math> and therefore <math> \sin C</math> is <math>\sqrt{\dfrac{1}{8}}</math>.  
+
\end{align*}</cmath> so <math>\cos C</math> is <math>\sqrt{\dfrac{7}{8}}</math> and therefore <math> \sin C</math> is <math>\sqrt{\dfrac{1}{8}}</math>.  
  
 
Similarily, we have <math>\sin A =\dfrac{2}{3}</math> and <math>\cos A=\sqrt{\dfrac{14}{9}-1}=\sqrt{\dfrac{5}{9}}</math> and the rest of the solution proceeds as above.
 
Similarily, we have <math>\sin A =\dfrac{2}{3}</math> and <math>\cos A=\sqrt{\dfrac{14}{9}-1}=\sqrt{\dfrac{5}{9}}</math> and the rest of the solution proceeds as above.
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===Solution 3===
 +
 +
Let
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<cmath> \begin{align*}
 +
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{    ------ (1)}\\
 +
\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \text{    ------ (2)}\\
 +
\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= x \text{                        ------ (3)}\\
 +
\end{align*} </cmath>
 +
 +
Adding (1) and (3) we get:
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<cmath> 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C +  \sin C \cos B)  = \frac{15}{8} + x</cmath>  or
 +
<cmath> 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C)  = \frac{15}{8} + x</cmath>    or
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<cmath> 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x</cmath>    or
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<cmath> \cos^2 B + \cos^2 C =  x - \frac{1}{8}    \text{    ------ (4)}</cmath>
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 +
Similarly adding (2) and (3) we get:
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<cmath> \cos^2 A + \cos^2 B =  x - \frac{4}{9}    \text{    ------ (5)} </cmath>
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Similarly adding (1) and (2) we get:
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<cmath> \cos^2 A + \cos^2 C =  \frac{14}{9}  - \frac{1}{8}    \text{    ------ (6)} </cmath>
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 +
And (4) - (5) gives:
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<cmath> \cos^2 C - \cos^2 A =  \frac{4}{9}  - \frac{1}{8}    \text{    ------ (7)} </cmath>
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 +
Now (6) - (7) gives:
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<math> \cos^2 A  =  \frac{5}{9} </math> or
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<math>\cos A = \sqrt{\dfrac{5}{9}}</math> and <math>\sin A = \frac{2}{3} </math>
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so <math>\cos C</math> is <math>\sqrt{\dfrac{7}{8}}</math> and therefore <math> \sin C</math> is <math>\sqrt{\dfrac{1}{8}}</math>
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Now <math>\sin B = \sin(A+C)</math> can be computed first and then <math>\cos^2 B</math> is easily found.
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Thus <math>\cos^2 B</math> and <math>\cos^2 C</math> can be plugged into (4) above to give x =  <math>\dfrac{111-4\sqrt{35}}{72}</math>.
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Hence the answer is = <math>111+4+35+72 = \boxed{222}</math>.
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Kris17
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=== Solution 4 ===
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Let's take the first equation <math>\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C = \frac{15}{8}</math>. Substituting <math>180 - A - B</math> for C, given A, B, and C form a triangle, and that <math>\cos C = \cos(A + B)</math>, gives us:
 +
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<math>\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos (A+B) = \frac{15}{8}</math>
 +
 +
Expanding out gives us <math>\cos^2 A + \cos^2 B + 2 \sin^2 A \sin^2 B - 2 \sin A \sin B \cos A \cos B = \frac{15}{8}</math>.
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 +
Using the double angle formula <math>\cos^2 k = \frac{\cos (2k) + 1}{2}</math>, we can substitute for each of the squares <math>\cos^2 A</math> and <math>\cos^2 B</math>. Next we can use the Pythagorean identity on the <math>\sin^2 A</math> and <math>\sin^2 B</math> terms. Lastly we can use the sine double angle to simplify.
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<math>\cos^2 A + \cos^2 B + 2(1 - \cos^2 A)(1 - \cos^2 B) - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}</math>.
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Expanding and canceling yields, and again using double angle substitution,
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<math>1 + 2 \cdot \frac{\cos (2A) + 1}{2} \cdot \frac{\cos (2B) + 1}{2} - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}</math>.
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Further simplifying yields:
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<math>\frac{3}{2} + \frac{\cos 2A \cos 2B - \sin 2A \sin 2B}{2} = \frac{15}{8}</math>.
 +
 +
Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation <math>2</math> yields:
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<math>\cos (2A + 2B) = \frac{3}{4}</math> and <math>\cos (2B + 2C) = \frac{1}{9}</math>.
 +
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Substituting the identity <math>\cos (2A + 2B) = \cos(2C)</math>, we get:
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<math>\cos (2C) = \frac{3}{4}</math> and <math>\cos (2A) = \frac{1}{9}</math>.
 +
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Since the third expression simplifies to the expression <math>\frac{3}{2} + \frac{\cos (2A + 2C)}{2}</math>, taking inverse cosine and using the angles in angle addition formula yields the answer, <math>\frac{111 - 4\sqrt{35}}{72}</math>, giving us the answer <math>\boxed{222}</math>.
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===Solution 5===
 +
We will use the sum to product formula to simply these equations. Recall <cmath>2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}} = \cos{\alpha}+\cos{\beta}.</cmath> Using this, let's rewrite the first equation: <cmath>\cos^2(A) + \cos^2(B) + 2 \sin(A) \sin(B) \cos(C) = \frac{15}{8}</cmath> <cmath>\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))\cos(C).</cmath> Now, note that <math>\cos(C)=-\cos(A+B)</math>. <cmath>\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))(-\cos(A+B))</cmath> <cmath>\cos^2(A) + \cos^2(B) - \cos(A+B)\cos(A-B)+cos^2(A+B)=\frac{15}{8}.</cmath> We apply the sum to product formula again. <cmath>\cos^2(A) + \cos^2(B) - \frac{\cos(2A)+\cos(2B)}{2}+cos^2(A+B)=\frac{15}{8}.</cmath> Now, recall that <math>\cos(2\alpha)=2\cos^2(\alpha)-1</math>. We apply this and simplify our expression to get: <cmath>\cos^2(A+B)=\frac{7}{8}</cmath> <cmath>\cos^2(C)=\frac{7}{8}.</cmath> Analogously, <cmath>\cos^2(A)=\frac{5}{9}.</cmath> <cmath>\cos^2(A+C)=\frac{p-q\sqrt{r}}{s}-1.</cmath> We can find this value easily by angle sum formula. After a few calculations, we get <math>\frac{111 - 4\sqrt{35}}{72}</math>, giving us the answer <math>\boxed{222}</math>.
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 +
 +
~superagh
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=14|after=Last Problem}}
 
{{AIME box|year=2013|n=II|num-b=14|after=Last Problem}}
 +
{{MAA Notice}}

Revision as of 18:45, 29 November 2020

Problem 15

Let $A,B,C$ be angles of an acute triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$.

Solutions

Solution 1

Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \sin{A}$.

By the Law of Sines, we must have $CA = \sin{B}$ and $AB = \sin{C}$.

Now let us analyze the given:

\begin{align*} \cos^2A + \cos^2B + 2\sin A\sin B\cos C &= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C \\ &= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C) \end{align*}

Now we can use the Law of Cosines to simplify this:

\[= 2-\sin^2C\]


Therefore: \[\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.\] Similarly, \[\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.\] Note that the desired value is equivalent to $2-\sin^2B$, which is $2-\sin^2(A+C)$. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\dfrac{111-4\sqrt{35}}{72}$. Thus, the answer is $111+4+35+72 = \boxed{222}$.

Note that the problem has a flaw because $\cos B < 0$ which contradicts with the statement that it's an acute triangle. Would be more accurate to state that $A$ and $C$ are smaller than 90. -Mathdummy

Solution 2

Let us use the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1$ .

Add \begin{align*} \cos^2 C+2(\cos A\cos B-\sin A \sin B)\cos C \end{align*} to both sides of the first given equation.



Thus, as \begin{align*} \cos A\cos B-\sin A\sin B=\cos (A+B)=-\cos C ,\end{align*} we have \begin{align*} \dfrac{15}{8}-2\cos^2 C +\cos^2 C=1,  \end{align*} so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$.

Similarily, we have $\sin A =\dfrac{2}{3}$ and $\cos A=\sqrt{\dfrac{14}{9}-1}=\sqrt{\dfrac{5}{9}}$ and the rest of the solution proceeds as above.

Solution 3

Let \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{     ------ (1)}\\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \text{     ------ (2)}\\ \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= x \text{                        ------ (3)}\\ \end{align*}

Adding (1) and (3) we get: \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C +  \sin C \cos B)  = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C)  = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x\] or \[\cos^2 B + \cos^2 C =  x - \frac{1}{8}     \text{     ------ (4)}\]

Similarly adding (2) and (3) we get: \[\cos^2 A + \cos^2 B =  x - \frac{4}{9}     \text{     ------ (5)}\] Similarly adding (1) and (2) we get: \[\cos^2 A + \cos^2 C =  \frac{14}{9}  - \frac{1}{8}     \text{     ------ (6)}\]

And (4) - (5) gives: \[\cos^2 C - \cos^2 A =  \frac{4}{9}  - \frac{1}{8}    \text{     ------ (7)}\]

Now (6) - (7) gives: $\cos^2 A  =  \frac{5}{9}$ or $\cos A = \sqrt{\dfrac{5}{9}}$ and $\sin A = \frac{2}{3}$ so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$

Now $\sin B = \sin(A+C)$ can be computed first and then $\cos^2 B$ is easily found.

Thus $\cos^2 B$ and $\cos^2 C$ can be plugged into (4) above to give x = $\dfrac{111-4\sqrt{35}}{72}$.

Hence the answer is = $111+4+35+72 = \boxed{222}$.

Kris17

Solution 4

Let's take the first equation $\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C = \frac{15}{8}$. Substituting $180 - A - B$ for C, given A, B, and C form a triangle, and that $\cos C = \cos(A + B)$, gives us:

$\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos (A+B) = \frac{15}{8}$

Expanding out gives us $\cos^2 A + \cos^2 B + 2 \sin^2 A \sin^2 B - 2 \sin A \sin B \cos A \cos B = \frac{15}{8}$.

Using the double angle formula $\cos^2 k = \frac{\cos (2k) + 1}{2}$, we can substitute for each of the squares $\cos^2 A$ and $\cos^2 B$. Next we can use the Pythagorean identity on the $\sin^2 A$ and $\sin^2 B$ terms. Lastly we can use the sine double angle to simplify.

$\cos^2 A + \cos^2 B + 2(1 - \cos^2 A)(1 - \cos^2 B) - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$.

Expanding and canceling yields, and again using double angle substitution,

$1 + 2 \cdot \frac{\cos (2A) + 1}{2} \cdot \frac{\cos (2B) + 1}{2} - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$.

Further simplifying yields:

$\frac{3}{2} + \frac{\cos 2A \cos 2B - \sin 2A \sin 2B}{2} = \frac{15}{8}$.

Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation $2$ yields:

$\cos (2A + 2B) = \frac{3}{4}$ and $\cos (2B + 2C) = \frac{1}{9}$.

Substituting the identity $\cos (2A + 2B) = \cos(2C)$, we get:

$\cos (2C) = \frac{3}{4}$ and $\cos (2A) = \frac{1}{9}$.

Since the third expression simplifies to the expression $\frac{3}{2} + \frac{\cos (2A + 2C)}{2}$, taking inverse cosine and using the angles in angle addition formula yields the answer, $\frac{111 - 4\sqrt{35}}{72}$, giving us the answer $\boxed{222}$.

Solution 5

We will use the sum to product formula to simply these equations. Recall \[2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}} = \cos{\alpha}+\cos{\beta}.\] Using this, let's rewrite the first equation: \[\cos^2(A) + \cos^2(B) + 2 \sin(A) \sin(B) \cos(C) = \frac{15}{8}\] \[\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))\cos(C).\] Now, note that $\cos(C)=-\cos(A+B)$. \[\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))(-\cos(A+B))\] \[\cos^2(A) + \cos^2(B) - \cos(A+B)\cos(A-B)+cos^2(A+B)=\frac{15}{8}.\] We apply the sum to product formula again. \[\cos^2(A) + \cos^2(B) - \frac{\cos(2A)+\cos(2B)}{2}+cos^2(A+B)=\frac{15}{8}.\] Now, recall that $\cos(2\alpha)=2\cos^2(\alpha)-1$. We apply this and simplify our expression to get: \[\cos^2(A+B)=\frac{7}{8}\] \[\cos^2(C)=\frac{7}{8}.\] Analogously, \[\cos^2(A)=\frac{5}{9}.\] \[\cos^2(A+C)=\frac{p-q\sqrt{r}}{s}-1.\] We can find this value easily by angle sum formula. After a few calculations, we get $\frac{111 - 4\sqrt{35}}{72}$, giving us the answer $\boxed{222}$.


~superagh

See Also

2013 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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