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

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Kris17
 
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>
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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>.
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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 + \cos 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>.
  
 
==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}}
 
{{MAA Notice}}

Revision as of 20:30, 7 July 2018

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 $111+4+35+72 = \boxed{222}$.

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

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Problem
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All AIME Problems and Solutions

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