Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solutions)
m
 
(8 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==Problem 15==
 
==Problem 15==
−
Let <math>A,B,C</math> be angles of an obtuse triangle with
+
Let <math>A,B,C</math> be angles of a triangle with
 
<cmath> \begin{align*}
 
<cmath> \begin{align*}
 
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
 
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
Line 112: Line 112:
 
===Solution 5===
 
===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>.
 
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>.
 +
~superagh
  
 +
===Solution 6===
 +
According to LOC <math>a^2+b^2-2ab\cos{\angle{c}}=c^2</math>, we can write it into <math>\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}=\sin^2{\angle{C}}</math>. <math>\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}+cos^2A+cos^2B+2sinAsinBcosC=\frac{15}{8}+sin^2C</math> We can simplify to <math>2=sin^2C+\frac{15}{8}</math>. Similarly, we can generalize <math>2=sin^2A+\frac{14}{9}</math>. After solving, we can get that <math>sinA=\frac{2}{3}; cosA=\frac{\sqrt{5}}{3}; sinC=\frac{\sqrt{2}}{4}; cosC=\frac{\sqrt{14}}{4}</math>
 +
Assume the value we are looking for is <math>x</math>, we get <math>sin^2B+x=2</math>, while <math>sinB=sin(180^{\circ}-A-C)=sin(A+C)</math> which is <math>\frac{2\sqrt{14}+\sqrt{10}}{12}</math>, so <math>x=\frac{111 - 4\sqrt{35}}{72}</math>, giving us the answer <math>\boxed{222}</math>.~bluesoul
  
−
~superagh
+
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/_wB0WyhNoQE?si=wjjJtQ_rxi2dsDbo
 +
 
 +
~MathProblemSolvingSkills.com
 +
 
 +
 
 +
==Video Solution by The Power Of Logic==
 +
 
 +
https://youtu.be/9TVhH2bFjT0
 +
 
 +
~Hayabusa1
  
 
==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}}

Latest revision as of 16:13, 28 June 2024

Problem 15

Let $A,B,C$ be angles of a 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

Solution 6

According to LOC $a^2+b^2-2ab\cos{\angle{c}}=c^2$, we can write it into $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}=\sin^2{\angle{C}}$. $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}+cos^2A+cos^2B+2sinAsinBcosC=\frac{15}{8}+sin^2C$ We can simplify to $2=sin^2C+\frac{15}{8}$. Similarly, we can generalize $2=sin^2A+\frac{14}{9}$. After solving, we can get that $sinA=\frac{2}{3}; cosA=\frac{\sqrt{5}}{3}; sinC=\frac{\sqrt{2}}{4}; cosC=\frac{\sqrt{14}}{4}$ Assume the value we are looking for is $x$, we get $sin^2B+x=2$, while $sinB=sin(180^{\circ}-A-C)=sin(A+C)$ which is $\frac{2\sqrt{14}+\sqrt{10}}{12}$, so $x=\frac{111 - 4\sqrt{35}}{72}$, giving us the answer $\boxed{222}$.~bluesoul


Video Solution

https://youtu.be/_wB0WyhNoQE?si=wjjJtQ_rxi2dsDbo

~MathProblemSolvingSkills.com


Video Solution by The Power Of Logic

https://youtu.be/9TVhH2bFjT0

~Hayabusa1

See Also

2013 AIME II (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_15&oldid=221230"