Difference between revisions of "2013 AIME II Problems/Problem 15"
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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>. | 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}} | {{MAA Notice}} |
Revision as of 21:45, 5 October 2020
Contents
Problem 15
Let be angles of an obtuse triangle with There are positive integers , , , and for which where and are relatively prime and is not divisible by the square of any prime. Find .
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 .
By the Law of Sines, we must have and .
Now let us analyze the given:
Now we can use the Law of Cosines to simplify this:
Therefore: Similarly, Note that the desired value is equivalent to , which is . All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of . Thus, the answer is .
Note that the problem has a flaw because which contradicts with the statement that it's an acute triangle. Would be more accurate to state that and are smaller than 90. -Mathdummy
Solution 2
Let us use the identity .
Add to both sides of the first given equation.
Thus, as
we have so is and therefore is .
Similarily, we have and and the rest of the solution proceeds as above.
Solution 3
Let
Adding (1) and (3) we get: or or or
Similarly adding (2) and (3) we get: Similarly adding (1) and (2) we get:
And (4) - (5) gives:
Now (6) - (7) gives: or and so is and therefore is
Now can be computed first and then is easily found.
Thus and can be plugged into (4) above to give x = .
Hence the answer is = .
Kris17
Solution 4
Let's take the first equation . Substituting for C, given A, B, and C form a triangle, and that , gives us:
Expanding out gives us .
Using the double angle formula , we can substitute for each of the squares and . Next we can use the Pythagorean identity on the and terms. Lastly we can use the sine double angle to simplify.
.
Expanding and canceling yields, and again using double angle substitution,
.
Further simplifying yields:
.
Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation yields:
and .
Substituting the identity , we get:
and .
Since the third expression simplifies to the expression , taking inverse cosine and using the angles in angle addition formula yields the answer, , giving us the answer .
Solution 5
We will use the sum to product formula to simply these equations. Recall Using this, let's rewrite the first equation: Now, note that . We apply the sum to product formula again. Now, recall that . We apply this and simplify our expression to get: Analogously, We can find this value easily by angle sum formula. After a few calculations, we get , giving us the answer .
~superagh
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.