Difference between revisions of "2013 AIME II Problems/Problem 15"
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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> | 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 | 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 | ||
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+ | ==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}} |
Revision as of 12:26, 2 June 2022
Contents
Problem 15
Let be angles of an acute 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
Solution 6
According to LOC , we can write it into
.
We can simplify to
. Similarly, we can generalize
. After solving, we can get that
Assume the value we are looking for is
, we get
, while
which is
, so
, giving us the answer
.~bluesoul
Video Solution by The Power Of Logic
~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 |
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