Difference between revisions of "1989 AIME Problems/Problem 10"
The 76923th (talk | contribs) (→Solution 1) |
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&= \frac{a^2 + b^2 - c^2}{2c^2} \\ | &= \frac{a^2 + b^2 - c^2}{2c^2} \\ | ||
&= \frac{1989 c^2 - c^2}{2c^2} \\ | &= \frac{1989 c^2 - c^2}{2c^2} \\ | ||
− | &= \frac{1988}{2} = | + | &= \frac{1988}{2} = \boxed{994}. \end{align*} </cmath> |
=== Solution 2 === | === Solution 2 === |
Revision as of 14:15, 18 December 2022
Problem
Let ,
,
be the three sides of a triangle, and let
,
,
, be the angles opposite them. If
, find
![$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$](http://latex.artofproblemsolving.com/8/5/8/858c49b569dc76c503dbdbfa8d0b78fd2c04fe1d.png)
Contents
Solution
Solution 1
We draw the altitude to
, to get two right triangles.
![[asy] size(170); pair A = (0,0), B = (3, 0), C = (1, 4); pair P = .5*(C + reflect(A,B)*C); draw(A--B--C--cycle); draw(C--P, dotted); draw(rightanglemark(C,P, B, 4)); label("$A$", A, S); label("$B$", B, S); label("$C$", C, N); label("$a$", (B+C)/2, NE); label("$b$", (A+C)/2, NW); label("$c$", (A+B)/2, S); label("$h$", (C+P)/2, E);[/asy]](http://latex.artofproblemsolving.com/c/a/7/ca7f7ca9366088b0274daace75e09ea38876ae3f.png)
Then , from the definition of the cotangent.
Let be the area of
Then
, so
.
By identical logic, we can find similar expressions for the sums of the other two cotangents:
Adding the last two equations, subtracting the first, and dividing by 2, we get
Therefore
Solution 2
By the law of cosines,
So, by the extended law of sines,
Identical logic works for the other two angles in the triangle. So, the cotangent of any angle in the triangle is directly proportional to the sum of the squares of the two adjacent sides, minus the square of the opposite side. Therefore
We can then finish as in solution 1.
Solution 3
We start as in solution 1, though we'll write instead of
for the area. Now we evaluate the numerator:
From the Law of Cosines and the sine area formula,
Then .
Solution 4
By the Law of Cosines,
Now
Solution 5
Use Law of cosines to give us or therefore
. Next, we are going to put all the sin's in term of
. We get
. Therefore, we get
.
Next, use Law of Cosines to give us . Therefore,
. Also,
. Hence,
.
Lastly, . Therefore, we get
.
Now, . After using
, we get
.
Solution 6
Let be
WLOG, assume that and
are legs of right triangle
with
and
By Pythagorean theorem, we have , and the given
. Solving the equations gives us
and
. We see that
, and
.
We see that our derived equation equals to as
approaches infinity.
Evaluating
, we get
.
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.