2001 AMC 12 Problems/Problem 24
Contents
Problem
In , . Point is on so that and . Find
Solution 2
Draw a good diagram! Now, let's call , so . Given the rather nice angles of and as you can see, let's do trig. Drop an altitude from to ; call this point . We realize that there is no specific factor of we can call this just yet, so let . Notice that in we get . Using the 60-degree angle in , we obtain . The comparable ratio is that . If we involve our , we get:
. Eliminating and removing radicals from the denominator, we get . From there, one can easily obtain . Now we finally have a desired ratio. Since upon calculation, we know that can be simplified. Indeed, if you know that or even take a minute or two to work out the sine and cosine using , and perhaps the half- or double-angle formulas, you get .
Solution 3
Without loss of generality, we can assume that and . As above, we are able to find that and .
Using Law of Sines on triangle , we find that Since we know that we can compute to equal and to be .
Next, we apply Law of Cosines to triangle to see that Simplifying the right side, we get , so .
Now, we apply Law of Sines to triangle to see that After rearranging and noting that , we get
Dividing the right side by , we see that so is either or . Since is not a choice, we know .
Note that we can also confirm that by computing with Law of Sines.
Solution 4(FAST)
Note that and . Seeing these angles makes us think of 30-60-90 triangles. Let be the foot of the altitude from to . This means and . Let and . This means and since we know that . This means . This gives . Note that . Looking that the answer options we see that . This means the answer is . ~coolmath_2018
Solution 5 (Law of Sines)
, , , let ,
By the Law of Sines, we have
By the Triple-angle Identities,
，so
Suppose , and
, ,
,
Two possible values of are and . However we can rule out because is positive, while is negative.
Therefore ,
Solution 6
For starters, we have Dropping perpendiculars and from and to gives since
Without loss of generality, let and This tells us that Using trigonometric identities, we find that
Thus, which gives Thus,
Now, note that is a triangle, so Thus, we have
Additionally, note that Applying the Pythagorean Theorem to triangle then tells us that By the trigonometric formula for area,
Setting this equal to our other area and solving gives so ~vaporwave
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.