2001 AMC 12 Problems/Problem 24
Problem
In , . Point is on so that and . Find
Solution 1
We start with the observation that , and .
We can draw the height from onto . In the triangle , we have . Hence .
By the definition of , we also have , therefore . This means that the triangle is isosceles, and as , we must have .
Then we compute , thus and the triangle is isosceles as well. Hence .
Now we can note that , hence also the triangle is isosceles and we have .
Combining the previous two observations we get that , and as , this means that .
Finally, we get .
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.
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.