2001 AMC 12 Problems/Problem 24
Contents
Problem
In , . Point is on so that and . Find .
Solution
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 .
Trig Bash
WLOG, 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 , , and , we can compute to equal and to be .
Next, we apply Law of Cosines to triangle to see that . Simplifying the RHS, we get , so .
Now, we apply Law of Sines to triangle to see that . After rearranging and noting that , we get .
Dividing the RHS through 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 |
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