Difference between revisions of "2001 AMC 12 Problems/Problem 24"
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pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); | pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); | ||
pair ortho=rotate(-90)*(D-A); | pair ortho=rotate(-90)*(D-A); | ||
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label("$A$",A,SW); | label("$A$",A,SW); | ||
label("$B$",B,SE); | label("$B$",B,SE); | ||
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Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}</math>. | Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}</math>. | ||
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==Trig Bash== | ==Trig Bash== |
Revision as of 19:05, 27 May 2017
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 |
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.