Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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− | <math>\frac{AB}{EB}=\frac{AD}{CD}=\frac{7}{21/5}=7\cdot\frac{5}{21}=\frac{5}{3} | + | <math>\frac{AB}{EB}=\frac{AD}{CD}=\frac{7}{21/5}=7\cdot\frac{5}{21}=\frac{5}{3} = \boxed{\text{(B)}}</math> |
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==Solution 3== | ==Solution 3== |
Revision as of 18:55, 6 August 2024
Contents
[hide]Problem
In triangle we have , , . Point is on the circumscribed circle of the triangle so that bisects angle . What is the value of ?
Solution 1
Set 's length as . 's length must also be since and intercept arcs of equal length (because ). Using Ballemy's Theorem, . The ratio is
Solution 2
Let . Observe that because they both subtend arc
Furthermore, because is an angle bisector, so by similarity. Then . By the Angle Bisector Theorem, , so . This in turn gives . Plugging this into the similarity proportion gives: .
Alternatively:
Solution 3
We know that bisects , so . Additionally, and subtend the same arc, giving . Similarly, and .
These angle relationships tell us that by AA Similarity, so . By the angle bisector theorem, . Hence,
--vaporwave
P.S We get by the numbers given in the problem. We get by setting up a systems of equations. Using the Angle Bisector Theorem:
We also know that and add up to 9 (using the numbers given in the problem)
We then solve.
--rosebuddy_vxd
See Also
2004 AMC 10B (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 10 Problems and Solutions |
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