Difference between revisions of "2004 AMC 10B Problems/Problem 24"
(→Solution 1) |
(→Solution 3) |
||
Line 51: | Line 51: | ||
We know that <math>\overline{AD}</math> bisects <math>\angle BAC</math>, so <math>\angle BAD = \angle CAD</math>. Additionally, <math>\angle BAD</math> and <math>\angle BCD</math> subtend the same arc, giving <math>\angle BAD = \angle BCD</math>. Similarly, <math>\angle CAD = \angle CBD</math> and <math>\angle ABC = \angle ADC</math>. | We know that <math>\overline{AD}</math> bisects <math>\angle BAC</math>, so <math>\angle BAD = \angle CAD</math>. Additionally, <math>\angle BAD</math> and <math>\angle BCD</math> subtend the same arc, giving <math>\angle BAD = \angle BCD</math>. Similarly, <math>\angle CAD = \angle CBD</math> and <math>\angle ABC = \angle ADC</math>. | ||
− | These angle relationships tell us that <math>\triangle ABE\sim \triangle ADC</math> by AA Similarity, so <math>AD/CD = AB/BE</math>. By the angle bisector theorem, <math>AB/BE = AC/ | + | These angle relationships tell us that <math>\triangle ABE\sim \triangle ADC</math> by AA Similarity, so <math>AD/CD = AB/BE</math>. By the angle bisector theorem, <math>AB/BE = AC/CD</math>. Hence, |
− | <cmath>\frac{AB}{BE} = \frac{AC}{ | + | <cmath>\frac{AB}{BE} = \frac{AC}{CD} = \frac{7}{21/5} = 7\cdot\frac{5}{21} = \frac{35}{21} = \frac{5}{3}}.</cmath> |
--vaporwave | --vaporwave | ||
+ | |||
+ | P.S | ||
+ | We get AB by the numbers given in the problem. | ||
+ | We get EB by setting up a systems of equations. | ||
+ | Using the Angle Bisector Theorem: | ||
+ | 7/8=EB/EC | ||
+ | |||
+ | We also know that EB and EC add up to 9 (using the numbers given in the problem) | ||
+ | EB+EC=9. | ||
+ | |||
+ | We then solve. | ||
+ | |||
+ | --rosebuddy_vxd | ||
== See Also == | == See Also == |
Revision as of 18:48, 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: .
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,
\[\frac{AB}{BE} = \frac{AC}{CD} = \frac{7}{21/5} = 7\cdot\frac{5}{21} = \frac{35}{21} = \frac{5}{3}}.\] (Error compiling LaTeX. Unknown error_msg)
--vaporwave
P.S We get AB by the numbers given in the problem. We get EB by setting up a systems of equations. Using the Angle Bisector Theorem: 7/8=EB/EC
We also know that EB and EC add up to 9 (using the numbers given in the problem) EB+EC=9.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.