Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC \cong \angle ADC</math> because they both subtend arc <math>\overarc{AC}.</math> | Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC \cong \angle ADC</math> because they both subtend arc <math>\overarc{AC}.</math> | ||
− | Furthermore, <math>\angle BAE \cong \angle EAC</math> because <math>\overline{AE}</math> is an angle bisector, so <math>\triangle ABE \sim \triangle ADC</math> by <math>\text{AA}</math> similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>. By the [[Angle Bisector Theorem]], <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math>, so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math>. This in turn gives <math>BE = \frac{21}{5}</math>. Plugging this into the similarity proportion gives: <math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>. | + | Furthermore, <math>\angle BAE \cong \angle EAC</math> because <math>\overline{AE}</math> is an angle bisector, so <math>\triangle ABE \sim \triangle ADC</math> by <math>\text{AA}</math> similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>. By the [[Angle Bisector Theorem]], <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math>, so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math>. This in turn gives <math>BE = \frac{21}{5}</math>. Plugging this into the similarity proportion gives: |
+ | <math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>. | ||
+ | alternatively: | ||
+ | <math>\frac{AB}{EB}=\frac{AD}{CD}=\frac{7}{21/5}=7\cdot\frac{5}{21}=\frac{5}{3}</math> | ||
==Solution 3== | ==Solution 3== |
Revision as of 18:53, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.