Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>\overline{CD}</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length (because <math>\angle BAD=\angle DAC</math>). Using [[Ballemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <math>\frac{5}{3}\implies\boxed{\text{(B)}}</math> | Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>\overline{CD}</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length (because <math>\angle BAD=\angle DAC</math>). Using [[Ballemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <math>\frac{5}{3}\implies\boxed{\text{(B)}}</math> | ||
− | ==Solution 2== | + | ==Solution 2 - Similarity Proportion== |
<asy> | <asy> | ||
import graph; | import graph; |
Revision as of 19:56, 6 August 2024
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 - Similarity Proportion
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,
--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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