Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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== Solution 1== | == Solution 1== | ||
− | Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>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 [[Ptolemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <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 [[Ptolemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <math>\frac{5}{3}\implies\boxed{\text{(B)}}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 01:02, 11 November 2017
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 Ptolemy's Theorem,
. The ratio is
Solution 2
Let
. Observe that
because they subtend the same arc. Furthermore,
, so
is similar to
by AAA similarity. Then
. By angle bisector theorem,
so
which gives
. Plugging this into the similarity proportion gives:
.
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.