2004 AMC 10B Problems/Problem 24: Difference between revisions
Rachanamadhu (talk | contribs) |
Rachanamadhu (talk | contribs) |
||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
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>\boxed{\frac{5}{3}}\implies(B)</math> | 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>\boxed{\frac{5}{3}}\implies(B)</math> | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 21:26, 19 December 2015
In triangle 
we have 
, 
, 
. Point 
is on the circumscribed circle of the triangle so that 
bisects angle 
. What is the value of 
?
Solution
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
