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>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> | ||
Revision as of 15:59, 7 January 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
![[asy] import graph; import geometry; import markers; unitsize(0.5 cm); pair A, B, C, D, E, I; A = (11/3,8*sqrt(5)/3); B = (0,0); C = (9,0); I = incenter(A,B,C); D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C)); E = extension(A,D,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(D--A); draw(D--B); draw(D--C); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, NE); markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true))); markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true))); markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); [/asy]](http://latex.artofproblemsolving.com/7/3/0/730d9e00b576f80d8652e97bc3bc2ad3f358297b.png)
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. 
