Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
+ | <asy> | ||
+ | import graph; | ||
+ | import geometry; | ||
+ | import markers; | ||
− | Let <math> | + | 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> | ||
+ | Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC = \angle ADC</math> because they subtend the same arc. Furthermore, <math>\angle BAP = \angle EAC</math>, so <math>\triangle ABE</math> is similar to <math>\triangle ADC</math> by AAA similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>. By angle bisector theorem, <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math> so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math> which gives <math>BE = \dfrac{21}{5}</math>. Plugging this into the similarity proportion gives: <math>\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}</math>. | ||
== See Also == | == See Also == |
Revision as of 16:36, 28 April 2016
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 |
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