Difference between revisions of "2004 AMC 10B Problems/Problem 24"
(→Solution 1) |
|||
Line 6: | Line 6: | ||
== Solution 1== | == Solution 1== | ||
− | 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 | + | 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== | ||
+ | <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> | ||
+ | Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC \cong \angle ADC</math> because they subtend the same arc, <math>\overarc{AC}</math>. Furthermore, <math>\angle BAE \cong \angle EAC</math> because <math>\overline{AE}</math> is an angle bisector, so <math>\triangle ABE \sim \triangle ADC</math> by <math>\text{AA}</math> similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>. By the [[Angle Bisector Theorem]], <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math>, so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math>. This in turn gives <math>BE = \frac{21}{5}</math>. Plugging this into the similarity proportion gives: <math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>. | ||
== See Also == | == See Also == |
Revision as of 12:26, 7 January 2021
Contents
[hide]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 Ptolemy's Theorem, . The ratio is
Solution 2
Let . Observe that because they subtend the same 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: .
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.