Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 10B Problems/Problem 24"

(→‎Solution)
m (Updated incorrect LATEX)
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
  
−
Set <math>\segment 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. 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. 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 23:06, 1 August 2015

In triangle $ABC$ we have $AB=7$, $AC=8$, $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $AD/CD$?

$\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$


Solution

Set $\overline{BD}$'s length as $x$. $CD$'s length must also be $x$ since $\angle BAD$ and $\angle DAC$ intercept arcs of equal length. Using Ptolemy's Theorem, $7x+8x=9(AD)$. The ratio is $\boxed{\frac{5}{3}}\implies(B)$

Solution 2

Let $P = \overline{BC}\cap \overline{AD}$. Observe that $\angle ABC = \angle ADC$ because they subtend the same arc. Furthermore, $\angle BAP = \angle PAC$, so $\triangle ABP$ is similar to $\triangle ADC$ by AAA similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BP}$. By angle bisector theorem, $\dfrac{7}{BP} = \dfrac{8}{CP}$ so $\dfrac{7}{BP} = \dfrac{8}{9-BP}$ which gives $BP = \dfrac{21}{5}$. Plugging this into the similarity proportion gives: $\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_24&oldid=71433"