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

Revision as of 12:03, 9 January 2025

Problem

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 $\frac{AD}{CD}$ ?

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

Solution 1 - (Ptolemy's Theorem)

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

Solution 2 - Similarity Proportion

$[asy] import graph; import geometry; import markers; unitsize(0.5 cm); pair A, B, C, D, E, I; A = (11/3,8*sqrt(5)/3); B = (0,0); C = (9,0); I = incenter(A,B,C); D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C)); E = extension(A,D,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(D--A); draw(D--B); draw(D--C); label("$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 $E = \overline{BC}\cap \overline{AD}$ . Observe that $\angle ABC \cong \angle ADC$ because they both subtend arc $\overarc{AC}.$

Furthermore, $\angle BAE \cong \angle EAC$ because $\overline{AE}$ is an angle bisector, so $\triangle ABE \sim \triangle ADC$ by $\text{AA}$ similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BE}$ . By the Angle Bisector Theorem, $\dfrac{7}{BE} = \dfrac{8}{CE}$ , so $\dfrac{7}{BE} = \dfrac{8}{9-BE}$ . This in turn gives $BE = \frac{21}{5}$ . Plugging this into the similarity proportion gives: $\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}$ .

Solution 3 (Angle Bisector Theorem)

Similar to solution 1, let $E$ be the intersection of diagonals $AC and$ BD $. We know that$ \overline{AD} $bisects$ \angle BAC $, so$ \angle BAD = \angle CAD $. Additionally,$ \angle BAD $and$ \angle BCD $subtend the same arc, giving$ \angle BAD = \angle BCD $. Similarly,$ \angle CAD = \angle CBD $and$ \angle ABC = \angle ADC$.

These angle relationships tell us that$ (Error compiling LaTeX. Unknown error_msg)\triangle ABE\sim \triangle ADC $by AA Similarity, so$ AD/CD = AB/BE $. By the angle bisector theorem,$ AB/BE = AC/CE$. Hence, <cmath>\frac{AB}{BE} = \frac{AC}{CE} = \frac{AB + AC}{BE + CE} = \frac{AB + AC}{BC} = \frac{7 + 8}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}.</cmath>

~vaporwave

== Solution 4 (Ptolemy's Theorem)

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

@@ Line 50: / Line 50: @@
 ==Solution 3 (Angle Bisector Theorem)==
-We know that <math>\overline{AD}</math> bisects <math>\angle BAC</math>, so <math>\angle BAD = \angle CAD</math>. Additionally, <math>\angle BAD</math> and <math>\angle BCD</math> subtend the same arc, giving <math>\angle BAD = \angle BCD</math>. Similarly, <math>\angle CAD = \angle CBD</math> and <math>\angle ABC = \angle ADC</math>.
+Similar to solution 1, let <math>E</math> be the intersection of diagonals <math>AC and </math>BD<math>. We know that </math>\overline{AD}<math> bisects </math>\angle BAC<math>, so </math>\angle BAD = \angle CAD<math>. Additionally, </math>\angle BAD<math> and </math>\angle BCD<math> subtend the same arc, giving </math>\angle BAD = \angle BCD<math>. Similarly, </math>\angle CAD = \angle CBD<math> and </math>\angle ABC = \angle ADC<math>.
-These angle relationships tell us that <math>\triangle ABE\sim \triangle ADC</math> by AA Similarity, so <math>AD/CD = AB/BE</math>. By the angle bisector theorem, <math>AB/BE = AC/CE</math>. Hence,
+These angle relationships tell us that </math>\triangle ABE\sim \triangle ADC<math> by AA Similarity, so </math>AD/CD = AB/BE<math>. By the angle bisector theorem, </math>AB/BE = AC/CE<math>. Hence,
 <cmath>\frac{AB}{BE} = \frac{AC}{CE} = \frac{AB + AC}{BE + CE} = \frac{AB + AC}{BC} = \frac{7 + 8}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}.</cmath>
-(Where did E come from?????)
+~vaporwave
---vaporwave
 == Solution 4 (Ptolemy's Theorem)
-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)$
 == 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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