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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>\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>
 
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>
  
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Note: Call CD <math>x</math> and BD <math>9-x</math>. By the Angle Bisector Theorem, we obtain the relation <math>\frac{8}{x} = \frac{7}{9-x}</math>. Solving for <math>x</math>, we find <math>72 - 8x = 7x</math>, which means <math>x = 72/15 = 24/5</math>. Thus, <math>CD = 24/5</math> and <math>BD = 21/5</math>, and they are not equal, so this solution is flawed.
+
Note: Call <math>CD x</math> and <math>BD 9-x</math>. By the Angle Bisector Theorem, we obtain the relation <math>\frac{8}{x} = \frac{7}{9-x}</math>. Solving for <math>x</math>, we find <math>72 - 8x = 7x</math>, which means <math>x = 72/15 = 24/5</math>. Thus, <math>CD = 24/5</math> and <math>BD = 21/5</math>, and they are not equal, so this solution is flawed.
  
 
==Solution 2 (Similarity Proportion) ==
 
==Solution 2 (Similarity Proportion) ==

Revision as of 17:35, 26 March 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 $\frac{5}{3}\implies\boxed{\text{(B)}}$

Note: Call $CD x$ and $BD 9-x$. By the Angle Bisector Theorem, we obtain the relation $\frac{8}{x} = \frac{7}{9-x}$. Solving for $x$, we find $72 - 8x = 7x$, which means $x = 72/15 = 24/5$. Thus, $CD = 24/5$ and $BD = 21/5$, and they are not equal, so this solution is flawed.

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 2, let $E$ be the intersection of diagonals $AD$ and $BC$. 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 $\triangle ABE\sim \triangle ADC$ by AA Similarity, so $AD/CD = AB/BE$. By the angle bisector theorem, $AB/BE = AC/CE$. Hence, \[\frac{AB}{BE} = \frac{AC}{CE} = \frac{AB + AC}{BE + CE} = \frac{AB + AC}{BC} = \frac{7 + 8}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}.\]

--vaporwave

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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