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

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(→‎Solution 2)
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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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==Solution 2==
 
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<asy>
 
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import graph;
 
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import geometry;
 
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import markers;
 
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unitsize(0.5 cm);
 
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pair A, B, C, D, E, I;
 
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A = (11/3,8*sqrt(5)/3);
 
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B = (0,0);
 
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C = (9,0);
 
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I = incenter(A,B,C);
 
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D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C));
 
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E = extension(A,D,B,C);
 
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draw(A--B--C--cycle);
 
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draw(circumcircle(A,B,C));
 
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draw(D--A);
 
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draw(D--B);
 
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draw(D--C);
 
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label("$A$", A, N);
 
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label("$B$", B, SW);
 
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label("$C$", C, SE);
 
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label("$D$", D, S);
 
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label("$E$", E, NE);
 
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markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true)));
 
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markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true)));
 
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markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true)));
 
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markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true)));
 
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markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true)));
 
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</asy>
 
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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 11:25, 2 January 2021

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 $\frac{AD}{CD}$?

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

Solution 1

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 $\frac{5}{3}\implies\boxed{\text{(B)}}$

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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