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

m (→‎Solution)
(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
 +
<asy>
 +
import graph;
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import geometry;
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import markers;
  
−
Let <math>P = \overline{BC}\cap \overline{AD}</math>.  Observe that <math>\angle ABC = \angle ADC</math> because they subtend the same arc.  Furthermore, <math>\angle BAP = \angle PAC</math>, so <math>\triangle ABP</math> is similar to <math>\triangle ADC</math> by AAA similarity.  Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BP}</math>.  By angle bisector theorem, <math>\dfrac{7}{BP} = \dfrac{8}{CP}</math> so <math>\dfrac{7}{BP} = \dfrac{8}{9-BP}</math> which gives <math>BP = \dfrac{21}{5}</math>.  Plugging this into the similarity proportion gives:  <math>\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}</math>.
+
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)));
 +
</asy>
 +
Let <math>E = \overline{BC}\cap \overline{AD}</math>.  Observe that <math>\angle ABC = \angle ADC</math> because they subtend the same arc.  Furthermore, <math>\angle BAP = \angle EAC</math>, so <math>\triangle ABE</math> is similar to <math>\triangle ADC</math> by AAA similarity.  Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>.  By angle bisector theorem, <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math> so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math> which gives <math>BE = \dfrac{21}{5}</math>.  Plugging this into the similarity proportion gives:  <math>\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 15:36, 28 April 2016

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(because $\angle BAD =\angle DAC$). Using Ptolemy's Theorem, $7x+8x=9(AD)$. The ratio is $\boxed{\frac{5}{3}}\implies(B)$

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 $E = \overline{BC}\cap \overline{AD}$. Observe that $\angle ABC = \angle ADC$ because they subtend the same arc. Furthermore, $\angle BAP = \angle EAC$, so $\triangle ABE$ is similar to $\triangle ADC$ by AAA similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BE}$. By angle bisector theorem, $\dfrac{7}{BE} = \dfrac{8}{CE}$ so $\dfrac{7}{BE} = \dfrac{8}{9-BE}$ which gives $BE = \dfrac{21}{5}$. 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

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