Difference between revisions of "1998 AHSME Problems/Problem 28"

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==Solution 2==
 
 
Let <math>AC=2</math> and <math>AD=3</math>. By the Pythagorean Theorem, <math>CD=\sqrt{5}</math>. Let point <math>P</math> be on segment <math>CD</math> such that <math>AP</math> bisects <math>\angle CAD</math>. Thus, angles <math>CAP</math>, <math>PAD</math>, and <math>DAB</math> are congruent. Applying the angle bisector theorem on <math>ACD</math>, we get that <math>CP=\frac{2\sqrt{5}}{5}</math> and <math>PD=\frac{3\sqrt{5}}{5}</math>. Pythagorean Theorem gives <math>AP=\frac{\sqrt{5}\sqrt{24}}{5}</math>.
 
 
Let <math>DB=x</math>. By the Pythagorean Theorem, <math>AB=\sqrt{(x+\sqrt{5})^{2}+2^2}</math>. Applying the angle bisector theorem again on triangle <math>APB</math>, we have <cmath>\frac{\sqrt{(x+\sqrt{5})^{2}+2^2}}{x}=\frac{\frac{\sqrt{5}\sqrt{24}}{5}}{\frac{3\sqrt{5}}{5}}</cmath>
 
The right side simplifies to<math>\frac{\sqrt{24}}{3}</math>. Cross multiplying, squaring, and simplifying, we get a quadratic: <cmath>5x^2-6\sqrt{5}x-27=0</cmath> Solving this quadratic and taking the positive root gives <cmath>x=\frac{9\sqrt{5}}{5}</cmath> Finally, taking the desired ratio and canceling the roots gives <math>\frac{CD}{BD}=\frac{5}{9}</math>. The answer is <math>\fbox{(B) 14}</math>.
 
 
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1998|num-b=27|num-a=29}}
 
{{AHSME box|year=1998|num-b=27|num-a=29}}

Revision as of 11:23, 30 July 2017

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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