Difference between revisions of "1998 AHSME Problems/Problem 28"
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<math> \mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26</math> | <math> \mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26</math> | ||
− | + | == Solution 1== | |
− | |||
− | |||
Let <math>\theta = \angle DAB</math>, so <math>2\theta = \angle CAD</math> and <math>3 \theta = \angle CAB</math>. Then, it is given that <math>\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}</math> and | Let <math>\theta = \angle DAB</math>, so <math>2\theta = \angle CAD</math> and <math>3 \theta = \angle CAB</math>. Then, it is given that <math>\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}</math> and | ||
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and <math>\frac{CD}{BD} = \frac{5}{9} \Longrightarrow m+n = 14 \Longrightarrow \mathbf{(B)}</math>. (This also may have been done on a calculator by finding <math>\theta</math> directly) | and <math>\frac{CD}{BD} = \frac{5}{9} \Longrightarrow m+n = 14 \Longrightarrow \mathbf{(B)}</math>. (This also may have been done on a calculator by finding <math>\theta</math> directly) | ||
+ | |||
+ | == Solution 2 == | ||
+ | By the application of ratio lemma for <math>\frac{CD}{BD}</math>, we get <math>\frac{CD}{BD} = 2\cos{3A}\cos{A}</math>, where we let <math>A = \angle{DAB}</math>. We already know <math>\cos{2A}</math> hence the rest is easy | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>AC = 2</math>, <math>AD = 3</math>. <math>\cos \angle CAD = \frac23</math> | ||
+ | |||
+ | By the pythagorean theorem <math>CD = \sqrt{3^2-2^2} = \sqrt{5}</math> | ||
+ | |||
+ | <math>\sin \angle BDA = \sin (180^{\circ} - \angle BDA) = \sin \angle CDA = \cos \angle (90^{\circ} - CDA) = \cos \angle CAD = \frac23</math> | ||
+ | |||
+ | <math>\sin \angle BAD = \sqrt{ \frac{1-cos (2\angle BAD)}{2} } = \sqrt{ \frac{1-\cos \angle CAD}{2} } = \sqrt{ \frac{1-\frac23}{2} } = \frac{\sqrt{6}}{6}</math> | ||
+ | |||
+ | By the Law of Sine, <math>\frac{ \sin \angle BDA }{AB} = \frac{ \sin \angle BAD }{BD}</math> | ||
+ | |||
+ | <math>\frac{ \frac23 }{ \sqrt{2^2 + ( \sqrt{5} + BD)^2} } = \frac{ \frac{\sqrt{6}}{6} }{BD}</math> | ||
+ | |||
+ | <math>8BD^2 = 3(9+ 2BD \sqrt{5} + BD^2)</math> | ||
+ | |||
+ | <math>5BD^2 - 6 BD \sqrt{5} -27=0</math> | ||
+ | |||
+ | As <math>BD>0</math>, <math>BD = \frac{6 \sqrt{5} + \sqrt{ (6 \sqrt{5})^2 - 4 \cdot 5 (-27) } }{10} = \frac{9\sqrt{5}}{5}</math> | ||
+ | |||
+ | <math>\frac{CD}{BD} = \frac{\sqrt{5}}{\frac{9\sqrt{5}}{5}} = \frac59</math>. | ||
+ | |||
+ | <math>5+9=\boxed{\textbf{(B) } 14}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:51, 2 October 2023
Problem
In triangle , angle is a right angle and . Point is located on so that angle is twice angle . If , then , where and are relatively prime positive integers. Find .
Solution 1
Let , so and . Then, it is given that and
Now, through the use of trigonometric identities, . Solving yields that . Using the tangent addition identity, we find that , and
and . (This also may have been done on a calculator by finding directly)
Solution 2
By the application of ratio lemma for , we get , where we let . We already know hence the rest is easy
Solution 3
Let and . By the Pythagorean Theorem, . Let point be on segment such that bisects . Thus, angles , , and are congruent. Applying the angle bisector theorem on , we get that and . Pythagorean Theorem gives .
Let . By the Pythagorean Theorem, . Applying the angle bisector theorem again on triangle , we have The right side simplifies to. Cross multiplying, squaring, and simplifying, we get a quadratic: Solving this quadratic and taking the positive root gives Finally, taking the desired ratio and canceling the roots gives . The answer is .
Solution 4
Let , .
By the pythagorean theorem
By the Law of Sine,
As ,
.
.
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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