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Difference between revisions of "1983 AHSME Problems/Problem 19"

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\textbf{(E)} \ 4  </math>   
 
\textbf{(E)} \ 4  </math>   
  
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==Solution==
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==Solution 1 (Law of Cosines)==
  
 
Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so let <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{(A)} \ 2}</math>.
 
Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so let <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{(A)} \ 2}</math>.
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 +
==Solution 2 (Similar Triangles)==
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 +
Let <math>E</math> be a point on line <math>AC</math> such that <math>AE = 6</math> and <math>A</math> is between <math>E</math> and <math>C</math>.  Then <math>\triangle ABE</math> must be equilateral, <math>BE = 6</math>, and <math>\angle BEC = 60^\circ</math>.  So <math>BE \parallel DA</math> and <math>\triangle CDA \sim \triangle CBE</math>.  So <math>\frac{AD}{6} = \frac{3}{9}</math>, so <math>AD = \boxed{(\mathbf{A})\ 2}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:12, 2 July 2025

Problem

Point $D$ is on side $CB$ of triangle $ABC$. If $\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$, then the length of $AD$ is

$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ 2.5 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 3.5 \qquad \textbf{(E)} \ 4$

Solution 1 (Law of Cosines)

Let $AD = y$. Since $AD$ bisects $\angle{BAC}$, the Angle Bisector Theorem gives $\frac{DB}{CD} = \frac{AB}{AC} = 2$, so let $CD = x$ and $DB = 2x$. Applying the Law of Cosines to $\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$, and to $\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$. Subtracting $4$ times the first equation from the second equation therefore yields $0 = 6y - 3y^2 \Rightarrow y(y-2) = 0$, so $y$ is $0$ or $2$. But since $y \neq 0$ ($y$ is the length of a side of a triangle), $y$ must be $2$, so the answer is $\boxed{\textbf{(A)} \ 2}$.

Solution 2 (Similar Triangles)

Let $E$ be a point on line $AC$ such that $AE = 6$ and $A$ is between $E$ and $C$. Then $\triangle ABE$ must be equilateral, $BE = 6$, and $\angle BEC = 60^\circ$. So $BE \parallel DA$ and $\triangle CDA \sim \triangle CBE$. So $\frac{AD}{6} = \frac{3}{9}$, so $AD = \boxed{(\mathbf{A})\ 2}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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