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# AMC 10B 2019 Question #16

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I also had trouble with this question.

16. In ABC with a right angle at C, point D lies in the interior of AB and point E lies in the interior of BC so that AC=CD, DE=EB, and the ratio AC:DE = 4:3. What is the ratio AD:DB?

A) 2:3

B) 2: $$\sqrt5$$

C) 1:1

D) 3:$$\sqrt5$$

E) 3:2

Feb 15, 2019

#1
+21848
+6

16.

In ABC with a right angle at C,

point D lies in the interior of AB and

point E lies in the interior of BC so that

AC=CD, DE=EB, and the ratio AC:DE = 4:3.

$$\text{Let \overline{AC} = \overline{CD} = {\color{red}x}  } \\ \text{Let \overline{DE} = \overline{EB} = {\color{red}y}  } \\ \text{Let \overline{AD} = {\color{red}s}  } \\ \text{Let \overline{DB} = {\color{red}t}  }$$

$$\text{Let \angle{CAD} = \angle{ADC} = \alpha  } \\ \text{Let \angle{ABC} = \angle{EDB} = 90^{\circ}-\alpha  } \\ \text{Let \angle{DCA} = 180^{\circ}-2\alpha  } \\ \text{Let \angle{ECD} = 90^{\circ}-\angle{DCA} = 90^{\circ} -(180^{\circ}-2\alpha) = 2\alpha-90^{\circ}  } \\ \text{Let \angle{BED} = 180^{\circ}-2(90^{\circ}-\alpha)=2\alpha  } \\ \text{Let \angle{DEC} = 180^{\circ}-\angle{BED}=180^{\circ}-2\alpha  }$$

$$\begin{array}{|rcll|} \hline \angle{CDE} &=& 180^{\circ}- \angle{ECD}-\angle{DEC} \\ &=& 180^{\circ}-(2\alpha-90^{\circ})- (180^{\circ}-2\alpha ) \\ &\mathbf{=}& \mathbf{90^{\circ} \ !} \\ \hline \end{array}$$

$$\mathbf{\text{\cos-rule in \triangle ACD}}$$

$$\begin{array}{|rcll|} \hline x^2 &=& x^2+s^2-2xs\cos(\alpha) \\ 2xs\cos(\alpha) &=& s^2 \\ \mathbf{2x\cos(\alpha)} & \mathbf{=} & \mathbf{s} \quad & (1) \\ \hline \end{array}$$

$$\mathbf{\text{\cos-rule in \triangle DEB}}$$

$$\begin{array}{|rcll|} \hline y^2 &=& y^2+t^2-2yt\cos(90^{\circ}-\alpha) \\ 2yt\sin(\alpha) &=& t^2 \\ \mathbf{2y\sin(\alpha)} & \mathbf{=} & \mathbf{t} \quad & (2) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \dfrac{(1)}{(2)}: & \dfrac{ 2x\cos(\alpha) } {2y\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \dfrac{x}{y} \cdot \dfrac{ \cos(\alpha) } {\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad & (3) \\ \hline \end{array}$$

$$\mathbf{\text{In the right-angled \triangle DCE}} \\ \mathbf{ \tan(\alpha) =\ ?}$$

$$\begin{array}{|rcll|} \hline \tan(180^{\circ}-2\alpha) &=& \dfrac{x}{y} \\\\ -\tan(2\alpha) &=& \dfrac{4}{3} \quad | \quad \text{Formula:} \ \boxed{ \tan(2\alpha)=\dfrac{2\tan{\alpha}}{1-\tan^2(\alpha)} } \\\\ \dfrac{-2\tan{\alpha}}{1-\tan^2(\alpha)} &=& \dfrac{4}{3} \\\\ \dfrac{2\tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{4}{3} \\\\ \dfrac{ \tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{2}{3} \\\\ 3\tan{\alpha} &=& 2(\tan^2(\alpha)-1) \\ 3\tan{\alpha} &=& 2\tan^2(\alpha)-2 \\ 2\tan^2(\alpha)-3\tan(\alpha)-2 &=& 0 \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{9-2\cdot4\cdot(-2)}} {2\cdot 2} \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{25}} {4} \\\\ \tan(\alpha)&=& \dfrac{3\pm 5} {4} \\\\ \tan(\alpha)&=& \dfrac{3 {\color{red}+} 5} {4} \quad | \quad \tan(\alpha) > 0\ ! \\\\ \tan(\alpha)&=& \dfrac{8} {4} \\ \mathbf{ \tan(\alpha) }& \mathbf{=} & \mathbf{2} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad | \quad \mathbf{\tan(\alpha)=2},\ \quad \dfrac{x}{y}=\dfrac{4}{3} \\\\ \dfrac{4}{3} \cdot \dfrac{1} {2} & = & \dfrac{ s }{t} \\\\ \dfrac{4}{6} & = & \dfrac{ s }{t} \\\\ \dfrac{2}{3} & = & \dfrac{ s }{t} \\\\ \mathbf{\dfrac{ s }{t}} &\mathbf{=}& \mathbf{\dfrac{2}{3}} \\ \hline \end{array}$$

$$(A) \ 2:3$$

Feb 15, 2019

#1
+21848
+6

16.

In ABC with a right angle at C,

point D lies in the interior of AB and

point E lies in the interior of BC so that

AC=CD, DE=EB, and the ratio AC:DE = 4:3.

$$\text{Let \overline{AC} = \overline{CD} = {\color{red}x}  } \\ \text{Let \overline{DE} = \overline{EB} = {\color{red}y}  } \\ \text{Let \overline{AD} = {\color{red}s}  } \\ \text{Let \overline{DB} = {\color{red}t}  }$$

$$\text{Let \angle{CAD} = \angle{ADC} = \alpha  } \\ \text{Let \angle{ABC} = \angle{EDB} = 90^{\circ}-\alpha  } \\ \text{Let \angle{DCA} = 180^{\circ}-2\alpha  } \\ \text{Let \angle{ECD} = 90^{\circ}-\angle{DCA} = 90^{\circ} -(180^{\circ}-2\alpha) = 2\alpha-90^{\circ}  } \\ \text{Let \angle{BED} = 180^{\circ}-2(90^{\circ}-\alpha)=2\alpha  } \\ \text{Let \angle{DEC} = 180^{\circ}-\angle{BED}=180^{\circ}-2\alpha  }$$

$$\begin{array}{|rcll|} \hline \angle{CDE} &=& 180^{\circ}- \angle{ECD}-\angle{DEC} \\ &=& 180^{\circ}-(2\alpha-90^{\circ})- (180^{\circ}-2\alpha ) \\ &\mathbf{=}& \mathbf{90^{\circ} \ !} \\ \hline \end{array}$$

$$\mathbf{\text{\cos-rule in \triangle ACD}}$$

$$\begin{array}{|rcll|} \hline x^2 &=& x^2+s^2-2xs\cos(\alpha) \\ 2xs\cos(\alpha) &=& s^2 \\ \mathbf{2x\cos(\alpha)} & \mathbf{=} & \mathbf{s} \quad & (1) \\ \hline \end{array}$$

$$\mathbf{\text{\cos-rule in \triangle DEB}}$$

$$\begin{array}{|rcll|} \hline y^2 &=& y^2+t^2-2yt\cos(90^{\circ}-\alpha) \\ 2yt\sin(\alpha) &=& t^2 \\ \mathbf{2y\sin(\alpha)} & \mathbf{=} & \mathbf{t} \quad & (2) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \dfrac{(1)}{(2)}: & \dfrac{ 2x\cos(\alpha) } {2y\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \dfrac{x}{y} \cdot \dfrac{ \cos(\alpha) } {\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad & (3) \\ \hline \end{array}$$

$$\mathbf{\text{In the right-angled \triangle DCE}} \\ \mathbf{ \tan(\alpha) =\ ?}$$

$$\begin{array}{|rcll|} \hline \tan(180^{\circ}-2\alpha) &=& \dfrac{x}{y} \\\\ -\tan(2\alpha) &=& \dfrac{4}{3} \quad | \quad \text{Formula:} \ \boxed{ \tan(2\alpha)=\dfrac{2\tan{\alpha}}{1-\tan^2(\alpha)} } \\\\ \dfrac{-2\tan{\alpha}}{1-\tan^2(\alpha)} &=& \dfrac{4}{3} \\\\ \dfrac{2\tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{4}{3} \\\\ \dfrac{ \tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{2}{3} \\\\ 3\tan{\alpha} &=& 2(\tan^2(\alpha)-1) \\ 3\tan{\alpha} &=& 2\tan^2(\alpha)-2 \\ 2\tan^2(\alpha)-3\tan(\alpha)-2 &=& 0 \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{9-2\cdot4\cdot(-2)}} {2\cdot 2} \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{25}} {4} \\\\ \tan(\alpha)&=& \dfrac{3\pm 5} {4} \\\\ \tan(\alpha)&=& \dfrac{3 {\color{red}+} 5} {4} \quad | \quad \tan(\alpha) > 0\ ! \\\\ \tan(\alpha)&=& \dfrac{8} {4} \\ \mathbf{ \tan(\alpha) }& \mathbf{=} & \mathbf{2} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad | \quad \mathbf{\tan(\alpha)=2},\ \quad \dfrac{x}{y}=\dfrac{4}{3} \\\\ \dfrac{4}{3} \cdot \dfrac{1} {2} & = & \dfrac{ s }{t} \\\\ \dfrac{4}{6} & = & \dfrac{ s }{t} \\\\ \dfrac{2}{3} & = & \dfrac{ s }{t} \\\\ \mathbf{\dfrac{ s }{t}} &\mathbf{=}& \mathbf{\dfrac{2}{3}} \\ \hline \end{array}$$

$$(A) \ 2:3$$

heureka Feb 15, 2019
#2
+53
-1

dang, I got it wrong, I circled C

Feb 15, 2019
#3
+99301
0

Mathcourts24,

How about saying thankyou to heueka for all his time consuming effort which he afforded you!

Don'r just grumble about getting it wrong!

You tried to vote your question up about 10 times but you have not even attempted to give Heureka a point.

That is incredibly rude of you!  Do you think that will encourage anyone to help you next time?

Melody  Feb 15, 2019
edited by Melody  Feb 15, 2019
edited by Melody  Feb 15, 2019
edited by Melody  Feb 15, 2019