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Can someone please help me?

 

In \(\triangle ABC\)\(AB = BC\), and BC is an altitude. \(BE=10\), and The values of \(\tan \angle CBE\)\(\tan \angle DBE\), and \tan \angle ABE form a geometric progression and the values of \(\cot \angle DBE\)\(\cot \angle CBE\), and \cot \angle DBC form an arithmetic progression. What is the area of \triangle ABC?

 Jul 31, 2019
 #1
avatar+23339 
+3

In  \(\triangle ABC\) , \(AB = BC\), and \(\overline{BD}\) is an altitude.
Point \(E\) is on the extension of \(\overline{AC}\)$ such that \(BE = 10\).
The values of \(\tan \angle CBE\), \(\tan \angle DBE\), and \(\tan \angle ABE\) form a geometric progression,
and the values of \(\cot \angle DBE\), \(\cot \angle CBE\), \(\cot \angle DBC\) form an arithmetic progression.
What is the area of \(\triangle ABC\)?

 

\(\text{Let $\angle DBC=\angle ABD = \alpha $}\)

 

\(1.\ \mathbf{ GP}:\)

\(\begin{array}{|rcll|} \hline \tan (\angle CBE) &=& a \\ \tan (\angle DBE) &=& ar \\ \tan (\angle ABE) &=& ar^2 \\\\ a\cdot ar^2 &=& (ar)^2 \\ \tan (\angle CBE)\cdot \tan (\angle ABE) &=& \tan^2(\angle DBE) \quad | \quad \angle CBE=\angle DBE-\alpha,\ \angle ABE=\angle DBE+\alpha\\ \tan (\angle DBE-\alpha)\cdot \tan (\angle DBE+\alpha) &=& \tan^2(\angle DBE) \\ && \boxed{ \text{Formula}\\ \tan(x+y)=\dfrac{\tan(x)+\tan(y)} {1-\tan(x)\tan(y)}\\ \tan(x-y)=\dfrac{\tan(x)-\tan(y)} {1+\tan(x)\tan(y)} \\ \tan(x-y)\tan(x+y)= \left(\dfrac{\tan(x)-\tan(y)} {1+\tan(x)\tan(y)}\right) \left(\dfrac{\tan(x)+\tan(y)} {1-\tan(x)\tan(y)}\right) \\ =\dfrac{\tan^2(x)-\tan^2(y)}{1-\tan^2(x)\tan^2(y)} } \\\\ \dfrac{\tan^2(\angle DBE)-\tan^2(\alpha)} {1-\tan^2(\angle DBE)\tan^2(\alpha)} &=& \tan^2(\angle DBE) \\ \tan^2(\angle DBE)-\tan^2(\alpha) &=& \left(1-\tan^2(\angle DBE)\tan^2(\alpha) \right)\tan^2(\angle DBE) \\ -\tan^2(\alpha) &=& -\tan^4(\angle DBE)\tan^2(\alpha) \\ \tan^4(\angle DBE)\tan^2(\alpha) -\tan^2(\alpha) &=& 0 \\ \underbrace{ \tan^2(\alpha)}_{\neq 0}\underbrace{(\tan^4(\angle DBE)-1)}_{=0} &=& 0 \\ \tan^4(\angle DBE)-1 &=& 0 \\ \tan^4(\angle DBE) &=& 1 \\ \tan (\angle DBE) &=& 1 \\ \angle DBE &=& \arctan(1) \\ \mathbf{\angle DBE} &=& \mathbf{45^\circ} \\ \hline \end{array} \)

 

\(2.\ \mathbf{ AP}:\)

\(\begin{array}{|rcll|} \hline \cot (\angle DBE) &=& b \quad | \quad \angle DBE = 45^\circ \\ \cot (45^\circ) &=& b \\ 1 &=& b \\\\ \cot (\angle CBE)=\cot (\angle DBE-\alpha)=\cot (45^\circ-\alpha) = b+d&=&1+b \\ \cot (\angle DBC)=\cot(\alpha) = b+2d=1+2d \\\\ \cot (\angle DBE) &=& 1 \\ \cot (45^\circ-\alpha) &=& 1+d \\ \cot (\alpha) &=& 1+2d \\\\ 2(1+d) &=& 1+(1+2d) \\ 2\cot (\angle 45^\circ-\alpha) &=& 1+ \cot (\alpha) \\ && \boxed{ \text{Formula}\\ \cot(x-y)=\dfrac{\cot(x)\cot(y)+1}{\cot(y)-\cot(x)}\\ \cot(45^\circ-\alpha)=\dfrac{\cot(45^\circ)\cot(\alpha)+1}{\cot(\alpha)-\cot(45^\circ)}\\ =\cot(45^\circ-\alpha)=\dfrac{1*\cot(\alpha)+1}{\cot(\alpha)-1}\\ =\cot(45^\circ-\alpha)=\dfrac{\cot(\alpha)+1}{\cot(\alpha)-1} } \\\\ 2\left(\dfrac{\cot(\alpha)+1}{\cot(\alpha)-1}\right) &=& 1+ \cot (\alpha) \\ (1+ \cot (\alpha))(\cot(\alpha)-1)&=& 2 (\cot(\alpha)+1)\\ (1+ \cot (\alpha))((\cot(\alpha)-1)-2 (\cot(\alpha)+1)&=& 0\\ (1+ \underbrace{\cot(\alpha)}_{\neq-1} ) ( \underbrace{\cot(\alpha)-3}_{=0} ) &=& 0\\ \cot(\alpha)-3 &=& 0 \\ \mathbf{\cot(\alpha)} &=& \mathbf{3} \\ \hline \end{array}\)

 

\(\mathbf{BD=\ ?}\)

\(\begin{array}{|rcll|} \hline \cos(\angle DBE) &=& \dfrac{BD}{BE} \\ \cos(45^\circ) &=& \dfrac{BD}{10} \\ BD &=& 10\cos(45^\circ) \\ BD &=& 10\dfrac{\sqrt{2}}{2} \\ \mathbf{BD} &=& \mathbf{5 \sqrt{2} } \\ \hline \end{array}\)

 

\(\mathbf{AC=\ ?}\)

\(\begin{array}{|rcll|} \hline \cot(\alpha)&=& \dfrac{BD}{DC}\times \dfrac{2}{2} \\\\ \cot(\alpha)&=& \dfrac{2BD}{2DC} \quad | \quad 2DC = AC \\\\ \cot(\alpha)&=& \dfrac{2BD}{AC} \\\\ AC&=& \dfrac{2BD}{\cot(\alpha)} \quad | \quad \cot(\alpha) = 3 \\\\ AC&=& \dfrac{2BD}{3} \quad | \quad \mathbf{BD = 5 \sqrt{2} }\\ AC&=& \dfrac{2 *5 \sqrt{2}}{3} \\ \mathbf{AC} &=& \mathbf{\dfrac{10 \sqrt{2}}{3}} \\ \hline \end{array}\)

 

\(\mathbf{\text{area of $\triangle ABC$ }}\)

\(\begin{array}{|rcll|} \hline \text{area of $\triangle ABC$ } &=& \dfrac{AC\cdot DB}{2} \\ \text{area of $\triangle ABC$ } &=& \dfrac{\dfrac{10 \sqrt{2}}{3}\cdot 5 \sqrt{2}}{2} \\ \mathbf{\text{area of $\triangle ABC$ }} &=& \mathbf{\dfrac{50}{3}} \\ \hline \end{array}\)

 

source: https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_24

 

laugh

 Jul 31, 2019
 #2
avatar+104937 
0

Nicely done, heureka!!!!!!

 

 

 

cool cool cool

CPhill  Aug 2, 2019

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