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Let ABCD be a square with side length 1. Points E and F lie on line BC and line CD, respectively, in such a way that angle EAF=45 degrees If [CEF]=1/8, what is the value of [AEF]?

 
 Jun 11, 2021
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https://web2.0calc.com/questions/geometry_41255

 
 Jun 11, 2021
 #2
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Let ABCD be a square with side length 1.
Points E and F lie on line BC and line CD,
respectively, in such a way that angle EAF=45 degrees.

If [CEF]=1/8, what is the value of [AEF]?

 

\(\text{Let $CE=x$ } \\ \text{Let $CF=y$ } \\ \text{Let $BE=1-x$ } \\ \text{Let $DF=1-y$ } \\ \text{Let $EF=l=\sqrt{x^2+y^2}$ } \\ \text{Let $AE=m=\sqrt{1+(1-x)^2}$ } \\ \text{Let $AF=n=\sqrt{1+(1-y)^2}$ }\)

 

\(\begin{array}{|rcll|} \hline [CEF]&=&\dfrac{xy}{2} \quad | \quad [CEF]=\dfrac{1}{8} \\\\ \dfrac{1}{8} &=& \dfrac{xy}{2} \quad | \quad * 2 \\\\ \dfrac{1}{4} &=& xy \\\\ \mathbf{xy} &=& \mathbf{\dfrac{1}{4}} ~\text{or} ~\mathbf{y=\dfrac{1}{4x}} \\ \hline \end{array}\)

 

cos - rule
\(\small{ \begin{array}{|rcll|} \hline \text{In }\triangle~ [AEF] \\ \hline l^2 &=& m^2+n^2-2mn\cos(45^\circ) \quad | \quad \cos(45^\circ) = \dfrac{\sqrt{2}}{2} \\\\ l^2 &=& m^2+n^2-2mn\dfrac{\sqrt{2}}{2} \\\\ l^2 &=& m^2+n^2-\sqrt{2}mn \\ \sqrt{2}mn &=& m^2+n^2-l^2 \\ && \boxed{ m^2 = 1+(1-x)^2\\ n = 1-(1-y)^2\\l^2=x^2+y^2 } \\ \sqrt{2}mn &=& 1+(1-x)^2+1+(1-y)^2-x^2-y^2 \\ \sqrt{2}mn &=& 1+1-2x+x^2+1+1-2y+y^2-x^2-y^2 \\ \sqrt{2}mn &=& 4-2x-2y \\ \sqrt{2}mn &=& 4-2(x+y) \\ \sqrt{2}mn &=& 2\left(2-(x+y)\right) \quad | \quad \text{square both sides}\\ 2m^2n^2 &=& 4\left(2-(x+y)\right)^2 \quad | \quad : 2 \\ m^2n^2 &=& 2\left(2-(x+y)\right)^2 \\ m^2n^2 &=& 2\left(4-4(x+y)+(x+y)^2\right) \\ m^2n^2 &=& 8-8(x+y)+2(x+y)^2 \\ && \boxed{ m^2 = 1+(1-x)^2\\ n = 1-(1-y)^2\\l^2=x^2+y^2 } \\ \left(1+(1-x)^2\right) \left(1-(1-y)^2\right) &=& 8-8(x+y)+2(x+y)^2 \\ (2-2x+x^2)(2-2y+y^2) &=& 8-8(x+y)+2(x^2+2xy+y^2) \\ \small{4-4y+2y^2-4x+4xy-2xy^2+2x^2-2x^2y+x^2y^2} &=& \small{8-8(x+y)+2x^2+4xy+2y^2)} \\ \small{4-4y-4x-2xy^2-2x^2y+x^2y^2} &=& \small{8-8(x+y))} \\ \small{4-4(x+y)-2xy*y-2xy*x+(xy)^2} &=& \small{8-8(x+y))} \\ \small{4-4(x+y)-2xy(x+y)+(xy)^2} &=& \small{8-8(x+y))} \quad | \quad xy=\dfrac{1}{4} \\\\ \small{4-4(x+y)-2\dfrac{1}{4}(x+y)+\dfrac{1}{16}} &=& \small{8-8(x+y))} \\\\ \small{4-4(x+y)-\dfrac{1}{2}(x+y)+\dfrac{1}{16}} &=& \small{8-8(x+y))} \\\\ \small{8(x+y)-4(x+y)-\dfrac{1}{2}(x+y)} &=& \small{8-4-\dfrac{1}{16})} \\\\ \dfrac{7}{2}(x+y) &=& \dfrac{63}{16} \quad | \quad * 2 \\\\ 7(x+y) &=& \dfrac{63}{8} \\\\ x+y &=& \dfrac{63}{56} \\\\ \mathbf{x+y} &=& \mathbf{\dfrac{9}{8}} \quad | \quad y = \dfrac{1}{4x} \\\\ x+\dfrac{1}{4x} &=& \dfrac{9}{8} \quad | \quad * x \\\\ x^2+\dfrac{1}{4} &=& \dfrac{9x}{8} \\\\ x^2- \dfrac{9x}{8} +\dfrac{1}{4} &=& 0 \\ \hline x &=& \dfrac{ \dfrac{9}{8}\pm \sqrt{\dfrac{81}{64}-4*\dfrac{1}{4} } }{2} \\\\ x &=& \dfrac{ \dfrac{9}{8}\pm \sqrt{\dfrac{81}{64}-1 } }{2} \\\\ x &=& \dfrac{ \dfrac{9}{8}\pm \sqrt{\dfrac{81-64}{64} } }{2} \\\\ x &=& \dfrac{ \dfrac{9}{8}\pm \sqrt{\dfrac{17}{64} } }{2} \\\\ x &=& \dfrac{ \dfrac{9}{8}\pm \dfrac{\sqrt{17}}{8} }{2} \\\\ x &=& \dfrac{ \dfrac{9+\sqrt{17}}{8} }{2} \\\\ \mathbf{x} &=& \mathbf{\dfrac{9+\sqrt{17}}{16}} \quad | \quad (x =0.8201941016)\\ \hline y &=& \dfrac{1}{4x} \\ \Rightarrow \mathbf{y} &=& \mathbf{\dfrac{9-\sqrt{17}}{16}} \quad | \quad (y =0.3048058984)\\ \hline \end{array} }\)

 

\(\begin{array}{|rcll|} \hline [AEF] &=& [ABCD] -[CEF] -[ABE] -[ADF] \\ [AEF] &=& 1^2 - \dfrac{1}{8} -\dfrac{1*(1-x)}{2} -\dfrac{1*(1-y)}{2} \\\\ [AEF] &=& 1^2 - \dfrac{1}{8} -\dfrac{1}{2}+\dfrac{x}{2} -\dfrac{1}{2}+\dfrac{y}{2} \\\\ [AEF] &=& -\dfrac{1}{8} +\dfrac{x}{2} +\dfrac{y}{2} \\\\ [AEF] &=& \dfrac{x+y}{2}- \dfrac{1}{8} \quad | \quad \mathbf{x+y=\dfrac{9}{8}} \\\\ [AEF] &=& \dfrac{ \dfrac{9}{8}}{2}- \dfrac{1}{8} \\\\ [AEF] &=& \dfrac{9}{16}- \dfrac{1}{8} \\\\ [AEF] &=& \dfrac{9}{16}- \dfrac{2}{16} \\\\ [AEF] &=& \dfrac{9-2}{16} \\\\ \mathbf{[AEF]} &=& \mathbf{ \dfrac{7}{16} } \\ \hline \end{array}\)

 

laugh

 
 Jun 11, 2021
edited by heureka  Jun 11, 2021

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