+0

# For a triangle XYZ, we use [XYZ] to denote its area.

0
371
3

For a triangle XYZ, we use [XYZ] to denote its area.

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]=1/9, what is the value of [AEF].

Sep 16, 2019

#1
+25532
+2

For a triangle XYZ, we use [XYZ] to denote its area.
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]=1/9, what is the value of [AEF].

$$\text{Let CE=x } \\ \text{Let BE=1-x } \\ \text{Let CF=y } \\ \text{Let DF=1-y } \\ \text{Let AB=AD=1 }$$

$$\begin{array}{|lrcll|} \hline & [CEF] = \dfrac{xy}{2} &=& \dfrac{1}{9} \\\\ \text{or}& xy &=& \dfrac{2}{9} \\\\ \text{or}& y &=& \dfrac{2}{9x} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline &[AEF] &=& [ABCD] - [CEF]-[ABE]-[ADF] \\\\ &[AEF] &=& 1 - \dfrac{xy}{2}-\dfrac{(1-x)}{2}-\dfrac{(1-y)}{2} \quad | \quad \cdot 2 \\\\ &2[AEF] &=& 2 - xy- (1-x)-(1-y) \\ &2[AEF] &=& 2 - xy- 1+x-1+y \\ (1) &\mathbf{2[AEF]} &=& \mathbf{(x+y) - xy} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline &2[AEF] &=& AE\cdot AF\cdot \sin(45^\circ) \quad | \quad AE=\sqrt{1^2+(1-x)^2},\ AF=\sqrt{1^2+(1-y)^2} \\\\ &2[AEF] &=& \sqrt{1^2+(1-x)^2}\sqrt{1^2+(1-y)^2} \sin(45^\circ) \quad | \quad \sin(45^\circ) = \dfrac{\sqrt{2}} {2} \\\\ (2) &\mathbf{2[AEF]} &=& \mathbf{\sqrt{1^2+(1-x)^2}\sqrt{1^2+(1-y)^2}\cdot \dfrac{\sqrt{2}} {2} } \\ \hline \end{array}$$

(1) = (2):

$$\small{ \begin{array}{|rcll|} \hline \mathbf{(x+y) - xy} &=& \mathbf{\sqrt{1^2+(1-x)^2}\sqrt{1^2+(1-y)^2}\cdot \dfrac{\sqrt{2}} {2} } \quad | \quad \text{square both sides} \\\\ \Big((x+y) - xy\Big)^2 &=& \left(\sqrt{1^2+(1-x)^2}\sqrt{1^2+(1-y)^2}\cdot \dfrac{\sqrt{2}} {2} \right)^2 \\\\ \Big((x+y) - xy\Big)^2 &=& \left(1^2+(1-x)^2 \right) \left(1^2+(1-y)^2 \right)\cdot \dfrac{1} {2} \\\\ 2\Big((x+y) - xy\Big)^2 &=& \left(1^2+(1-x)^2 \right) \left(1^2+(1-y)^2 \right) \\ 2\Big( (x+y)^2-2(x+y)xy+x^2y^2 \Big) &=& 1+(1-y)^2+(1-x)^2+(1-x)^2(1-y)^2 \\ 2\Big( x^2+y^2 +2xy-2(x+y)xy+x^2y^2 \Big) &=& 1+1-2y+y^2+1-2x+x^2+(1-2x+x^2)(1-2y+y^2) \\ 2x^2+2y^2 +4xy-4(x+y)xy+2x^2y^2 &=& 3-2y+y^2-2x+x^2+1-2y+y^2 -2x+4xy-2xyy +x^2-2xyx+x^2y^2 \\ -4(x+y)xy+2x^2y^2 &=& 4-2y -2x -2y -2x -2xyy -2xyx+x^2y^2 \\ -4(x+y)xy+2x^2y^2 &=& 4-4(x+y) -2xy(x+y)+x^2y^2 \\ 4(x+y)-4xy(x+y)+2xy(x+y)+x^2y^2-4&=& 0 \\ (x+y)(4-4xy+2xy)+x^2y^2-4 &=& 0 \\ (x+y)(4-2xy)+x^2y^2-4 &=& 0 \quad | \quad y = \dfrac{2}{9x} \\ (x+\dfrac{2}{9x})(4-2xy)+x^2y^2-4 &=& 0 \\ \left(\dfrac{9x^2+2}{9x}\right)(4-2xy)+x^2y^2-4 &=& 0 \quad | \quad \cdot 9x \\\\ (9x^2+2)(4-2xy)+ 9x(x^2y^2-4) &=& 0 \quad | \quad \cdot 9x \\ 9x^2(4-2xy)+9x(x^2y^2-4)+2(4-2xy)&=& 0 \quad | \quad xy = \dfrac{2}{9} \\ 9x^2(4-\dfrac{4}{9})+9x(\dfrac{4}{81}-4)+2(4-\dfrac{4}{9})&=& 0 \\ 9x^2(\dfrac{36-4}{9})+9x(\dfrac{-320}{81})+2(\dfrac{36-4}{9})&=& 0 \\ 9x^2(\dfrac{32}{9})-x(\dfrac{-320}{9})+2(\dfrac{32}{9})&=& 0 \\ 32x^2-\dfrac{320}{9}x+ \dfrac{64}{9} &=& 0 \quad | \quad \cdot 9 \\ 288x^2-320x+64&=& 0 \quad | \quad :4 \\ \mathbf{ 72x^2-80x+16 } &=& \mathbf{0} \\ \hline \end{array} }$$

$$\begin{array}{|lrcll|} \hline & \mathbf{ 72x^2-80x+16 } &=& \mathbf{0} \\ & x &=& \dfrac{5}{9} + \dfrac{\sqrt{7}}{9} \\ \text{or}& x &=& \dfrac{5}{9} - \dfrac{\sqrt{7}}{9} \\\\ & y &=& \dfrac{5}{9} - \dfrac{\sqrt{7}}{9} \\ \text{or}& y &=& \dfrac{5}{9} + \dfrac{\sqrt{7}}{9} \\\\ & 2[AEF] &=& (x+y) -xy \\ & 2[AEF] &=& \left(\dfrac{5}{9} + \dfrac{\sqrt{7}}{9}+\dfrac{5}{9} - \dfrac{\sqrt{7}}{9} \right) - \left(\dfrac{5}{9} + \dfrac{\sqrt{7}}{9}\right) \left(\dfrac{5}{9} - \dfrac{\sqrt{7}}{9}\right) \\ & 2[AEF] &=& \dfrac{10}{9} - \left(\dfrac{5^2}{9^2} - \dfrac{(\sqrt{7})^2}{9^2}\right) \\ & 2[AEF] &=& \dfrac{10}{9} - \dfrac{2}{9} \\ & 2[AEF] &=& \dfrac{8}{9} \\ & \mathbf{ [AEF] } &=& \mathbf{ \dfrac{4}{9} } \\ \hline \end{array}$$

Sep 16, 2019
edited by heureka  Sep 17, 2019
edited by heureka  Sep 17, 2019
#2
+111433
+2

Very nice, heureka!!!

I want to look at this one, again......!!!

Sep 16, 2019
#3
+25532
+2

Thank you, CPhill !

heureka  Sep 17, 2019