An interesting relationship in a square

           Starting with a post on Facebook:

https://www.facebook.com/groups/646056952197053/permalink/2475000499302680/ 

          In translation (thanks to Miss Google):

          "Given the square $ABCD$ and a point $E$ on the side $(BC)$. $AF$ is the bisector of the  angle   $\angle DAE,\;\;F \in  (DC)$. Prove that $AE=DF+BE$."

Solution CiP

    

          Let's rotate clockwise the square around the  point $A$ by $90^\circ$ degree.

Point $D$ turns into $D'=B$. Point $C$ turns into $C'$ on the extension of the half-line $[CB$ and point $F$ will be in the new position $F' \in (D'C')=(BC')$ so that $BF'=DF$.

     If we denote the measure of the angle $\angle BAE=2x^\circ$, then 

$\angle EAF=\angle FAD=45^\circ-x^\circ$.     

 Then $\angle AFD=45^\circ+x^\circ$ and so $\angle AF'D'=\angle AF'E=45^\circ+x^\circ$.

     This rotation allows us to say again $\angle D'AF'=\angle BAF'=\angle DAF=45^\circ-x^\circ$ and then it's easy to see that 

$\angle F'AE=\angle F'AB+\angle BAE=(45^\circ-x^\circ)+2x^\circ=45^\circ+x^\circ$.

     We obtained that the triangle $\Delta AEF'$ has equal angles $\angle AF'E$ and $\angle EAF'$, so it is an isosceles triangle with $EA=EF'$. 

     Now it turns out immediately

 $AE=EF'=EB+BF'=EB+D'F'=EB+DF.$

$\blacksquare$

  

*******

 A very interesting solution that uses relations between the areas of some triangles can be found here 

https://www.facebook.com/photo.php?fbid=7507678955923909&set=p.7507678955923909&type=3

***************

          Related to this problem we mention  Problems #1334, #1363, #1366 from http://www.gogeometry.com/index.html

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Problem E:15900, not from AMERICAN MATHEMATICAL MONTHLY

It is from the Romanian magazine "Gazeta Matematica", No. 2, 2021, at page 105, proposed for 8th grade.

 In the translation of Miss Google:

          "Solve the equation in the set of integers

$$\left\{\frac{10x+4}{x+1}\right\}=\left\{\frac{7x+8}{x+2}\right\}.$$

     We have noted $\{a\}$ the fractional part of the real number $a$."

Answer CiP

$$x\in \{-4,\;-3,\;0,\;1\;\}$$

Solution CiP

     We are looking for the solutions of the equation in the set $\mathbb {Z} \setminus\{-2,\;-1\;\}.$  

      If we denote by $[ a ]$ the integer part of the real number $a$, we have

$$a=[ a ]+\{ a \},\;\;\;[ a ] \in \mathbb{Z},\;\{ a \} \in [ 0,\;1).$$

      The equation is written successively

$$\frac{10x+4}{x+1}-\left [ \frac{10x+4}{x+1} \right ]=\frac{7x+8}{x+2}-\left [ \frac{7x+8}{x+2} \right ]$$

$\Leftrightarrow \;\;\frac{10x+4}{x+1}-\frac{7x+8}{x+2}=\left [ \frac{10x+4}{x+1} \right ]-\left [ \frac{7x+8}{x+2} \right ]$

$ \Leftrightarrow \;\;\left ( 10-\frac{6}{x+1} \right )-\left (7-\frac{6}{x+2} \right )=\left [ \frac{10x+4}{x+1} \right ]-\left [ \frac{7x+8}{x+2} \right ]$

$\Leftrightarrow \;\;\frac{6}{x+1}-\frac{6}{x+2}=3-\left [ \frac{10x+4}{x+1} \right ]+\left [ \frac{7x+8}{x+2} \right ]$

$\Leftrightarrow \;\;\frac{6}{(x+1)(x+2)}=3-\left [ \frac{10x+4}{x+1} \right ]+\left [ \frac{7x+8}{x+2} \right ]. \tag{1}$

In the relation (1) the right side is an integer. So we will have solutions only if 

$$\frac{6}{(x+1)(x+2)} \in \mathbb{Z}. \tag{2}$$

The integers $x$ that satisfy relation (2) are determined from the condition  

$(x_+1)(x+2)\in \{\pm 1,\;\pm 2,\;\pm 3,\;\pm 6\}$ - the integer divisors of the number 6.

 Moreover, since $(x+1)(x+2)$ is an even number, we only have to consider the cases

$(x+1)(x+2)=\pm 2$ and $(x+1)(x+2)=\pm 6.$

 The first equation provides the integer solutions $0,\;-3$ and the second - $1,\;-4$.

     All the found values check the equation:

$\left \{\frac{10 \cdot 0+4}{0+1}\right \}=\{4\}=0;\;  \left \{\frac{7\cdot 0+8}{0+2} \right \}=\left \{\frac{8}{2}\right \}=\{4\}=0$

 $\left \{ \frac{10\cdot (-3)+4}{-3+1} \right \}=\left \{ \frac{-26}{-2} \right \}=\{ 13 \}=0;\; \left \{ \frac{7 \cdot(-3)+8}{-3+2} \right \}=\{13 \}=0$

$\left \{ \frac{10\cdot 1 +4}{1+1} \right \}=\left \{ \frac{14}{2} \right \}=\{7\}=0;\;\left \{ \frac{7 \cdot 1+8}{1+2} \right \}=\left \{ \frac{15}{3} \right \}=\{5\}=0$

$\left \{ \frac{10 \cdot (-4)+4}{-4+1} \right \}=\left \{ \frac{-36}{-3} \right \}=\{12\}=0;\;\left \{ \frac{7\cdot (-4)+8}{-4+2} \right \}=\left \{ \frac{-20}{-2} \right \}=\{10 \}=0.$

$\blacksquare$

 

 

 

 

 

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