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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Problem E:2718, not from AMERICAN MATHEMATICAL MONTHLY

           It is from the Romanian magazine "Gazeta Matematica"

 

Proposed for 7th grade on page 224. The statement is (thanks to Miss Google for the translation):

           "Let the polynomial be given $P(x,y)=by(x^2+a^2)-ax(y^2+b^2)+xy.$ Show that:

$$P(b,a)=P(-\frac{a}{4},-4b)=P(\frac{b}{a^2-b^2},\frac{a}{a^2-b^2})=P(-\frac{5ab+1}{5b},\frac{4-5ab}{5a})."$$

 

Solution CiP

          It's easy to calculate $P(b,a)=b \cdot a(a^2+b^2)-a \cdot b(a^2+b^2)+b \cdot a=a \cdot b.$

           To avoid laborious calculations, we will write the polynomial in a different way.

$$P(x,y)=\underline{bx^2y}+\underline{\underline{a^2by}}\; \underline{-axy^2}\;\underline{\underline{-ab^2x}}+xy,$$

now we group the identically-underlined terms,

$$P(x,y)=xy(bx-ay)+ab(ay-bx)+xy,$$ 

$$P(x,y)=(xy-ab)\cdot (bx-ay)+xy. \tag{1}$$ 

      For the couple  $(x,y)=(-\frac{a}{4},-4b)$ we see that $xy=ab$ so the first paranthesis in the formula $(1)$ is equal to zero, finally obtaining the result $P(x,y)=ab.$

      For the couple $(x,y)=(\frac{b}{a^2-b^2}, \frac{a}{a^2-b^2})$ we see that

$$bx-ay=\frac{b^2}{a^2-b^2}-\frac{a^2}{a^2-b^2}=-1.$$

 So for these two values of $x$ and $y$ we have from $(1)$ that

$$P(x,y)=(xy-ab)(-1)+xy=ab.$$
      The same goes for the couple $(x,y)=(-\frac{5ab+1}{5b},\frac{4-5ab}{5a})$; here again

 $$bx-ay=-\frac{5ab+1}{5}-\frac{4-5ab}{5}=\frac{-5ab-1-4+5ab}{5}=-1.$$

     All values we calculated are equal to $ab$.

$\blacksquare$

G M B #5 - 1967 (contine articolul ARIA TRIUNGHIULUI IN FUNCTIE DE MEDIANE - pag. 205-206)

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          Articolul de la pp. 205-206 l-am prezentat in anul 1976 la Cercul de Matematica  al scolii (Lic Energetic  Timisoara), sub conducerea prof. Vasile BIVOLARU....

       Am "generalizat"  Aplicatia prezentata la sfarsit, aratand ca:

          Daca medianele unui triunghi verifica relatia $m_a^2 + m_b^2 =m_c^2$ atunci aria triunghiului este egala cu

$A_{\Delta ABC}=\frac{2}{3}\cdot m_a \cdot m_b$.

     Nu mai stiu cum am demonstrat; ar merge cam asa:

     Avem formula generala

$$A_{\Delta ABC}=\frac{1}{3}\cdot \sqrt{2 \cdot \sum m_a^2 \cdot m_b^2 - \sum m_a^4}.$$

Calculam expresia de sub radical, inlocuind $m_c^2$ cu $m_a^2+m_b^2$ si obtinem

$2 \cdot (m_a^2 \cdot m_b^2+(m_a^2+m_b^2) \cdot m_c^2))-m_a^4-m_b^4-(m_a^2+m_b^2)^2=$

$=2(m_a^2 m_b^2 +(m_a^2 +m_b^2)^2)-m_a^4-m_b^4-(m_a^2+m_b^2)^2=2m_a^2 m_b^2+(m_a^2+m_b^2)^2-m_a^4-m_b^4=4m_a^2 m_b^2$

Deci $A_{\Delta ABC}=\frac{1}{3}\cdot \sqrt{4m_a^2m_b^2}=\frac{2}{3}m_a m_b$.

$\blacksquare$

 

G M B #2-1965 Румънско математическо списание

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