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duminică, 30 ianuarie 2022

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

miercuri, 26 ianuarie 2022

GAZETA MATEMATICĂ Seria B N0 3/2021

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 Vezi ERATA



 

SUPLIMENTUL cu EXERCIȚII al GMB N0 3/2021

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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

 

 

 

 

 

joi, 20 ianuarie 2022

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


miercuri, 19 ianuarie 2022

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

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Vezi ERATA (necorectat inca)



          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