About the PROBLEM #2 ("икки") from International Olympiad "TUYAMAADA" 2021 - Junior League

           See link, obtained from www.molympiad.net.

          We will formulate the following related problem:

 

"The bisector of angle $B$ of a parallelogram $ABCD$ meets itd diagonal $AC$ at $E$, and the external bisector of angle $B$ meets line $AD$ at $F$.

Prove that the right segments $CE$ and $FM$ are parallel and congruent  if and only if $BC=2\cdot AB$."

 

Proof CiP of Original Problem 

          First of all, notice that from the equal angles:

$\angle B'BF\equiv \angle BFA$, and $\angle CBE \equiv \angle AGC$ (as angles formed by the parallel lines $BC$ and $AD$ with the transversals $BF$ and $BG$ respectively). And because the bisectors form equal angles, we have $\angle B'BF \equiv \angle FAB$, $\angle CBG \equiv \angle ABG$, so we get some isosceles triangles, from where it come

$$ AB=AF=AG\;.$$

          We will solve the original problem by the vector method.

 Denote with $L=BC,\;l=AB$. The bisector theorem $BE$ in the triangle $ABC$

$$\frac{AE}{EC}=\frac{AB}{BC}=\frac{l}{L}$$ 

allows us to find out  

$$\overrightarrow{BE}=\frac{L\cdot \overrightarrow{BA}+l\cdot \overrightarrow{BC}}{L+l}.\tag{1}$$

(Indeed, $L\cdot \overrightarrow{AE}=l\cdot \overrightarrow{EC}$, or $L(\overrightarrow{BE}-\overrightarrow{BA})=l(\overrightarrow{BC}-\overrightarrow{BE})$...)

We also obtain from the bisector theorem $AE=\frac{l}{L+l}\cdot AC$, so $\overrightarrow{AE}=\frac{l}{L+l}\overrightarrow {AC}=\frac{l}{L+l}(\overrightarrow{BC}-\overrightarrow{BA})$, hence

$$\overrightarrow {AE}=\frac{l}{L+l}\overrightarrow{BC}-\frac{l}{L+l}\overrightarrow{BA}.$$

      Further, because $AF=AB$, we have $\overrightarrow{AF}=-\frac{l}{L}\cdot \overrightarrow{AD}=-\frac{l}{L}\overrightarrow{BC}$, and now we calculate $\overrightarrow{EF}=\overrightarrow{AF}-\overrightarrow{AE}=-\frac{l}{L}\overrightarrow{BC}-(\frac{l}{L+l}\overrightarrow{BC}-\frac{l}{L+l}\overrightarrow{BA}),$ so 

$$\overrightarrow{EF}=-\frac{2Ll+l^2}{L(L+l)}\overrightarrow{BC}+\frac{l}{L+l}\overrightarrow{BA}.\tag{2}$$

 Also $\overrightarrow{CM}=\overrightarrow{BM}-\overrightarrow{BC}=\frac{1}{2}\overrightarrow{BE}-\overrightarrow{BC}\;\overset{(1)}{=}\frac{L}{2(L+l}\overrightarrow{BA}+\frac{l}{2(L+l)}\overrightarrow{BC}-\overrightarrow{BC}$, getting

$$\overrightarrow{CM}=-\frac{2L+l}{2(L+l)}\overrightarrow{BC}+\frac{L}{2(L+l)}\overrightarrow{BA}.\tag{3}$$

           From relations $(2)$ and $(3)$ we see that

$\overrightarrow{EF}=\frac{l}{L}[-\frac{2L+l}{L+l}\overrightarrow{BC}=\frac{l}{L+l}\overrightarrow{BA}]=\frac{l}{L}\cdot 2\overrightarrow{CM}$, so

$$\overrightarrow{EF}=\frac{2l}{L}\overrightarrow{CM}.\tag{4}$$

Relation $(4)$ states that  $EF\;\parallel\;CM$.

$\blacksquare$

     REMARK CiP

      Also from the formula $(4)$ we deduce that 

$EF \;\overset{\parallel}{=}\;CM\;\;\Leftrightarrow\;\;2l=L\;\;\Leftrightarrow\;\;EFMC-parallelogram\;\;\Leftrightarrow\;\;CE\;\overset{\parallel}{=}\;MF$

and we get our reformulation of the problem.

$\blacksquare\;\blacksquare$

 

 

 

 

 

 

 

 

 

Крайні значення раціонального дробу

           We are looking for the extreme values of some rational fractions of form 

$$\frac{Ax^2+2Bx+C}{ ax^2+2bx+c}.$$

     We find in the book Daniel SITARU - Fenomen Algebric

     PROBLEM #1, page 27

In translation (thanks to Miss Google for this):

     " Find the minimum of the expression 

$$E=\frac{4ab-11a^2-14b^2}{3(a^2+b^2)}\; ;\;a,b \in \mathbb{R^*}."$$ 

 

ANSWER CiP

           We will find out more, both the minimum and the maximum of the expression.

$$-5=E(a,-2a)\leqslant  E(a,b) \leqslant E(2b,b)=-\frac{10}{3}\;\;,a,b \in \mathbb{R^*}.$$ 

 

Solution CiP

      The  number $\lambda$ is a value of the fraction $E$ if, and only if

" there are two numbers $a$ and $b$ such that $\lambda =E(a,b)"

$$\Leftrightarrow \; \exists \;a,b \in \mathbb{R^*}\;such\; that\;\lambda =\frac{4ab-11a^2-14b^2}{3(a^2+b^2)}$$

 $$\Leftrightarrow\;\exists\;a,b \in \mathbb{R^*}\;such\;that\;4ab-11a^2-14b^2=3\lambda a^2+3 \lambda b^2$$

 $$\Leftrightarrow\;\exists\;a,b \in \mathbb{R^*}\;such\;that\;(3\lambda+11)a^2-4ab+(3\lambda+14)b^2=0$$

$$\Leftrightarrow\; t=\frac{a}{b}\;is\;a\;root\;of\;the\;equation\;(3\lambda+11)t^2-4t+(3\lambda+14)=0 \tag{1}$$

$\Leftrightarrow\;\Delta _t\;\geqslant 0,\tag{2}$

where $\Delta _t$ is the half-discriminant to the equation (1),

$\Delta _t=4-(3\lambda+11)(3\lambda+14).$

But $(2)\;\Leftrightarrow\;9\lambda ^2+75 \lambda+150 \leqslant 0\;\Leftrightarrow\;3\lambda^2+25\lambda+50\leqslant 0\;\Leftrightarrow\;\lambda \in [-5;-\frac{10}{3}]$.

Here, the ends of the range $[-5;-\frac{10}{3}]$ are the roots of the equation $3\lambda^2+25\lambda+50=0.$

           Result that we have  $\lambda _{min}=-5,\;\lambda _{max}=-\frac{10}{3}.$

Further, 

$E(a,b)=-5\;\Leftrightarrow\;-4a^2-4ab-b^2=0\;\Leftrightarrow\;-(2a+b)^2=0\;\Leftrightarrow\;b=-2a$,

 and $E(a,b)=-\frac{10}{3}\;\Leftrightarrow\;a^2-4ab+4b^2=0\;\Leftrightarrow\;(a-2b)^2=0\;\Leftrightarrow\;a=2b.$

 We got the answer.

$\blacksquare$

 

Solution #2

(the authors' solution, only for the minimum, completed by CiP for the maximum)

 See page 72.

           Completed by CiP: $E=\frac{-10a^2-10b^2-a^2+4ab-4b^2}{3(a^2+b^2)}=-\frac{10}{3}-\frac{(a-2b)^2}{3(a^2+b^2)}\leqslant -\frac{10}{3}$,  $max\;E=-\frac{10}{3}$ for $a-2b=0.$

$\blacksquare$