[KOUSHIK SOM]

we will get back to you within 24 hours

[KOUSHIK SOM]

Let D, E, F be the midpoints of BC, CA, AB respectively.
Area of $\frac{[DEF]}{[ABC]}$=$\frac{1}{4}$

we know that any median is divided at the centroid $2:1$
Now  $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G1F$
Similarly,$GG_2 = 2G2D$ and$GG_3 = 2G3E$
Thus, From  homothetic transformation  $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$
Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$
Therefore we say that $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]} $= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$

So $p+q$=9+1=10

[KOUSHIK SOM]

Join A,D .Note that $AD \perp BC $.Also let $AF \cap BE $ be ${G}$ and $H$ be the mid point of the segment  $EC$.It suffices to prove that A,B,D,G are concyclic(because that implies $\angle ADB=\angle AGB$.This is equivalent to proving that $\angle FAD=\angle EBD$.

Join D,H .Note that $\triangle ADE |||\triangle CDE$ and the sides $DE $ and $EC$ correspond. As AF and DH are medians DE and EC respectively,they correspond too.Hence $\triangle DAF ||| \triangle DCH$

Hence $\angle FAD =\angle HDC$ As $D,H$ are mid points of $BC,CE$ respectively.we have $DH ||BE$.Hence $\angle HDC =\angle EBD$.Hence $\angle FAD=\angle EBD$

[KOUSHIK SOM]

 

Let A BE THE POINT (0,0) and AH the positive X axis .If C is $(x_0,y_0)$ then B has to be $(x_0,-y_0)$ and H,$(x_0,0)$

Let E=$(e_1,e_2)$,E must satisfy the equation $(\frac{e_1}{x_0}=\frac{e_2}{y_0})$ and $(\frac{e_2 -0}{e_1 -x_0})(\frac{y_0}{x_0})$=-1

i.e$ (e_1y_0 = e_2 x_0)$ and $e_2y_0 = {x_2}^2 x_0 e_1$

From these we get $(e_1 x_0)y_0 = e_2{x_0}^2$

$\Rightarrow y_0({x_0}^2- e_2 y_0)=e_2 {x_0}^2$

$\Rightarrow e_2 = \frac{y_0{x_0}^2}{{x_0}^2+{y_0}^2}$

$\Rightarrow e_1 = \frac{e_2 x_0}{y_0}=\frac{{x_0}^3}{{x_0}^2+{y_0}^2}$

Hence, O=$(\frac{1}{2} (\frac{{x_0}^3}{{x_0}^2+{y_0}^2} +x_0)$,$\frac{1}{2} \frac{y_0{x_0}^2}{{x_0}^2+{y_0}^2})$

To prove that AO perpendicular BE we must check $(\frac {y_0 {x_0}^2}{2{x_0}^2+x_0{y_0}^2})(\frac {\frac {y_0 {x_0}^2}{{x_0}^2+{y_0}^2} +y_0}{\frac {{x_0}^3}{{x_0}^2+{y_0}^2 }-x_0})=-1$

L.H.S,

$(\frac{y_0 x_0}{2{x_0}^2+{y_0}^2})(\frac{ y_0{x_0}^2+y_0 {x_0}^2+{y_0}^3}{{x_0}^3-{x_0}^3-x_0{y_0}^2})$

=$(\frac {y_0 x_0}{2{x_0}^2 +{y_0}^2})(\frac{2{x_0}^2 +{y_0}^2}{-x_0 y_0})=-1$

[KOUSHIK SOM]

Observe that $\angle A F E=\angle A E F=90^{\circ}-A / 2$ and $\angle F D E=\angle A E F=90^{\circ}-A / 2 .$ Again
$\angle E I_{1} F=90^{\circ}+A / 2 .$ Thus
$$
\angle E I_{1} F+\angle F D E=180^{\circ}
$$
Hence $I_{1}$ lies on the incircle. Also
$$
\angle I_{1} F E=(1 / 2) \angle A F E=(1 / 2) \angle A E F=\angle I_{1} E F
$$
Thus $I_{1} E=I_{1} F .$ But then they are equal chords of a circle and so they must subtend equal angles at the circumference. Therefore $\angle I_{1} D F=\angle I_{1} D E$ and so $I_{1} D$ is the internal bisector
of $\angle F D E .$ Similarly we can show that $I_{2} E$ and $I_{3} F$ are internal bisectors of $\angle D E F$ and $\angle D F E$ respectively. Thus the three lines $I_{1} D, I_{2} E, I_{3} F$ are concurrent at the incentre of triangle $D E F$

[Author]

Posts