Question

A $33$-gon $P_1$ is drawn in the Cartesian plane. The sum of the $x$-coordinates of the $33$ vertices equals $99$. The midpoints of the sides of $P_1$ form a second $33$-gon, $P_2$. Finally, the midpoints of the sides of $P_2$ form a third $33$-gon, $P_3$. Find the sum of the $x$-coordinates of the vertices of $P_3$.

Answer 1

1.
Given any 2 points P(a, b) and Q(c, d), the midpoint $M_P_Q$ of PQ is given by $( \frac{a+c}{2}, \frac{b+d}{2} )$.

2.
Let the x coordinates of the vertices of P_1 be : 

${x_1, x_2, x_3....x_3_3}$

the x coordinates of P_2 be :

${z_1, z_2, z_3....z_3_3}$

and the x coordinates of P_3 be:

${w_1, w_2, w_3....w_3_3}$

3.
we are given that 

$x_1+ x_2+ x_3....+x_3_3=99$

and we want to find the value of $w_1+ w_2+ w_3....+w_3_3$.

4.

According to the midpoint formula:

$z_1= \frac{x_1+x_2}{2}$

$z_2= \frac{x_2+x_3}{2}$

$z_3= \frac{x_3+x_4}{2}$
.
.
$z_3_3= \frac{x_3_3+x_1}{2}$

and 


$w_1= \frac{z_1+z_2}{2}$

$w_2= \frac{z_2+z_3}{2}$

$w_3= \frac{z_3+z_4}{2}$
.
.
$w_3_3= \frac{z_3_3+z_1}{2}$

5.

$w_1+ w_2+ w_3....+w_3_3=\frac{z_1+z_2}{2}+\frac{z_2+z_3}{2}+...\frac{z_3_3+z_1}{2}= \frac{2(z_1+z_2+ z_3....+z_3_3)}{2}$

$=(z_1+z_2+ z_3....+z_3_3)=\frac{x_1+x_2}{2}+\frac{x_2+x_3}{2}+...\frac{x_3_3+x_1}{2}$

$=\frac{2(x_1+x_2+ x_3....+x_3_3)}{2}=(x_1+x_2+ x_3....+x_3_3)=99$


Answer: 99