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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 $P_3$.

User Bsabiston
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1 Answer

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1.
Given any 2 points P(a, b) and Q(c, d), the midpoint
M_P_Q of PQ is given by
( (a+c)/(2), (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= (x_1+x_2)/(2)


z_2= (x_2+x_3)/(2)


z_3= (x_3+x_4)/(2)
.
.

z_3_3= (x_3_3+x_1)/(2)

and



w_1= (z_1+z_2)/(2)


w_2= (z_2+z_3)/(2)


w_3= (z_3+z_4)/(2)
.
.

w_3_3= (z_3_3+z_1)/(2)

5.


w_1+ w_2+ w_3....+w_3_3=(z_1+z_2)/(2)+(z_2+z_3)/(2)+...(z_3_3+z_1)/(2)= (2(z_1+z_2+ z_3....+z_3_3))/(2)


=(z_1+z_2+ z_3....+z_3_3)=(x_1+x_2)/(2)+(x_2+x_3)/(2)+...(x_3_3+x_1)/(2)


=(2(x_1+x_2+ x_3....+x_3_3))/(2)=(x_1+x_2+ x_3....+x_3_3)=99


Answer: 99

User Ian Nato
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8.3k points