Question

A 33-gon P1 is drawn in the Cartesian plane. The sum of the x-coordinates of the 33 vertices equals 99. The midpoints of the sides of P1 form a second 33-gon, P2. Finally, the midpoints of the sides of P2 form a third 33-gon, P3. Find the sum of the x-coordinates of the vertices of P3.

Answer 1

Take an isosceles triangle with apex angle 360/33

the apothem is cos(theta/2) shorter than the edge

rotating the 33gon twice means you scale it by

cos (theta/2)^2

all the vretices will move towards the centre by this amount.

I THINK (but not 100% sure) that the sum of x vertices will therfora also scale by cos (theta/2)^2

Hmmmm - not sure about that - it comes out at 98

(but this is close to a circle so it could be such a small scale factor...)
to be continued..

Answer 2

IF this were an even number of sides you could argue that for ach movement of a vertex in positive direction there would be an opposite vertex movement negative, so the sum would remain the same.

Not sure this is true for an odd number of vertices..

Answer 3

Nood - the question is from Algebra ... just to use midpoint, slope intercept form, standard linear equation..the topic just studied.. so it is to be much simpler solution than this.

Answer 4

OK - I have confirmed by drawing that the scale factor IS cos^2 (theta/2)

(=.991 in this case)

@danyboy9169
Not sure what you mean from your last post
I don't see a linear equation there - but maybe I haven't seen th simple way forward...

Answer 5

@Nood so the Ans is .991

Answer 6

NO - definitely not - that is not what I said.

This does not initially seem trivial and I was merely opening a discussion (you can see that I said I was not sure in my first post)

Answer 7

So with further thought how about this for an argument:

The sum of x vertices = 99 so th AVERAGE x value is 3

IF we translate the polygon -3 in x then the average becomes 0

THEN carry out the 2 rotations

Question??? IS the average STILL zero? (Not 100% sure myself)

THEN translate +3 in x give and average of 3 so a sum of 99

i.e. the sum does not change (as suggested previously for an even polygon)

Answer 8

ALL this assumes we are talking about a REGULAR polygon which is not stated in the question.
IF irregular then all my discussion is wasted.

Answer 9

Suppose we translate so that it's centre is at the origin lets say dx

So the sum is now 99-33dx

now convert all vertices to polar cords

x= r cos theta (where r is the 'radius' of the polygon to its vertex)

so sum of x = r (sum cos theta)

After the rotations the r = .991 r

take it from there... for now

Answer 10

so the sum is .991 times the original

when we translate back to original then the answer is sum = 98.1

So- I have answered two options - with different results.

Guess there needs another set of brain cells on this one.....

Answer 11

Let the x-coordinates of the vertices of \[P_1\] be \[x_1,x_2,\ldots,x_{33}\]
Then, by the midpoint formula, the x-coordinates of the vertices of \[P_2 are \frac{x_1+x_2}2,\frac{x_2+x_3}2,\ldots,\frac{x_{33}+x_1}2 \\\]

Answer 12

The sum of these equals \[\frac{2x_1+2x_2+\cdots +2x_{33}}2=x_1+x_2+\cdots+x_{33}.\]
Similarly, the sum of the x-coordinates of the vertices of \[P_3\] equals the sum of the x-coordinates of the vertices of \[P_2\]. Thus the desired answer is 99

Answer 13

@londoncat