Suppose there is a pentagon, 4 of whose vertices are directly above the midpoint of the side opposite them...

Suppose there is a pentagon, 4 of whose vertices are directly above the midpoint of the side opposite them. How can one prove that the same is true for the other vertex?

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homework

it's not homework.
I'm curious about how the problem works and would like to discuss it.

Fucking add up the angles or some shit. Is it a regular pentagon? If so, should be easy

lol
This is the whole problem.
How can I prove that if 4 of the vertices are over the opposite midpoints, then the other does too, and whole pentagon is regular.
I don't know that it's regular.
I basically know that 4 sides are the same length though. maybe?

Actually. Let me change the problem slightly.
For 4 of the vertices, the lines from the vertex to the midpoint of the opposide side pass through the point P, how can I prove that the line from the last vertex to the midpoint of the opposite side also passes through P?

You know, all pentagons have the same sum of interior angles. If you know 4 of the angles, you know the 5th

I don't know 4 of the angles.

Aight motherfucker, I'm getting the whiteboard marker out.

please do user

Well I tried

I think I proved that angle dpe is 72, and if the angle is the same as all the others, it should have the same properties. Not a very rigorous proof

You don't know that the other angles are 72.
Where are you getting that from?

Ok fuck, I can prove the angles are the same by 3 congruent sides of the triangles, but I can't figure out how to prove line pa=pc.

What do you have so far, we can try to work together

Pic

Here is my entire whiteboard, closeups of each part are in the thread

You're still pulling 72 degrees out of thin air.

Yes, until the triangles are proven to be similar. Once I prove they're similar, their angles will be proven to be the same

Well PC = PB.

How?

Replying to myself. I can see how you would prove pa=pb with law of sines, but I don't see how you can prove pc=pb

you're right, PA = PB, but we can't yet prove that PC=PB

Aight, new approach. Is there a way to prove the triangles are similar using Side Angle Side?

If you can prove that PA1 is perpendicular to CD then the problem becomes simpler, I think

>Suppose there is a pentagon, 4 of whose vertices are directly above the midpoint of the side opposite them.
Looks like it's given information

I did a sketch in GeoGebra.

If four vertices are above the midpoint of their opposite side in a pentagon, it implies regularity and each angle is 108 degrees. It MUST follow that the same is true for the remaining vertice (E) since 108*5 - 108*4 = 108

Ignore that. (I'm OP. The real problem is )

Alright

see

Well, obviously it's true. the question is how the hell do we prove it?

this problem is impossible

Another way we can prove the pentagon is regular, which would prove is if we can prove the length of a diagonal is phi*sidelength
en.wikipedia.org/wiki/Golden_ratio
probably not very feasible though

This is the problem

Still the same problem and still trivial solution. Look at the image: I drew the midpoints mCE and mDE. If you were to draw all of CE and DE, at their intersection the 5th vertex (E) would be at (1.5388..., 0.5) , which obviously is intersected by the line y = 0.5, which has already been shown to be the y value of P and the midpoint AB.

is there a way to prove that P is the midpoint of the line between A and the midpoint of CD

...

a lot of assumptions were made here

Where did you get those 90 degree angles from?

The proof is trivial if it's a regular pentagon but it's not true in general

Who says it's not true in general?

Me. Because I can easily pull the fifth point to any arbitrary point.

which breaks the conditions listed for the 4 other points
great going

No, just take op's original figure and move a single point. Moving that point does not change the relative location of the other 4.

It's not that hard to see.

it changes the length of the line segments the point is connected to
which changes the midpoint
which changes the location of P

Only sometimes.

That line of thought is close though.

?

You can use that argument formalized to prove it's true. However, there are an infinite number of configurations in which the fifth bisects the opposite line segment into equal parts.

How?

Bump