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/sci/ - Science & Math

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File: 47 KB, 503x509, 503px-Regular_polygon_5_annotated.svg.png [View same] [iqdb] [saucenao] [google]
8492916 No.8492916 [Reply] [Original]

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?

>> No.8492926

homework >>>/hm/

>> No.8492934

>>8492926
it's not homework.
I'm curious about how the problem works and would like to discuss it.

>> No.8492939

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

>> No.8492944

>>8492939
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?

>> No.8492978

>>8492916
>>8492944
>>8492939
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?

>> No.8492995

>>8492978
You know, all pentagons have the same sum of interior angles. If you know 4 of the angles, you know the 5th

>> No.8492997

>>8492995
I don't know 4 of the angles.

>> No.8493004

>>8492997
Aight motherfucker, I'm getting the whiteboard marker out.

>> No.8493013

>>8493004
please do anon

>> No.8493050
File: 52 KB, 800x1165, received_173954369735560.jpg [View same] [iqdb] [saucenao] [google]
8493050

Well I tried

>> No.8493055
File: 37 KB, 800x541, received_173954903068840.jpg [View same] [iqdb] [saucenao] [google]
8493055

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

>> No.8493060

>>8493050
>>8493055
You don't know that the other angles are 72.
Where are you getting that from?

>> No.8493077

>>8493060
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.

>> No.8493081

>>8493077
What do you have so far, we can try to work together

>> No.8493083
File: 52 KB, 800x1165, received_173958646401799.jpg [View same] [iqdb] [saucenao] [google]
8493083

>>8493077
Pic

>> No.8493086
File: 47 KB, 1165x800, received_173959153068415.jpg [View same] [iqdb] [saucenao] [google]
8493086

>>8493081
Here is my entire whiteboard, closeups of each part are in the thread

>> No.8493090

>>8493083
>>8493086
You're still pulling 72 degrees out of thin air.

>> No.8493094

>>8493090
Yes, until the triangles are proven to be similar. Once I prove they're similar, their angles will be proven to be the same

>> No.8493103

>>8493094
Well PC = PB.

>> No.8493109

>>8493103
How?

>> No.8493114

>>8493109
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

>> No.8493118

>>8493103
>>8493109
>>8493114
you're right, PA = PB, but we can't yet prove that PC=PB

>> No.8493169

>>8493118
Aight, new approach. Is there a way to prove the triangles are similar using Side Angle Side?

>> No.8493172

>>8493169
If you can prove that PA1 is perpendicular to CD then the problem becomes simpler, I think

>> No.8493178

>>8493172
>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

>> No.8493182
File: 11 KB, 413x451, pentagon.png [View same] [iqdb] [saucenao] [google]
8493182

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

>> No.8493183

>>8493178
Ignore that. (I'm OP. The real problem is >>8492978 )

>> No.8493189

>>8493183
Alright

>> No.8493190

>>8493182
see
>>8493183

>> No.8493194

>>8493182
Well, obviously it's true. the question is how the hell do we prove it?

>> No.8493200

this problem is impossible

>> No.8493201

Another way we can prove the pentagon is regular, which would prove >>8492978 is if we can prove the length of a diagonal is phi*sidelength
https://en.wikipedia.org/wiki/Golden_ratio
probably not very feasible though

>> No.8493212
File: 20 KB, 594x87, cZ3CpGp.png [View same] [iqdb] [saucenao] [google]
8493212

This is the problem

>> No.8493217
File: 17 KB, 467x397, pentagon2.png [View same] [iqdb] [saucenao] [google]
8493217

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.

>> No.8493222

>>8493217
is there a way to prove that P is the midpoint of the line between A and the midpoint of CD

>> No.8493265
File: 77 KB, 700x624, pentagon3.png [View same] [iqdb] [saucenao] [google]
8493265

>>8493212

>> No.8493271

>>8493265
a lot of assumptions were made here

>> No.8493341

>>8493265
>>8493271
Where did you get those 90 degree angles from?

>> No.8493393

>>8492916
>>8492944

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

>> No.8493394

>>8493393
Who says it's not true in general?

>> No.8493397

>>8493394
Me. Because I can easily pull the fifth point to any arbitrary point.

>> No.8493402

>>8493397
which breaks the conditions listed for the 4 other points
great going

>> No.8493406

>>8493402
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.

>> No.8493409

>>8493406
it changes the length of the line segments the point is connected to
which changes the midpoint
which changes the location of P

>> No.8493417

>>8493409
Only sometimes.

That line of thought is close though.

>> No.8493424

>>8493417
?

>> No.8493556

>>8493424
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.

>> No.8493784

>>8493556
How?

>> No.8493796

Bump