Lets help ViperMagic do some Geometry!

[ViperMagic]

Link to picutre
HELP!

 

[Legendary]

Is that a parallelogram or just a quadrilateral?

 

[GoodToGo]

Is this a parallelogram? Because if it is, then the solution is real simple.

 

[minendo]

Originally posted by: GoodToGo
Is this a parallelogram? Because if it is, then the solution is real simple.

You must solve the problem using only the given and theorems/postulates.

 

[RbSX]

Can't be a parallelogram or the oppisite sides would be marked as equal. But yeah I think some parts are missing.

 

[ViperMagic]

that is exactly whats in the book. Though this is chapter on triangle inequalitys... (two sides of a triabgle must be greater than the thrid) if that helps

 

[josphII]

edit: misread the problem

 

[Yomicron]

k, I haven't done geometry in a long time, but:

if MN == 0 then ON = OM = PO

so, PO + MN = ON , therefore PO + MN !> ON

so it is disproved by example

 

[fatbaby]

There has to be more given because ON can have infinitely many sizes

 

[GoodToGo]

You cant prove anything without being given some characteristics of this shape. If it is a parrallelogram, then I can tell the solution pretty easily:

PO = MN = OM
OM + MN > ON (sums of two sides of a trianlge is greater than the third side)
But since PO = OM,
PO + MN > ON

 

[SgtStedenko]

Since PO=MO then angle p=angle m(definition of an isosceles triangle).
That's all I can get from it.

 

[josphII]

OM+MN>ON

PO=OM

PO+MN>ON

proved

 

[Yomicron]

Originally posted by: josphII
OM+MN>ON

PO=OM

PO+MN>ON

proved

how do you know OM + MN > ON? MN and OM do not have set lengths

 

[Darien]

Well, you're given a Triangle MNO.

The length of MO is set and is Equal to PO.

|MO| = |PO|

You know that the sum of length of the smaller sides of a triangle is greater than the length of the largest side, and that the sum of the length of any two sides is greater than the third side.

Since you don't know the length of MN and ON, you have 3 situations

|MN| = |ON|, |MN| > |ON|, |MN| < |ON|



if |MN| = |ON|, then |x| + |MN| > |ON|

if |MN| > |ON|, then no matter what |x| you add, |x| + |MN| > |ON|

if |MN| < |ON|, you know that |MO| + |MN| > |NO|. Substitute |MO| for |PO|, then |PO| + |MN| > |NO|



I'll leave it up to you to construct a more formal proof.

 

[tweakmm]

didn't we help you do your geometry homework last weekend?

 

[Bobomatic]

just skip it and move on to the next problem, and if it needs to be done for class, copy the asian kids homework before class.

 

[Darien]

Originally posted by: Bobomatic
just skip it and move on to the next problem, and if it needs to be done for class, copy the asian kids homework before class.

Words of wisdom