# Lets help ViperMagic do some Geometry!

#### Legendary

##### Diamond Member
Is that a parallelogram or just a quadrilateral?

#### GoodToGo

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

#### minendo

##### Elite Member
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

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

#### ViperMagic

##### Platinum Member
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

##### Banned
edit: misread the problem

#### Yomicron

##### Golden Member
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

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

#### GoodToGo

##### Diamond Member
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

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

OM+MN>ON

PO=OM

PO+MN>ON

proved

#### Yomicron

##### Golden Member
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

##### Platinum Member
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

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

#### Bobomatic

##### Senior member
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

##### Platinum Member
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