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**NeoXtremeX****Member**- Registered: 2005-08-09
- Posts: 1

The sum of the lengths of any two sides is greater than the lenght of the third side. in triangle ABC, BC = 4 and AC = 8 - AB. Write an inequality for AB.

Help?

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,566

The Triangle inequality is

AB + BC > AC

AC + BC > AB

AB + AC > BC

You have given BC=4 and AC = 8 - AB

Therefore, the required inequality is

AB < BC + AC

AB < BC + 8 - AB

2 AB < BC + 8

AB < (BC + 8)/2

Character is who you are when no one is looking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Ganesh is right, but as BC=4, it can be continued further:

AB < (4+8)/2

AB < 12/2

AB < 6

Why did the vector cross the road?

It wanted to be normal.

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

How do we prove that the sum of the lengths of any two sides is greater than the lenght of the third side in the first place?

Just interested.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Try disproving it and fail. Just try drawing a triangle with lengths of 2, 3 and 6. The 2 and 3 are too short and too far apart to be able to join up.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,566

Lets take the length of sides Mathsy has given.

The Hero's formula for area of a triangle is

Area = √(s(s-a)(s-b)(s-c)

where a, b, c are the sides and s = (a+b+c)/2

Therefore, Area = √[(11/2)(11/2 - 2)(11/2 - 3) (11/2 - 6)]

11/2 - 6 is a negative number,

So the Area of the triangle is an imaginary number

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

Ooohh ... so you CAN have one side of a triangle longer than the other two, it just somehow slips into imaginary space ...

Or not ...

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

here is a proof for right angled triangle:

by pythagoras theorem,

a²+b²=c²

c=√(a²+b²) ----(1)

(a+b)²=a²+2ab+b²

a+b=√(a²+2ab+b²) ----(2)

comparing (1) and (2),

one can see that

a+b>c

*Last edited by wcy (2005-08-10 00:07:00)*

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

The sum of the lengths of any two sides is greater than the lenght of the third side.

For all triangles,

by cosine rule,

c²=a²+b²-2abcosC

(a+b)²=a²+2ab+b²

a+b=√(a²+2ab+b²)

c=√(a²+b²-2abcosC)

now, a+b is definitely larger than c, as the maximum possible value of -2abcosC is less than 2ab, as angle C is smaller than 180 but larger than 0.

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