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## #1 2005-12-10 14:45:59

cevilian
Member
Registered: 2005-12-10
Posts: 3

### Triangle related

Hi guys,
I am new to the forum and this is my first post. I need some help with the following problem:

*********************************************************************************
ABC is a triangle, with the length of sides:
AB = h
BC = i
CA = j

If ((h+j-i) / (h+j)) > (1/2), then the angle between the AB and AC is:
1) Less than 60 degrees
(Also: Smaller than the other two angles)
*********************************************************************************
Is the above right? If yes, How do we prove it?

Similarly, for ((h+j-i) / (h+j)) > (1/3, or 1/4 or any other value x such that 0 < x < 1), can we establish an upper and a lower limit on the angle between the AB and AC?

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## #2 2005-12-10 17:46:14

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: Triangle related

Welcome to the forum!
The folks in Europe will be waking up pretty soon, so they'll probably
give you some insight into this.
I just did a few examples and got no where, though, sorry.

igloo myrtilles fourmis

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## #3 2005-12-11 11:11:25

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

### Re: Triangle related

((h+j-i)) / (h+j)) > 1/2

Inequalities are exactly like equations, except that you can't multiply or divide by negative numbers. However, all of these letters mean lengths, so that is irrelevent in this case and we can treat it like an equation.

h+j-i > 0.5(h+j)
0.5(h+j)-i > 0
0.5(h+j) > i

The angle can be worked out by using the cosine equation:

i² = h² + j² - 2hj cos I, where I is angle BAC.

Rearranging: I = arccos ((h² + j² - i²) / (2hj))

I is at its biggest possible when the expression inside the arccos is at its smallest. This happens when h = j and i is at its maximum, so i = 0.5(h+j) = h, because h = j.

I_max = arccos ((h²) / (2h²) = arccos (1/2) = 60°

And, because we were told that i < (instead of ≤) 0.5(h+j), that means that I < 60°.

The other part is to follow.

Why did the vector cross the road?
It wanted to be normal.

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## #4 2005-12-11 11:26:55

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

### Re: Triangle related

The extension starts in the same way, but then we get to here:

h+j-i > x(h+j)
(1-x)(h+j)-i > 0
(1-x)(h+j) > i

I = arccos ((h² + j² - i²) / (2hj)) still applies, as does the rule that h = j and i is at its maximum to find I_max.

So this time, i would be 2h(1-x).

I_max = arccos (-2x² + 4x - 1), after simplification.

And, again, because it is < instead of ≤, angle I is  < arccos (-2x² + 4x - 1)

The lower limit of the angle is independent of x because x just restricts the maximum length of i.
The angle is at its lowest when i is at its lowest, which is when it is almost 0. It can't be 0, because then ABC would be a line instead of a triangle, but it can be very close and in this case, angle I ≈ 0.

And John, I'm amazed that you couldn't do this considering what you did with the curvy calculus in the other topic. I barely even understood that!

Why did the vector cross the road?
It wanted to be normal.

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## #5 2005-12-11 13:43:59

cevilian
Member
Registered: 2005-12-10
Posts: 3

### Re: Triangle related

Mathsy,
Many thanks indeed for your response. I think, I kinda lost touch with my math fundamentals (makes me sad ). It has been a while since I worked on math problems, and I encountered this problem while working on a location based service issue. I did not quite follow a small part of your response, I greatly appreciate if you can help me understand.

"I is at its biggest possible when the expression inside the arccos is at its smallest. This happens when h = j and i is at its maximum, so i = 0.5(h+j) = h, because h = j."

I looked at the same expression, agree with you that i should be at its maximum, but dont understand the following part:
---------------------------------------------------------------------------------------------------------
"This happens when h = j"

looking at ((h² + j² - i²) / (2hj)), how could you conclude that? I would say h² + j² should be small and i should be as big as possible, but h = j ?
---------------------------------------------------------------------------------------------------------

Also, here is an example triangle (H=30, I = 59, J = 91) with sides (h = 10.00, i = 17.14, j = 20.00) which satisfies the condition that I < 60, but ((h+j-i)) / (h+j)) < 1/2

Thanks again

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## #6 2005-12-12 05:16:46

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

### Re: Triangle related

It's just a generally accepted thing that (a² + b²) / ab is smallest when a = b. I'll have a go at proving it.

Let's get rid of i for the purpose of this, as we just want to know when the expression is smallest for h and j.

(h² + j²) / 2hj

It wouldn't matter if we swapped the positions of h and j, because (j² + h²) / 2jh is the same thing, so we can assume that h ≥ j. If it isn't, we can swap the positions of h and j so that it is.

That way, we can write h as (j+k), where k is an unknown constant that is at least 0.

So now the expression is [(j+k)² + j²] / 2(j+k)j

This multiplies out to give (2j² + 2jk + k²) / (2j² + 2jk)

Separating the k² from the rest of the numerator allows us to cancel out:

1 + [k² / (2j² + 2jk)]

The new fraction will obviously be smallest when k = 0.

Therefore, (h² + j²) / 2hj is smallest when h = j.

--------------------------------------------------------------------------

It's perfectly alright that triangles exist where the angle is < 60° and ((h+j-i)) / (h+j)) < 1/2. All that means is that the converse of the original statement is false and it doesn't have any connection with the original proof.

It's like trying to disprove that every even number is a whole number by saying that 3 is whole, but not even.

Why did the vector cross the road?
It wanted to be normal.

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## #7 2005-12-12 05:59:20

cevilian
Member
Registered: 2005-12-10
Posts: 3

### Re: Triangle related

Mathsy,
Thank you very much for the explanation, it really helped!

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## #8 2005-12-12 11:33:08

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: Triangle related

Hi mathsyperson,
Yeah.  I think in retrospect, I just got extremely lucky in the curvy calculus stuff.
And also it occurs to me that maybe I was taught it back in 1985 or 6, but my
memory is so bad, I may have been reinventing something I'd seen before.
My major was Electrical Engineering, but I took a lot of Calculus courses, but
just barely passed somehow.  Now I am starting all over from nearly scratch again.

igloo myrtilles fourmis

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