Question 1207717
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Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.


AB, BC, and AC are the three sides of triangle ABC.


The Triangle Inequality Theorem generates these 3 inequalities
AB + BC > AC
AB + AC > BC
AC + BC > AB
I'll refer to them as inequality (1) through (3).


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Problem 2


We're given that
AC = 2 + AB


Let's solve for AB
AC = 2 + AB
AB = AC - 2


Then apply this to inequality (1)
AB + BC > AC
AC - 2 + 7 > AC
AC + 5 > AC
5 > 0
The last inequality is always true, so the first inequality is always true.
We don't really generate anything interesting here, so let's move on.


Now move to inequality (2)
AB + AC > BC
AC-2 + AC > 7
2AC-2 > 7
2AC > 7+2
2AC > 9
AC > 9/2
AC > 4.5
Inequality (2) is only true when AC is larger than 4.5


Lastly inequality (3)
AC + BC > AB
AC + 7 > AC-2
7 > -2
This is always true.
Like with the first case shown above, nothing interesting is here.



To wrap up problem 2, if triangle ABC has sides BC = 7 and AC = 2+AB, then we conclude that <font color=red>AC > 4.5</font>



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Problem 3


AB + AC = 5 solves to AB = 5 - AC
AC + BC = 4 solves to BC = 4 - AC


Plug them into inequality (1)
AB + BC > AC
5-AC + 4-AC > AC
9-2AC > AC
9 > 3AC
AC < 9/3
AC < 3


Also, plug them into inequality (2)
AB + AC > BC
5-AC + AC > 4-AC
5 > 4-AC
5-4 > -AC
1 > -AC
AC > -1
This will always be true since AC is a positive length.
This subcase isn't useful so we move on.


Lastly inequality (3)
AC + BC > AB
AC + 4-AC > 5-AC
4 > 5-AC
4+AC > 5
AC > 5-4
AC > 1
1 < AC


Combine 1 < AC with AC < 3 to conclude <font color=red>1 < AC < 3</font> is the answer to problem 3.



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Problem 4


BC + AC = 22 solves to BC = 22 - AC


Plug this into the other equation so we get something in terms of AC.
AB + BC = 12
AB + 22-AC = 12
AB - AC = 12-22
AB - AC = -10
AB = AC-10


The key equations we'll be using are 
BC = 22-AC and AB = AC-10
Like with the previous problem, they are useful for substitutions.


Let's revisit inequality (1)
AB + BC > AC
AC-10 + 22-AC > AC
12 > AC
AC < 12


Now focus on inequality (2)
AB + AC > BC
AC-10 + AC > 22-AC
2AC-10 > 22-AC
2AC+AC > 22+10
3AC > 32
AC > 32/3
AC > 10.6667 approximately


And now inequality (3)
AC + BC > AB
AC + 22-AC > AC-10
22 > AC-10
22+10 > AC
32 > AC
AC < 32


Combine or overlap AC < 12 with AC < 32, so we determine overall that AC < 12.
If some length of AC makes AC < 12 true, then it autometically makes AC < 32 true as well (but not always the other way around).


Combine 32/3 < AC with AC < 12 to determine <font color=red>32/3 < AC < 12</font> 
In other words, <font color=red>10.6667 < AC < 12</font> when going with the approximate form.


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Answers:
Problem 2)  <font color=red>AC > 4.5</font>
Problem 3)  <font color=red>1 < AC < 3</font>
Problem 4)  <font color=red>32/3 < AC < 12</font>  (where 32/3 = 10.6667 approximately)
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