# 6th Grade (12 Years Old) Mathematics - Triangle Inequality Question

My son's math homework has this question which I find to be extremely confusing. Neither my wife or I can wrap our heads around what they are expecting as an answer:

Two sides of a triangle have lengths $7$ and $9$. The third side has length $x$. Tell whether each of the following statements is always, sometimes but not always or never true.

a) $x = 12$

b) $x = 2$

c) $x < 2$

d) $x < 16$

e) $x < 15$

To my mind the answer is sometimes to all of them. For $x = 12$ and $x = 2$, when considering all the possible values of $x$ it will sometimes be those values. For $x < 2$, $x < 15$, and $x < 16$, it's sometimes because all values less than those make it true, until you hit $0$ which forms a line and no longer a triangle.

It looks like I've found the the teachers edition of the book (scroll to page 6 in the pdf; question 12), that just confuses me even further!

• "it's sometimes because all values less than those make it true, until you hit 0 which forms a line and no longer a triangle." - Can you show me a triangle with side lengths 7, 9 and 1? You should be able to, if 1 is a possible value of x. – immibis Jan 30 '17 at 3:21
• Who or what is "sixth grade"? I don't know the intricacies of your school system (which you haven't even bothered to specify). If the information is relevant, give an age instead. – TRiG Jan 30 '17 at 17:28

I think you and your wife and your son can have fun figuring this out.

Even before you start thinking about triangles, you should be clear about distinguishing among "sometimes", "always" and "never". Each requires a different kind of argument. That's what I'll focus on in this answer.

I'd also recommend cutting out some strips of paper $7$ and $9$ inches long, and and a strip $16$ inches long marked off in inches. Then play around a bit before trying to think about the arithmetic. You'll notice that if you put the two shorter strips together at one end, the other ends will be at most $16$ inches apart and at least $2$ inches apart – the extreme cases when they're lined up.

You will have discovered something that you already know: the sum of any two sides of a triangle must be strictly greater than the third side. If the sum is equal, the "triangle" lies on one line.

Now you're ready for a careful reading, one statement at a time.

a) $x=12$.  There is a triangle with a third side of $12$, but the third side does not have to be $12$.  So the answer is that it's sometimes $12$.

My original answer was wrong. I'll leave it here since it's instructive to think about why.

a) $\require{enclose}\enclose{horizontalstrike}{x=12}$.  Since there's just one number to think about, the answer can't be "sometimes".
It has to be "always" or "never".  Your strips of paper will tell you it's "always".  The arithmetic is $\enclose{horizontalstrike}{9-7 < 12 < 9 + 7}$.

b) and c) Clearly "never", since $9-7 = 2$.

d) $x < 16$. Well, every triangle you managed to build had a third side less than $16$. The question starts with an actual triangle, so the answer is "always". That's not the same as "for every $x < 16$ you have a triangle". That's false, as you observed in a comment, since $2 < 16$ and there's no triangle with third side of $2$.

e) $x < 15$. OK, we have a triangle with third side $x$. Can it be less than $15$? Well, yes, obviously. Must it be less than $15$? No. It must be less than $16$, but it might be $15.5$.  So the answer to this one is "sometimes".  (Even if $x$ had to be an integer (not specified in the question), it could be exactly $15$.)

Note: I've deliberately written a rather long winded answer. Thinking things out this way is the best way to learn, even if what your son ends up turning in to his teacher is closer to "just the answer".

• For part (a), how can $x = 12$ always? It is sometimes $x = 12$. Sometimes it is $x = 5$. The most restrictive property of $x$ that is always true is $2 < x < 16$. (ETA: Note that the teacher's manual has an answer of "sometimes" for (a).) – Brian Tung Jan 29 '17 at 20:42
• @BrianTung I realized that mistake while I was out with my dogs and have fixed it now. Glad someone else notices. In the sixth grade I suspect triangles shouldn't be degenerate. – Ethan Bolker Jan 29 '17 at 21:42
• The answer for b) is "sometimes" not "never". A degenerate triangle will have a length of 9-7=2 – Dale M Jan 30 '17 at 2:26
• @DaleM That depends on whether a degenerate triangle counts as a triangle. I would assume not in 6th grade math. – DRF Jan 30 '17 at 7:32
• @DRF: It appears that the book linked by the OP agrees with you: it states the triangle inequality in two parts, as "Part 1: If A, B, and C are any three points, then AB + BC ≥ AC. Part 2: If A, B, and C are vertices of a triangle, then AB + BC > AC." I actually rather like that formulation; it manages to provide both forms of the inequality, while avoiding confusing terms like "degenerate triangle". – Ilmari Karonen Jan 30 '17 at 11:15

The important thing here is to understand the bounds on the third side of the triangle. The triangle inequality tells us that the third side must have length $x\in (2,16)$, i.e. $2<x<16$ . This makes sense, because the sum of the lengths of the first two sides $A+B$ should never exceed the length $x$ of side $C$. Similarly, the difference between the lengths of the sides: $A-B$ should never be less than the length $x$ of the side $C$.

$(a)$ This answer is sometimes true, since $2<12<16$.

$(b)$ This answer is never true, since $2\not<2<16.$

$(c)$ This answer is also never true, for the same reason as $(b)$.

$(d)$ This answer is always true, because any $x$ satisfying the constraint will satisfy $x<16$.

$(e)$ This is sometimes true, because a side of length $x=15.5$ could work, for example.

• Makes sense, d and e are still confusing. "because any x satisfying the constraint", seems a little weaseling to me. I'm think I'm going to chalk it up to a bad question. – Mark Jan 29 '17 at 20:08
• Suppose $x$ is the length of the side $C$ (the third side), then $x<16$. Because $2<x<16$. That's the logic behind $(d)$. This should be unambiguous. Now, in the case of $(e)$, I am showing that it is sometimes true, because lengths $2<x<15$ will work, but there are some allowable lengths with $15<x<16$. More precisely, if $x$ is the length of $C$, it could be the case that $2<x<15$, but it also could be the case that $15<x<16$, hence the sometimes. – Antonios-Alexandros Robotis Jan 29 '17 at 20:10
• I got it now. If I build all the possible triangles with all the possible lengths, x will always be less than 16, but sometimes it'll be less than 15. Eureka! – Mark Jan 29 '17 at 20:14
• That's the idea. – Antonios-Alexandros Robotis Jan 29 '17 at 20:14
• Thanks so much, I'm torn between accepting your answer or @ethanbolker. I'm think I'm going with him as his verbosity is what made it click in my head. Not to take away from your excellent answer, though! – Mark Jan 29 '17 at 20:16

From your pdf, just use the following:

If a triangle has sides of lengths $x$, $y$, and $z$, the following three inequalities must be true: \begin{align} x+y&>z,\\ x+z&>y,\\ y+z&>x. \end{align}

Here, $y=7$ and $z=9$.

So, for example, b. and c. can never happen since it would violate the first inequality.

• Ok, thanks, that makes sense, but why is d always. If x < 16, x could be 2, which violates the rule. – Mark Jan 29 '17 at 19:51
• These three become: \begin{align} x&>7-9\\ x&>9-7\\ x&<7+9. \end{align} when plugging in the values for $y$ and $z$. The first one becomes superfluous in this case, and we end with: $$2<x<16$$ – Jeppe Stig Nielsen Jan 29 '17 at 20:08
• @mark $d$ and $e$ are statements that will be true (or sometimes true) about a valid $x$. They are not full statements about how to find a valid $x$, – Joffan Jan 29 '17 at 20:14

Think of a "hinge" of two sides with fixed lengths 7 and 9, respectively. You can put it at any opening between $0^\circ$ and $180^\circ$, but not including the extreme values $0^\circ$ and $180^\circ$.

The third side will be given once you fix the degree of opening of your hinge. In the limit where the angle is extremely close to $0^\circ$, the third side tends to $9-7=2$. In the other limit, $180^\circ$, the third side tends to $9+7=16$.

So the third side can have all lengths between 2 and 16, not including those two extreme values.

The answer is $d)$ as $x < 7+9 = 16$ is always true.
• That's only answers $(d)$. What about \$a, b, c, e, f? – Namaste Jan 29 '17 at 19:49