Pythagorean Triples

[Fl Engineer]

Hi all,

I was curious after helping my daughter with a geometry problem tonight....

Say I have a right triangle with integer sides a,b,c (c = hypotenuse)

a, b= unknown; c= k(integer) multiple of a primitive Pythagorean triple.

Can anyone provide a counterexample that disproves:

If a and b are unknown, and c=k*(Pythagorean Triple value), then necessarily if all lengths are assumed to be integers, that the triangle must be a Pythagorean triple, or integer multple thereof.

I know this is a strange question, and if necessary I can post the problem that generates it.

Thanks

 

[Denis]

c= k(integer) multiple of a primitive Pythagorean triple.

Whadda heck d'you mean :wink:

c = k, k being an integer multiple of THE HYPOTENUSE of a primitive Pythagorean TRIANGLE ?

Suggest you show an example...

 

[mmm4444bot]

?

Hi Fl Engineer:

Let's say that a right-triangle with legs A and B and with hypotenuse C is a primitive Pythagorean triple.

If we enlarge this triangle by an integer factor k, then the resulting triangle has legs k*A and k*B and has hypotenuse k*C.

We could label this result as triangle abc.

a = k*A

b = k*B

c = k*C ? Is this what you're talking about ?

If so, then you won't be able to find an example of some triangle abc (generated from some triangle ABC as described above) that is not a Pythagorean triple because triangle sides increase proportionately and the set of Integers is closed under multiplication.

If this is not the situation in your mind, then I'm at least as confused as Denis.

Cheers ~ Mark

 

[Fl Engineer]

mmm,

I should have emphasised more that legs a and b are unknown, and the hypotenuse is an integer multiple of a primitive Triple. I don't know how to post diagrams here, so let me try to describe the problem.

Find area of trapezoid:

base 1 = 24, base 2 = 10, legs of length 13 and 15. No angles given.

My first look, I immediately recognized the 15 as the hypotenuse of a 3 multiple of a 3,4,5 triple. Used that as a check, and it worked. I didn't want to tell her that every right triangle with a hypotenuse that is an integer multiple of a primitive triple (in this case the 3,4,5) can be "solved" with no info other than the hypotenuse length and a 90 degree angle, without looking into it further.

I did a few calcs to try and find a counterexample, but couldn't. I also tried to find another method of solving the problem without using the Pythagorean Triple recognition. I couldn't come up with a solution that didn't involve 2 or more variables.

I'm just curious if anyone thinks that as a general rule, I can tell her that it is at least highly likely that when dealing with lengths that are integers, a Pythagorean Triple multiple recognized in a right triangle, should at least be checked as a first possibility.

I'm not sure if that makes it any clearer, but thanks for your input.

 

[mmm4444bot]

Hi Engineer:

I was hoping that you would answer the question in my first response. (The same question appears in red below.)

?

I should have emphasised more that legs a and b are unknown, and the hypotenuse is an integer multiple of a primitive Triple.

Emphasis was not needed; clarification was. Your statement above (in blue) is not correctly worded. It does not make sense.

A hypotenuse is a length. We can certainly say that a given hypotenuse is an integer multiple of some other length, but a primitive Pythagorean triple is not another length; it's a set of three integers.

c = hypotenuse in your original triangle abc

C = hypotenuse in some other triangle that is a primitive Pythagorean triple

k = an integer

c = k*C ? Is this what you're trying to say?

?

Find area of trapezoid:

base 1 = 24, base 2 = 10, legs of length 13 and 15. No angles given.

My first look, I immediately recognized the 15 as the hypotenuse of a 3 multiple of a 3,4,5 triple. Used that as a check, and it worked.

Your usage above with both pronouns "that" and "it" is ambiguous. I would need to see your steps, to be confident that I understand this "check".

15 is not a "3 multiple of 3,4,5".

15 is a "3 multiple of 5".

Okay. Perhaps this is enough to confirm that your answer to my question in red is "yes", but I still only have partial confidence that I know what you're thinking. I understand how you could form a right triangle in the trapezoid, using the 15-unit side as the hypotenuse. I understand how you could interpret this right triangle as an enlargement of the primitive triple 3-4-5. I understand how you could thus conclude that the legs on the right-triangle with hypotenuse 15 have measure 9 and 12.

But I do not understand how you concluded that this right triangle (with hypotenuse 15) is oriented such that the 9-unit leg is a section of the trapezoid's 24-unit base — versus the 9-unit leg being the trapezoid altitude — because there are two possibilities.

If you don't understand what I just wrote, please let me know, and I will upload some labeled diagrams to illustrate what I mean.

?

I did a few calcs to try and find a counterexample, but couldn't. I also tried to find another method of solving the problem without using the Pythagorean Triple recognition. I couldn't come up with a solution that didn't involve 2 or more variables.

You won't be able to find an example of some enlarged primitive-Pythagorean-triple triangle, where the enlargment has an integer hypotenuse, whose sides are not all corresponding integer multiples because triangle sides increase proportionately and the set of Integers is closed under multiplication.

The method that I used to find the area of your trapezoid involves three variables; does it matter? There's many ways to skin this cat!

(I don't know in which course your daughter is currently enrolled, and I obviously don't know her math background and skills, but I used an algebraic approach to find the trapezoid's altitude, followed by applying a standard formula for the area of a trapezoid in terms of the base lengths and altitude.)
?

I'm just curious if anyone thinks that as a general rule, I can tell her that it is at least highly likely that when dealing with lengths that are integers, a Pythagorean Triple multiple recognized in a right triangle, should at least be checked as a first possibility.

Sure, you can tell her that. Any time we can break an object into sub-pieces, and one of those pieces is a right triangle with a hypotenuse that's an integer multiple of another hypotenuse in some primitive triple, we know that the legs on this sub-piece are integer mutliples of the legs in the triple. Always!

But, one needs sufficient information to know how such a primitive-triple-multiple is oriented, in the exercise at hand, because there are two possibilities for the legs.

Keep your questions coming, if you like. Especially if I wrote anything that you do not understand.

I'm happy to continue this discussion.

Cheers ~ Mark

 

[Denis]

base 1 = 24, base 2 = 10, legs of length 13 and 15. No angles given.
My first look, I immediately recognized the 15 as the hypotenuse of a 3 multiple of a 3,4,5 triple.

OK, assume that "the 15" was 25;
then you would have recognized it as hypotenuse of a 5 multiple of a 3,4,5 triple...right?
In other words, a 15,20,25.
But 7,24,25 is also one with hypotenuse = 25. Get my drift?

As Mark says, "multiple" ain't the right "word" here...but we know what you mean :wink:

 

[Fl Engineer]

Guys,

You'll have to excuse my mathspeak, I'm an engineer, and to be honest, it's been a few years since I've had to strictly define these terms concisely. I understand the frustration, I do it to colleagues that don't use concise language.

What I am refering to is a hypotenuse that is an integer multiple of the hypotenuse length of a primitive Pythagorean Triple right triangle.

Yes, c=k*C

But I do not understand how you concluded that this right triangle (with hypotenuse 15) is oriented such that the 9-unit leg is a section of the trapezoid's 24-unit base — versus the 9-unit leg being the trapezoid altitude — because there are two possibilities.

The length of the bases allows for only one to be true, therefore process of elimination.

I'm interested in seeing your algebraic solution, I'll admit I haven't had the time to revisit it with my schedule. If it's not too much trouble, I'd like to see it.

Denis,

I think you hit where I was going, that just because the hypotenuse length is an integer multiple of a primitive Triple, doesn't necessarily mean that it's a Pythagoran triple. It may have hit me that way just because it was a "small" multiple, and by testing it in the problem it proved out.

I understand that I should have stated that I recognized the 15 length as the hypotenuse of 3(3,4,5).

I'll clean up my semantics.

Thanks for the help.

 

[mmm4444bot]

The length of the bases allows for only one to be true, therefore process of elimination.

Ah, there was a "process" involved. Fair enough.

As I mentioned, there are different ways to approach this exercise. I did not invest any time trying to be efficient; I just plugged away with a mechanical approach that I knew was sufficient, unconcerned about relative efforts.

I started by recalling the standard formula for the area (A) of a trapezoid in terms of base lengths (b[sub3l8t99p]1[/sub3l8t99p] and b[sub3l8t99p]2[/sub3l8t99p]) and height (h).

\(\displaystyle A \;=\; h \cdot \frac{b_1 \;+\; b_2}{2}\)

In other words, the area is a product of the height and the average base length.

If I knew the value of h, I could use this formula, since b[sub3l8t99p]1[/sub3l8t99p] and b[sub3l8t99p]2[/sub3l8t99p] are given. So, I tried to find h.

I drew a picture, in order to organize the known and unknown pieces of information.

[attachment=03l8t99p]junk444.JPG[/attachment3l8t99p]

I wrote down some obvious relationships.

\(\displaystyle h^2 \;+\; y^2 \;=\; 13^2\)

\(\displaystyle h^2 \;+\; x^2 \;=\; 15^2\)

\(\displaystyle y + 10 + x \;=\; 24\)

Solved each of the first two equations for h^2, giving me two expressions that must be equal (Transitive Property of Equality).

\(\displaystyle 169 \;-\; y^2 \;=\; 225 \;-\; x^2\)

Rearranged.

\(\displaystyle x^2 \;-\; y^2 \;=\; 56\)

Next, I chose to express y^2 in terms of x, so that I could solve this equation for x (and use that value to calculate h).

To express y^2 in terms of x, I returned to the relationship between x and y.

\(\displaystyle y \;+\; 10 \;+\; x \;=\; 24\)

Rearranged.

\(\displaystyle y \;=\; 14 \;-\; x\)

Determined y^2.

\(\displaystyle y^2 \;=\; (14 \;-\; x)^2\)

\(\displaystyle y^2 = x^2 \;-\; 28x \;+\; 196\)

Substituted this expression for y^2 in the equation x^2 - y^2 = 56, then solved for x.

\(\displaystyle x^2 \;-\; (x^2 \;-\; 28x \;+\; 196) \;=\; -56\)

\(\displaystyle 28x \;=\; 252\)

\(\displaystyle x \;=\; 9\)

Next, found h using x = 9 in the prior relationship h^2 = 15^2 - x^2.

\(\displaystyle h^2 \;=\; 225 - 81 \;=\; 144\)

\(\displaystyle h \;=\; 12\)

Finally, I found the area.

\(\displaystyle A \;=\; 12 \cdot \frac{10 \;+\; 24}{2}\)

The area is 204 square units.

If your process involved recognizing 15 as the hypotenuse of a 9-12-15 triangle, and recognizing 13 as the hypotenuse of a 5-12-13 triangle, and realizing that the common leg of 12 must be the height of the trapezoid, then that's a pretty slick way to go straight to the formula with h=12.