Thanks to TC's inspired challenge to our readers in a comment on the Medians of a Triangle post, I've decided to expand it into an investigation for our readers and students (geometry with some trig needed).
Consider a parallelogram whose sides have lengths 39 and 25 and with one diagonal of length 34.
(a) Explain why this parallelogram is unique, i.e., all parallelograms with these characteristics are congruent. Why was it not necessary to specify that the 'shorter' diagonal was given?
(b) A parallelogram has sides of lengths a and b and diagonals of lengths c and d. Use the Law of Cosines to show that
c2 + d2 = 2(a2 + b2).
(c) Determine the length of the other diagonal. As an alternative, how would you do it without the formula in (b)?
(d) Determine the area of this parallelogram.
(e) So what makes this parallelogram unusual?
Comments:
(1) From the comments on the Medians post (Cotton Blossom and others), we know that we can construct rectangles and rhombuses whose sides and diagonals have integer lengths, but the above demonstrates a parallelogram that is neither of these special cases.
(2) The formula in (b) is not too difficult to prove, however, finding solutions to this Diophantine equation or a general solution is far more challenging!
(3) Note that the parallelogram in this challenge also has integral area. Finding other such parallelograms is not a simple exercise!
Thursday, March 6, 2008
A Parallelogram Has Sides of Lengths 39 and 25 and a Diagonal of Length 34. So, What Makes It So Special!
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Consider a parallelogram whose sides have lengths 39 and 25 and with one diagonal of length 34.
(a) Explain why this parallelogram is unique, i.e., all parallelograms with these characteristics are congruent. Why was it not necessary to specify that the 'shorter' diagonal was given?
The longer diagonal has to be longer than either side. Or, use the formula in (b).
(b) A parallelogram has sides of lengths a and b and diagonals of lengths c and d. Use the Law of Cosines to show that
c2 + d2 = 2(a2 + b2).
You don't need the law of cosines, just Pythag. Drop an altitude, and maybe there are some special cases to worry about, but in the usual sort of parallelogram drawing you end up with (with x labeling the piece of b cut off by the altitude)
x^2 + h^2 = a^2
(b-x)^2 + h^2 = d^2
(b+x)^2 + h^2 = c^2
Square out those bottom two equations, add, and substitute a^2 in wherever you can, and you're done. (Of course, this is almost equivalent to the proof of the law of cosines, but I never had to say "cosine" in it, which makes it more accessible to first-year geometry students.)
(c) Determine the length of the other diagonal. As an alternative, how would you do it without the formula in (b)?
Well, the 34 diagonal makes me guess that there'll be a 16-30-34 right triangle in there, especially given all the clues about the special integerness of this one. Then the 25 side looks like a 15-20-25, but that wouldn't fit together quite right, would it? So I guess the height turns out to be yucky after all even though the area is nice, and the other diagonal is nice.
Well, one of many ways to find the other diagonal would be to say that the area of a 39-25-d triangle must be the same as the area of a 39-25-34 triangle. You can use Heron's formula from there and get a not-too-awful equation to solve, but I wouldn't recommend that approach.
How about the law of cosines? You can say that 34^2 = 25^2 + 39^2 - 2*25*29*cos(theta),
so the 2*25*29*cos(theta) part must equal 990. Now the other diagonal is made by swapping theta to its supplement, hence switching the sign of that term, so I get
d^2 = 25^2 + 39^2 + 990,
which luckily enough is a nice perfect square, so d = 56.
But that really amounts to a long way of using the formula in (b), doesn't it? Hm.
OK, I'm stumped for now. Are there any really slick ways of finding that diagonal?
(d) Determine the area of this parallelogram.
Heron's formula is pretty easy, or you can find the height using some of my Pythag equations above. Either way I think you get 420, or wait, that's the area of the 25-34-39 triangle, so the whole parallelogram is 840.
(e) So what makes this parallelogram unusual?
Integer sides, diagonals, and area.
By the way, messing around a bit in answer to your further challenge, I found:
17 28 sides, 25 39 diagonals, 420 area (look at the cool relationship between this one and your parallelogram!)
21 89 sides, 82 100 diagonals, 1680 area (look at all these multiples of 420! this can't be a coincidence)
26 51 sides, 35 73 diagonals, 840 area
26 73 sides, 51 97 diagonals, 840 area
33 58 sides, 41 85 diag, 1320 area (multiple of 60 but not 420!?)
34 56 sides, 50 78 diag, 1680 area
35 73 sides, 52 102 diag, 1680 area
41 50 sides, 21 89 diag, 840 area
41 85 sides, 66 116 diag, 2640 area
50 78 sides, 68 112 diag, 3360 area
51 84 sides, 75 117 diag (note gcd 3 here), 3780 area
51 97 sides, 52 146 diag, 1680 area
So apparently they're not THAT rare, or maybe I'm making some mistakes of some sort here.
There is a 3-dimensional version of this problem:
Does a rational box exist? (MathWorld calls it the perfect cuboid problem.)
That is, is there a box with integer sides a, b, and c, with each face diagonal integral, and with body diagonal integral?
So, a, b, and c are integers, and these four equations are solvable in integers:
a² + b² = d²,
a² + c² = e²,
b² + c² = f²,
a² + b² + c² = g².
People have come one equation short.
An associated challenge:
If a parallelogram has adjacent sides a&b, and diagonals c &d, show that
(a+b+c)(a+b-c)(a+c-b)(b+c-a) = (a+b+d)(a+b-d)(a+d-b)(b+d-a)
TC
Eric,
Thank you for the fascinating 3-D extension. Students often believe every equation can be solved in 'closed' form, never mind Diophantine equations. It's a surprise to them that there is no general solution or method for solving all Diophantine equations and in many cases, of course, no non-trivial solution exists at all.
TC-- The formula in part (b) is the parallelogram law that the sum of the squares of the 4 sides equals the sum of the squares of the diagonals. Your new identity looks familiar. Something more for me to play with!
BTW, the parallelogram formula reminded of Ptolemy's Theorem for quadrilaterals inscribed in a circle, but that involves the products of the diagonals and products of opposite sides not the sum of their squares. I guess they're equivalent for rectangles...
Joshua,
Thank you for the insights and the detailed explanations! I would agree that there does not appear to be any elegant or satisfying alternative to finding the other diagonal.
Stewart's Theorem, applied to the median on the '34' side of the 25-34-39 triangle leads to (1/2)√(2(25^2+39^2)-34^2) = 28, so the longer diagonal is 56. However, this is just another version of the formula I provided, and it is normally proved using the Law of Cosines anyway.
Interestingly, the cos of the base angle θ included between the '25' and '39' sides turns out to be 33/65. This leads to a 33-56-65 Pythagorean triple and the sin θ = 56/65. That '56' is just coincidental of course! Since the area is also integral, we would expect that the cos and sin of the base angles would be rational, therefore related to some triple.
Using the sin and cos of θ we can also represent the sides of the parallelogram in vector form. Then the longer diagonal is the SUM of these vectors and its length of course is the magnitude of that vector sum.
Details: Place the origin of the vectors at the intersection of the '25' and '39' sides. Then the vectors representing these sides are
(165/13,280/13) and (39,0).
You can work out the details of the magnitude of the sum of these two vectors but you will get 56!
Still not elegant since it depends on the trig, but that Pythagorean triple still fascinates me.
I'm impressed by all of the solutions you found. I still feel these integer triangles or parallelograms are pretty special, but that's just my take.
Students think that all equations can be solved, and in closed form, because teachers always give examples and problems that can be solved in closed form. Frankly, it would do students and teachers a world of good to take a problem we know is unsolvable, and try to solve it.
A calculus class could try to find a closed form for the logarithmic integral by using substitution and integration by parts. They won't find any, of course, but they'd learn of the wider world of mathematics. They might also discover asymptotic series while doing their research.
Number theory problems are less amenable to that sort of experimentation because they rapidly lead to arithmetic with numbers large enough as to require calculator and computer support.
You should look at the article on Hilbert's Tenth Problem in the current issue of the Notices of the AMS; it describes the problem's history and consequences.
You can have an interesting discussion with your students on what a closed-form solution really is!
Eric Jablow wrote:
'You should look at the article on Hilbert's Tenth Problem in the current issue of the Notices of the AMS;'
My question (taken from an essay titled 'Andy Rooney, Ph. D.' that I saw in the late 80's: Just how messed up was this guy Hilbert?
Another observation from the same essay: Can you imagine a mathematician writing Moby Dick>? It would begin, 'Let my name be Ishmael, let the captain's name be Ahab, let the ship's name be Pequod, and let the whale's name be Moby Dick.'
TC
"If a parallelogram has adjacent sides a&b, and diagonals c &d, show that
(a+b+c)(a+b-c)(a+c-b)(b+c-a) = (a+b+d)(a+b-d)(a+d-b)(b+d-a)"
This is REALLY easy if you know Heron's formula. It says the area of one half of the parallelogram equals the area of the other half ... I used Heron a bunch as a quick way of calculating the areas in my examples without having to mess with sines and cosines.
Joshua,
Bingo! Your reference to Heron's formula in your earlier post was what prompted this problem.
It is one of those problems that looks complicated and the algebra could get really hairy if you try to expand it, but there is a simple solution.
It is one more of those proofs where mathematicians just say 'Consider.....' without going much into the thinking that went into arriving at that step.
TC
Amazingly enough, finding solutions to these diophantine equations is related to Pell's Equation.
c² + d² = 2a² + 2b²,
c² - 2a² = - (d² - 2b²).
Remember, the Pell's Equation for index D is the problem of finding Diophantine solutions (a, b) of
a² - Db² = 1.
(A different a and b in this case.)
This is solvable for any D. The equation with 1 replaced by -1 is solvable for some D. In any case, both are solvable for D = 2. So, let e and f be rational numbers satisfying e² - 2f² = -1
Now, go back to the original problem, and write things as follows:
c² - 2a² = (c + a√2) (c - a√2).
d² - 2b² = (d + b√2) (d - b√2).
-1 = (e + f√2) (e - f√2).
Multiply the last two equations, and get that
c² - 2a² = - (d² - 2b²) if
(c + a√2) (c - a√2) = (d + b√2) (d - b√2) (e + f√2) (e - f√2).
Now, reorder the terms on the right side as:
(d + b√2) (e + f√2) (d - b√2) (e - f√2)
Now, show that the first equation below is equivalent to the second:
(c + a√2) = (d + b√2) (e + f√2)
(c - a√2) = (d - b√2) (e - f√2).
It's just like complex conjugation. (Do you think the students would understand Galois theory?)
Unfortunately, some values of e and f lead to values of a, b, c, and d that violate the triangle inequality. The original problem came from a = 39, b = 25, c = 56, d = 34, e = 23/47, f = 37/47.
Joshua--
Nice solution. You are my 'Hero'! When one solves TC's 4th degree equation for d, using the known values of a,b, and, c we obtain d = +-56; d = +-34. No surprises there!
Eric--
Thanks for the Pell's! I'm working on an activity that would develop solutions in certain special cases - this is not that easy. I'd love to introduce middle schoolers to continued fractions and convergents. I believe it can be done at any introductory level.
More specifically, I'm working on an investigation of integer-valued triangles having a 60 degree or 120 degree angle. These lead to some interesting Diophantine equations:
x^2 + y^2 - xy = z^2
or
x^2 + y^2 + xy = z^2.
There are some easily guessed solutions (like 3,7,8), but the general cases is of course related to our old friend, Pell (don't you love it - Euler may have mistakenly given credit to Pell for this!).
Ironically, one of our recent Mystery Mathematicians, Ito, was also credited with solutions in this area.
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