As promised...
In a triangle, two sides and the median to the third side have respective lengths 5, 13, and x. List all possible integer values of x.
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Comments:
- There is an elegant solution/proof here! I found a far more complicated coordinate argument.
- Some students might make an educated guess based on insight but the real challenge is to explain one's conjecture!
- Challenge yourselves with this - if you see how to do it, suggest an approach but don't give it all away for at least a few hours!
- We generally think of these as only appropriate for our best and brightest but there's no harm in throwing it out to all of our geometry students- amazing things happen when we open up these kinds of problems to everyone!
Posts
19 comments:
Would it be more interesting to ask for solutions such that the length of the third side and the median to the third side are both integers?
Unless I have made a mistake in my calculations (which is eminently possible), I think there are two solutions to the problem above.
TC
tc--
I believe the wording of the question is unchanged from the original contest.
I also believe there are 4 integer values for x, one of which is 6. I don't want to give away more than that right now!
It appears at first blush that one needs some considerable machinery to obtain relationships about the median to a side of a triangle, however, remember I said there is an elegant solution.
Consider a special case of a median, namely the median to the hypotenuse of a right triangle. This doesn't seem to help in our problem but I strongly urge our readers to consider how that theorem is proved!
Hi Dave,
I also now believe there are 4 integer values for x, and also that none of them is such that the length of the third side is an integer.
A slightly simpler (maybe?) offshoot: How many integer values can the altitude to the third side have?
TC
Say a=5, b=13, w = Angle(a,b).
Then for w we have: 0 < w < 180.
(I'm omitting the cases w=0,180
because then the triange disaapears,
all sides overlap)
Since the median originates the
vertex (a,b) and divides the 3rd
side of the triangle in the middle
we get:
For w -> 0 => x -> 0
For w -> 180 => x -> 5 + |13-5| / 2 = 9
So the possible solutions for x
are: 0 < x < 9.
And the integer values:
x=1,2,3,4,5,6,7,8
Your comments indicate that this is
wrong, what's the mistake I made?
Hi Florian,
I made the same mistake the first time too. I believe some of the values you have do not generate valid triangles.
My approach was slightly different: Assuming the 3rd side was y, I used the cosine formula in the two triangles to get
5^2 = x^2 +(y/2)^2 -xy cos(t)
13^2 = x^2 +(y/2)^2 -xy cos(pi-t)
Adding, we get
97-(y/2)^2 = x^2, which gives the possible values of x as 1..8. However, for some of these, y is such that the sides 5,13,y do not form a triangle.
TC.
florian--
I like your analysis of the 'extreme' cases but I drew slightly different conclusions. As the angle approaches 180, I felt that the median approaches |13+5|/2 - 5 = 4, so the least integer value of x would be 5.
As the angle approaches 0, I believe the median approaches 5 + |13-5|/2 = 9, so the largest possible value of x would be 8. I think students would be able to follow this boundary value approach but it is a bit tricky, since the student would have to carefully focus on the correct vertex and where the midpoint would be in each case. This is not that obvious, IMO.
As I suggested, there is another approach that 'blew me away.' However, I like your argument for instructional purposes!
tc--
I missed your comment as I was posting my reply to Florian. Thanks for that explanation, a bit different from mine.
BTW, I like the altitude problem, an easier version of the median problem. I would probably give yours first as a lead-in, although the medians require a lot more visualization and reasoning IMO.
My approach:
Virtually draw the triangle on the coordinate plane, with the first side along the positive x axis, and the second side in the positive y side of the plane. The set of possible triangles will be represented by a point at (5,0) for the first side, followed by a semicircle of radius 13 centered on the origin for the second.
The resulting median will be represented by the locus of midpoint of the line drawn between the previous point & curve - a semi circle of radius 6 1/2 centered at (2 1/2, 0).
The length of the median is the distance from the origin to this semicircle. The two degenerate cases are at x = -4 (d = 4) and x = 9 (d = 9) , where the triangle collapses to a line.
The resultant possible integer values for the length of the median are then in between those two cases: the consecutive integers 5 through 8.
(note - this is a visual equivalent of using the cosine formulas. with a whiteboard, the explanation would be more suited for middle school, however).
My "discovery" about problems like this is that whenever you have the median of a triangle, you can think about it as half a parallelogram.
So, my rephrasing of the question:
In a parallelogram with side lengths of 5 and 13, what are all possible integer lengths for half the diagonal?
Since the diagonal (by the triangle inequality) has to be between 8 and 18 (not inclusive), the median has to be between 4 and 9 (not inclusive), so 5,6,7,8 are the possibilities.
This approach also yields some nice formulas for the lengths of the medians in terms of the sides, since for any parallelogram with sides a and b and diagonals d and e, you can use Pythag to prove
2a^2 + 2b^2 = d^2 + e^2
so if you have three sides of a triangle, say a, b, and e, then d is twice the median, so you can find the length of the median from
2a^2 + 2b^2 = (2m)^2 + e^2
where e is the length of the side the median is to.
I really like (Mr.?) "Joshua Zucker"'s discovery! I've now seen the light--
I drew the problem as one parallelogram embodying FOUR of such 5-13-x triangles, with two sets of two congruent triangles. That means that the median of one triangle is half of the "x" side of the other! Then I simply used the two Triangle Inequality Theorem pieces Mr. Marain has been trying to drill into my brain: the third side has to be greater than 8 (13-5) and less than 18 (13+5). Taking the even numbers, 10, 12, 14, and 16 and dividing them to get whole numbers, I get 5, 6, 7, and 8 as the solutions to the problem, since these half-sides are also medians for the overlapping triangles in the parallelogram! HOORAY!!!!!!!!
-Melissa
Here is another mental visualization
of that leads to a
solution with only little
math:
Consider the following:
If we had a vector A and a vector B and X
representing the sides of the triange
and the median we could obtain X simply
by adding A and B and divideing the sum by 2.
X = (A+B)/2
With this visual picture we can easily see how the
length of X is dependent on A and B.
The two extreme cases are when the two vectors are
parallel and
1. have the same direction
2. have opposite directions
We can and formulate the bounds for the
length x of X (lowercase means length).
1. (a+b)/2 < x
2. x < (b-a)/2
Add the two inequalities and the result is:
(b-a)/2 < x < (a+b)/2
With a=5 and b=13 we get:
4 < x < 9
A related problem:
Prove or disprove: A parallelogram with integer side lengths has integer diagonal lengths only if it is a rectangle (i.e., the sides are two parts of a Pythagorean triplet).
TC
Wow. I spent a long time trying to figure out the Law of Cosines to try to prove TC's statement-to-be-proven. But at last, I've discovered how simple it is to disprove!
Take a look at a parallelogram that has perpendicular diagonals, that is, it is a rhombus, but not necessarily a rectangle. Now draw in the two diagonals, forming four congruent right triangles. If you just make sure these right triangles have integral sides, viola! You have a parallelogram that has two different integral diagonals, but that does not necessarily have right angles--adjacent angles just have to be supplementary!
So, TC's to-be-proven statement is FALSE.
Nice work, Cotton Blossom! You've seen the light! Isn't it amazing how one good geometry problem gives life to others, thanks to TC!
I've been keeping quiet for a day or so to allow our readers to share their insights, creative solutions and other ideas. TC, your 'integer diagonal' problem is formidable. I believe I could probably come up with a specific numerical example, but I'll need to play around with it. Cotton Blossom found a nice counterexample, but I'd love to see the actual lengths for both diagonals. Describing such parallelograms in general is probably far more challenging!
Joshua, your approach is that 'elegant' solution to which I referred. It certainly didn't occur to me immediately - I actually used Mr. K's approach, which is more complicated but highly instructive for students.
Florian's vector approach is also very powerful. Vector geometry doesn't get enough attention (if any) in most precalculus classes and here it provides a more efficient method than coordinates. Nice job, Florian!
Again, this problem and TC's variations are taking on a life of their own. Cotton Blossom is showing us how motivated students might react to challenges and how such challenges are critical in their development.
Ok, to follow up on TC and Cotton Blossom with a specific numerical example, here's a simple Pythagorean triple that would produce such a rhombus:
5-12-13
Obviously, one diagonal would have length 10 and the other 24.
This was easy but finding the dimensions of a non-rhombus parallelogram with integer length diagonals will take some more time.
The rhombus example certainly provides a quick counterexample.
Great job, Cotton Blossom. I like your approach.
My reasoning was as follows: The problem is in some sense, equivalent to finding a triangle with integer sides (one half of the parallelogram) and an integer (or half integer) median. With this formulation, and the low of cosines, you can find that if the sides are a & b, and the first diagonal is c, the other diagonal d is given by
d^2=2(a^2+b^2) - c^2.
It was easy to find integers such that d^2 was a perfect square.
Of course, that raises the obvious question: Are the only possible cases when either the sides or the diagonals are perpendicular? My guess is not, but I haven't had time to verify this.
TC
TC's comments continue to make my brain dance...
I've been trying to get an example of a parallelogram 1) that does not have congruent diagonals and 2) whose diagonals are not perpendicular, either. I've come up with a drawing that's a little hard to explain, but here's a try:
Draw a right triangle whose sides are integers (a Pythagorean triplet). Pick a vertex (NOT the right angle!) and rotate the entire triangle about that point 180 degrees, making another triangle. Now you have two congruent right triangles. Then connect the dots--make two lines, each from a right angle to the third vertex (the one that isn't right and isn't the point of rotation.) Now you have a parallelogram. By now, the parallelogram has integer diagonals and a pair of integer sides. Now you just have to make the remaining pair of sides integer lengths. I then picked a diagonal and drew a perpendicular line (perpendicular bisector, going through the point of rotation). Since we have right triangles, we know the sines of one pair of vertical angles in the middle of the parallelogram. The sine of that angle (that I've named x) is also the cosine of the angle it is supplementary to (the angle that is across from the side we are trying to find.) (This is all inverse trig functions, which is great because these angles are quite messy.) Now use the law of cosines--the denominator of the sine of the known angle should cancel out with either "a" or "b" in the "-2ab cos x " part of the law (that should be the length of the hypotenuses of the original right triangles). So now all we have to do is try to find a Pythagorean triplet that, when plugged into the law of cosines, will produce a perfect square. And THAT would take a long time. Perhaps someone should write a computer program?
-Melissa
Holey Moley! I just had a revelation:
In my previous comment, my final step in finding the magic parallelogram was plugging in certain lengths having to do with a pythagorean triple into the law of cosines. I also mentioned that in the 2ab cos x part, (well, I used sine of the supplementary angle), the denominator of x (hypotenuse of right triangle) would cancel out with either a or b (because one of them is the hypotenuse). Take a look at what that gives you:
c^2 = a^2 + b^2 - 2ac
which looks stunningly similar to a^2 - 2ab +b^2 , or (a-b)^2. Here a-b could be the length of side c, except it can't, because 1) you'd have a flat triangle and 2) c cannot equal b. So this "2ac" thing is a little more or a little less than a perfect square, which leads me to believe that it may be true that such a magic parallelogram does not exist. But I'm still counting on that computer program to figure it out.
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