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I probably should start a new word game called find all the typos! 'Simpel' and rightt' are pretty impressive variations. I guess if you can't spell write rite, then you're not alright! Well, I'm trying to correct these but once they go to a reader, the mistakes are out there for all to see! Haste makes --------------.
Just another one of those square and right triangle problems. Yes, I do have a passion for these types of geometry questions (I've published similar ones previously) and you probably will recognize this one from your own experience. The difference of course is that an investigation takes students from particular cases to a general formula which is related to the harmonic mean of the legs of the right triangle.
Part I
Find the length of the side of the square in Fig. I.
Comments:
This one is eminently guessable and doesn't require the use of similar triangles. Encourage students to justify their conjecture that D and F are midpoints.
Part II
Do the same for Fig. II.
Comments: Ok, the answer is 12/7. You may find students trying a Pythagorean approach, guessing midpoints or assuming special angles. Most students do not look for similar triangle solutions unless they have considerable experience with this or the problem is assigned in that unit!
Part III
Of course, we will now ask students to generalize the result:
If the legs of the right triangle have lengths a and b, show that the side of the indicated square (inscribed square, largest inscribed square with sides parallel to the legs, however you want to describe it!) has length ab/(a+b).
Friday, October 31, 2008
Inscribed Square in Right Triangle Investigation
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12 comments:
Taking it one step further, when is the side of the square maximized?
One way is to use the AM:HM (rather than AM:GM) inequality.
TC
tc--
Am I missing something here? There is only one 'inscribed'square for each right triangle. Are you referring to finding the rectangle of maximum area for a given right triangle? OR??
Dave
Ah,
I think I was trying to say something like 'What is the ratio of the sides a to b that maximizes the side of the square?'
Do we need to put some kind of a normalization condition on a and b? Not clear to me, but I don't have the time to think right now.
TC
Related questions? Hmm...
1) Find the side length of the biggest square you can fit in an equilateral triangle
2) Prove that the biggest right-angled triangle you can fit in a square is the obvious one
3) What's the smallest triangle you can draw that contains a 1x1 square?
Nice, Alex!
Your first question reminds me of a post I published last year. Look
here.
#2 looks straightforward using bases and heights. #3 I really like. I'm assuming you mean the triangle of smallest area.
For #3, Alex, I'm leaning toward the isos. right triangle of area 2 as the 'smallest'. I think I can prove it but I'm wondering if my conjecture is right and if you have a proof in mind before I suggest an argument for it.
Hi Dave,
I don't think that the isosceles right triangle is the unique solution for #3, though its area might be equal to the area of the smallest triangle.
Nice problem, Alex!
TC
tc--
I'm not sure if it's unique, but if we are looking for the triangle of smallest area, I think I have a proof that this area is 2. The other issue is orientation of the square inside the triangle. I am assuming, perhaps incorrectly, that we are restricting our attention to triangles in which at least one side of the square lies on a side of the triangle. With this condition, my proof demonstrates that the minimum area is 2. Let me echo tc's comment: Nice question! It deserves to be published as a separate challenge.
Some further generalizations:
1. Consider triangle ABC with angles A and B acute. Construct a square that has its base on side AB of length c and two other vertices on sides AC and BC. Let the altitude from C to AB be of length h. Find the length of the side of the square.
2. Now, triangle ABC is isosceles, with angles A & B acute, but C is obtuse. Let the length of side AB be equal to c, and let the altitude from C to AB be of length h. You can have square 1 like in the previous problem. Alternatively, you can have square 2 within the triangle whose diagonal is the altitude from C to AB. Find the relationship between c & h so that the area of square 1 is greater than the area of square 2.
TC
Nice generalizations, tc...
The first one is straightforward assuming you want the result in terms of c and h.
If one inscribes the 1x1 square in an isosceles right triangle so that a side of the square lies on the hypotenuse, then the area of the square is 9/4 or 2.25. This is actually a nice problem for students to try, although not difficult.
On the other hand, if the 1x1 square is placed in the "corner" of the isos. right triangle so that only one vertex of the square is on the hypotenuse, the area of the triangle is 2 as previously noted.
Alex, tc--
Could you describe for us another triangle containing this square whose area is 2?
tc, is that what you were trying to do with your questions?
Well, I see it now that I've worked through more of the details and my brain is actually functioning in the AM! Yes, there are 'many' such triangles whose area is 2. One can show that they all have a base of 2 and a height of 2. Of course the isos. rt. triangle is one special case of this, indeed a 'limiting' case of the side of the square lying on a side of the triangle, That is, we can slide the side of the square toward one vertex of the triangle until it reaches the 'corner'.
Alex and tc --
Thank you for enriching the original problem.
Hi Dave,
Yes, My problem (1) was trying to show that there were an infinity of triangles of area two that inscribe the square of area 1.
I found the solution to my problem (2) somewhat surprising. According to my calculations, square 1 is larger than square 2 if h < (sqrt(2)-1)*c
TC
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