[After writing several thousand SAT-type problems over the years and authoring many math contests here in my state, I should be able to come up with a few of these. Now if I could only locate those 250 file folders with the materials...]
Some geometry today...This question is easily guessed by those with a passing knowledge of the Pythagorean Theorem, but its usefulness may go far beyond this. Sorry, but if you were looking for a quickie, this isn't it. You could still use part of it and limit the discussion for now to making conjectures and doing some verification (not a full-blown proof).
Determine the least possible length of the hypotenuse of a right triangle whose legs have a sum of 12. Also, what would be true about the area of this triangle compared to other right triangles with the same condition on its legs?
Comments: Target audience -- Grades 7-12. Again, I apologize. This is far more than a 5-minute class opener. I usually use this kind of problem to encourage discovery and an inductive approach to problem-solving before getting into the issue of proof. Middle schoolers who have learned the Pythag rule can be shown how to begin the investigation by setting up a table of values with 3 or 4 columns (the last one could be area), computing various hypotenuses, areas and making conjectures. I'll leave the issue of calculator use up to you! Lot of meat here but most will accurately guess that the isosceles right triangle solves it. No calculus needed to PROVE that the isosceles case minimizes the hypotenuse and maximizes the area! There's a wonderful algebraic approach that proves both results with one broad stroke! What approaches will your students devise??? Pls share if you can or email me and I'll post a few.
Friday, January 5, 2007
'Warm Up' #2 (not a 5-minute one, perhaps)
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4 comments:
And the least possible hypotenuse if the perimeter were 12?
(I grabbed a pencil and started muttering, before I realized I had misread!)
Understanding referred me to this site. Interesting. I am adding you to my blogroll.
jonathan,
i do that all the time even with problems i write myself! unfortunately, your misreading has created a new problem that we can now pose, but i believe this one is for the real math devotees...
i'd love to post this on my main page today, but then the number of visitors i have will approach zero asymptotically!
Here's what I came up with -- let me know if you find an easier way:
If the perimeter is 12, the relationship among the sides is more sophisticated and I think we have to be creative to avoid calculus. here goes...
Since the 3-4-5 triangle occurred to me first (I usually examine special cases to get a feel for the problem), I had a suspicion that 5 was the least possible hypotenuse since something less than 5 in my mind would produce a smaller perimeter than 12 (this seemed obvious to me but I needed to verify it). To prove this, I labeled the legs p,q and the hypotenuse r. I obtained the following:
p^2 + q^2 = r^2 < 25
Then (p+q)^2 = p^2 + q^2 + 2pq < 25 + 2pq
Using the Arithmetic-Geom Mean Inequality we know that 2pq <= (p^2+q^2)/2 = 25/2.
Thus, (p+q)^2 < 25 + 25/2, so p+q < (37.5)^.5 and p+q+r < (37.5)^.5 + 5 < 7+5. QED!!!
I'm sure i made this way too hard but that's all i could come up with before 6 AM!
Dave
Do we need to show that an isosceles right triangle maximizes c/(a + b) ??
Assuming not, let c be the hypotenuse:
c[sqr(2)]/2 + c[sqr(2)]/2 + c = 12
c[sqr(2)/2 + sqr(2)/2 + 1] = 12
c[sqr(2) + 1] = 12
c = 12/[sqr(2) + 1]
or, rationalized and simplified, 12(sqr(2) - 1)
about 4.97
Yuch. Ugly. That's what I was muttering before.
thanks, jonathan!
i found my careless error:
should be 2pq <= p^2+q^2
The isosceles right triangle does it! (by calculus) I should have known.
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