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PLS NOTE: BLOGGER HAS APPARENTLY BEEN DOWN FOR MOST OF THE DAY. I APOLOGIZE FOR ANY INCONVENIENCE THIS MAY HAVE CAUSED. I WILL TRY TO PUT UP NEW PROBLEMS FOR 2-2-07 BUT NO GUARANTEES AS TO IF OR WHEN...
Just 3! Have fun!!
Note for #1: Assume k is a positive integer.
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12 comments:
1. E
2. 21, 22
3. 40
I like the last one a lot! Thanks for these, they are fun to do after a long day...
On (1), what is a clown quiz?
(2) is nice. Is there a way of knowing whether to choose the first number as odd or even before doing the problem rather than inferring it after things work out or don't?
On these kinds of geometry problems like (3), you rock, man!
Totally_clueless
no need for an answer key - see mrc! nice job
TC --
a clown quiz is one-third of a clown test!
(2) has several purposes and can be approached in a variety of ways; beyond the sequence notation and piecewise function, I want students to use number sense and estimate: well, the square should be the dominant term and 20^2 = 400, but that's too samll, so let's go to 21,22 - bingo! Of course, I would have used algebra in school and tested both cases: odd,even and even,odd...
(3) is an original from just playing around with triangles in squares and rectangles.
So here's the 'what if':
What if AE = 3 instead of 4, all other conditions the same. No congruent triangles this time...
For the what-if, no congruent triangles, but still similar triangles. The numbers are much messier, but I'm coming up with 73*9/16 which is, um, not an integer. :) But it is rational, assuming I got it right.
On 2, if I wanted estimation, I would need to make the algebra tougher
x^2 + 2(x + 1) = 485
(or x-1, it is not necessary to assume 1st even or 1st odd)
Maybe a quarter the square of evens and two less than double the aquare of the odds... (sum is 1177)
I don't know.
Hi Dave,
Here is a problem suggestion:
The legs of a right triangle are 6 and 7 units long. Find the radius of the circle that passes through the three vertices of the triangle.
Nice one, totally! I hadn't run into it, and wasn't sure what to do. I ended up throwing it onto coordinates, and it dropped out nicely (although I thought it was going to be a 'nice' answer)
totally, is it sqrt(85)/2?
I'm thinking you use the "inscribed angle is half the central angle" theorem to show that the hypotenuse is also a diameter of the circle. But my circle geometry is weak, so I might have that wrong. :)
Are your students comfortable with the idea of a function being defined by two rules on different subsets of the domain? If they are, you might want to introduce the 3n+1 problem to them.
Mathmom, your answer and approach are correct, and along the lines of what I was thinking.
Jonathan, as an alternative approach, I tried the co-ordinate thing, it works out much easier if you choose the points wisely.
A third approach is to actually find the perperdicular bisectors of the sides. Using the theorem that says that a line that bisects a side and is parallel to another side bisects the third side, it can be shown that the perpendicular bisector of any leg and the hypotenuse meet at the midpoint. Thus, the midpoint is the center of the circumcircle, and the answer follows.
TC --
your problem motivated me to write the right triangle problem for 2-2-07. The other piece is that the midpoint of the hypotenuse is equidistant from the 3 vertices, a theorem that is non-trivial IMO! Thanks!
I gave problem 3 as a challenge today to my Geometry classes, on an otherwise slow day due to most students having taken the CAHSEE math test all morning. Surprisingly, a good number of them were willing to take up with the problem and put some effort into it even though I allowed them to do any other work they wanted to. They chose the challenge! I gave very small hints, but none of them managed it all the way through.
I will use this problem tomorrow as the launching point for a review of quadrilaterals, congruence theorems, and right triangles. The more I play with this problem, the more I like it!
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