Well, the June SATs have arrived today so these problems come too late for that, but these kinds of questions can be used to review basic ideas while strengthening thinking skills. Both questions below are appropriate for both middle and secondary students, although the second requires knowledge of a fundamental geometry principle regarding the sides of triangles.
There are other important principles embedded in these problems as well. In the end, I believe that students need to be exposed to many of these "contest-type" challenges to improve reading skill, learn how to pay attention to detail and think clearly. As a separate issue, performing well under testing conditions requires extensive training. You may not feel this is an important objective for math teaching in the classroom, but testing is a reality for the student...
These questions may appear fairly straightforward at first but be careful! I believe the second is more challenging than the first. These are not so different from the "gotcha" problem on our latest online contest.
1) The dimensions of a rectangle are odd integers and its perimeter is 100. How many different values are possible for its area?
2) The perimeter of an isosceles triangle is 96 and the lengths of its sides are even integers. How many noncongruent triangles satisfy these conditions?
For my "unofficial" answers, click on Read more...
Unofficial Answers (no solutions):
1) 13
2) 11
Feel free to challenge these answers or express agreement!
Comments
Which of the following do you believe would cause the most difficulty for students?
- The wording/terminology (e.g., noncongruent); general reading comprehension issues
- The sheer number of details (e.g., odd vs. even, perimeter vs. area, integer values)
- A precise counting/listing strategy vs. an abstract or commonsense approach
- The "square is also a rectangle", "equilateral is also isosceles" traps
- The issue of different areas for #1
- The triangle inequality for #2
- Other concerns?
Posts
9 comments:
Now I'm concerned ...
#2) I got 22 and a possible 23rd if you count the one special case as isosceles - what the heck am I missing?
#1) I agree with the 13.
I may have messed up but I believe the "base" of the triangle must be a multiple of 4. E.g., if the base were 2, then the 'legs' would be 47 and 47. The conditions required that all sides be even integers! Thus the 'base' could be any positive integer multiple of 4 up to and including 44, hence 11 possibilities. This includes the equilateral case.
Holy Guacamole.
I actually had a triangle that was 2,2,92 ...
I am really wiped out. This year really needs to end and I've got two weeks left.
I know that feeling and I'm retired!
Compared to some of the errors I've made....
At least yours can be attributed to end of year brain freeze!
Seriously, when you get a different answer from mine, I assume I made the error!
For part 2, let the triangle have sides a, a, and b. B the triangle inequality and the the sides being even,
2 ≤ b ≤ 2a-2.
So, 2a+2 ≤ 2a+b ≤ 4a - 2.
2a+2 ≤ 96 ≤ 4a - 2.
This is where students will have the most problems.
2a + 2 ≤ 96 means a ≤ 47, and so a ≤ 46 by the problem conditions.
96 ≤ 4a - 2 means 98 ≤ 4a, 24.5 ≤ a and so 26 ≤ a by the problem conditions.
26 ≤ a ≤ 46. This sort or inequality reversal always seems to give students fits. Since a must be even, there are 11 solutions in this range.
Beautiful analysis, Eric, and some nice manipulations.
I would agree with you about about the inequalities for those students who approach this problem analytically. However, when confronted with this type of question, most students I've worked with plug in values for the sides. They might start with the largest possible value of the equal sides or, equivalently, as I did, the smallest possible value of the 'base'. If they start with a base of 2, they have to remember to keep the other sides even. My gut feeling is that it's one or more of the following which would trip them up:
the even condition OR
the triangle inequality OR
the equilateral case.
As an instructor, I would certainly want to show them the algebraic analysis as well. I also think it's a good exercise in simple number theory for them to understand why the 'base' has to be a multiple of 4.
Problems like these seems to be much more algebraic than geometric. Even though they speak of areas and perimeters, the restrictions on the sides to integers of specific parities mean that their geometric contents are minimal. Only the most basic techniques of geometry show up in their solutions. Last week's problem on the length of the tangent to a circle in the first quadrant was much more geometric.
Whether the student counts the number of solutions or applies algebra does not really matter. I would prefer an algebraic solution, but the time pressure of the SAT makes counting plausible.
Agreed, Eric, on all counts...
The bigger issue for me is whether there should be a place for these kinds of problems in a regular curriculum. I think you know how I feel but a natural question is how and when does one bring these into the classroom.
I belive student would spend time trying to find all the solutions by using a listing strategy. Few student wold take an abstract approach. For example in the first question what do we really need. we need two odd numbers that add to fifty, and we have our answer. there are 25 combinations that add to fifty, twelve of which are two even numbers, and thirteen are two odd numbers. I do not think most students would think of that initially.
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