Exactly two of the sides of a triangle are congruent and one of the equal angles is known to have degree measure greater than 50. How many integer values are possible for the measure of the remaining angle (the one that is different from the base angles)?

Comments:

(1) This is not intended to be a significant challenge. Rather it is meant as a warmup or for a slightly more extended discussion.

(2) Since middle school students know the sum of the angles of a triangle and can be told the basic fact about the base angles of an isosceles triangle, this problem is appropriate for them too.

(3) How would you expect most students to approach this? Do you think the majority would start by plugging in 51, 52, 53, etc.?

(4) Does this type of question promote important problem-solving skills and strategies? Do students recognize the significance of the 'boundary values' 50 and 90, values that are not in the domain of the base angles yet can be critical for the analysis?

(5) There are at least two 'traps' set in this problem that are intended to help students become more critical thinkers and not jump too quickly to conclusions. After all, what is a trap? If one is circumspect, details are not so easily overlooked.

## Wednesday, December 3, 2008

### Geometry Problem Requiring Critical Reading & Thinking

Posted by Dave Marain at 2:39 PM

Labels: geometry, inequalities, SAT-type problems, triangles

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## 4 comments:

I'm not a middle-schooler, so I probably approached this differently from the way they would.

I recognized the first trap right away, which is that just because the remaining angle has an integer value doesn't mean that the two equal angles are integral. So the first triangle is 50.5, 50.5, 79.

The smallest integral remaining angle is 1 degree, so the last triangle is 89.5, 89.5, 1.

So the first thought is that there are 79 different integer values possible for the remaining angle, from 1 to 79 inclusive.

Then I reread the sentence about the TWO traps, reread the problem description, and said AHA.

Back to "exactly two sides ... are congruent", which rules out the 60, 60, 60 solution.

So there are only 78 possible integers.

I bet if you gave this question without the hint about traps, the most common answers would be 38, 39, 78, and 79.

Of course, if there are MORE than two traps, then I've missed something, and will have egg on my face when I figure out what.

I knew I couldn't fool you, Susan!

My goal of course is to provide examples for teachers that improve critical thinking for students. Telling them in advance to look out for at least two 'traps' is not necessarily a bad thing to do when they are developing these skills. Actually, I know I would have appreciated being told that at any age! Over time, students learn to read more critically and reason more carefully when exposed to these kinds of problems. Of course, people like you and me actually find these to be 'fun' activities. I'm not sure all of my students from the past would agree. However, I do know they felt very good about themselves when they solved them.

Students and adults alike need to have their brain challenged. That's how children's minds develop and adults' minds keep from atrophying!

Ha! I fell for the equilateral, too. Curse me and my horrible reading skills.

Sean,

Considering how careless I've been lately, don't give it a second thought. Then again, giving it a second and even third thought is exactly what we want our students to do! In the end, we humans only learn from our mistakes -- which is why I'm a lifelong learner!!

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