## Saturday, September 15, 2007

### A Special Case of the Random Triangle Problem

jd2718's fascinating discussion regarding selection of a random triangle a couple of weeks ago led me to consider the strategy of particularization, aka, 'consider a special case' or 'make it simpler', approaches I typically suggested to students when faced with a problem that was confusing, vague or overly general. Instead of phrasing it in terms of geometric probability, I've reduced it to a much simpler problem, followed by a locus problem, a topic perhaps currently underemphasized.

First a 'simpler problem' fro your geometry students:

Consider points A(-3,0) and B(3,0) in the coordinate plane. If C(x,y) is a point in the plane such that angle ACB is a right angle, determine the value of x2 + y2.
Note: Many capable students would rush into the distance formula and the Pythagorean Theorem, but there is another approach that is less algebraically cumbersome!

Now for more generalization...

The following questions regard points in a single plane. If A and B are 2 arbitrary points in a plane, d units apart, determine the locus of all points C in that plane such that
(a) Angle ACB is right
(b) Angle ACB is acute
(c) Angle ACB is obtuse

Notes:
(i) Why do you think I chose to focus on angle ACB rather than triangle ACB as in Jonathan's discussion?
(ii) If students are unfamiliar with the term locus, rephrase as "Describe the set of all points C such that..."
(iii) Will this question lead directly to solving the general probability question raised by zac or Jonathan? Probably not, but I was thinking more in terms of accessibility for most geometry students. The open-endedness of the original triangle problem is highly instructive and can lead to profound considerations but that can be tackled later...

Anonymous said...

But extension to right triangle ABC in your first example is easy. I would consider asking that after discussion of the problem, although it is quite basic.

But look at this: if students worked through your generalized (second) question, then asking about right triangle ABC might be fun, and asking the acute vs obtuse questions bring us close to that flag...

Anonymous said...

By the way, the idea that we can pick up and modify each other's problems is an important one.

Borrow, modify, credit, share. This is a perfect example. And there should be much more.

In research mathematics this would be normal. In secondary math? Publishers and the bottom line get in the way.

But there can be something remarkably democratic, remarkably productive about math teachers talking on the internet...

Anonymous said...

It always helps if students take a deep breath and think first. In the original problem, AB is the hypotenuse. It's easy to find the altitude from C to AB. Just use the standard area formula to get 3|y|.

Part a of the second problem is easily guessable, though that doesn't make a real proof. Remember that a triangle inscribed on a semicircle is a right triangle. So, draw the circle with diameter AB, providing the school security police hasn't confiscated the students' compasses. Then any point on that circle other than A or B satisfies the result. A further guess would be that the exterior gives part b and the interior part c.

mathmom said...

I love how these blogs help teachers share, modify and think about great questions for their students.

The internet is a powerful resource that I think teachers are, for the most part, just beginning to see the power of and take advantage of. I already emailed my son's geometry (though they're studying counting and probability first, and haven't actually seen any geometry) teacher some pointers, for which she thanked me, but I don't know if she was just being polite. ;-)

As to this question, I see that the area of triangle ABC is 3|y| but I don't see how that particularly gets us closer to finding x^2 + y^2.

Anonymous said...

Hi Mathmom,

As Eric has pointed out, ACB is a right angle, so the locus of C forms a circle with diameter AB. (so the center is (0,0)). Thus x^2+y^2 = 9.

I actually don't see where Dave has asked for the area of the triangle, so 3|y| is an interesting answer, though to a different question.

Actually, the locus problem for the triangle, not just the angle, is straightforward too, and is discussed in the comments section of this topic on Jonathan's blog.

TC

mathmom said...
This comment has been removed by the author.
mathmom said...

Aha, thanks, I always forget about the fact that the hypotenuse of a right angled triangle is always the diameter of a circle whose circumference contains the third point. Or as Eric said much more mathematically, "a triangle inscribed on a semicircle is a right triangle."

Actually, Eric, I think you do have a proof for part (a) as long as the rule you gave is an "if and only if" which I believe it is.

My intuition for parts (b) and (c) is the same as yours, but I don't immediately see how to prove it either.

(And I had to look up locus.) ;-)

Anonymous said...

Sorry, mathmom. I must have been hallucinating.

Dave Marain said...

jonathan--
(2) Yes, I agree, that there is much fertile ground from sharing each other's ideas, provided we give proper attribution! My focus is usually less on solving opn your classic puzzlers, more on how to modify it for students in both honors and regular classes.

mathmom, eric, anonymous--
My preferred method of solution is the one anonymous gave. I gave this question to a group of students (SAT) yesterday and a few employed a test-taking 'strategy', but one that's worthy of discussion. I had drawn a diagram for the problem, placing point C in the 2nd quadrant, near the y-axis. Two students made the assumption it was actually ON the y-axis and figured it was the point (0,3) by symmetry and 45-45-90 considerations. This trivialized the problem but one commented that since I didn't really specify the location of C, it probably meant that C could be placed anywhere and the value of the expression would not change. This is more than an SAT 'trick'- I saw it as a teachable moment and a subtle point about WHY we often CAN assume special cases even though we frequently tell students to NEVER ASSUME! No one thought about the fact that C had to be on a circle of radius 3, centered at (0,0). A couple of the stronger Algebra 2 students actually worked through the distance formula and the Pythagorean Theorem to obtain the same result. It really wasn't that messy and I'll show it here because it's useful for the obtuse and acute cases:
(x+3)^2 + y^2 + (x-3)^2 + y^2 represents the sum of the squares of sides AC and BC respectively. This simplifies to
2x^2 + 2y^2 + 18.
Angle ACB is obtuse if
(AC)^2 + (BC)^2 < (AB)^2. This is a less well-known variation on the Pythagorean Thm and can be proved easily by the Law of Cosines.
Thus, for the obtuse case,
2x^2 + 2y^2 +18 < 36 or
x^2 + y^2 < 9 which is equivalent to stating that point C is INSIDE the circle. Similar argument in the acute case. I did not have time to discuss this aspect with the students.
The locus problem is a straightforward generalization using the circle whose radius is d/2 and whose center is the midpoint of segment AB. Coordinatizing the problem makes it easier to prove results.

eric--
Could you explain how you would use the altitude, |y|, or the area expression 3|y| to obtain the value of x^2+y^2? Did you choose a special case for y, namely y = 3?

Anonymous said...

Dave,

Please ignore the first half of my first comment--I must have had a brain cramp.

Unknown said...

Is there something like an 'Inverse Locus' problem, i.e., I give you the locus, and you try to find what it is the locus for.

Example (1) The locus is a straight line. This can be the set of points equidistant from two given points.

I believe the answer is non-unique, though, as in Example (2): The locus is a circle. Two possible answers: (a) The set of points equidistant from a given point, or (b) the set of points, when chosen as a vertex along with two given points as end points, will give a right angle. ( I guess I could work on the language for (b))

Food for thought/discussion?

TC

Anonymous said...

T.C.,

There's a funny anecdote along those lines. When topologist R.L. Moore held his long-lasting seminar at U. texas, students would challenge each other to find spaces that satisfied various sets of properties. Quite often, Moore would announce that the student had found yet another characterization of the 1-point topological space. They eventually collected 200 such example.

Anonymous said...

Hi