## Monday, February 13, 2012

### the best I could do after much toiling

[REMINDER: Anonymous comments are never published.]

Ah, the things I do for my goddaughter. After hours and hours of struggle, I finally came up with what I think is a proof that will work (see previous post for the original problem). What you'll see below is called an indirect proof, also known as a proof by contradiction. The idea is to assume something that will lead to a contradiction, then make deductions therefrom. Because the hour is late and I've been working on this thing all evening, there's a good chance that the reasoning may be sloppy. I invite you to check the proof out and notify me of any holes you may find.

_

Kevin Kim said...

An anonymous commenter (comment deleted because it was anonymous, per the commenting policy written directly above the comment box) claimed that lines C and D cannot be parallel if angles 10 and 16 are 90 degrees.

Imagine the above lines drawn in a tic-tac-toe configuration, with everything orthogonal. In that scenario, angles 10 and 16 are both 90 degrees and thus equal to each other. That configuration is a possible way for 10 and 16 to be equal.

Assume C and D are parallel, then imagine a scenario in which transversal N does NOT hit lines C and D at 90 degrees. 10 and 16 cannot be equal, but it's given that they are. Assuming parallel lines, the transversal MUST be perpendicular.

But the commenter's insight does bring an uncomfortable fact to light: if we assume C and D are not parallel, with transversal N passing through them, then it becomes obvious that, in such a situation, it's conceivable for angles 10 and 16 to be equal.

In other words: my proof is wrong because it seems to insist on parallelism for C and D, but C and D don't have to be parallel to satisfy the given (i.e., that angles 10 and 16 are congruent).

So: I thank my commenter for leading me to the correct solution, and for helping me see my own failure of imagination in this instance, but I'll also ask that s/he identify him-/herself from now on. I don't publish anonymous comments, no matter how intelligent.

David Wester said...

The claim that c and d can not be parallel if angle 10 and angle 16 are equal is just plain wrong.

It is in fact, the only condition where c and d CAN be parallel.

The issue with your proof is that you assume angle 10 HAS so be greater than OR less than 90. The third possible value is 90, which is indeed the solution.

Kevin Kim said...

Mr. Wester:

"The claim that c and d can not be parallel if angle 10 and angle 16 are equal is just plain wrong."

I completely agree. See my previous comment, just above yours.

"The issue with your proof is that you assume angle 10 HAS so be greater than OR less than 90. The third possible value is 90, which is indeed the solution."

I think you've misunderstood the proof. A proof by contradiction is SUPPOSED to lead to absurdities, which is what it did-- twice in that proof-- to indicate exactly what you're contending, i.e., that 90 degrees is the only possible solution for which angles 10 and 16 can be equal IF lines C and D are parallel.

Putting it more concisely: my proof basically did this--

1. Let's ask ourselves what happens if we assume C and D are parallel and angle 10 is less than 90 degrees. Absurdity results.

2. Let's ask ourselves what happens if we assume C and D are parallel and angle 10 is greater than 90 degrees. Absurdity results.

Unfortunately, the proof didn't do a very good job of explicitly stating that angles 10 and 16 must be 90 degrees, respectively. But that idea is there by strong implication (see line 15 of the proof). The only three options are less than, greater than, and equal; the proof shows that inequalities lead to absurdity, given my assumptions.

However, as I noted in a subsequent post, my proof is fundamentally incorrect because it fails to prove that lines C and D must be parallel (which was the original goal: proving whether any of the lines are parallel, given 10/16 congruence). There are other scenarios in which angles 10 and 16 can be equal without C and D being parallel (see the above link).

And thus do I fall on my sword.

Dave Wester said...

Kevin,
You are too hard on yourself. But the proof, as you have it, only covered 2 cases out of a possibility of 3. The third possibility, being the correct one, is untouched by the absurdities of invalid cases. At least, thats my gut reaction - which is always a spotty way to "know" anything, as I have come to learn.

Kevin Kim said...

Thanks, but I must respectfully disagree with your gut (which is apparently more mathematically inclined than mine is; my gut is fixated on the Food Network).

While the proof should have been more explicit, it does cover the third possibility by strong implication, as there are only three possibilities (<, >, =). By eliminating two of them, only the third was left (cf. Sherlock Holmes on "eliminating the impossible"). So I plead "not guilty" to neglecting the third possibility. If it was neglect, it was, at worst, a benign neglect. Heh.

However, the real problem, as we both agree, was my attempt to shoehorn the congruent angles into a scenario involving parallel lines. A classic blunder: assuming what you want to prove. And that assumption is, unfortunately, written right into my proof on line 2.