GR0568 problem #39

 Posts: 7
 Joined: Mon Sep 21, 2009 7:31 am
GR0568 problem #39
Anyone know how to do this?
I'll do my best to describe the problem without the use of a diagram.
We are given a triangle, with sides of length 1,s and r. The angle opposite the side of length s measures 110 degrees. As r and s increase, the side whose length is 1 remains fixed at 1, and the 110 degree angle remains fixed at 110 degrees. What is the limit (as both r AND s go to infinity) of the difference sr?
(A) 0
(B) A positive number less than 1
(C) 1
(D) A finite number greater than 1
(E) Infinity
I'll do my best to describe the problem without the use of a diagram.
We are given a triangle, with sides of length 1,s and r. The angle opposite the side of length s measures 110 degrees. As r and s increase, the side whose length is 1 remains fixed at 1, and the 110 degree angle remains fixed at 110 degrees. What is the limit (as both r AND s go to infinity) of the difference sr?
(A) 0
(B) A positive number less than 1
(C) 1
(D) A finite number greater than 1
(E) Infinity
Re: GR0568 problem #39
Law of Cosines gives s^2 = r^2 + 1  2rcos(110degrees)
s^2  r^2 = 1 + kr , k < 1 but close to 1 (cos110 is close to cos120 = .5)
factor (s^2r^2)
sr = 1/(s+r) + kr/(s+r).
as r, s go to infinity together, 1/(s+r) certainly goes to 0. kr/(s+r) would seem to go to k*1/2, implying (B) A positive number less than 1. Nonrigorous though. Any ideas on how to make that precise? I mean, you can say: let r=s, then just take a single variable going to infinity, to get this result, but this isn't satisfactory I don't think. especially since that would indicate the left side is 0.
s^2  r^2 = 1 + kr , k < 1 but close to 1 (cos110 is close to cos120 = .5)
factor (s^2r^2)
sr = 1/(s+r) + kr/(s+r).
as r, s go to infinity together, 1/(s+r) certainly goes to 0. kr/(s+r) would seem to go to k*1/2, implying (B) A positive number less than 1. Nonrigorous though. Any ideas on how to make that precise? I mean, you can say: let r=s, then just take a single variable going to infinity, to get this result, but this isn't satisfactory I don't think. especially since that would indicate the left side is 0.

 Posts: 7
 Joined: Mon Sep 21, 2009 7:31 am
Re: GR0568 problem #39
That seems to work. I guess for the term
$$\frac{kr}{s+r}$$
we can maybe use the fact that
$$s>r\Rightarrow \frac{kr}{s+r} < \frac{kr}{r+r} = \frac{k}{2},$$
giving (B).
Thanks prong, this was doing my head in!
$$\frac{kr}{s+r}$$
we can maybe use the fact that
$$s>r\Rightarrow \frac{kr}{s+r} < \frac{kr}{r+r} = \frac{k}{2},$$
giving (B).
Thanks prong, this was doing my head in!
Re: GR0568 problem #39
Ahh! That is the missing step. The Law of Cosines always holds, and s > r always holds (because of the law of sines, or just visualize it) so kr/(s+r) < kr/(r+r) holds for every s,r. Nice.

 Posts: 7
 Joined: Mon Sep 21, 2009 7:31 am
Re: GR0568 problem #39
Yes, I just hope this kind of stuff occurs to us within 2.5 minutes during the exam!

 Posts: 9
 Joined: Tue Sep 29, 2009 1:07 pm
Re: GR0568 problem #39
There's a much simpler solution to this problem. It's rather informal, but with a little work (and time, which we don't have during the test) it can be made rigorous. Here it goes:
Let b the inner angle to the right and a the one above. A simple application of the sine law shows that b goes to zero when r and s go to infinity. In particular, a converges to 70 degrees. So, the picture we have after taking limits is two parallel halflines joined together by a segment whose angle with the bottom halfline is 110 degrees and to the top one is 70 degrees.
If we lift a perpendicular segment at the point of intersection between the bottom halfline and the segment described previously, we form a right triangle whose inner angles are 70 degrees on the top left and 20 degrees on the bottom. The quantity we seek is precisely the length of the base of this triangle, which can be easily seen to be cos(70 degrees), a real number greater than zero but less than one.
It took me about 20 minutes to rigorously prove all these claims, and if anyone wants a detailed proof just ask.
Let b the inner angle to the right and a the one above. A simple application of the sine law shows that b goes to zero when r and s go to infinity. In particular, a converges to 70 degrees. So, the picture we have after taking limits is two parallel halflines joined together by a segment whose angle with the bottom halfline is 110 degrees and to the top one is 70 degrees.
If we lift a perpendicular segment at the point of intersection between the bottom halfline and the segment described previously, we form a right triangle whose inner angles are 70 degrees on the top left and 20 degrees on the bottom. The quantity we seek is precisely the length of the base of this triangle, which can be easily seen to be cos(70 degrees), a real number greater than zero but less than one.
It took me about 20 minutes to rigorously prove all these claims, and if anyone wants a detailed proof just ask.
Re: GR0568 problem #39
Interesting. Although I wouldn't necessarily call that simpler, the Law of Cosines is pretty simple. Maybe that's just me being in love with my own solution.
Re: GR0568 problem #39
You guys think to much.
This was what I had on my paper:
_________
\_________
the diagonal segment is 1, so obviously the difference is less than 1.
This was what I had on my paper:
_________
\_________
the diagonal segment is 1, so obviously the difference is less than 1.
Re: GR0568 problem #39
Agreed. This type of qualitative reasoning saves valuable time on the exam.origin415 wrote:You guys think to much.
This was what I had on my paper:
_________
\_________
the diagonal segment is 1, so obviously the difference is less than 1.