jump to navigation

Some unsolved problems January 3, 2010

Posted by Damien Jiang in Problem-solving.
Tags: , , , , ,
trackback

Happy New Year!

Since we have been too lazy to post lately (and the so-not-lazy Akhil posts mostly elsewhere now), I’m going to post some problems that I probably should be able to solve, but haven’t.

First, an interesting (and apparently classic) problem given to me by Tim Chu, USA IMO alternate:

Find all such that f is multiplicative (for relatively prime positive integers), monotonically increasing, and f(2) = 2.

A second from an old IMO longlist:

Show that the union of all segments with endpoints in includes every point in .

EDIT: Well apparently this is wrong. However, part b of the problem asks if the convex hull of is equal to .

And a third from Russia 2005:

A quadrilateral ABCD without parallel sides is circumscribed around a circle with center O. Show that O is the barycenter of ABCD iff (OA)(OC) = (OB)(OD).

(Looks quite vector/complex-number friendly, but… I probably didn’t try hard enough.)

About these ads

Comments»

1. Akhil Mathew - January 3, 2010

In the second one, you mean the union, no?

2. Damien Jiang - January 3, 2010

of course. I am silly.

Akhil Mathew - January 3, 2010

Actually I just realized this can’t be the case—a line segment has 2-dimensional measure zero, and a countable union of them can’t fill \mathbb{R}^2.

Damien Jiang - January 3, 2010

lol maybe this is why I couldn’t solve it? Now I’m hoping I didn’t misread the problem. Time to go take another look…

3. Damien Jiang - January 3, 2010

Well, that’s what the problem says; I guess it’s wrong.

4. Maurizio Monge - January 3, 2010

There also is an “algebraic” proof that the union of segments with rational endpoints cannot be all R^2: if (x,y) is in a segment from (a,b) to (c,d), than it is straightforward to see that x,y,1 must be linearly dependent over Q. But if the union was all of R^2, than it would not be possible to find three real numbers that are independent over Q, and R would be an algebraic extension of degree at most 2 over Q, which is absurd.

5. Sam - January 10, 2010

There’s also a more elementary approach just considering what the slope of the segment would be, on a case by case basis.

6. ks - January 10, 2010

A quadrilateral ABCD without parallel sides is circumscribed around a circle with center O. Show that O is the barycenter of ABCD iff (OA)(OC) = (OB)(OD).

choose the coordinate system so that a represents A. and let the angles
at A, B , C, D be 2t1, 2t2, 2t3 and 2t4, modulus of B, C, D are b, c, d.
t’s are constrained by t1 + t2 + t3 + t4 = 180 and sin(t1) = 1/a
cos(t1) = a1/a where a1 = \sqrt(a^2-1) and similar expressions
for t2, t3 and t4.

the arguments of B, C, D are 180 – t1 -t2, 180 -t2 + t4 and -t2 -t4. let g be the center of mass, then
4g = a + be^i(180-t1-t2) + ce^i(180-t2+t4)+de^i(-t2-t3)
4ge^it2=ae^it2 – be^-it1-ce^it4+de^-it3
re(4ge^it2) = ab1/b – ba1/a -cd1/d + dc1/c
= (acb1+bdc1)/bc -(acd1+bda1)/ad —-(01)
im(4ge^it2) = a/b + b/a -c/d -d/c
= (ac-bd)/bc -(ac-bd)/ad ————-(02)

now we use sin(t2+t3) = sin(180 – t1 -t4) which gives
1/b c1/c + 1/c b1/c = 1/a d1/d + 1/d a1/a which simplifies to
(b1+c1)/bc = (a1+d1)/ad ——————–(03)

ac = bd and (03) gives g = 0.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 39 other followers

%d bloggers like this: