AIME Questions (I have to make USAMO)

[tongos]

ABC is a paper triangle with AB = 36, AC = 72 and ∠B = 90o. Find the area of the set of points P inside the triangle such that if creases are made by folding (and then unfolding) each of A, B, C to P, then the creases do not overlap.

My Ideas:
?

R is a 6 x 8 rectangle. R' is another rectangle with one vertex on each side of R. R' can be rotated slightly and still remain within R. Find the smallest perimeter that R' can have.

My ideas: How can this be? If it was rotated then it would seem to me that the vertices of the inscribed rectangle would pop out of the rectangle it is inscribed in.

thanks in advance

 

[tongos]

can someone help? thanks

 

[tongos]

anyone? please?

 

[devious_]

Hmm.. I can't get my mind around question #1, but here are my thoughts for question #2:

Let the points where the diagonals of R intersect be O, and the vertices of R and R' be ABCD and WXYZ respectively. Draw a diagram with R, and on each side of R let draw a point to represent one of the vertices of R'. If you picked suitable positions for the vertices of R', you can get some nice trig action. Say R has lengths AB and CD and widths BC and AD. Let W be on AB, X on AD, Y on DC, Z on CB. Now draw lines and complete R'. There are four equal angles, namely AWX, DXY, CYZ and BZX. Since WX=YZ, then AX=CZ, and thus XZ passes through O. (It might help to note that triangles AWX and CYZ are similar.) You can also do the same for WY.

Now XZ and WY are radii of a circle with center O (since that's where they intersect!). Previously I said that there were four equal angles, so let those angles equal T. You can use some trig to find values for AB and BC (and hence CD and AD). Some more manipulation (and maybe pythogras?) and I believe you can get the perimeter of R' in terms of T. I'm going to go ahead and guess, but if you do find the equation, you'd probably note that the perimeter would decrease as T increases, so it'll remain inside R. Try to find the largest possible value of T to find the smallest possible perimeter of R'. Unfortunately I have no idea how to do that, as I'm having a tought time trying to visualize the movement of the vertices of R'. Actually.. Will R' remain a rectangle throughout the movement?

Hrm. I've managed to confuse myself.

Oh well, I tried.

 

[devious_]

Wait.. Is the question asking for a complete revolution? Because it's impossible for a rectangle of fixed sides to remain inside another rectangle if its rotated like that.

If it's simply a revolution such that the vertices of R' are always in contact with the sides of R, then it's possible. Maybe you can try drawing diagrams and seeing when R' would be of really small perimeter, and then you could compare all the small ones and see where that leads you. i.e. what does it tell you about the vertices and angle T?

What a horrible question...

How hard is it to get to USAMO after taking the AIME? Do you need to answer ALL the questions? Personally I'd skip questions like this.

 

[DoubleMike]

I was thinking about this today, and it would seem that one diagonal has to be in the exact center of one of the larger rectangle's sides in order for it to rotate... It's freedom of rotation is then dictated by its other sides. That's all I got.

Is there a book with sample AIME questions that teaches you some of the algorithms or whatever?

 

[Hurkyl]

It seems to me that the obvious first steps are:

#1: Figure out how to find the crease when folding paper

#2: Determine a criterion for a vertex of the inscribed rectangle to remain inside the outer rectangle after a slight rotation.

 

[Hurkyl]

P.S.: I'm highly confident that there's a straightforward proof that the center of the two rectangles must coincide (whether or not the inner rectangle remains inside the outer after rotation)

 

[devious_]

I think I agree with DoubleMike. If, say, X bisected AD, then we can calculate areas of varios triangles and get that the area of R' is 24. Now set up an equation:
XY . YZ = 24 ... (1)

Now since X is on the midpoint of AD, draw a line from X perpendicular to BC (i.e. it intersects Z). The line would have length 8 (since it's of the same length as the longer sides of R). Now construct a triangle XYZ with a right-angle at Y. The hyp of this triangle is XZ, which equals 8 as we just found out. Now by pythagoras:
XY² + YZ² = XZ² = 64 ... (2)

Solve (1) and (2) simultaneously to get XY and YZ (you should get some surds) and then you should be able to find the perimeter of R'.

Edit:
You don't really need to solve for XY and YZ (that would take a lot of time and some boring manipulation, and I believe the AIME has a time limit, right?). You could notice that XY² + YZ² + 2(XY)(YZ) = (XY+YZ)² = 64 + 2(24) = 112. Then XY+YZ=sqrt{112} => 2(XY+YZ)=2sqrt{112}, which is the perimeter of R'.

 

[tongos]

devious, that's right! thanks. Both these questions are fifteens from different years (which means that their very difficult on the AIME)