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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1631.0. "1992 American Invitiational Math Exam" by BEING::EDP (Always mount a scratch monkey.) Fri Jun 19 1992 09:47

    Here are four more questions from the exam.
    
    
    				-- edp
    
    
    9. Trapezoid ABCD has sides AB=92, BC=50, CD=19, and AD=70, with AB
    parallel to CD.  A circle with center P on AB is drawn tangent to BC
    and AD.  Given that AP=m/n, where m and n are relatively prime positive
    integers, find m+n.
    
    10. Consider the region A in the complex plane that consists of all
    points z such that both z/40 and 40/conjugate(z) have real and
    imaginary parts between 0 and 1 inclusive.  What is the integer that is
    nearest the area of A?
    
    11. Lines l1 and l2 both pass through the origin and make
    first-quadrant angles of pi/70 and pi/54 radians, respectively, with
    the positive x-axis.  For any line l, the transformation R(l) produces
    another line as follows:  l is reflected in l1, and the resulting line
    is then reflected in l2.  Let R^1(l) = R(l), and for integer n>=2,
    define R^n(l) = R(R^(n-1)(l)).  Given that l is the line y = 19/92 x,
    find the smallest positive integer m for which R^m(l)=l.
    
    12. In a game of Chomp, two players alternately take bytes from a
    5-by-7 grid of unit squares.  To take a bite, the player chooses one of
    the remaining squares, then removes ("eats") all squares found in the
    quadrant defined by the left edge (extended upward) and the lower edge
    (extended rightward) of the chosen square.  For example, the bite
    determined by the B in the diagram would remove the B square and the
    squares marked by X.  (The - squares have already been moved; the O
    squares remain.)
    
    	OOX----
    	OOXX---
    	OOBX---
    	OOOOOOO
    	OOOOOOO
    
    The object of the game is to make one's opponent take the last bite. 
    The diagram shows one of the many subsets of the set of 35 unit squares
    that can occur during games of Chomp.  How many different subsets are
    there in all?  Include the full board and the empty board in your
    count.
T.RTitleUserPersonal
Name		DateLines
1631.1TRACE::GILBERTOwnership ObligatesTue Jun 23 1992 12:079
> 12. In a game of Chomp, ...

Easy.  Suppose a squares have been removed from the first row; let b <= a
be removed from the second row; let c <= b be removed from the third row, etc.
Then the total number of diagrams is:

	 7   a   b   c   d
	Sum Sum Sum Sum Sum 1
	a=0 b=0 c=0 d=0 e=0
1631.2Number 9CRONIC::CRONIC::MCINTYREFri Jun 26 1992 12:3646
Wow!  The AIME!  This was my favorite exam in high school, probably 
because its the one I did best on, and won some money from.  It wasn't 
much, but $40 to a high school student with no job seems like a lot of 
money, especially for doing something you enjoy!

This will be fun trying to figure these out.  If I can find my copies of
the first 2 AIME exams back in 1983 and 1984 I'll post the problems from
them. 

>    9. Trapezoid ABCD has sides AB=92, BC=50, CD=19, and AD=70, with AB
>    parallel to CD.  A circle with center P on AB is drawn tangent to BC
>    and AD.  Given that AP=m/n, where m and n are relatively prime positive
>    integers, find m+n.
    
This one's easy once you see how to do it.  

Of course, you want to find where the center P of the circle is on AB.

A simple way to do this is to find the point on AB where 2 equal length
line segments can be drawn to AD and BC, respectively, which will 
meet them at right angles.  The point where they meet will be the 
tangent points of the circle mentioned in the problem; call these points 
T  and T  .
 ad     bc

The line segments will be radii of the circle, so call their length r.

Now draw lines segments from AB and perpendicular to AB up to D and C
and call the perpendicular points on AB, D' and C', respectively.  The
length of these lines will be the height of the trapezoid, h.

Now notice that the right triangle APT   is similar to ADD', with a 
                                      ad
ratio in lengths of sides of r/h.  Also notice that BPT   is similar 
                                                       bc
to BCC', with the same r/h ratio in lengths.  

So  AP =  (r/h) AD  =  (r/h) 70
and PB =  (r/h) BC  =  (r/h) 50

So  AP = (7/5) PB
and AP = (7/12) AB = (7/12) 92 = 161/3

m=161, n=3, m+n = 164

Jon
1631.3CRONIC::CRONIC::MCINTYREFri Jun 26 1992 14:4538
> > 12. In a game of Chomp, ...
>
> Easy.  Suppose a squares have been removed from the first row; let b <= a
> be removed from the second row; let c <= b be removed from the third row, etc.
> Then the total number of diagrams is:
>
> 	 7   a   b   c   d
>	Sum Sum Sum Sum Sum 1
>	a=0 b=0 c=0 d=0 e=0

Well, that's halfway to the answer.  For those unfamiliar with the AIME,
all the answers are integers between 0 and 999.  So it's kind of like a 
multiple choice, with 1000 choices on each problem.  They don't care how 
you solve it, just that you get the right integer.

So the other half of this problem is to figure out how to count up all 
those sums.

There's a short cut to do this.  Using the observation above that the set 
of possible diagrams is the set of all diagrams for which each row has 
at least as many squares as the one below it, you can make a further 
observation that each possible diagram can be generated by simply starting 
at the top of the grid and limiting yourself to downward and rightward 
movements until you reach the bottom right corner of the grid.  If you 
start at the top left of the grid and generate every possible "path" 
along the edges of the squares to the bottom right corner, moving only 
right or down, you will generate every possible diagram.

Each of these paths will have length 12 (5 down, 7 to the right).  
Counting the paths can be done by simply counting the number of ways 
you can choose when to take your 5 "down steps" out of your 12 total 
steps (or your 7 "right steps" out of your 12 total).

This is "12 choose 5", or   12!
                          -------  =  11 * 9 * 8  =  792
                           5! 7!

Jon
1631.4HANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Thu Jul 23 1992 17:432
What about the strategy for chomp ?  Is that an interesting topic ?