## December 2003

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Solution:

(Part 1):

If |AB|=|DC| then P is a parallelogram, which we already solved.
Assume |AB| > |DC| (AB is longer than DC).
Let h denote the height, the distance between the parallel
lines AB and DC.
Continue sides AD and BC until they meet at point E,
at height g=h*(|AB|)/(|AB|-|DC|) above AB.
Draw a line parallel to AB, at distance
√{(g-h)*g} = g*√{|DC|/|AB|} from E.
This line partitions P into two trapezoids, similar to each other,
one larger than the other by a factor √{|AB|/|DC|}.

(Part 2):

Let P be a convex quadrilateral whose angles are linearly
independent over the rationals; that is, no nontrivial integer linear
combination of the four angles gives 0. For concreteness,
suppose their measures (in degrees) are

α_{1}=180/√(5), α-2=180/√(7),
α_{3}=180/√(11), α_{4}=360-α_{1}-α_{2}-α_{3}.

Suppose P admits a "similar dissection" into n tiles, each
having k sides. Let the angles of the tile be β_{1},...,β_{k},
where each β_{i} is strictly between 0 and 360, and none is 180.
(A k-gon with an angle of 180 is really a k-1-gon.)

The dissected polygon has (say) a total of m+r+4 vertices:
m interior vertices where the angles of the various tiles sum to 360;
r vertices where the angles of the various tiles sum to 180 (these
vertices can be on a side of P or on a side of another tile, which
would contribute the other 180 degrees); and the four vertices A,B,C,D
of the original polygon. Since each tile contributes 180(k-2) degrees,
we have

360*m+180*r+(α_{1}+α_{2}+α_{3}+α_{4})=180*(k-2)*n,
2*m+r+2=(k-2)*n.

We have m+r+4 equations, each expressing an angle (α_{i}, 360 or 180)
as a nonnegative integer combination of β_{j}. When β_{j} appears
in one of these equations, we say that β_{j} is a "component" of
the appropriate angle (α_{i}, 360 or 180).

Suppose first that one of these m interior vertices meets only two
tiles, so that two angles sum to 360: β_{a}+β_{b}=360, with
β_{a}>β_{b}. So β_{a}>180, and since P is convex, β_{a}
is not a component of any α_{i}, nor of any of the r instances
of 180. So β_{a} appears n times in our m+r+4 equations, each time
as a component of 360. In each of these appearances, if it is not
exactly β_{a}+β_{b}=360, but rather
β_{a}+(∑ c_{j} β_{j}), we will find another of our equations
that contains β_{b} but not β_{a}, and exchange the
(∑ c_{j} β_{j}) with the β_{b}, to produce an equation of the
form β_{a}+β_{b}=360. Our equations no longer correspond to the
dissection, but that's all right.

If we eliminate the n equations of the form β_{a}+β_{b}=360,
and eliminate the two angles β_{a} and β_{b} from our tiles,
we come up with a set of (m-n)+r+4 equations relating the k-2
remaining tile-angles. It is as if we had started with (k-2)-gons,
m-n vertices of 360, r vertices of 180, and the four angles α_{i}.

Continue until there is no equation of the form β_{i}+β_{j}=360.

So we can assume that each of the m equations for 360 involves at
least three tile angles (counting multiplicity), and each of the
r equations for 180 involves at least two.
Among the m+r+4 equations, the number of tile angles (including
multiplicity) is then

n*k ≥ 3*m + 2*r + 4.

Combine with the earlier equation

n*(k-2) = 2*m + r + 2,

(subtracting 3 times the equation from 2 times the inequality)

to obtain

n*(6-k) ≥ r+2.

So k is at most 5.

Because no linear relation exists among α_{i}, the four equations
expressing α_{i} as sums of β_{j} must involve at least four
distinct β_{j}. So k must be at least 4.

If k=4, then from
α_{1}+α_{2}+α_{3}+α_{4}=360=β_{1}+β_{2}+β_{3}+β_{4},
we must have α_{i}=β_{i}, i=1,2,3,4. Then r must be 0 (no
combination of β_{i} gives 180). Our transformations (to get
rid of β_{a}+β_{b}=360) did not change r, nor did they decrease
the number of tile angles composing α_{i}, so that in the original
dissection we must have had r=0 (no tile vertices along the sides
of the large polygon) and each angle α_{i} was just one tile angle.
So the dissection was a single tile.

Suppose k=5. If α_{i}=β_{i} for each i=1,2,3,4, then from
∑ α_{i}=360 and ∑ β_{i}=540
we would have β_{5}=180, impossible.
Not all β_{j} are components of α_{i}, since otherwise
∑ α_{i}=360 would be at least ∑ β_{i}=540.
So exactly four of the β_{j}
are components of α_{i}, and we can write
360=∑ α_{i}=c_{1}*β_{1}+...+c_{4}*β_{4},
with all c_{j} positive integers.
One of them, say c_{1}, is at least 2.

Take one of the occurrences of angle β_{5},
either as one of the m equations expressing 360,
or as one of the r equations expressing 180;
in the latter case, multiply by 2 to get 360 on the right hand side.
We get an equation giving 360
as a nonnegative integer linear combination of β_{1},...,β_{5}:
say 360=d_{1}*β_{1}+...+d_{5}*β_{5}, with d_{5}>0.
So we have three equations on the β_{i}:

c_{1}*β_{1} + ... + c_{4}*β_{4} = 360,
d_{1}*β_{1} + ... + d_{5}*β_{5} = 360,
β_{1} + ... + β_{5} = 540,
with c_{i} and d_{j} nonnegative integers and
c_{1} ≥ 2, c_{2} ≥ 1, c_{3} ≥ 1, c_{4} ≥ 1, d_{5} ≥ 1.

Compute (3*d_{5}-2) times the first equation, plus 2 times the second
equation, minus (2*d_{5}) times the third equation.
The right-hand sides cancel, as do the β_{5}.
We are left with the equation
f_{1}*β_{1} + .. + f_{4}*β_{4} = 0,
with f_{i} integers.

But since (α_{1},...,α_{4}) are linearly independent over the
integers, (β_{1},...,β_{4}) are also linearly independent,
so that f_{i}=0.

In particular,

0 = f_{1} = (3*d_{5}-2)*c_{1} + 2*d_{1} - 2*d_{5}.

The only solution satisfying d_{5} ≥ 1, c_{1} ≥ 2, d_{1} ≥ 0 is
d_{5}=1, c_{1}=2, d_{1}=0.

So we learn that d_{5}=0.

Now looking at f_{i}, i=2,3,4:

0 = f_{i} = (3*d_{5}-2)*c_{i} + 2*d_{i} - 2*d_{5}
0 = c*i + 2*d_{i} - 2.

The only integer solution with c_{i} ≥ 1 and d_{i} ≥ 0 is
c_{i}=2, d_{i}=0.

Then β_{5}=360, again impossible.

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