aime26-17

Proposed solution

I need to find the number of possible values of AB for the given pentagon with specific angle conditions and area constraints.

Let me set up a coordinate system to analyze this problem systematically.

Setting up coordinates:
Let me place point A at the origin (0, 0) and point E on the positive x-axis at (20, 0) since AE = 20.

Since ∠A = 90°, side AB is vertical. Let B be at (0, h) for some positive value h.
Since ∠E = 90°, side ED is vertical. Let D be at (20, k) for some value k.

Since ∠B = 45° and B is at (0, h), the angle between AB and BC is 45°.
Since AB points upward from A, BC must make a 45° angle with the vertical.
This means BC makes a 45° angle with the positive y-axis.

Since ∠D = 45° and D is at (20, k), the angle between ED and DC is 45°.
Since ED points downward from E, DC must make a 45° angle with the vertical.
This means DC makes a 45° angle with the negative y-axis.

Let me reconsider this more carefully.

Actually, let me think about the angles properly:
- At vertex A: ∠A = 90°, so the angle between AB and AE is 90°
- At vertex E: ∠E = 90°, so the angle between ED and EA is 90°  
- At vertex B: ∠B = 45°, so the angle between BA and BC is 45°
- At vertex D: ∠D = 45°, so the angle between DC and DE is 45°

Let me place A at (0, 0) and E at (20, 0).

Since ∠A = 90°, AB is perpendicular to AE. So B is at (0, h) for some h > 0.

Since ∠E = 90°, ED is perpendicular to EA. So D is at (20, k) for some k.

Now, since ∠B = 45°, the angle between BA and BC is 45°.
BA points from B to A: BA = (0, -h)
BC makes a 45° angle with BA.

Since we're looking at the interior of the pentagon, and B is at (0, h), BC should point toward the right (since we want the pentagon to be nonconvex in a reasonable way).

If BA points downward, then BC pointing at 45° from BA means BC points toward the upper-right.
So if BA = (0, -h), then BC = (h, h) or BC = (-h, h) depending on which direction.

But we also know that BC = 14√2.

Let's say B is at (0, h) and C is at (x, y).
Then BC = (x, y-h) and |BC| = 14√2.
Also, the angle between BA = (0, -h) and BC = (x, y-h) should be 45°.

The angle between vectors u and v is given by cos θ = (u·v)/(|u||v|).

BA·BC = (0, -h)·(x, y-h) = 0·x + (-h)(y-h) = -h(y-h)

|BA| = h
|BC| = 14√2

So cos(45°) = -h(y-h)/(h·14√2) = -(y-h)/(14√2) = √2/2

This gives us: -(y-h) = 14√2 · √2/2 = 14

So y - h = -14, which means y = h - 14.

Since BC = 14√2 and y = h - 14:
x² + (h-14-h)² = (14√2)²
x² + 196 = 392
x² = 196
x = ±14

Let's say x = 14, so C is at (14, h-14).

Similarly, for the angle at D:
D is at (20, k).
DE points from E to D: DE = (0, k)
DC makes a 45° angle with DE.

Since ∠D = 45°, and we're looking at the interior of the pentagon, DC should point toward the left.
If DE = (0, k), then DC should make a 45° angle with this vector.

The angle between DE = (0, k) and DC is 45°.
If DC = (a, b), then:
cos(45°) = (0·a + k·b)/(k·|DC|) = kb/(k·14√2) = b/(14√2) = √2/2

So b = 14.

Also, |DC| = 14√2, so a² + b² = 392.
a² + 196 = 392
a² = 196
a = ±14

If DC points toward the left (negative x-direction), then a = -14.

So D is at (20, k) and C is at (14, h-14).
The vector DC = (14-20, h-14-k) = (-6, h-14-k)

But we said DC = (-14, 14), so:
-6 = -14 → This is wrong.

Let me reconsider the constraint that points B, C, D are on the same side of line AE.

Since AE is along the x-axis from (0,0) to (20,0), and we want B, C, D on the same side, let's say they're all above the x-axis.

So h > 0 and k > 0.

Let me reconsider the angle at D more carefully.
D is at (20, k).
DE = (0, k) points upward from E to D.
DC makes a 45° angle with DE.

If we consider the interior angle at D, then DC should point toward the left (since we're going around the pentagon in the right direction).

Actually, let me reconsider what "same side of line AE" means and be more systematic.

Let's place A = (0, 0), E = (20, 0).
B = (0, h) since ∠A = 90°.
D = (20, k) since ∠E = 90°.

For the angle at B:
BA = (0, -h)
BC makes 45° with BA.
If we go cou