This beautifully neat solution was sent in by Johnny Chen.

If we rotate the figure by 90 degrees about B (clockwise). This should result in a new square A¢BC¢D¢ with a point P¢ inside the new square such that, ÐPBP¢=90o and BP=BP¢=2. By Pythagoras, PP¢ = Ö8. Now consider triangle AP¢P. The side lengths are 3, Ö8, and 1which satisfies Pythagoras again: 32 - 12 = (Ö8)2.

So this triangle has 90 degrees at ÐP¢PA. Since we know that ÐBPP¢=45o therefore ÐBPA is 135 degrees.

This alternative method comes from Yatir from Maccabim-Reut High School, Israel.


AB=BC=CD=AD=x, BP=2, CP=3, AP=1. Name the angles this way: ÐABP = c, ÐCBP = 90-c, ÐAPB = a, ÐCPB = b.

We need to find angle a. By the Sine Law:
x
sinb
 = 3
sin(90-c)
 = 3
cosc
and
x
sina
 = 1
sinc
Hence sinc = sina /x, cosc = 3sinb /x and we use the identity sin2 c + cos2 c = 1 to eliminate c. So
sin2 (a/x2) + 9sin2 (b / x2)
= 1
sin2 b
= (x2 - sin2 a)/9.
Using the Cosine Law we find cosb and then eliminate b:
x2 = 9 + 4 - 12 cosb
so
cos2 b = [(13-x2)/12]2.

Using the identity sin2 b + cos2 b = 1 we get
(x2 - sin2 a)/9 + [(13 - x2)2]/144 = 1     (1).
Finally we eliminate x. By the Cosine Law x2 = 4 + 1 - 4cos a so
x2 = 5 - 4 cosa     (2).

Combine (1) and (2) and simplify and we get:
2 cos2 a /9 + 8 /9 = 1
hence cos2 a = 1/2.

This yields solutions: a=45 or a=135 degrees.

Now this angle cannot be 45 degrees because angle c is smaller (AP is the shortest side of triangle APB) and the third angle of the triangle is less than 90 degrees. Hence angle a is 135 degrees.