MHB Big Square composed of Small Squares ?

[Albert1]

View attachment 580

 

[soroban]

Hello, Albert!

Prove that it is impossible for a square to be composed
of five smaller square as shown.

         a       b
      *-----*---------*
      |     |         |
    a |     |         | b
      |     |   Q     |
      |    P*---*-----*
      |     |   |     |
      *-----*---*R    |
      |     S   |     |
      |         |     | c
    d |         |     |
      |         |     |
      |         |     |
      *---------*-----*
           d       c

The four outer squares have sides $a,b,c,d$ as shown.

The inner square is $PQRS$.We find that: .$\begin{Bmatrix}PQ \:=\:b-c \\ SR \:=\:d-a \end{Bmatrix} \quad \begin{Bmatrix}QR \:=\:c-d \\ PS \:=\:a-b \end{Bmatrix}$Since $PQ = SR\!:\:b-c \:=\:d-a \quad\Rightarrow\quad a+b-c-d \:=\:0\;\;[1]$

Since $PS =QR\!:\:a-b \:=\:c-d \quad\Rightarrow\quad a-b-c+d \:=\:0\;\;[2]$Add [1] and [2]: .$2a-2c\:=\:0 \quad\Rightarrow\quad a \:=\:c$

Subtract [1] and [2]: .$2b-2d \:=\:0 \quad\Rightarrow\quad b \:=\:d$Hence, the large square is divided into four congruent squares.

The inner square has zero area.

 

[Nono713]

The inner square has zero area.

Still a square! (Tongueout)

 

[Albert1]

Hence, the large square is divided into four congruent squares.
The inner square has zero area. (Tongueout)

soroban :well done !