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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

- Thread starter Albert
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- Jan 25, 2013

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Prove that it is impossible for a square to be composed

of five smaller square as shown.

Code:`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.

- Jan 26, 2012

- 644

Still a square!The inner square has zero area.

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- Jan 25, 2013

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soroban :well done !Hence, the large square is divided into four congruent squares.

The inner square has zero area.