Originally Posted by

**Dans** The old-style Kit-Kat bar allowed for a saving of resources and costs in the size of the foil wrapping. The bar was offset on an angle of approx 21 degrees.

A given rectangular bar of chocolate is *a* cm by *b* cm, placed in a background rectangle ( *c* cm by *d* cm).

a= 12cm, b=8cm

Find the angle of offset, and find the length and width of the rectangular wrapping paper.

EDIT: For the purpose of this question, assume that the chocolate bar is flat.

I take it that the picture is meant to look like this:

$\displaystyle \setlength{\unitlength}{4.5mm}

\begin{picture}(10,12)

\put(0,0){\line(1,0){9}}

\put(0,12){\line(1,0){9}}

\put(0,0){\line(0,1){12}}

\put(9,0){\line(0,1){12}}

\put(5,0){\line(-5,2){5}}

\put(5,0){\line(2,5){4}}

\put(9,10){\line(-5,2){5}}

\put(0,2){\line(2,5){4}}

\put(7,4){$a$}

\put(6,10.3){$b$}

\put(2.8,0.1){$\theta$}

\end{picture}

$

I also assume that "assume that the chocolate bar is flat" means that we disregard its thickness and treat it as two-dimensional. The idea is that if you fold over the four triangular areas then they just cover the inner rectangle. If that is the case, then the area *cd* of the outer rectangle (width *c*, height *d*) must be twice the area *ab* of the bar of chocolate. If the offset angle is $\displaystyle \theta$, then $\displaystyle c = a\cos\theta + b\sin\theta$ and $\displaystyle d = b\cos\theta + a\sin\theta$. Thus $\displaystyle cd = (a\cos\theta + b\sin\theta)(b\cos\theta + a\sin\theta) = ab + (a^2+b^2)\sin\theta\cos\theta$. If this is equal to 2ab then $\displaystyle \sin(2\theta) = \frac{2ab}{a^2+b^2}$. From that, given a and b you can find $\displaystyle \theta$, and hence c and d.