Sangaku problems from Contours magazine 2014-15 Edition

Problem 1

Show that \(4T=9t\).

Problem 2

Show that \(3B=8b\).

## Solutions

### Problem 1

Show that \(4T=9t\).

Let the square have length \(\ell\). Consider the three right triangles in the picture.

The blue triangle has sides \(\frac\ell2\), \(T\) and hypotenuse \(\ell -T\), hence Pythagoras says that \[ \frac{\ell^2}{4} + T^2 = (\ell - T)^2 \implies T = \frac{3\ell}8~. \]

The red and green triangles have common height \(h\). The other side of the red triangle is \(\ell - t\) and the hypotenuse is \(\ell + t\), hence Pythagoras says that \[ h^2 = (\ell +t)^2 - (\ell -t)^2 = 4 \ell t~. \]

On the other hand, the other side of the green triangle is \(t\) and its hypotenuse is \(\ell -t\), and thus Pythagoras says that \[ h^2 = (\ell-t)^2 - t^2 = \ell(\ell - 2 t)~. \] Equating the heights, we have \[ 4 \ell t = \ell (\ell - 2 t) \implies t = \frac\ell6~. \]

### Problem 2

Show that \(3B=8b\).

Let the square have length \(\ell\). Then consider the two right triangles in the picture.

The blue triangle has sides \(\frac\ell2\) and \(\frac\ell2 + B\) and hypotenuse \(\ell - B\). Pythagoras says that \[ \frac{\ell^2}{4} + \left(\frac\ell2 + B\right)^2 = (\ell - B)^2 \implies B = \frac\ell6~. \] The red triangle has sides \(\frac\ell2\) and \(\ell-b\) and hypotenuse \(\ell +b\), whence Pythagoras says that \[ \frac{\ell^2}{4} + \left(\ell - b\right)^2 = (\ell + b)^2 \implies b = \frac\ell{16}~. \]