# Sangaku problems

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 $\frac{\ell^2}{4} + \left(\ell - b\right)^2 = (\ell + b)^2 \implies b = \frac\ell{16}~.$