# Sunday Times Teaser 2653 – Tough Guys

by H Bradley and C Higgins

A ship has two identical vertical masts on the centre line of its deck. These masts are a whole number of feet tall and are seven feet apart horizontally.

The tops of these masts mast are each attached by straight guy ropes to the same anchor point on the centre line of the deck.

One rope is two feet longer than the other and their combined length is a whole number of feet.

What is the height of the masts?

1. This teaser can be solved analytically as follows. First let the height of the masts be $$h$$, the length of the guy ropes be $$l_1$$ and $$l_2$$ with $$l$$ and $$s$$ being the sum and difference of these lengths respectively. Then let the distance of the anchor point from the closest mast be $$a$$ and the distance between the masts be $$d$$. We hence have $$l_1=(l+s)/2$$ and $$l_2=(l-s)/2$$. Now using the two pythagorean triangles formed by the deck, the masts and the guy ropes we have:$l_1^2=h^2+(a+d)^2=(l+s)^2/4$ $l_2^2=h^2+a^2=(l-s)^2/4$ Taking the difference of these two equations gives $$2ad+d^2=ls$$ and hence: $a=(ls-d^2)/(2d)$ We can now substitute for $$a$$ in the above equations, which gives: $(2dl_1)^2=(2dh)^2+(ls+d^2)^2=d^2(l+s)^2$ $(2dl_2)^2=(2dh)^2+(ls-d^2)^2=d^2(l-s)^2$ Finally by adding these two equations and simplifying we obtain:$(2dh)^2=(d^2-s^2)(l^2-d^2)$

Substituting the given values $$d=7$$ and $$s=2$$ and simplifing the result gives the quadratic diophantine equation $x^2 – 5y^2=-5$ where $$x=2h/3$$ and $$y=l/7$$. This is a variant of Pell’s equation and has multiple solutions, the first being the trivial one $$x=0,y=1$$, the n’th solution being given by expanding$x_n+\sqrt{5}y_n=(9+4\sqrt{5})^n$ and equating the terms on either side. These solutions can also be generated recursively using: $x_{n+1}=9x_n+20y_n$ $y_{n+1}=4x_n+9y_n$ which provides the basis for this Python solution:

with the output:

including the intended solution of 30 feet masts.

• • 2. Let $$x$$ be the height of the masts, $$y$$ and $$y+2$$ be the lengths of the guy ropes with the anchor point on the left side of the mast holding the shorter rope. Using Pythagorean Theorem we can get: $\sqrt{(y+2)^2-x^2}-\sqrt{y^2-x^2}=7$ giving us $x=(3/14)\sqrt{5(2y-5)(2y+9)}$ whereby by trial and error and using the fact that $$y>7$$ and $$y=a/2$$ where $$a$$ is an integer, we get $$y=(61/2)$$ ft and $$x=30$$ ft.