A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Squares are placed successively inside the right angled triangle thus formed as in the figure below. What is the area enclosed by all squares when their number goes to infinity?

  1. Find the points of intersection of the line $y= x+6$ with the parabola $y=x^2.$
  2. Evaluate the sum $\sum_{n=0}^{a}{2^n}.$
  3. What is the equation of the line with gradient $5$ that passes through the point $(2,5)?$

A line equation may help.

What can you say about the $x$ and $y$ coordinates of the top right corner of the largest square?

… does that help to compute its side length?

Is there a relationship between the large triangle and the one above the largest square?

What about the successive triangles?

What is then the sum of such numbers that are in such proportion, i.e. constant ratio?

The equation of the line is $y=1-\frac{x}{a}.$ At the top right hand corner of the first square $y=x.$ If we substitute this into the first equation we get $x = 1 - \frac{x}{a},$ which we can solve to get the side length of the first square which is $\frac{a}{a+1}.$

We notice that the triangle bounded by the lines $x=\frac{a}{a+1},$ $y = 1 - \frac{x}{a}$ and $y=0$ is similar to the larger triangle but with a height of $\frac{a}{a+1}.$ The heights of successive nested triangles follow a geometric progression with ratio $\frac{a}{a+1}.$

The squares contained within the triangles have side lengths equal to the height of the next triangle and the total area is a geometric series.$$ \begin{aligned} \sum_{n=1}^{\infty}\bigg(\frac{a}{a+1}\bigg)^{2n} &= \frac{\big(\frac{a}{a+1}\big)^2}{1-\big(\frac{a}{a+1}\big)^2} \\[7pt]&= \frac{a^2}{2a+1} \end{aligned}$$