Area of infinite number of squares inside triangle

Question

This question is from the third sample paper for the CSAT (Computer Science Admissions Test) at Cambridge. My working is as follows:

First, I worked out the equation of the diagonal line touching the top right corner of each square as being: $y+\frac{1}{a}x=1 \to ay+x=a$

I then worked out the x-coordinate of the intersection point between the diagonal and the top right corner of the first square. I did this using simultaneous equations with the square's diagonal which is clearly $y=x$

$$ay+x=a$$ $$y=x$$ Sub in: $$ax+x=a$$ $$x(a+1)=a$$ $$x_1=\frac{a}{a+1}$$ Then clearly the area of the first square $A_1 = (\frac{a}{a+1})^2$

I then did the exact same thing to work out the top right corner of the second square. I used the equation $y-y_1=m(x-x_1)$ to work out the equation of the diagonal of the second square as $y=x-\frac{a}{a+1}$. Again I did simultaneous equations as follows:

$$ay+x=a$$ $$y=x-\frac{a}{a+1}$$ Sub in: $$a(x-\frac{a}{a+1})+x=a$$ $$ax-\frac{a^2}{a+1}+x=a$$ $$ax+x=a+\frac{a^2}{a+1}$$ $$x(a+1)=\frac{2a^2+a}{a+1}$$ $$x_2=\frac{a(2a+1)}{(a+1)^2}$$

Then, to work out the length of the second square, I did $x_2-x_1$ which yielded $(\frac{a}{a+1})^2$. I then saw a pattern between the two lengths so I repeated the working for the third fraction and the pattern held. The pattern I found is that the length of the square is always $(\frac{a}{a+1})^n$ and so the area of each square is $(\frac{a}{a+1})^{2n}$.

$\therefore$ the total area is $\sum_{n=1}^{\infty} (\frac{a}{a+1})^{2n}$

I have 2 questions:

Is my working correct?
Are there any better/more efficient ways of doing this? (I'm pre-calculus and haven't learnt summations etc.)

@LordSharktheUnknown has a nice answer. Yours may in fact agree with his after you sum the geometric series you found. (I haven't checked.) — Ethan Bolker, Jul 22 '18 at 18:31

score 3 · Accepted Answer · answered Jul 22 '18 at 18:23

3

Look at the left-most trapezium. It consists of a square of sidelength $h$, say, surmounted by a right-angled triangle whose legs must measure $h$ and $h/a$. Therefore $h+h/a=1$, that is $h=a/(1+a)$. The area of the trapezium is $(h/2)(h+1)$, and so the proportion of its area contained in the square is $$\frac{h^2}{(h/2)(h+1)}=\frac{2h}{h+1}=\frac{2a}{1+2a}.$$ But that is the proportion of the whole triangle occupied by the squares. The squares' area must add up to $$\frac{2a}{1+2a}\frac{a}{2}=\frac{a^2}{1+2a}.$$

answered Jul 22 '18 at 18:23

Angina Seng

158,341

Thank you, I understand all of that. I did actually think about using the trapezium at the beginning but decided against it :-( But where's the mistake in my working then? – Dan Jul 22 '18 at 18:29
Very nice way to avoid having to sum a geometric series – Ethan Bolker Jul 22 '18 at 18:29
1

@dan Your summation should be $\sum_{n=1}^\infty(a/(1+a))^{2n}$. – Angina Seng Jul 22 '18 at 18:33
@LordSharktheUnknown Yep sorry. I copied my working wrong. – Dan Jul 22 '18 at 18:35

Area of infinite number of squares inside triangle

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