Mixing
Area of infinite number of squares inside triangle
Clash Royale CLAN TAG#URR8PPP
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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: $y+frac1ax=1 to ay+x=a$
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:
$$ax+x=a$$
$$x(a+1)=a$$
$$x_1=fracaa+1$$
Then clearly the area of the first square $A_1 = (fracaa+1)^2$
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 $y=x-fracaa+1$. Again I did simultaneous equations as follows:
$$ay+x=a$$
$$y=x-fracaa+1$$
Sub in:
$$a(x-fracaa+1)+x=a$$
$$ax-fraca^2a+1+x=a$$
$$ax+x=a+fraca^2a+1$$
$$x(a+1)=frac2a^2+aa+1$$
$$x_2=fraca(2a+1)(a+1)^2$$
Then, to work out the length of the second square, I did $x_2-x_1$ which yielded $(fracaa+1)^2$. 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 $(fracaa+1)^n$ and so the area of each square is $(fracaa+1)^2n$.
$therefore$ the total area is $sum_n=1^infty (fracaa+1)^2n$
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.)
proof-verification
asked Jul 22 at 18:16
Dan
25517
add a comment |Â
up vote
1
down vote
favorite
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: $y+frac1ax=1 to ay+x=a$
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:
$$ax+x=a$$
$$x(a+1)=a$$
$$x_1=fracaa+1$$
Then clearly the area of the first square $A_1 = (fracaa+1)^2$
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 $y=x-fracaa+1$. Again I did simultaneous equations as follows:
$$ay+x=a$$
$$y=x-fracaa+1$$
Sub in:
$$a(x-fracaa+1)+x=a$$
$$ax-fraca^2a+1+x=a$$
$$ax+x=a+fraca^2a+1$$
$$x(a+1)=frac2a^2+aa+1$$
$$x_2=fraca(2a+1)(a+1)^2$$
Then, to work out the length of the second square, I did $x_2-x_1$ which yielded $(fracaa+1)^2$. 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 $(fracaa+1)^n$ and so the area of each square is $(fracaa+1)^2n$.
$therefore$ the total area is $sum_n=1^infty (fracaa+1)^2n$
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.)
proof-verification
asked Jul 22 at 18:16
Dan
25517
@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 at 18:31
@EthanBolker Ah yes they are equal. Thank you.
– Dan
Jul 22 at 18:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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: $y+frac1ax=1 to ay+x=a$
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:
$$ax+x=a$$
$$x(a+1)=a$$
$$x_1=fracaa+1$$
Then clearly the area of the first square $A_1 = (fracaa+1)^2$
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 $y=x-fracaa+1$. Again I did simultaneous equations as follows:
$$ay+x=a$$
$$y=x-fracaa+1$$
Sub in:
$$a(x-fracaa+1)+x=a$$
$$ax-fraca^2a+1+x=a$$
$$ax+x=a+fraca^2a+1$$
$$x(a+1)=frac2a^2+aa+1$$
$$x_2=fraca(2a+1)(a+1)^2$$
Then, to work out the length of the second square, I did $x_2-x_1$ which yielded $(fracaa+1)^2$. 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 $(fracaa+1)^n$ and so the area of each square is $(fracaa+1)^2n$.
$therefore$ the total area is $sum_n=1^infty (fracaa+1)^2n$
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.)
proof-verification
asked Jul 22 at 18:16
Dan
25517
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: $y+frac1ax=1 to ay+x=a$
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:
$$ax+x=a$$
$$x(a+1)=a$$
$$x_1=fracaa+1$$
Then clearly the area of the first square $A_1 = (fracaa+1)^2$
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 $y=x-fracaa+1$. Again I did simultaneous equations as follows:
$$ay+x=a$$
$$y=x-fracaa+1$$
Sub in:
$$a(x-fracaa+1)+x=a$$
$$ax-fraca^2a+1+x=a$$
$$ax+x=a+fraca^2a+1$$
$$x(a+1)=frac2a^2+aa+1$$
$$x_2=fraca(2a+1)(a+1)^2$$
Then, to work out the length of the second square, I did $x_2-x_1$ which yielded $(fracaa+1)^2$. 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 $(fracaa+1)^n$ and so the area of each square is $(fracaa+1)^2n$.
$therefore$ the total area is $sum_n=1^infty (fracaa+1)^2n$
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.)
proof-verification
asked Jul 22 at 18:16
Dan
25517
edited Jul 22 at 18:34
asked Jul 22 at 18:16
Dan
25517
asked Jul 22 at 18:16
Dan
25517
asked Jul 22 at 18:16
Dan
25517
25517
@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 at 18:31
@EthanBolker Ah yes they are equal. Thank you.
– Dan
Jul 22 at 18:37
add a comment |Â
@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 at 18:31
@EthanBolker Ah yes they are equal. Thank you.
– Dan
Jul 22 at 18:37
@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 at 18:31
@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 at 18:31
@EthanBolker Ah yes they are equal. Thank you.
– Dan
Jul 22 at 18:37
@EthanBolker Ah yes they are equal. Thank you.
– Dan
Jul 22 at 18:37
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
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
$$frach^2(h/2)(h+1)=frac2hh+1=frac2a1+2a.$$
But that is the proportion of the whole triangle occupied by the
squares. The squares' area must add up to
$$frac2a1+2afraca2=fraca^21+2a.$$
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
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 at 18:29
Very nice way to avoid having to sum a geometric series
– Ethan Bolker
Jul 22 at 18:29
1
@dan Your summation should be $sum_n=1^infty(a/(1+a))^2n$.
– Lord Shark the Unknown
Jul 22 at 18:33
@LordSharktheUnknown Yep sorry. I copied my working wrong.
– Dan
Jul 22 at 18:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
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
$$frach^2(h/2)(h+1)=frac2hh+1=frac2a1+2a.$$
But that is the proportion of the whole triangle occupied by the
squares. The squares' area must add up to
$$frac2a1+2afraca2=fraca^21+2a.$$
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
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 at 18:29
Very nice way to avoid having to sum a geometric series
– Ethan Bolker
Jul 22 at 18:29
1
@dan Your summation should be $sum_n=1^infty(a/(1+a))^2n$.
– Lord Shark the Unknown
Jul 22 at 18:33
@LordSharktheUnknown Yep sorry. I copied my working wrong.
– Dan
Jul 22 at 18:35
add a comment |Â
up vote
3
down vote
accepted
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
$$frach^2(h/2)(h+1)=frac2hh+1=frac2a1+2a.$$
But that is the proportion of the whole triangle occupied by the
squares. The squares' area must add up to
$$frac2a1+2afraca2=fraca^21+2a.$$
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
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 at 18:29
Very nice way to avoid having to sum a geometric series
– Ethan Bolker
Jul 22 at 18:29
1
@dan Your summation should be $sum_n=1^infty(a/(1+a))^2n$.
– Lord Shark the Unknown
Jul 22 at 18:33
@LordSharktheUnknown Yep sorry. I copied my working wrong.
– Dan
Jul 22 at 18:35
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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
$$frach^2(h/2)(h+1)=frac2hh+1=frac2a1+2a.$$
But that is the proportion of the whole triangle occupied by the
squares. The squares' area must add up to
$$frac2a1+2afraca2=fraca^21+2a.$$
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
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
$$frach^2(h/2)(h+1)=frac2hh+1=frac2a1+2a.$$
But that is the proportion of the whole triangle occupied by the
squares. The squares' area must add up to
$$frac2a1+2afraca2=fraca^21+2a.$$
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 18:23
Lord Shark the Unknown
85.2k950111
85.2k950111
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 at 18:29
Very nice way to avoid having to sum a geometric series
– Ethan Bolker
Jul 22 at 18:29
1
@dan Your summation should be $sum_n=1^infty(a/(1+a))^2n$.
– Lord Shark the Unknown
Jul 22 at 18:33
@LordSharktheUnknown Yep sorry. I copied my working wrong.
– Dan
Jul 22 at 18:35
add a comment |Â
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 at 18:29
Very nice way to avoid having to sum a geometric series
– Ethan Bolker
Jul 22 at 18:29
1
@dan Your summation should be $sum_n=1^infty(a/(1+a))^2n$.
– Lord Shark the Unknown
Jul 22 at 18:33
@LordSharktheUnknown Yep sorry. I copied my working wrong.
– Dan
Jul 22 at 18:35
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 at 18:29
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 at 18:29
Very nice way to avoid having to sum a geometric series
– Ethan Bolker
Jul 22 at 18:29
Very nice way to avoid having to sum a geometric series
– Ethan Bolker
Jul 22 at 18:29
1
1
@dan Your summation should be $sum_n=1^infty(a/(1+a))^2n$.
– Lord Shark the Unknown
Jul 22 at 18:33
@dan Your summation should be $sum_n=1^infty(a/(1+a))^2n$.
– Lord Shark the Unknown
Jul 22 at 18:33
@LordSharktheUnknown Yep sorry. I copied my working wrong.
– Dan
Jul 22 at 18:35
@LordSharktheUnknown Yep sorry. I copied my working wrong.
– Dan
Jul 22 at 18:35
add a comment |Â
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@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 at 18:31
@EthanBolker Ah yes they are equal. Thank you.
– Dan
Jul 22 at 18:37