Interesting way to solve

function area = triangle_sequence(n)

y(1)=3;
y(2)=4;

for i=1:n
y(i+2)=sqrt(y(i)^2+y(i+1)^2)
end
area = y(n+2)^2;
end

C=[3,4,5];
for i=2:n
C(end+1,1,1)=C(end,2,1);
C(end,2,1)=C(end-1,3,1);
C(end,3,1)=sqrt(C(end,1,1)^2+C(end,2,1)^2);
end

C= [C(end,:,1)]
area=C(3)^2

Dear Cody Team,
solutions like mine shall be not allowed.

Elegant brain teaser and a good reminder to always read the question carefully...

And now, nobody will ever see it. You should have, at least, posted it.

Seems broken, when using Fibonacci ratio it won't work so I have botched it instead

So Exciting !

Finally made it :D ... Fibonacci sequence !

please help me out in understanding the problem

Didn't understand question

very hard example
R_T=[3 4 5];
if (n>1)
for i=1:(n-1)
R_T=[R_T(2) R_T(3) (R_T(2)^2+R_T(3)^2)^0.5]

Attention to the last line!

Double fibonacci sequence, subtle! I love this problem!

If anyone cares to point out why this is wrong, I'm open to hearing.

Note that the ratio of consecutive areas is asymptotically the golden ratio, i.e., phi=(1+sqrt(5))/2.

Weak! My recursion solution should work just fine, but the server is unable to evaluate it properly. I'm going to count it.

Its the best way that I found :)

I think it is the only way to solve this Problem. We Need to get the sides of nth triangle and ist done.

I consider my answer as a cheat

Yeah, you have just copied the required values of the test suits.

Interesting... (not a reflection on this particular solution but cody in general) - most of the lowest scored answers could be entries in an "obfuscated code" contest.

I'm curious how all these "Size 10" solutions look like. I don't get even close to this value. I terms of education, it would be great to see example codes which are able to go this low.

Brilliant !

I join you @Eugen, it would be great to share some "Size 10" solutions samples.

yes, this is probably cheating