A sequence of triangles is constructed in the following way:
1) the first triangle is Pythagoras' 3-4-5 triangle
2) the second triangle is a right-angle triangle whose second longest side is the hypotenuse of the first triangle, and whose shortest side is the same length as the second longest side of the first triangle
3) the third triangle is a right-angle triangle whose second longest side is the hypotenuse of the second triangle, and whose shortest side is the same length as the second longest side of the second triangle etc.
Each triangle in the sequence is constructed so that its second longest side is the hypotenuse of the previous triangle and its shortest side is the same length as the second longest side of the previous triangle.
What is the area of a square whose side is the hypotenuse of the nth triangle in the sequence?
Solution Stats
Problem Comments
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25 Comments
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22 older comments
Jean
on 12 Dec 2013
good for brain !
Jean-Marie Sainthillier
on 13 Dec 2013
The best problem of the CUP challenge.
Jean-Marie Sainthillier
on 13 Dec 2013
I think that thematic challenge is a wonderful idea. I solved this one with a great pleasure.
Matt
on 4 Feb 2014
Great problem. So many ways to solve it. It was fun trying different methods to see which scored the best.
kwijibo28
on 8 Feb 2014
The connection of this problem with the Fibonacci sequence is most interesting.
Philipp
on 5 Mar 2014
indeed: like!
Guilherme Coco Beltramini
on 26 Jun 2014
Very interesting problem!
Vishal
on 8 Sep 2014
Interesting problem....!
Vidushi Jain
on 10 Jun 2015
Interesting!!
Pooja Narayan
on 23 Feb 2016
Good one :)
Giovanni Mottola
on 7 Jun 2016
Combining the Fibonacci sequence and the Pythagorean theorem is a nice idea
Dhaval
on 11 Sep 2016
I thought I had the right code and it didn't work. Then I realized that the area of the first triangle (n=1 case) should be 0.5*4*3 = 6. But is given in the solution as 25. Am I understanding this wrong? I thought the three sides of the first triangle (n=1) was 3,4,5 and we go on from there. Please let me know if my understanding is wrong and then I can proceed Otherwise I have a general code.
Milan Petrovic
on 21 Oct 2016
@ Dhaval: Try reading the last sentence of the problem very carefully.
Dhaval
on 10 Dec 2016
@Milan Petrovic: Thank you, I completely missed that; silly me. Now I got it to work :)
Joao Vitor Monteiro Lopes
on 17 Jun 2017
Good problem. Kinda fun too.
Chris Eio
on 13 Oct 2017
Definitely needed to read that last line carefully. Like at least one other here, I misread it.
Anthony Cafeo
on 28 Nov 2017
can someone help me with this problem?
Wycliff Dembe
on 26 Aug 2018
tough but fun... had issues with assertion at first through
Zhilin Ye
on 6 Jan 2019
Does Cody not recognize the Fibonacci function?
Anthony Zúñiga
on 16 Jan 2019
Can anybody PLEASE send me the solution of this problem? My e-mail is: anthonydavid72@hotmail.com
Dani Péter
on 20 Apr 2019
so interesting!
Christian Schröder
on 8 Jul 2019
Nice one! Makes you think, and then the solution is simple and elegant.
Parv Verma
on 6 Dec 2019
It was a really great problem to solve. For me doing the steps as mentioned on paper helped a lot for proper visualization of the problem, and to understand the flow.
Thanks Tanya Morton for the problem exercise.
Ahmet Burak Aydogdu
on 18 Apr 2020
Very elegantly done!
carlos fuentes
on 25 Apr 2020
Very very interesting problem. It introduces the importance of integer sequences.
Solution Comments
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1 Comment
Piero Cimule
on 21 Jun 2020 at 1:09
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
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1 Comment
Chris Guenther
on 24 Sep 2019
Dear Cody Team,
solutions like mine shall be not allowed.
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1 Comment
Chris Coote
on 10 Apr 2019
Elegant brain teaser and a good reminder to always read the question carefully...
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1 Comment
Rafael Siqueira Telles Vieira
on 3 Jun 2020
And now, nobody will ever see it. You should have, at least, posted it.
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1 Comment
James Best
on 21 Jan 2019
Seems broken, when using Fibonacci ratio it won't work so I have botched it instead
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1 Comment
reyhane mohebali
on 14 Oct 2018
So Exciting !
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1 Comment
Othmane ELMOUATAMID
on 9 Aug 2018
Finally made it :D ... Fibonacci sequence !
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1 Comment
durga sadasivuni
on 19 Jul 2018
please help me out in understanding the problem
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1 Comment
durga sadasivuni
on 7 Jun 2018
Didn't understand question
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1 Comment
Rachel Wedemeyer
on 6 Apr 2018
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]
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1 Comment
Thiago César
on 10 Jan 2018
Attention to the last line!
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1 Comment
ccljohn123
on 16 Sep 2017
Double fibonacci sequence, subtle! I love this problem!
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1 Comment
Thomas Blackwood
on 7 Sep 2016
If anyone cares to point out why this is wrong, I'm open to hearing.
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1 Comment
John D'Errico
on 21 Aug 2016
Note that the ratio of consecutive areas is asymptotically the golden ratio, i.e., phi=(1+sqrt(5))/2.
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1 Comment
Daniel Zimmermann
on 16 Jul 2016
Weak! My recursion solution should work just fine, but the server is unable to evaluate it properly. I'm going to count it.
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2 Comments
Carlos Zúñiga
on 9 Mar 2016
Its the best way that I found :)
Suraj Gurav
on 3 Sep 2018
I think it is the only way to solve this Problem. We Need to get the sides of nth triangle and ist done.
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1 Comment
Raphael
on 20 Mar 2014
I consider my answer as a cheat
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3 Comments
Bruce
on 1 Apr 2015
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.
Eugen Zimmermann
on 23 May 2018
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.
Othmane ELMOUATAMID
on 9 Aug 2018
Brilliant !
I join you @Eugen, it would be great to share some "Size 10" solutions samples.
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1 Comment
Jon
on 3 Feb 2014
yes, this is probably cheating
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