Question

Jeremy loves right triangles that have side length 1. In fact, he loves to combine them so that the hypotenuse of one becomes the leg of the next. He does this over and over to create beautiful pictures like this:

[asy]
pair A=(0,0),B=(0,1),C=(0.4404,1.957), D=(1.2,2.9), E=(3.316,0);
draw(A--B--C--D--E--A);
draw(B--E--C);
draw(D--E);
draw(rightanglemark(E,A,B));
draw(rightanglemark(E,B,C));
draw(rightanglemark(E,C,D));
label("1",midpoint(A--B),W);
label("1",midpoint(C--B),NW);
label("1",midpoint(C--D),NW);
label("... (pattern continues here)",midpoint(D--E),E);
[/asy]


If the longer leg of the first triangle is $\sqrt{11}$, what will be the hypotenuse of the 2014th triangle?

Answer 1

Let the longer leg of triangle k be L_k and its hypotenuse H_k. Each triangle has the other leg 1 and the next triangle's longer leg equals the previous hypotenuse: L_1 = √11, L_{k+1} = H_k, and H_k^2 = L_k^2 + 1.

So H_1^2 = 11 + 1 = 12, and inductively H_k^2 = 11 + k. Thus
H_{2014} = √(11 + 2014) = √2025 = 45.