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Cut a given straight line so that the sum of the square of one part and twice the square of the other part equals a given size.

Given a line of length $L$ which is cut in two pieces, lengths $x$ and $L-x$, I first thought of constructing the squares with sides $x$ and $\sqrt{2}(L-x)$ but I cannot relate the total area to a given length.

If $M$ is the given length the length to construct is $$\frac{2L\pm\sqrt{3M-2L^2}}{3}$$ but I assume there is an easier way to solve this problem using only straight-edge and compass.

Any idea how to solve this problem? TIA.

mf67
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  • It is a mistake to treat L and M as separate things. As for using a square and a compass, you need to divide a line segment into 3 parts and I don't know if that can be done. – poetasis Jan 30 '21 at 04:35

2 Answers2

1

Let the segment lengths be $(x,L-x)$ for total length $L.$

Total area $$ A=x^2+2(L-x)^2= 3x^2-4xL+2L^2 \tag 1 $$

should be a constant or minimum / maximum when segmented at some $x$.

Differentiating to find location of partition,

$$ 6x-4L=0, x=\dfrac{2L}{3} \to A_{min}= \dfrac{2L^2}{3};$$

which is a function of $L.$ It is easy to divide a line length L = OB into three equal parts. (A line drawn through any one end parallel lines from three equal segments as shown, take line trisected through D).

enter image description here

However in this problem the minimum is a priori specified to have a given size =S, but not derivative driven $2 L^2/3$ minimum. This is not possible without any modification of area expression in (1) accordingly, like in ..

$$ A=x^2+2(L-x)^2- \dfrac{2L^2}{3} +S \tag2 $$

A numerical example for $ L=100, S=10000 $ gives desired area $=S$ at the calculated two thirds length.

Geometric construction for physical Area Check

Total area $ S= 10,000$ square units is fully accounted for as one square of $2L/3$ side and another five of $L/3$ side in rough sketch.

enter image description here

enter image description here

Narasimham
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Given the square of one part plus twice the square of the other part equals the whole.

$$L=(x)^2+2(L-x)^2=2 L^2 - 4 L x + 3 x^2\\ \implies 3 x^2 - 4 L x + (2 L^2 - L) = 0$$

$$x=\frac{4L\pm\sqrt{16L^2-4(3)(2 L^2 - L)}}{2(3)} =\frac{2L\pm \sqrt{L(3 - 2 L)}}{3}\quad x\in\mathbb{R}\iff 0\le L\le \frac{3}{2}$$

Any value of $L$ in this range will work but the only (L,x) solutions that "contain" integers are $$ (0,\space0)\qquad (1,\space1)\qquad (3/2,\space1)$$

So $$0^2+2(0-0)^2=0\qquad 1^2+2(1-1)^2=1\qquad 1^2+2(1.5-1)^2=1+2(.25)=1.5$$

Outside of this range, solutions are complex (containing imaginary parts), i.e. $x\in\mathbb{C}$

$\textbf{Update}$ WolframAlpha shows the solution to be different from mine where I stayed in the limits.

$$(x)^2+2(L-x)^2-L=0,\quad L=414\\ \implies x = 276 + 5 i \sqrt{1518}\qquad L-x=138 + 5 i \sqrt{1518}$$

Multiplied back, it yields an enormous number for $L$, even with plus-minus the imaginary part here, and here.

poetasis
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  • The solution should only involve straight-edge and compass. It is not about finding particular solution, but a general method for any two lines. – mf67 Jan 30 '21 at 07:19
  • If you just want to trisect a line segment, this video shows one easy way to do that with a straight edge and compass. If you want it to meet your criteria, however, the line segment length $;L;$ must be measured in units where $0\le L \le \frac{3}{2}$ as indicated. in my answer – poetasis Jan 30 '21 at 15:14
  • I'm not sure I fully understand, but any rational number can be created using straight-edge and compass. E.g. $L=24$, $M=414$ gives $x=(48\pm3\sqrt{10})/3$. It is possible to construct $x$ and $x^2+2(L-x)^2=M$ as requested. The problem is to construct $x$ using tools, but no algebra. – mf67 Jan 30 '21 at 16:52
  • You can go outside the limits but the results are complex as I said in my answer. $$x^2$+(414-x)^2-414=0\ \implies 2 x^2 - 828 x + 170982=0$$

    $$x=\frac{828\pm\sqrt{828^2-4(2)(170982)}}{2(2)}=\frac{828\pm\sqrt{-682272}}{2(2)}\ x=207\pm 3i\sqrt{4738}$$

     – poetasis Jan 30 '21 at 17:57
  • I mentioned a link for trisecting a line segment if that's all you want to do but you may or may not meet the criteria of summing squares. If you can ingnore the imaginary multiplier, I am glad. I simply have not explored it. – poetasis Jan 30 '21 at 18:05