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Find the number of positive integer solutions such that $a+b+c\le 12$ and the shown square can be formed.

enter image description here

$a \perp b$ and $b\perp c$.
the segments $a,b,c$ lie completely inside the square as shown.


Here is my attempt but I am pretty sure this is not the efficient method
Let the angle between left edge of square and segment $a$ be $\alpha$. To form a square we need the horizontal projections equal the vertical projections. Using similar triangles it is easy to get to below equation $$\langle \cos\alpha ,~\sin \alpha\rangle \cdot \langle b-a-c,~a \rangle = 0 $$

I feel stuck after this. Any help ?

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Here is a start.

Extend the line of length $a$ by an amount $c$ and then draw a line from the end of that to the corner of the square that ends the line of length $c$.

This forms a right triangle with sides $a+c$ and $b$ whose hypotenuse is the diagonal of the square. This length is $\sqrt{(a+c)^2+b^2}$, so the side of the square is $\sqrt{((a+c)^2+b^2)/2}$.

This does not take into account the condition that the lines line inside the square, but it is a start.

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  • $\begingroup$ That looks really neat! thank you so much. Does that mean there exists a square for every combination of $a,b,c$ ? $\endgroup$ – drae Apr 12 '15 at 5:54
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    $\begingroup$ @drae: If you ignore the condition that the segments are within the square, then yes of course. You just draw the red shape first and then you know where the square's diagonal must be, which determines the square. $\endgroup$ – user21820 Apr 12 '15 at 6:08
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We start off by noting that $$a\sin(\theta)+b\cos(\theta)+c\sin(\theta)=a\cos(\theta)-b\sin(\theta)+c\cos(\theta)$$ and that therefore all solutions will be of the form $$(a+b+c)^2\cos^2(\theta)=(a-b+c)^2\sin^2(\theta)$$

Which, in turn yields $$2(ab+bc)+(a^2+b^2+c^2+2c)(\cos^2(\theta)-\sin^2(\theta))\equiv2(ab+bc)+(a^2+b^2+c^2+2c)\cos(2\theta)=0$$ Which implies that for all $a+b+c \leq 12$ such that $(a^2+b^2+c^2+2ac) \leq 2b(a+c)$ we have a solution. It also implies that $\theta>\pi/4$ for all solutions, which makes sense if you look at your picture (otherwise $|b|<0$)

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