Puzzle of the week [15]

The puzzle in the last weekend edition of Le Monde is simple to state: given a rectangle with sides 1 and a<1, what is the condition on a for an equilateral triangle to be inscribed in it? (Meaning that one summit of the triangle must coincide with one summit of the rectangle and that both other summits must stand on opposite sides of the rectangle.)

The equality on the sides of the triangle leads to finding solutions in (b,c) to the equations

$a^2+b^2 = (1-b)^2 + (a-c)^2 = 1 + c^2$

Eliminating c leads to an equation in b of the form

$\left[ \sqrt{b^2-(1-a^2)} - a \right]^2 + 1-a^2 -2 b = 0$

which can only have a solution for $b^2>1-a^2$ . If we rewrite $b^2=(1-a^2)v^2$ and $\varphi^1=1/(1-a^2)$ , this equation turns into

$\left[ \sqrt{v^2-1} - \sqrt{\varphi^2-1} \right]^2 + 1 - 2v\varphi=0$

with the constraint that $1<v<\varphi$ . Since the right hand side is decreasing in v, the equation has a solution if the rhs takes different signs for $v=1$ and $v=\varphi$ . This is only the case when $\varphi>2$ , which is equivalent to

$a>\sqrt{3}/2$ .

This entry was posted on April 21, 2010 at 12:22 am and is filed under Statistics with tags Le Monde, mathematical puzzle. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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Puzzle of the week [15]

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