By G. Mauger
Booklet through Mauger, G.
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Additional resources for Cours de Langue et de Civilisation Françaises III
Sample text
We need to find a formula for the area A of the equilateral triangle and then try to maximize this formula in terms of θ. ) Since the base of the triangle is x, its height is ( 3/2) x. So A = ( 3/4) x2 , but we wanted it in terms of θ. Well, cos θ = 1/x, so x = 1/ cos θ = sec θ. Thus we have √ 3 sec2 θ. A= 4 Folding Equilateral Triangles in a Square 11 We could take the derivative of this and try to maximize it using calculus, but we don’t really need to. Since cos θ is a decreasing function on the interval 0 ≤ θ ≤ π/12 (we really should be working in radians, after all), we know that sec θ is an increasing function on this interval.
The key is that EDB is similar to ABC since they’re both right triangles and they share the angle at B. Therefore, √ √ √ √ √ 3 3 3 ( 2 − 3) x √ = = ⇒ x= = 2 3 − 3. 1−x 2 1 2+ 3 √ √ Thus our 15◦ -75◦ -90◦ triangle has long leg 3, short leg 2 3 − 2, and hypothenuse √ 2 6 − 3 3. These values are rather awkward, and normalizing the shortest leg to be 1 seems√ somewhat √ standard for √ this kind of thing. Doing this gives us a long leg of length 3/(2 3 − 2) = 2 +√ 3 once we rationalize the denominator.
Since the base of the triangle is x, its height is ( 3/2) x. So A = ( 3/4) x2 , but we wanted it in terms of θ. Well, cos θ = 1/x, so x = 1/ cos θ = sec θ. Thus we have √ 3 sec2 θ. A= 4 Folding Equilateral Triangles in a Square 11 We could take the derivative of this and try to maximize it using calculus, but we don’t really need to. Since cos θ is a decreasing function on the interval 0 ≤ θ ≤ π/12 (we really should be working in radians, after all), we know that sec θ is an increasing function on this interval.