Introduction While waiting for train- or bus connections, travellers pass time by reading a book.In most cases, these books have little value and there is no need to keep them in good shape. Also, the book will not be finished during one period of waiting. Therefore, the page to continue reading has to be remembered. Three ways are available to accomplish that: -
1. pages read are torn from the book and discarded. This method is recommended for walkers.
2. a bookmark is inserted after the page that was last read. Disadvantage : it may me be lost. 3. the page last read is folded such, that a small triangle sticks out. The problem Please refer to to picture right.After folding the page, a small triangle at the top sticks out, indicating the place to continue reading. How must we fold to maximize the area of this triangle? Or : how to know the length AF, given a certain width and height of the page. Note: EG is the folding line. Point B folds over F, C becomes H. BE = EF CG = GH What to do? This problem carries the danger of getting stuck in endless calculations.A good strategy is of major importance. We assume that the Pythagoras lemma is needed as well as the theory of similarity of triangles. We extend edge BC and folding line EG to make triangles. For convenience we define -
AB = a
BC = b AF = x BE = EF = p TC = h CG = GH = v (calculations left to the reader) Final goal is a formula that expresses the area of triangle GHI in x,a and b. The sign ~ means similarity. The area of triangle GHI is written as [GHI] The strategy p may be calculated from width a and x.DTBE ~ DBAF, so: h may be calculated DTCG ~ DTBE, so: v may be calculated DGHI ~ DEAF Define f = v : AE, then [GHI] = f ^{2}[EAF],because the ratio of areas of similar triangles is the squared ratio of similar edges. Calculation of p Pythagoras lemma in DEAFCalculation of h DTBE ~ DBAFCalculation of v DTCG ~ DTBEmultiply by x replace hx by ap - bx Calculation of f v substitution: p substitution and simplifying The area of triangle GHI [GHI] = f^{2}[EAF][GHI] = f ^{2}x(a-p)/2f, a-p substitution and simplifying If x=0 the area is zero, which is correct. x=a is not possible. If folding line is CE, we notice p:b = x:a, so ap = bx and f = 0 The maximum area Differentiation is not advisable, because a polynomial of degree 5 will result.We feed the formula to Graphics-Explorer, and call the area y -
y = x(x^2+a^2-2b*x)^2 / (4a(a^2-x^2))
and select autoplot and replace
The maximum occurs at x = 1,72 |
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