By Francis Hirsch, Gabriel Mokobodzki
ISBN-10: 3540172106
ISBN-13: 9783540172109
Read or Download Seminaire de Theorie du Potentiel Paris No 8 PDF
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Additional info for Seminaire de Theorie du Potentiel Paris No 8
Sample text
What is the least possible area of the region where the paper covers itself? Solution: Label the vertices of the triangle which covers itself A, B, C, where AB is the folding axis and ∠BAC is acute. Drop altitudes from A and B and label the points across the strip from them A and B , respectively. Note that ∠BAB folds onto ∠BAC, so these angles have the same measure, which we call x. We consider two cases. First, suppose 0 < x ≤ π/4. Then C is between A and B, ∠ACA = 2x and ∠ABA = x. Then [ABC] = [ABA ] − [ACA ] 1 1 (3)(3 cot x) − (3)(3 cot 2x) = 2 2 9 9 (cot x − cot 2x) = csc 2x.
Then x − y = (2k − 1)(2b − 1)(2a ), z − w = (2k − 1)(2d − 1)(2c ). Also x/y = 2b , z/w = 2d . Now x/y = z/w if and only if b = d if and only if x − y and z − w have the same largest odd divisor. 6 Czech and Slovak Republics 1. Let ABC be a triangle with sides a, b, c and corresponding angles α, β, γ. Prove that the equality α = 3β implies the inequality (a2 − b2 )(a − b) = bc2 , and determine whether the converse also holds. Solution: By the extended law of sines, a = 2R sin α, b = 2R sin β, c = 2R sin γ, where R is the circumradius of ABC.
Prove that no matter how the edges were colored initially, it is possible to make the number of blue segments at each vertex even. Prove also that the resulting coloring is uniquely determined by the initial coloring. Solution: All congruences are taken modulo 2. First, changing the order in which we choose the vertices does not affect the end coloring. Also, choosing a vertex twice has no net effect on the coloring. Then choosing one set of vertices has the same effect as choosing its “complement”: the latter procedure is equivalent to choosing the first set, then choosing all the vertices.
Seminaire de Theorie du Potentiel Paris No 8 by Francis Hirsch, Gabriel Mokobodzki
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