Applied Mathematicsematics

Mathematical Morsels by Ross Honsberger

By Ross Honsberger

Publication through Honsberger, Ross

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I n this case a(n) + cp(n) 2n n·d(n), = = "AMM, 1 %5, p. 1 86, Problem E 1 674, proposed by C A. Nicol, University of South Carolina; solution by Ivan Niven, University of Oregon (unpublished) 32 a(n) + cp(n)= n·d(n) 33 implying the condition is necessary. (ii) Suppose that a(n) + cp(n) = n-d(n) and also that n is not a prime number. The equation is not satisfied by n = I (I + I =1= I · I), implying n ;;;. 2. For n > I , cp(n) does not count the number n, itself, and we have cp(n) < n. Because n is a composite number it must have at least 3 divisors.

The desired conclusion follows the observation that the overlap in question is simply (the sum of the legs) -(the hypotenuse). x x y y FIG. 19. ----=:. C FIG. 20. The following relation is also easily deduced. If the altitude BD to the hypotenuse of right-angled L;A BC is drawn, then the sum of the three in-radii r, r,. r2, of L;A BC and the smaller triangles A SURPRISING PROPERTY OF RIGHT-ANGLED TRIANGLES 29 into which it is divided by BD, is simply the altitude B D, itself (FIG. 20). We have 2 r + 2r, + 2r2 = (A B + BC -A C ) + (A D + B D -A B ) + ( B D + DC-BC ) = (A D + DC ) -A C + 2 BD hence r + r, + r2 = BD.

The number of winning lines for a cube of edge k in n-dimensional space is (k+2)"-kn 2 "AMM, 1 948, p. 99, Problem E773, proposed by A. L. Rubinoff, University of Toronto, solved by Leo Moser, University of Manitoba. 26 PROBLEM 14 A SURPRISING PROPERTY OF RIGHT-ANGLED TRIANGLES* Prove that if each leg of a right-angled triangle is rotated about its vertex on the hypotenuse so as to lie along the hypotenuse, then the legs overlap in a segment whose length is the diameter of the circle inscribed in the triangle (FIG.

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