Suppose we are given the length of the hypotenuse of a right triangle, and we also know the sides of a rectangle inscribed in the triangle as shown below __ |\ | \ | \ |_____ \ H Y | | \ | | \ | b| \ __ |_____|________\ a |<-----X------>| Given a, b, and H, how can we determine the remaining dimensions of the right triangle? If Y denotes the height of the triangle, then the equation of the hypotenuse is y = Y - (Y/X)x. We also know Y = sqrt(H^2-X^2), and when we plug "a" into the equation for the line we must get "b". Therefore, b = sqrt(H^2 - X^2) (1 - a/X) Squaring both sides and rearranging terms gives the quartic X^4 - (2a)X^3 + (a^2+b^2-H^2)X^2 + (2aH^2)X - (a^2 H^2) = 0 For example, if a=11, b=5, and H=25 we have X^4 - 22 X^3 - 479 X^2 + 13750 X - 75625 = 0 The Galois group of this polynomial is the fully symmetric group S_4, so there's no hope of reducing the general problem to anything simpler than the general solution of the quartic. The roots of the above polynomial give the four solutions X Y ----------- ----------- 23.10702596 9.54281674 14.59544931 20.29711456 9.05666503 -23.30186298 -24.75914031 3.46193175 These correspond to the four lines that pass through the point (a,b) and such that the distance between the x-intercept and the y-intercept is H. Of course, the first two are the solutions that match the geometrical shown above. Interestingly, if we define Z = (Y - b)/a, we have the quartic Z^4 + (2b/a)Z^3 + (1+(b/a)^2-(H/a)^2)Z^2 + (2b/a)Z + (b/a)^2 = 0 which is palindromic if a=b. In The Fundamental Theorem of Palindromic Polynomials we showed how a palindromic quartic of this kind can be reduced to a resolvant quadratic, which explains why the problem is easy if the rectangle is a square. Returning to the original quartic X^4 - (2a)X^3 + (a^2+b^2-H^2)X^2 + (2aH^2)X - (a^2 H^2) = 0 (1) it occurs to me that the palindromic procedure can be applied to ANY quartic of this type, not just those with a=b, reducing it to, at most, a cubic. Thus, it can be added to the well known methods of Ferrari and Descartes for reducing quartics to cubics. Given any quartic Ax^4 + Bx^3 + Cx^2 + Dx + E = 0 (2) it was shown in the previous article that this can be converted to "palindromic" form if E = A(D/B)^2. Making the substitution x=z+k we have a quartic in z whose coefficients A',B'...,E' are a function of the parameter k. To ensure that this polynomial can be converted to palindromic form we need only impose the requirement that E' = A'(D'/B')^2, which gives a cubic for k p k^3 + q k^2 + r k + s = 0 (3) where p = (B^3 - 4ABC + 8DA^2) q = (2ABD + 16EA^2 + CB^2 - 4AC^2) r = (8ABE - 4ACD + DB^2) s = (EB^2 - AD^2) With k equal to any root of this cubic we can substitute into the general quartic (2), reducing it to a palindromic form with a resolvant quadratic. Of course, if E=A(D/B)^2 then s=0, and we have k=0 is a root of (3), meaning that no shifting of the roots is required to make the original equation (2) palindromic. Also, note that the expression for "p" is the same as the "discriminant" in the note on palindromic polynomials. The sign of "p" determines whether the roots are real or complex conjugates, and p=0 implies two equal real roots. In that case the equation for k reduces to a quadratic, so again the solution is easy. Finally, what about the case of equation (1) with a=b? In that case, none of the coefficients of the resolvant cubic (3) vanish, so it might seem strange that this case can be easily solved. However, when a=b it turns out that the resolvant cubic (3) factors as 4 (a^2 + H^2) (a-k) (2ak^2 - H^ k + aH^2) = 0 so this gives the simple transformation given previously.

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