ABC is right-angled at A. The angle bisector from A meets BC at D, so that DAB = 45°. If CD = 1 and BD = AD + 1, find the lengths of AC and AD.

With various triangles to which the law of sines, the law of cosines, and Pythagoras' Theorem may be applied, there are no doubt many solutions to this problem.

Let BD = d, so that AD = d − 1.

Applying the law of sines (also known as the sine rule) to:

CAD, (sin C) / (d − 1) = (sin 45°) / 1 = 1 /,

ABD, (sin B) / (d − 1) = (sin 45°) / d = 1 / (d).

Note that sin B = AC/BC = cos C.

Squaring and adding both sides, we obtain (since sin^{2}C + cos^{2}C = 1),

1/(d − 1)^{2} = 1/2 + 1/(2d^{2}).

Multiplying both sides of the equation by 2d^{2}(d − 1)^{2}, and simplifying, we get

d^{4} − 2d^{3} − 2d + 1 = 0.

Although the general quartic equation is difficult to solve, we can make a substitution that will halve the degree of this particular equation.

Notice that the sequence of coefficients, (1, −2, 0, −2, 1), forms a palindrome. Since d = 0 is not a root, we may divide by d^{2}, yielding

d^{2} − 2d − 2/d + 1/d^{2} = 0.

If we substitute u = d + 1/d, then u^{2} = d^{2} + 1/d^{2} + 2. We thereby obtain

u^{2} − 2u − 2 = (u − 1)^{2} − 3 = 0.

Rejecting the negative root, as it would lead to non-real values of d, we have

u = d + 1/d = 1 + . (1)

Multiplying by d, and rearranging, we get

d^{2} − (1 + )d + 1 = 0.

Solving this quadratic equation, we get d = ½ (1 + ± ).

We reject the smaller root as it is less than 1. (The length of AD = d − 1 must be positive.)

Therefore the length of AD is ½ (−1 + + ).

Now let AC = x and AB = y.

Applying the law of sines to:

CAD, 1 / sin 45° = x / sin ADC,

ABD, d / sin 45° = y / sin BDA.

Note that sin BDA = sin (180° − ADC) = sin ADC.

Hence x = sin ADC / sin 45°, y = d · sin ADC / sin 45°, and so y = dx.

Applying Pythagoras' Theorem to ABC, we obtain

x^{2} + y^{2} = (d + 1)^{2}.

Substituting y = dx, we get x^{2}(d^{2} + 1) = (d + 1)^{2}.

Hence x^{2} = (d^{2} + 2d + 1)/(d^{2} + 1) = 1 + 2d/(d^{2} + 1).

From (1), d + 1/d = (d^{2} + 1)/d = 1 + .

Therefore 2d/(d^{2} + 1) = 2/(1 + ).

Rationalizing the denominator of this fraction, we obtain

2d/(d^{2} + 1) = 2( − 1) / [(1 + )( − 1)] = − 1.

Hence x^{2} = 1 + 2d/(d^{2} + 1) = .

Therefore the length of AC is .

Having found AD, we could obtain the numerical value of AC by applying the law of cosines to ADC, leading to a quadratic equation in x. However, it would be difficult to obtain the exact solution from this equation.

Notice that, in proving AB/AC = BD/CD, we did not make use of the fact that ABC is a right triangle. Indeed, this is a general result, known as the Angle Bisector Theorem.

The polynomial p(x) = a_{n}x^{n} + ... + a_{1}x + a_{0}, with real coefficients, is said to be reciprocal if a_{i} = a_{n−i} for i = 0, ..., n. If r is a root of p(x) = 0, it is easy to show, by direct substitution, that 1/r is also a root.

It can be shown that any reciprocal polynomial p(x) of degree 2n can be written in the form p(x) = x^{n}q(u), where u = x + 1/x, and q(u) is a polynomial of degree n Further, every reciprocal polynomial of odd degree is divisible by x + 1 (since f(−1) = 0), and the quotient is a reciprocal polynomial of even degree. Hence a reciprocal equation of degree 2n or 2n + 1 can always be reduced to one equation of degree n and one quadratic equation.

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