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Solution to puzzle 112: Angle bisector

Skip restatement of puzzle.triangleABC is right-angled at A.  The angle bisector from A meets BC at D, so that angleDAB = 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.

Determine AD

Let BD = d, so that AD = d − 1.

Right triangle ABC, with angle bisector and lengths as described above.

Applying the law of sines (also known as the sine rule) to:
triangleCAD,  (sin C) / (d − 1) = (sin 45°) / 1 = 1 /root 2,
triangleABD,  (sin B) / (d − 1) = (sin 45°) / d = 1 / (droot 2).

Note that sin B = AC/BC = cos C.
Squaring and adding both sides, we obtain (since sin2C + cos2C = 1),
1/(d − 1)2 = 1/2 + 1/(2d2).

Multiplying both sides of the equation by 2d2(d − 1)2, and simplifying, we get
d4 − 2d3 − 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 d2, yielding
d2 − 2d − 2/d + 1/d2 = 0.

If we substitute u = d + 1/d, then u2 = d2 + 1/d2 + 2.  We thereby obtain
u2 − 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 + root 3. (1)
Multiplying by d, and rearranging, we get
d2 − (1 + root 3)d + 1 = 0.

Solving this quadratic equation, we get d = ½ (1 + root 3 ± root 2fourth root of 3).
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 + root 3 + root 2fourth root of 3).

Determine AC

Now let AC = x and AB = y.

Right triangle ABC.

Applying the law of sines to:
triangleCAD,  1 / sin 45° = x / sin ADC,
triangleABD,  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 triangleABC, we obtain
x2 + y2 = (d + 1)2.
Substituting y = dx, we get x2(d2 + 1) = (d + 1)2.
Hence x2 = (d2 + 2d + 1)/(d2 + 1) = 1 + 2d/(d2 + 1).

From (1),  d + 1/d = (d2 + 1)/d = 1 + root 3.
Therefore 2d/(d2 + 1) = 2/(1 + root 3).
Rationalizing the denominator of this fraction, we obtain
2d/(d2 + 1) = 2(root 3 − 1) / [(1 + root 3)(root 3 − 1)] = root 3 − 1.

Hence x2 = 1 + 2d/(d2 + 1) = root 3.

Therefore the length of AC is fourth root of 3.


Remarks

Having found AD, we could obtain the numerical value of AC by applying the law of cosines to triangleADC, 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.

Reciprocal equations

The polynomial p(x) = anxn + ... + a1x + a0, with real coefficients, is said to be reciprocal if ai = an−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) = xnq(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.

Source: Original

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