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Solution to puzzle 150: Isosceles apex angle

Skip restatement of puzzle.Triangle ABC is isosceles with AB = AC.  Point D on AB is such that angleBCD = 15° and BC = root 6AD.  Find, with proof, the measure of angleCAB.


Let angleABC = t and AC = x.
Then angleADC = t + 15° and angleDCA = t - 15°.

Isosceles triangle ABC, with AB = AC, and as described above.

Applying the law of sines (also known as the sine rule) to triangleADC, we obtain:

AD/AC = sin(t - 15°)/sin(t + 15°).  (1)

Dropping a perpendicular from A to BC, we have:

cos t = ½BC/AC = root 6/(2x).

Hence AD/AC = (2 cos t)/root 6(2)

By inspection, if t = 45°, then:

sin(t - 15°)/sin(t + 15°) = ½ / (½root 3) = 1/root 3.
(2 cos t)/root 6 = root 2/root 6 = 1/root 3.

Hence t = 45° is one solution.  To show it is the only solution, we first expand the right-hand side of (1) using standard trigonometric identities, and then divide by sin t sin 15°:

sin(t - 15°)/sin(t + 15°) = (sin t cos 15° - cos t sin 15°)/(sin t cos 15° - cos t sin 15°)
  = (cot 15° - cot t)/(cot 15° + cot t)

Clearly, 15° < t < 90°.
For this range of values of t, f(t) = (2 cos t)/root 6 is a strictly decreasing function.
As cot t is strictly decreasing, g(t) = (cot 15° - cot t)/(cot 15° + cot t) is a strictly increasing function.
Since both f(t) and g(t) are continuous functions, it follows that t = 45° is the only solution.

Therefore, angleCAB is a right angle.


Remarks

More generally, if BC/AD = r and angleBCD = u, then (2 cos t)/r = (cot u - cot t)/(cot u + cot t).
Setting x = cot t, we get cos t = x/(1 + x2)1/2.
Then, setting v = cot u, squaring, simplifying and collecting terms, we obtain a quartic equation in x:

(r2 - 4)x4 - 2v(r2 + 4)x3 + (v2(r2 - 4) + r2)x2 - 2vr2x + v2r2 = 0.

For given r and u, the (unique) positive root of this equation that is less than cot u (hence t > u) yields t, and hence angleCAB.

Source: Mathematical Fallacies, Flaws and Flimflam, by Edward J. Barbeau. See Chapter 3.

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