# Solution to puzzle 150: Isosceles apex angle

Triangle ABC is isosceles with AB = AC. Point D on AB is such that BCD = 15° and BC = AD. Find, with proof, the measure of CAB.

Let ABC = t and AC = x.

Then ADC = t + 15° and DCA = t - 15°.

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

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

Dropping a perpendicular from A to BC, we have:

cos t = ½BC/AC = /(2x).

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

By inspection, if t = 45°, then:

sin(t - 15°)/sin(t + 15°) = ½ / (½) = 1/.

(2 cos t)/ = / = 1/.

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)/ 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, CAB is a right angle.

## Remarks

More generally, if BC/AD = r and BCD = u, then (2 cos t)/r = (cot u - cot t)/(cot u + cot t).

Setting x = cot t, we get cos t = x/(1 + x^{2})^{1/2}.

Then, setting v = cot u, squaring, simplifying and collecting terms, we obtain a quartic equation in x:

(r^{2} - 4)x^{4} - 2v(r^{2} + 4)x^{3} + (v^{2}(r^{2} - 4) + r^{2})x^{2} - 2vr^{2}x + v^{2}r^{2} = 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 CAB.

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

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