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Geometry Difficulty 5.3 AIME, harder Find the answer

Example 9 As shown in Figure 12,
in △ABC\triangle A B C, ACA C =BC,∠ACB==B C, \angle A C B= 80∘80^{\circ}, take a point MM inside △ABC\triangle A B C, such that ∠MAB=10∘\angle M A B=10^{\circ}, ∠MBA=30∘\angle M B A=30^{\circ}. Find the degree measure of ∠AMC\angle A M C.
(1983, Former Yugoslavia Mathematical Olympiad)

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution 1: Since AC=BC,∠ACB=80∘A C=B C, \angle A C B=80^{\circ}, then ∠CAB=∠CBA=50∘\angle C A B=\angle C B A=50^{\circ}.
Since ∠MAB=10∘,∠MBA=30∘\angle M A B=10^{\circ}, \angle M B A=30^{\circ}, therefore,
∠CAM=40∘,∠CBM=20∘.  \angle C A M=40^{\circ}, \angle C B M=20^{\circ} \text {. }

Let ∠BCM=x\angle B C M=x, then, ∠ACM=80∘−x\angle A C M=80^{\circ}-x.
Applying the trigonometric form of Ceva's Theorem to △ABC\triangle A B C and point MM we have
1=sin⁡∠CAMsin⁡∠MAB⋅sin⁡∠ABMsin⁡∠MBC⋅sin⁡∠BCMsin⁡∠MCA=sin⁡40∘sin⁡10∘⋅sin⁡30∘sin⁡20∘⋅sin⁡xsin⁡(80∘−x). Then sin⁡(80∘−x)sin⁡x=sin⁡40∘⋅sin⁡30∘sin⁡10∘⋅sin⁡20∘=cos⁡20∘sin⁡10∘=sin⁡(80∘−10∘)sin⁡10∘. \begin{array}{l} 1=\frac{\sin \angle C A M}{\sin \angle M A B} \cdot \frac{\sin \angle A B M}{\sin \angle M B C} \cdot \frac{\sin \angle B C M}{\sin \angle M C A} \\ =\frac{\sin 40^{\circ}}{\sin 10^{\circ}} \cdot \frac{\sin 30^{\circ}}{\sin 20^{\circ}} \cdot \frac{\sin x}{\sin \left(80^{\circ}-x\right)} . \\ \text { Then } \frac{\sin \left(80^{\circ}-x\right)}{\sin x}=\frac{\sin 40^{\circ} \cdot \sin 30^{\circ}}{\sin 10^{\circ} \cdot \sin 20^{\circ}} \\ =\frac{\cos 20^{\circ}}{\sin 10^{\circ}}=\frac{\sin \left(80^{\circ}-10^{\circ}\right)}{\sin 10^{\circ}} . \end{array}

Since sin⁡(80∘−x)sin⁡x=sin⁡80∘⋅cot⁡x−cos⁡80∘\frac{\sin \left(80^{\circ}-x\right)}{\sin x}=\sin 80^{\circ} \cdot \cot x-\cos 80^{\circ}
is a strictly decreasing function of xx in (0,π)(0, \pi), therefore,
∠BCM=x=10∘.  \angle B C M=x=10^{\circ} \text {. }

Thus, ∠AMC=∠MAB+∠ABC+∠BCM=70∘\angle A M C=\angle M A B+\angle A B C+\angle B C M=70^{\circ}.
Solution 2: Since AC=BC,∠ACB=80∘A C=B C, \angle A C B=80^{\circ}, then
∠CAB=∠CBA=50∘.  \angle C A B=\angle C B A=50^{\circ} \text {. }

Since ∠MAB=10∘,∠MBA=30∘\angle M A B=10^{\circ}, \angle M B A=30^{\circ}, therefore,
∠CAM=40∘,∠CBM=20∘,∠AMB=140∘. \begin{array}{l} \angle C A M=40^{\circ}, \angle C B M=20^{\circ}, \\ \angle A M B=140^{\circ} . \end{array}

Let ∠AMC=x\angle A M C=x, then, ∠CMB=220∘−x\angle C M B=220^{\circ}-x. Applying the trigonometric form of Ceva's Theorem to △MAB\triangle M A B and point CC we have
1=sin⁡∠AMCsin⁡∠CMB⋅sin⁡∠MBCsin⁡∠CBA⋅sin⁡∠BACsin⁡∠CAM=sin⁡xsin⁡(220∘−x)⋅sin⁡20∘sin⁡50∘⋅sin⁡50∘sin⁡40∘. Then sin⁡(220∘−x)sin⁡x=12cos⁡20∘=sin⁡(220∘−70∘)sin⁡70∘. \begin{array}{l} 1=\frac{\sin \angle A M C}{\sin \angle C M B} \cdot \frac{\sin \angle M B C}{\sin \angle C B A} \cdot \frac{\sin \angle B A C}{\sin \angle C A M} \\ =\frac{\sin x}{\sin \left(220^{\circ}-x\right)} \cdot \frac{\sin 20^{\circ}}{\sin 50^{\circ}} \cdot \frac{\sin 50^{\circ}}{\sin 40^{\circ}} . \\ \text { Then } \frac{\sin \left(220^{\circ}-x\right)}{\sin x}=\frac{1}{2 \cos 20^{\circ}} \\ =\frac{\sin \left(220^{\circ}-70^{\circ}\right)}{\sin 70^{\circ}} . \end{array}

Since sin⁡(220∘−x)sin⁡x=sin⁡220∘⋅cot⁡x−\frac{\sin \left(220^{\circ}-x\right)}{\sin x}=\sin 220^{\circ} \cdot \cot x- cos⁡220∘(sin⁡220∘<0)\cos 220^{\circ}\left(\sin 220^{\circ}<0\right) is a strictly increasing function of xx in (0,π)(0, \pi), therefore,
∠AMC=x=70∘. \angle A M C=x=70^{\circ} .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.