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National Olympiad Final Round

Geometry Difficulty 5.8 AIME, harder Prove it Estonia

Triangle $ABC$ has $AC = BC$ . The bisector of angle $CAB$ meets side $BC$ at point $D$ . The difference of the sizes of some two internal angles of triangle $ABD$ is $40^\circ$ . Find all possibilities of what the size of angle $ACB$ can be.

Solution

Fig. 16

* If $\alpha - \frac{\alpha}{2} = 40^\circ$ then $\alpha = 80^\circ$ , whence $\angle ACB = 20^\circ$ .

* The case $\frac{\alpha}{2} - \alpha = 40^\circ$ is impossible since it would imply $\alpha < 0^\circ$ .

* If $(180^\circ - \frac{3}{2}\alpha) - \frac{\alpha}{2} = 40^\circ$ then $\alpha = 70^\circ$ , whence $\angle ACB = 40^\circ$ .

* If $\frac{\alpha}{2} - (180^\circ - \frac{3}{2}\alpha) = 40^\circ$ then $\alpha = 110^\circ$ , but the base angle of an isosceles triangle cannot be obtuse.

* If $(180^\circ - \frac{3}{2}\alpha) - \alpha = 40^\circ$ then $\alpha = 56^\circ$ , whence $\angle ACB = 68^\circ$ .

* If $\alpha - (180^\circ - \frac{3}{2}\alpha) = 40^\circ$ then $\alpha = 88^\circ$ , whence $\angle ACB = 4^\circ$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.