Competition problem — Geometry — Maths Olympiad Prep

Library / /7 of 11

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 $\angle ACB = 4^\circ$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.