Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it North Macedonia

Let $k$ is the circumscribed circle for the quadrangle $ABCD$ . The angle at the vertex $B$ is twice bigger than the angle at the vertex $A$ and for $40^\circ$ smaller than the angle at the vertex $D$ . Calculate the angles of $ABCD$ .

Solution

From the condition in the problem we have that $\beta = 2\alpha$ and $\beta = \delta - 40^\circ$ . From this we have $\delta = 2\alpha + 40^\circ$ . Because $ABCD$ can be inscribed in a circle we have that $\alpha + \gamma = \beta + \delta$ and because $\alpha + \gamma + \beta + \delta = 360^\circ$ we have that $\alpha + \gamma = \beta + \delta = 180^\circ$ . So $180^\circ = \beta + \delta = 2\alpha + 2\alpha + 40^\circ = 4\alpha + 40^\circ$ , from where we obtain $4\alpha = 140^\circ$ i.e. $\alpha = 35^\circ$ . Now we get that $\beta = 2\alpha = 70^\circ$ , $\gamma = 180^\circ - \alpha = 145^\circ$ and $\delta = 2\alpha + 40^\circ = 2 \cdot 35^\circ + 40^\circ = 110^\circ$ .

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