two intersecting circles inscribed in an angle with AB=AC=AD problem set

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two intersecting circles inscribed in an angle with AB=AC=AD problem set

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Source: 2020 Olympiad YuMSh Finals IX-XI p2 - Olympiad of Youth Mathematical School of St. Petersburg State University

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#1 h Mar 5, 2021, 4:51 PM

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Two circles inscribed in an angle with a vertex $R$ meet at points $A$ and $B$ . A line is drawn through A that intersects the smaller circle at point $C$ , and the larger one at point $D$ . It turned out that $AB = AC = AD$ .

1. Prove that the tangents to the circles at point $A$ are perpendicular.

2. Let $C$ and $D$ coincide with the points of tangency of the circles and the angle. Prove that the $\angle R$ is right.

3. Let $C$ and $D$ coincide with the points of tangency of the circles and the angle. Find the angle $ADR$ .

4. Prove that if $\angle R$ is right, then $C$ and $D$ coincide with the tangency points of the circles and the angle.

5. Let $\angle R = 135^o$ . The perpendicular from A to the nearest side of the angle intersects the smaller circle at point $P$ , the perpendicular from $A$ to the second side intersects $BP$ at point $Q$ . Finally, let $O_1$ and $O_2$ be the centers of the original circles, $O$ be the center of the circle circumscribed around $\vartriangle ABQ$ . Prove that $BO$ is the bisector of the angle $O_1BO_2$ .

6. What values can the angle $\angle RAO_1$ take, where $O_1$ is the center of the smaller circle?

Grade 9: 1-4 , Grade 10: 2-5 Grade 11: 2-4,6

This post has been edited 1 time. Last edited by parmenides51, Mar 5, 2021, 4:52 PM

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#2 h Mar 5, 2021, 5:19 PM • 1 Y

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1. Let the centers of the smaller and larger circle be $O_1,O_2$ , respectively. Note that the length condition implies that $C,D$ are the reflections of $B$ over the $AO_1,AO_2$ , respectively. Then $\angle CAO_1+\angle DAO_2 = \angle BAO_1+\angle BAO_2$ , and using the fact that $\angle CAO_1+\angle DAO_2 + \angle BAO_1+\angle BAO_2 = 180^{\circ}$ yields that $90^{\circ}=\angle BAO_1+\angle BAO_2 = \angle O_1AO_2$ , as desired.

2. $\angle DCR+\angle CDR = \angle ABC+\angle ABD = \frac{1}{2}(180^{\circ}-\angle BAC)+\frac{1}{2}(180^{\circ}-\angle BAD)=\frac{1}{2}(360^{\circ}-180^{\circ})=90^{\circ}$ , so $\angle CRD = 90^{\circ}$ , as desired.

This post has been edited 1 time. Last edited by kootrapali, Mar 5, 2021, 5:27 PM

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