answer:To demonstrate that triangle (AXY) is isosceles at (Y), we need to show that (AY = XY). 1. **Understand the configuration:** - ( Gamma ) is the circumcircle of triangle (ABC). - ( X ) is the foot of the angle bisector from (A) on (BC). - ( Y ) is the intersection of line (BC) and the tangent to ( Gamma ) at (A). 2. **Angles and Cyclic Properties:** - Since (Y) lies on the tangent to ( Gamma ) at (A), by the tangent-secant theorem, we know that ( angle BAY = angle YAC ) (both angles subtend the arc (BC) on ( Gamma )). 3. **Evaluate angles involving (AX):** - Since (X) is on (BC), we get: - (angle BAX = angle CAX) (angles bisected by (AX)). 4. **Transformation involving tangent properties:** - By the properties of the circumcircle and tangent, we use the identity involving directed angles: [ (AX, AY) - (XY, XA) = (AX, AB) + (AB, AY) + (AX, AB) + (AB, XY). ] 5. **Break down the angle calculations:** - Evaluating each angle difference separately: [ (AX, AB) + (AB, AY) = angle BAX + angle BAY = angle BAX + angle YAC = angle BAC ] since (angle BAY = angle YAC). 6. **Use cyclic and tangential properties:** - Similarly, the other set of angles: [ (AX, AB) + (AB, XY) = angle BAX + angle BAC = angle BAC + angle ACY, ] and noting ( angle BAC = angle ACY ) yields: [ (BC, CA) + (gamma, beta). ] 7. **Summarizing the directed angle differences:** - We combine the angle differences: [ (AX, AY) - (XY, XA) = 0^circ, ] indicating symmetrical angles around (Y). **Conclusion:** Since the angle calculations symmetrically satisfy zero degree difference, ( angle AXY = angle AYA ). Therefore, ( AY = XY ). Thus, triangle (AXY) is isosceles at (Y). [ boxed{AY = XY} ]

answer:1. **Define the intersection point:** Let ( Q ) be the intersection point of lines ( AM ) and ( DN ). We need to prove ( Q ) lies on line ( XY ). 2. **Establish relationship with point ( P ):** Connect ( MN ) and note that ( P ) lies on the radical axis ( XY ) of the circles defined by ( AC ) and ( BD ). 3. **Triangle similarity:** Since point ( P ) lies on ( XY ) and on the radical axes of the two circles (implying equal power to both circles), we have: [ PC cdot PM = PB cdot PN ] Therefore, triangles ( triangle PBC ) and ( triangle PMN ) are similar: [ triangle PBC sim triangle PMN ] 4. **Angle equivalence:** From the similarity of triangles, we get: [ angle PMN = angle PBC ] 5. **Right angles:** Because ( AC ) and ( BD ) are the diameters of the respective circles, the angles subtended by these diameters at the circumference are right angles: [ angle AMP = 90^circ ] Moreover, since ( XY ) intersects ( BC ) at ( Z ), forming a right angle with diameter ( BD ), we have: [ angle PBC + angle D = 90^circ ] 6. **Supplementary angles:** Utilize ( angle AMP ) and angles in triangle ( triangle PMN ): [ angle AMN + angle D = angle AMP + angle PMN + angle D = 180^circ ] Thus, points ( A, M, N, D ) are concyclic. 7. **Power of a point:** Given the concyclic nature of points ( A, M, N ) and ( D ), apply the power of a point theorem for ( Q ): [ AQ cdot QM = QD cdot QN ] Therefore, point ( Q ) is equidistant in terms of power from both circles, implying ( Q ) must lie on the radical axis ( XY ). # Conclusion: Thus, we have demonstrated that ( Q ) lies on line ( XY ). Hence, ( Q ) satisfies the condition of being on the radical axis, consistent with the properties of the geometric setup. [blacksquare]

answer:First, simplify 18 pmod{5}. Since 18 equiv 3 pmod{5}, the problem reduces to finding 3^{63} pmod{5}. At this point, recognize the cyclical pattern of powers of 3 modulo 5, which repeats every 4 terms: begin{align*} 3^1 &equiv 3 pmod{5}, 3^2 &equiv 4 pmod{5}, 3^3 &equiv 2 pmod{5}, 3^4 &equiv 1 pmod{5}, end{align*} Since 3^4 equiv 1 pmod{5}, by properties of exponents and modular arithmetic, for any integer k, 3^{4k} equiv 1 pmod{5}. For 3^{63}, we can express the exponent as 63 = 4*15 + 3, so: begin{align*} 3^{63} &= (3^4)^{15} cdot 3^3 &equiv 1^{15} cdot 3^3 pmod{5} &equiv 1 cdot 2 pmod{5} &equiv 2 pmod{5}. end{align*} Thus, the remainder when 18^{63} is divided by 5 is boxed{2}.

answer:1. Add the number of students in the drama club and the science club together: 90 + 140 = 230. 2. Substract the total number of students participating in either or both clubs from the sum obtained: 230 - 200 = 30. 3. According to the principle of inclusion and exclusion, this result suggests we have counted 30 students twice, ones who are members of both clubs. 4. Therefore, there are boxed{30} students who are in both the drama and science clubs.