answer:Given: - ( C_1 ) is a circle. - ( P ) is a fixed point outside the circle ( C_1 ). - Quadrilateral ( ABCD ) lies on the circle ( C_1 ) such that rays ( AB ) and ( CD ) intersect at ( P ). - ( E ) is the intersection of ( AC ) and ( BD ). We need to prove: (a) The circumcircle of triangle ( ADE ) and the circumcircle of triangle ( BEC ) pass through a fixed point. (b) Find the locus of point ( E ). # Part (a) 1. **Construct Tangents and Points:** - Let ( PK ) and ( PQ ) be two tangents from ( P ) to ( C_1 ). - Let ( O ) be the center of ( C_1 ). - Let ( H ) be the intersection of ( PO ) and ( KQ ). - Let ( CH ) intersect ( C_1 ) at ( J ). - Let ( BH ) intersect ( C_1 ) at ( F ). 2. **Cyclic Quadrilateral:** - Since ( HA cdot HO = HK cdot HQ = HC cdot HJ ), it follows that ( PJOC ) is cyclic. - Therefore, ( angle OJH = angle OPC ). 3. **Similar Triangles:** - Since ( OH cdot OP = R^2 = OD^2 ), we have ( frac{OH}{OD} = frac{OD}{OA} ). - Thus, ( triangle HOD sim triangle DOP ), implying ( angle OPC = angle ODH ). 4. **Perpendicularity and Parallelism:** - From the above, ( angle HJO = angle HDO ), so ( HO perp JD ). - Therefore, ( JD parallel KQ ). - Similarly, ( AF parallel KQ parallel JD ), implying ( text{arc } AD = text{arc } JF ). 5. **Cyclic Quadrilaterals:** - We have ( angle BHC = angle BEC ), so ( BHEC ) is cyclic. - Similarly, ( AHED ) is cyclic. 6. **Conclusion:** - The circumcircles of triangles ( ADE ) and ( BEC ) pass through a fixed point ( H ). # Part (b) 1. **Angle Calculation:** - We have ( 180^circ - angle BHE = angle BCE = frac{1}{2} (text{arc } BK + text{arc } KA) = frac{1}{2} (text{arc } BK + text{arc } QF) = angle BHK ). 2. **Collinearity:** - Therefore, ( K, H, E ) are collinear. 3. **Locus of ( E ):** - Hence, ( E in [KQ] ). The final answer is ( boxed{ E in [KQ] } ).

answer:Let N be the positive integer that satisfies all the given conditions. Since N leaves a remainder of 9 when divided by 10, N+1 must be divisible by 10. Similarly, N+1 should be divisible by 11, 12, and 13, as it leaves remainders of 10, 11, and 12 respectively when divided by these numbers. To find the least such N, we need to find the least common multiple (LCM) of 10, 11, 12, and 13. We determine the LCM by considering the highest powers of all prime factors: - 10 = 2 cdot 5 - 11 = 11 (prime) - 12 = 2^2 cdot 3 - 13 = 13 (prime) Thus, text{LCM}(10, 11, 12, 13) = 2^2 cdot 3 cdot 5 cdot 11 cdot 13. Calculating this, we get: - 2^2 = 4 - 4 cdot 3 = 12 - 12 cdot 5 = 60 - 60 cdot 11 = 660 - 660 cdot 13 = 8580 Therefore, N+1 = 8580. So, the smallest integer N satisfying all conditions is 8580 - 1 = 8579. Hence, the answer is boxed{8579}.

answer:To solve the given system of nonlinear equations: [ begin{aligned} & x^2 + y^2 = 1, & 4xy(2y^2 - 1) = 1, end{aligned} ] we will proceed step-by-step. 1. **Identify the constraints:** The first equation ( x^2 + y^2 = 1 ) represents a unit circle in the (xy)-plane. 2. **Parameterize the circle using trigonometric identities:** Since ( x^2 + y^2 = 1 ), we can use the parametrization: [ x = sin varphi quad text{and} quad y = cos varphi, ] where ( varphi ) is some angle. 3. **Substitute the parameterization into the second equation:** Substitute ( x = sin varphi ) and ( y = cos varphi ) into the second equation (4xy(2y^2 - 1) = 1): [ 4 (sin varphi) (cos varphi) left( 2 (cos^2 varphi) - 1 right) = 1. ] Simplify the equation: [ 4 sin varphi cos varphi left( 2 cos^2 varphi - 1 right) = 1. ] 4. **Recognize and utilize trigonometric identities:** Note that (4 sin varphi cos varphi) can be rewritten using the double-angle identity: [ 4 sin varphi cos varphi = 2 (sin 2varphi). ] Also, recognize that (2 cos^2 varphi - 1) is the double-angle identity for cosine: [ 2 cos^2 varphi - 1 = cos 2varphi. ] Thus, the equation becomes: [ 2 sin 2varphi cos 2varphi = 1. ] 5. **Simplify using a further trigonometric identity:** Apply the product-to-sum identity: ( sin 4varphi = 2 sin 2varphi cos 2varphi ): [ sin 4varphi = 1. ] 6. **Solve for ( varphi ):** The equation ( sin 4varphi = 1 ) has solutions: [ 4varphi = frac{pi}{2} + 2kpi quad text{for} quad k in mathbb{Z}. ] Solving for ( varphi ): [ varphi = frac{pi}{8} + frac{kpi}{2}. ] 7. **Determine specific solutions:** The specific solutions for ( varphi ) within one period ((0 leq varphi < 2pi)) are found by substituting ( k = 0, 1, 2, text{and} 3): [ varphi = frac{pi}{8}, ; frac{5pi}{8}, ; frac{9pi}{8}, ; frac{13pi}{8}. ] 8. **Back-substitute to find (x) and (y):** Find the corresponding ( x ) and ( y ) values for these ( varphi ) values: [ varphi = frac{pi}{8} Rightarrow (x, y) = left(sin frac{pi}{8}, cos frac{pi}{8}right), ] [ varphi = frac{5pi}{8} Rightarrow (x, y) = left(sin frac{5pi}{8}, cos frac{5pi}{8}right), ] [ varphi = frac{9pi}{8} Rightarrow (x, y) = left(sin frac{9pi}{8}, cos frac{9pi}{8}right), ] [ varphi = frac{13pi}{8} Rightarrow (x, y) = left(sin frac{13pi}{8}, cos frac{13pi}{8}right). ] 9. **Simplify the results:** Noting that (sin (varphi + pi) = -sin varphi) and (cos (varphi + pi) = -cos varphi), we get the corresponding points: [ left( pm sin frac{pi}{8}, pm cos frac{pi}{8} right), left( pm sin frac{5pi}{8}, pm cos frac{5pi}{8} right). ] Converting these to simpler forms using known identities, we get: [ left( pm frac{sqrt{2-sqrt{2}}}{2}, pm frac{sqrt{2+sqrt{2}}}{2} right), left( pm frac{sqrt{2+sqrt{2}}}{2}, mp frac{sqrt{2-sqrt{2}}}{2} right). ] # Conclusion: The solutions to the system are: [ boxed{left( pm frac{sqrt{2- sqrt{2}}}{2}, pm frac{sqrt{2+sqrt{2}}}{2} right), left( pm frac{sqrt{2+sqrt{2}}}{2}, mp frac{sqrt{2- sqrt{2}}}{2} right)}. ]

answer:Given: Quadrilateral ABCD with its diagonals AC and BD intersecting at E, making perpendicular lines from E to each side: - EP perpendicular to AB - EQ perpendicular to BC - ER perpendicular to CD - ES perpendicular to DA Points P', Q', R', and S' are respective intersections of these perpendiculars from E with the opposite sides, namely: - P' on CD - Q' on DA - R' on AB - S' on BC We aim to prove: - R'S' parallel Q'P' parallel AC - R'Q' parallel S'P' parallel BD. To prove first part, we will show: [ frac{A Q'}{D Q'} cdot frac{D P'}{C P'} =1. ] 1. Consider the triangles triangle AQE and triangle DQE. Their area ratio can be written using the base and height: [ frac{S_{triangle AQE}}{S_{triangle DQE}} = frac{AE cdot QEP}{DE cdot QEP} = frac{AE}{DE}. ] 2. Similarly, for triangles triangle DPE and triangle CPE: [ frac{S_{triangle DPE}}{S_{triangle CPE}} = frac{DE cdot PEP}{CE cdot PEP} = frac{DE}{CE}. ] 3. Now, multiply the area ratios: [ frac{S_{triangle AQE}}{S_{triangle DQE}} cdot frac{S_{triangle DPE}}{S_{triangle CPE}} = frac{AE}{DE} cdot frac{DE}{CE}. ] 4. Simplify the expression: [ frac{A Q'}{D Q'} cdot frac{D P'}{C P'} = frac{AE}{DE} cdot frac{DE}{CE} = frac{AE}{CE}. ] 5. Show this is equal to 1. We get: [ frac{A E cdot DE cdot CE}{C E cdot DE cdot AE} = 1 Rightarrow 1 = 1. ] Thus, P' Q' parallel AC. Similarly, by using same line of reasoning for other pair of triangles: 6. We have: [ frac{ S_{triangle S P E}}{ S_{triangle R P E}} = frac{ S_{triangle R Q'E}}{S_{triangle S Q' E}}. ] 7. Combining above relations, we get: [ frac{AR'}{BR'} cdot frac{BS'}{C S'} = 1. ] Thus, R'S' parallel Q'P' parallel AC. And so, boxed{R'S' parallel Q'P' parallel AC}. 8. Following the same steps, we can prove: [ R'Q' parallel P'S' parallel BD. ] Thus, boxed{R'Q' parallel S'P' parallel BD}.