answer:1. Consider the circles ( Gamma_1 ) and ( Gamma_2 ) intersecting at points ( M ) and ( N ). Let ( l ) be a common tangent to both circles closer to ( M ) than ( N ). Let ( l ) intersect ( Gamma_1 ) at point ( A ) and ( Gamma_2 ) at point ( B ). 2. Draw a line parallel to ( l ) that passes through ( M ). This line intersects ( Gamma_1 ) at another point ( C ) and ( Gamma_2 ) at another point ( D ). 3. Let lines ( CA ) and ( DB ) intersect at point ( E ). Let lines ( AN ) and ( CD ) intersect at point ( P ), and lines ( BN ) and ( CD ) intersect at point ( Q ). We need to prove that ( EP = EQ ). 4. To aid our proof, connect points ( N ) and ( M ), extending it to intersect line ( AB ) at point ( R ), and draw lines ( EM ), ( AM ), and ( BM ). 5. Using the Power of a Point Theorem, observe that: [ AR^2 = RM cdot RN = RB^2 ] This implies that: [ AR = RB ] Thus, ( R ) is the midpoint of ( AB ). 6. Because ( PQ parallel AB ) (both being parallel to the common tangent ( l )), it follows that: [ PM = MQ ] 7. To demonstrate that ( EM perp PQ ), we need to show ( EM perp AB ). 8. Observe that: [ angle EAB = angle ECQ = angle MAB ] Given the symmetry and the fact that ( R ) is the midpoint of ( AB ), the following holds: [ angle ABE = angle ABM ] Hence: [ angle EAB + angle ABE = angle MAB + angle ABM = 90^circ ] Therefore, ( EM ) is perpendicular to ( AB ), which implies ( EM perp PQ ). 9. Given that ( EM ) is the perpendicular bisector of ( PQ ), it follows by symmetry that ( EP = EQ ). Conclusion: [ boxed{EP = EQ} ]

answer:1. Calculate the units digit of 5^{11}. From the original problem, we know that the units digit of any power of 5 is 5. 2. Calculate the units digit of 2^3. Multiplying out, 2^3 = 8, so the units digit is 8. 3. Now multiply these results to find the units digit of 5^{11} times 8. Since the units digit of 5^{11} is 5, we multiply 5 by 8 to get 40. 4. The units digit of 40 is boxed{0}.

answer:To make one third of the cake, only one third of the 6 frac{1}{3} cups of flour are needed. First, convert 6 frac{1}{3} to an improper fraction: [ 6 frac{1}{3} = 6 + frac{1}{3} = frac{18}{3} + frac{1}{3} = frac{19}{3} ] Now, compute one third of frac{19}{3}: [ text{One third of } frac{19}{3} = frac{1}{3} times frac{19}{3} = frac{19}{9} ] Change frac{19}{9} to a mixed number: [ frac{19}{9} = 2 frac{1}{9} ] Therefore, you would need boxed{2 frac{1}{9}} cups of flour to make one third of the cake.

answer:Since the equation (a-1)x^2 + ax + 1 = 0 is a quadratic equation, it follows that a - 1 neq 0, thus a^2 + a = 2, solving this gives: a = -2, substituting a = -2 into the equation yields -3x^2 - 2x + 1 = 0, solving this equation gives: x_1 = -1, x_2 = frac{1}{3}, since x = -1 is a root of the equation (m^2 + m)x^2 + 3mx - 3 = 0, it follows that m^2 + m - 3m - 3 = 0, which simplifies to m^2 - 2m - 3 = 0, thus, m_1 = -1, m_2 = 3. Therefore, the values of a and m are boxed{a = -2, m = -1 text{ or } 3}.