answer：Alright, so I have this geometry problem here. Let me try to parse it step by step. We have triangle ( ABC ), and ( D ) is the foot of the internal bisector of angle ( angle A ). So, ( AD ) is the angle bisector, splitting ( angle BAC ) into two equal angles. That makes sense. Now, the problem mentions the straight line that joins the incenters of triangles ( ABD ) and ( ACD ). Okay, so first, I need to figure out where these incenters are. The incenter of a triangle is the intersection point of its angle bisectors, right? So, for triangle ( ABD ), its incenter ( I_1 ) would be where the angle bisectors of ( angle ABD ), ( angle BAD ), and ( angle ADB ) meet. Similarly, for triangle ( ACD ), its incenter ( I_2 ) is where the angle bisectors of ( angle ACD ), ( angle CAD ), and ( angle ADC ) meet.Once I have these two incenters, ( I_1 ) and ( I_2 ), the problem says that the line joining them intersects ( AB ) at ( M ) and ( AC ) at ( N ). So, if I draw a line from ( I_1 ) to ( I_2 ), this line will cross ( AB ) at some point ( M ) and ( AC ) at some point ( N ). The goal is to show that lines ( BN ) and ( CM ) meet on the bisector ( AD ). So, if I draw lines from ( B ) to ( N ) and from ( C ) to ( M ), their intersection point should lie somewhere on ( AD ). Hmm, okay. Let me try to visualize this. Maybe drawing a diagram would help, but since I can't draw here, I'll try to imagine it. Triangle ( ABC ), with ( AD ) as the angle bisector. Then, inside ( ABD ), the incenter ( I_1 ) is somewhere, and inside ( ACD ), the incenter ( I_2 ) is somewhere else. The line ( I_1I_2 ) cuts ( AB ) at ( M ) and ( AC ) at ( N ). Then, connecting ( B ) to ( N ) and ( C ) to ( M ), their intersection is supposed to be on ( AD ).I think I need to use some properties of angle bisectors and incenters. Maybe properties related to harmonic division or Ceva's theorem? Or perhaps using coordinate geometry? Let me think.First, let me recall that the incenter lies at the intersection of angle bisectors. So, in triangle ( ABD ), ( I_1 ) lies on the bisector of ( angle ABD ) and ( angle BAD ). Similarly, ( I_2 ) lies on the bisector of ( angle ACD ) and ( angle CAD ).Since ( AD ) is the angle bisector of ( angle BAC ), it splits ( BC ) into segments proportional to ( AB ) and ( AC ). That is, ( frac{BD}{DC} = frac{AB}{AC} ). That's the Angle Bisector Theorem. Maybe that will come into play.Now, considering the line ( I_1I_2 ), which intersects ( AB ) at ( M ) and ( AC ) at ( N ). I need to find some relationship between these points and the lines ( BN ) and ( CM ).Perhaps I can use Ceva's Theorem. Ceva's Theorem states that for concurrent lines from the vertices of a triangle, the product of certain ratios equals 1. In this case, if lines ( BN ) and ( CM ) intersect on ( AD ), then maybe Ceva's condition is satisfied.Wait, but Ceva's Theorem usually involves three lines, not two. Maybe I need to consider the third line as well, but since we're dealing with two lines, perhaps another approach is needed.Alternatively, maybe using Menelaus' Theorem could help, since we have a transversal cutting across a triangle. The line ( I_1I_2 ) is cutting across triangle ( ABC ), intersecting ( AB ) and ( AC ) at ( M ) and ( N ). So, Menelaus' Theorem might give a relation between the segments.Let me recall Menelaus' Theorem: for a triangle ( ABC ) and a transversal line that intersects ( AB ) at ( M ), ( BC ) at ( L ), and ( AC ) at ( N ), the product ( frac{AM}{MB} cdot frac{BL}{LC} cdot frac{CN}{NA} = 1 ). But in our case, the transversal is ( I_1I_2 ), which only intersects ( AB ) and ( AC ), not ( BC ). So, maybe Menelaus isn't directly applicable here.Hmm, perhaps I need to use properties of incenters. The incenter is equidistant from all sides of the triangle. So, maybe the distances from ( I_1 ) to ( AB ), ( BD ), and ( AD ) are equal, and similarly for ( I_2 ).Alternatively, maybe coordinate geometry would be a good approach. Assign coordinates to the triangle and compute the necessary points. Let me try that.Let me place triangle ( ABC ) in the coordinate plane. Let me set point ( A ) at the origin ( (0, 0) ), point ( B ) at ( (c, 0) ), and point ( C ) at ( (d, e) ). Then, point ( D ) is the foot of the angle bisector from ( A ) to ( BC ). Using the Angle Bisector Theorem, the coordinates of ( D ) can be determined.But this might get messy. Maybe it's better to use barycentric coordinates or another coordinate system. Alternatively, maybe using vectors.Wait, perhaps using mass point geometry could help, considering the ratios along the sides.Alternatively, maybe using projective geometry concepts, like harmonic conjugates or something.Wait, another idea: since ( I_1 ) and ( I_2 ) are incenters, the line ( I_1I_2 ) might have some special properties. Maybe it's related to the Gergonne line or something similar, but I'm not sure.Alternatively, perhaps considering the excenters. But I don't see a direct connection.Wait, let's think about the problem again. We have two incenters, ( I_1 ) and ( I_2 ), in triangles ( ABD ) and ( ACD ). The line connecting them intersects ( AB ) at ( M ) and ( AC ) at ( N ). Then, lines ( BN ) and ( CM ) meet on ( AD ).Maybe I can consider the properties of ( M ) and ( N ). Since ( M ) and ( N ) lie on ( AB ) and ( AC ), respectively, and are defined by the line through ( I_1 ) and ( I_2 ), perhaps there is a harmonic division or something similar.Wait, another thought: perhaps using Ceva's condition for concurrency. If I can show that the lines ( BN ), ( CM ), and ( AD ) are concurrent, then their intersection lies on ( AD ).So, to apply Ceva's Theorem, I need to compute the ratios ( frac{AM}{MB} ), ( frac{BN}{NC} ), and ( frac{CP}{PA} ) for some point ( P ) on ( BC ), but I'm not sure how that connects here.Alternatively, maybe using trigonometric Ceva's Theorem, which involves angles. But I'm not sure.Wait, perhaps I can use the fact that ( I_1 ) and ( I_2 ) lie on the angle bisectors of their respective triangles. So, in triangle ( ABD ), ( I_1 ) lies on the bisector of ( angle ABD ), and in triangle ( ACD ), ( I_2 ) lies on the bisector of ( angle ACD ).Since ( AD ) is the angle bisector of ( angle BAC ), maybe there is a relationship between these bisectors.Alternatively, maybe considering the inradius or distances from ( I_1 ) and ( I_2 ) to the sides.Wait, perhaps I can parametrize the coordinates. Let me try that.Let me set ( A ) at ( (0, 0) ), ( B ) at ( (1, 0) ), and ( C ) at ( (0, 1) ). Then, ( AD ) is the angle bisector of ( angle BAC ), which in this case is the line ( y = x ), since it's a 45-degree angle.Wait, no, in this coordinate system, ( AB ) is along the x-axis from ( (0, 0) ) to ( (1, 0) ), and ( AC ) is along the y-axis from ( (0, 0) ) to ( (0, 1) ). So, ( angle BAC ) is 90 degrees, and its bisector would be the line ( y = x ), but actually, in this case, the angle bisector would make equal angles with both axes, so yes, ( y = x ) is correct.Then, point ( D ) is the foot of the angle bisector from ( A ) to ( BC ). So, first, let me find the coordinates of ( D ).The line ( BC ) goes from ( (1, 0) ) to ( (0, 1) ), so its equation is ( x + y = 1 ).The angle bisector from ( A ) is ( y = x ). So, the intersection point ( D ) is where ( y = x ) meets ( x + y = 1 ). Solving these, we get ( x = y = 0.5 ). So, ( D ) is at ( (0.5, 0.5) ).Okay, so now, triangles ( ABD ) and ( ACD ) are both right triangles, since ( A ) is at the origin, ( B ) at ( (1, 0) ), ( D ) at ( (0.5, 0.5) ), and ( C ) at ( (0, 1) ).So, for triangle ( ABD ), which has vertices at ( A(0,0) ), ( B(1,0) ), and ( D(0.5, 0.5) ), the incenter ( I_1 ) can be found by finding the intersection of its angle bisectors.Similarly, for triangle ( ACD ), with vertices ( A(0,0) ), ( C(0,1) ), and ( D(0.5, 0.5) ), the incenter ( I_2 ) can be found.Let me compute the incenter ( I_1 ) of triangle ( ABD ). The incenter is located at the intersection of the angle bisectors. Alternatively, it can be found using the formula:If a triangle has vertices ( (x_A, y_A) ), ( (x_B, y_B) ), ( (x_C, y_C) ), and sides opposite these vertices have lengths ( a ), ( b ), ( c ), then the incenter ( (x, y) ) is given by:[x = frac{a x_A + b x_B + c x_C}{a + b + c}, quad y = frac{a y_A + b y_B + c y_C}{a + b + c}]So, for triangle ( ABD ), let's compute the lengths of the sides.First, side ( AB ): from ( A(0,0) ) to ( B(1,0) ). Length is ( sqrt{(1-0)^2 + (0-0)^2} = 1 ).Side ( BD ): from ( B(1,0) ) to ( D(0.5, 0.5) ). Length is ( sqrt{(0.5 - 1)^2 + (0.5 - 0)^2} = sqrt{(-0.5)^2 + (0.5)^2} = sqrt{0.25 + 0.25} = sqrt{0.5} = frac{sqrt{2}}{2} ).Side ( AD ): from ( A(0,0) ) to ( D(0.5, 0.5) ). Length is ( sqrt{(0.5)^2 + (0.5)^2} = sqrt{0.25 + 0.25} = sqrt{0.5} = frac{sqrt{2}}{2} ).So, the sides are ( AB = 1 ), ( BD = frac{sqrt{2}}{2} ), ( AD = frac{sqrt{2}}{2} ).Therefore, the incenter ( I_1 ) coordinates are:[x = frac{AB cdot x_A + BD cdot x_B + AD cdot x_D}{AB + BD + AD} = frac{1 cdot 0 + frac{sqrt{2}}{2} cdot 1 + frac{sqrt{2}}{2} cdot 0.5}{1 + frac{sqrt{2}}{2} + frac{sqrt{2}}{2}}]Similarly,[y = frac{AB cdot y_A + BD cdot y_B + AD cdot y_D}{AB + BD + AD} = frac{1 cdot 0 + frac{sqrt{2}}{2} cdot 0 + frac{sqrt{2}}{2} cdot 0.5}{1 + frac{sqrt{2}}{2} + frac{sqrt{2}}{2}}]Let me compute the denominator first:Denominator ( = 1 + frac{sqrt{2}}{2} + frac{sqrt{2}}{2} = 1 + sqrt{2} ).Now, compute the numerator for ( x ):( 0 + frac{sqrt{2}}{2} cdot 1 + frac{sqrt{2}}{2} cdot 0.5 = frac{sqrt{2}}{2} + frac{sqrt{2}}{4} = frac{2sqrt{2} + sqrt{2}}{4} = frac{3sqrt{2}}{4} ).Similarly, numerator for ( y ):( 0 + 0 + frac{sqrt{2}}{2} cdot 0.5 = frac{sqrt{2}}{4} ).Therefore,[x = frac{3sqrt{2}/4}{1 + sqrt{2}} = frac{3sqrt{2}}{4(1 + sqrt{2})}][y = frac{sqrt{2}/4}{1 + sqrt{2}} = frac{sqrt{2}}{4(1 + sqrt{2})}]To rationalize the denominators:Multiply numerator and denominator by ( 1 - sqrt{2} ):For ( x ):[frac{3sqrt{2}(1 - sqrt{2})}{4(1 + sqrt{2})(1 - sqrt{2})} = frac{3sqrt{2} - 6}{4(1 - 2)} = frac{3sqrt{2} - 6}{4(-1)} = frac{6 - 3sqrt{2}}{4} = frac{3(2 - sqrt{2})}{4}]Similarly, for ( y ):[frac{sqrt{2}(1 - sqrt{2})}{4(1 - 2)} = frac{sqrt{2} - 2}{4(-1)} = frac{2 - sqrt{2}}{4}]So, ( I_1 ) is at ( left( frac{3(2 - sqrt{2})}{4}, frac{2 - sqrt{2}}{4} right) ).Now, let's compute the incenter ( I_2 ) of triangle ( ACD ). Triangle ( ACD ) has vertices ( A(0,0) ), ( C(0,1) ), and ( D(0.5, 0.5) ).Compute the lengths of the sides:Side ( AC ): from ( A(0,0) ) to ( C(0,1) ). Length is ( 1 ).Side ( CD ): from ( C(0,1) ) to ( D(0.5, 0.5) ). Length is ( sqrt{(0.5 - 0)^2 + (0.5 - 1)^2} = sqrt{0.25 + 0.25} = sqrt{0.5} = frac{sqrt{2}}{2} ).Side ( AD ): same as before, ( frac{sqrt{2}}{2} ).So, sides are ( AC = 1 ), ( CD = frac{sqrt{2}}{2} ), ( AD = frac{sqrt{2}}{2} ).Therefore, the incenter ( I_2 ) coordinates are:[x = frac{AC cdot x_A + CD cdot x_C + AD cdot x_D}{AC + CD + AD} = frac{1 cdot 0 + frac{sqrt{2}}{2} cdot 0 + frac{sqrt{2}}{2} cdot 0.5}{1 + frac{sqrt{2}}{2} + frac{sqrt{2}}{2}}][y = frac{AC cdot y_A + CD cdot y_C + AD cdot y_D}{AC + CD + AD} = frac{1 cdot 0 + frac{sqrt{2}}{2} cdot 1 + frac{sqrt{2}}{2} cdot 0.5}{1 + frac{sqrt{2}}{2} + frac{sqrt{2}}{2}}]Compute the denominator: same as before, ( 1 + sqrt{2} ).Compute the numerator for ( x ):( 0 + 0 + frac{sqrt{2}}{2} cdot 0.5 = frac{sqrt{2}}{4} ).Numerator for ( y ):( 0 + frac{sqrt{2}}{2} cdot 1 + frac{sqrt{2}}{2} cdot 0.5 = frac{sqrt{2}}{2} + frac{sqrt{2}}{4} = frac{3sqrt{2}}{4} ).So,[x = frac{sqrt{2}/4}{1 + sqrt{2}} = frac{sqrt{2}}{4(1 + sqrt{2})}][y = frac{3sqrt{2}/4}{1 + sqrt{2}} = frac{3sqrt{2}}{4(1 + sqrt{2})}]Rationalizing:For ( x ):[frac{sqrt{2}(1 - sqrt{2})}{4(1 - 2)} = frac{sqrt{2} - 2}{4(-1)} = frac{2 - sqrt{2}}{4}]For ( y ):[frac{3sqrt{2}(1 - sqrt{2})}{4(1 - 2)} = frac{3sqrt{2} - 6}{4(-1)} = frac{6 - 3sqrt{2}}{4} = frac{3(2 - sqrt{2})}{4}]So, ( I_2 ) is at ( left( frac{2 - sqrt{2}}{4}, frac{3(2 - sqrt{2})}{4} right) ).Now, we have coordinates for ( I_1 ) and ( I_2 ). Let me write them down clearly:( I_1 left( frac{3(2 - sqrt{2})}{4}, frac{2 - sqrt{2}}{4} right) )( I_2 left( frac{2 - sqrt{2}}{4}, frac{3(2 - sqrt{2})}{4} right) )Now, the line ( I_1I_2 ) can be found by determining the equation of the line passing through these two points.Let me compute the slope first.Slope ( m = frac{y_2 - y_1}{x_2 - x_1} )Compute ( y_2 - y_1 ):( frac{3(2 - sqrt{2})}{4} - frac{2 - sqrt{2}}{4} = frac{3(2 - sqrt{2}) - (2 - sqrt{2})}{4} = frac{2(2 - sqrt{2})}{4} = frac{4 - 2sqrt{2}}{4} = 1 - frac{sqrt{2}}{2} )Compute ( x_2 - x_1 ):( frac{2 - sqrt{2}}{4} - frac{3(2 - sqrt{2})}{4} = frac{2 - sqrt{2} - 6 + 3sqrt{2}}{4} = frac{-4 + 2sqrt{2}}{4} = frac{-2 + sqrt{2}}{2} )So, slope ( m = frac{1 - frac{sqrt{2}}{2}}{frac{-2 + sqrt{2}}{2}} = frac{2 - sqrt{2}}{-2 + sqrt{2}} )Multiply numerator and denominator by ( -2 - sqrt{2} ) to rationalize:Numerator: ( (2 - sqrt{2})(-2 - sqrt{2}) = -4 - 2sqrt{2} + 2sqrt{2} + 2 = -4 + 2 = -2 )Denominator: ( (-2 + sqrt{2})(-2 - sqrt{2}) = 4 + 2sqrt{2} - 2sqrt{2} - 2 = 4 - 2 = 2 )So, slope ( m = frac{-2}{2} = -1 )So, the line ( I_1I_2 ) has a slope of ( -1 ). Now, let's find its equation.Using point ( I_1 ):( y - y_1 = m(x - x_1) )So,( y - frac{2 - sqrt{2}}{4} = -1 left( x - frac{3(2 - sqrt{2})}{4} right) )Simplify:( y = -x + frac{3(2 - sqrt{2})}{4} + frac{2 - sqrt{2}}{4} )Combine constants:( frac{3(2 - sqrt{2}) + (2 - sqrt{2})}{4} = frac{4(2 - sqrt{2})}{4} = 2 - sqrt{2} )So, the equation of line ( I_1I_2 ) is ( y = -x + 2 - sqrt{2} ).Now, we need to find where this line intersects ( AB ) and ( AC ).First, intersection with ( AB ). In our coordinate system, ( AB ) is the x-axis, so ( y = 0 ).Set ( y = 0 ) in the line equation:( 0 = -x + 2 - sqrt{2} )So, ( x = 2 - sqrt{2} )Therefore, point ( M ) is at ( (2 - sqrt{2}, 0) ).But wait, in our coordinate system, ( AB ) is from ( (0,0) ) to ( (1,0) ). So, ( x ) cannot be ( 2 - sqrt{2} ), which is approximately ( 2 - 1.414 = 0.586 ), which is within the segment ( AB ). Wait, no, ( 2 - sqrt{2} ) is approximately 0.586, which is between 0 and 1, so it's valid.Wait, actually, ( 2 - sqrt{2} approx 0.5857 ), which is less than 1, so it's on ( AB ). So, ( M ) is at ( (2 - sqrt{2}, 0) ).Similarly, find intersection with ( AC ). In our coordinate system, ( AC ) is the y-axis, so ( x = 0 ).Set ( x = 0 ) in the line equation:( y = -0 + 2 - sqrt{2} = 2 - sqrt{2} )So, point ( N ) is at ( (0, 2 - sqrt{2}) ).Now, we have points ( M(2 - sqrt{2}, 0) ) and ( N(0, 2 - sqrt{2}) ).Next, we need to find lines ( BN ) and ( CM ) and show that they intersect on ( AD ).First, find the equation of line ( BN ). Points ( B(1, 0) ) and ( N(0, 2 - sqrt{2}) ).Slope of ( BN ):( m = frac{(2 - sqrt{2}) - 0}{0 - 1} = frac{2 - sqrt{2}}{-1} = sqrt{2} - 2 )Equation of ( BN ):Using point ( B(1, 0) ):( y - 0 = (sqrt{2} - 2)(x - 1) )So,( y = (sqrt{2} - 2)x - (sqrt{2} - 2) )Simplify:( y = (sqrt{2} - 2)x + (2 - sqrt{2}) )Now, find the equation of line ( CM ). Points ( C(0, 1) ) and ( M(2 - sqrt{2}, 0) ).Slope of ( CM ):( m = frac{0 - 1}{(2 - sqrt{2}) - 0} = frac{-1}{2 - sqrt{2}} )Rationalize denominator:Multiply numerator and denominator by ( 2 + sqrt{2} ):( m = frac{-1(2 + sqrt{2})}{(2 - sqrt{2})(2 + sqrt{2})} = frac{-2 - sqrt{2}}{4 - 2} = frac{-2 - sqrt{2}}{2} = -1 - frac{sqrt{2}}{2} )Equation of ( CM ):Using point ( C(0, 1) ):( y - 1 = (-1 - frac{sqrt{2}}{2})(x - 0) )So,( y = (-1 - frac{sqrt{2}}{2})x + 1 )Now, we need to find the intersection point ( K ) of lines ( BN ) and ( CM ).Set the equations equal:( (sqrt{2} - 2)x + (2 - sqrt{2}) = (-1 - frac{sqrt{2}}{2})x + 1 )Bring all terms to left side:( (sqrt{2} - 2)x + (2 - sqrt{2}) + (1 + frac{sqrt{2}}{2})x - 1 = 0 )Combine like terms:( [(sqrt{2} - 2) + (1 + frac{sqrt{2}}{2})]x + (2 - sqrt{2} - 1) = 0 )Simplify coefficients:For ( x ):( sqrt{2} - 2 + 1 + frac{sqrt{2}}{2} = (sqrt{2} + frac{sqrt{2}}{2}) + (-2 + 1) = frac{3sqrt{2}}{2} - 1 )For constants:( 2 - sqrt{2} - 1 = 1 - sqrt{2} )So, equation becomes:( left( frac{3sqrt{2}}{2} - 1 right)x + (1 - sqrt{2}) = 0 )Solve for ( x ):( x = frac{-(1 - sqrt{2})}{frac{3sqrt{2}}{2} - 1} )Multiply numerator and denominator by 2 to eliminate fraction:( x = frac{-2(1 - sqrt{2})}{3sqrt{2} - 2} )Multiply numerator and denominator by ( 3sqrt{2} + 2 ) to rationalize denominator:Numerator:( -2(1 - sqrt{2})(3sqrt{2} + 2) = -2[ (1)(3sqrt{2}) + (1)(2) - sqrt{2}(3sqrt{2}) - sqrt{2}(2) ] )Simplify inside:( 3sqrt{2} + 2 - 3*2 - 2sqrt{2} = 3sqrt{2} + 2 - 6 - 2sqrt{2} = (3sqrt{2} - 2sqrt{2}) + (2 - 6) = sqrt{2} - 4 )So, numerator:( -2(sqrt{2} - 4) = -2sqrt{2} + 8 )Denominator:( (3sqrt{2} - 2)(3sqrt{2} + 2) = (3sqrt{2})^2 - (2)^2 = 18 - 4 = 14 )So, ( x = frac{-2sqrt{2} + 8}{14} = frac{8 - 2sqrt{2}}{14} = frac{4 - sqrt{2}}{7} )Now, substitute ( x ) back into one of the equations to find ( y ). Let's use the equation of ( BN ):( y = (sqrt{2} - 2)x + (2 - sqrt{2}) )So,( y = (sqrt{2} - 2)left( frac{4 - sqrt{2}}{7} right) + (2 - sqrt{2}) )Compute the first term:( (sqrt{2} - 2)(4 - sqrt{2}) = 4sqrt{2} - (sqrt{2})^2 - 8 + 2sqrt{2} = 4sqrt{2} - 2 - 8 + 2sqrt{2} = 6sqrt{2} - 10 )So,( y = frac{6sqrt{2} - 10}{7} + (2 - sqrt{2}) = frac{6sqrt{2} - 10}{7} + frac{14 - 7sqrt{2}}{7} = frac{6sqrt{2} - 10 + 14 - 7sqrt{2}}{7} = frac{-sqrt{2} + 4}{7} )So, point ( K ) is at ( left( frac{4 - sqrt{2}}{7}, frac{4 - sqrt{2}}{7} right) ).Wait, interesting, both coordinates are equal. So, ( x = y ), which means that ( K ) lies on the line ( y = x ), which is our angle bisector ( AD ). Therefore, ( K ) is on ( AD ).Thus, we've shown that the intersection point ( K ) of lines ( BN ) and ( CM ) lies on ( AD ).This completes the proof.

answer：First, I need to determine the total time John spent completing two tasks. He started at 8:30 AM and finished the second task at 11:10 AM. Calculating the time difference:- From 8:30 AM to 11:30 AM is 3 hours.- However, he finished at 11:10 AM, which is 20 minutes earlier than 11:30 AM.So, the total time spent on two tasks is 2 hours and 40 minutes.Next, I'll convert this time into minutes to make the calculations easier:- 2 hours = 120 minutes- Adding the 40 minutes gives a total of 160 minutes.Since John completed two tasks in 160 minutes, each task takes:- 160 minutes ÷ 2 = 80 minutes per task.Now, to find out when he finishes the third task, I'll add the duration of one task (80 minutes) to the time he finished the second task (11:10 AM):- Adding 80 minutes to 11:10 AM: - 11:10 AM + 60 minutes = 12:10 PM - Adding the remaining 20 minutes: 12:10 PM + 20 minutes = 12:30 PMTherefore, John finishes his third task at 12:30 PM.

answer：Okay, so I have this problem where a company is creating unique employee IDs. Each ID is five characters long, and they can use the letters C, A, T and the digits 2, 0, 1. But there's a catch: no character can appear in an ID more times than it does in "CAT" or "2021". So, in "CAT", each letter appears once, and in "2021", the digit '2' appears twice, while '0' and '1' appear once each. So, translating that into constraints for the IDs: I can use C, A, T each at most once, and the digit '2' at most twice, while '0' and '1' can be used at most once each. The question is asking for the total number of unique IDs possible under these constraints, which is M, and then to calculate M divided by 10.Alright, so first, I need to figure out all possible sequences of five characters that meet these constraints. Since each ID is a sequence, the order matters, so we're dealing with permutations here.Let me break this down. The available characters are C, A, T, 2, 0, 1. So that's six distinct characters, but with the caveat that '2' can be used up to two times, while the others can only be used once.So, the problem is similar to counting the number of permutations of a multiset, but with specific constraints on the number of times each element can appear.I think the way to approach this is to consider cases based on how many times '2' appears in the ID. Since '2' can appear at most twice, we can have IDs with zero '2's, one '2', or two '2's. Let me outline the cases:1. **Case 1: No '2's in the ID.** - So, all five characters must be chosen from C, A, T, 0, 1. Each of these can be used at most once. - So, we're selecting 5 distinct characters from 5 available (C, A, T, 0, 1). The number of permutations is 5! = 120.2. **Case 2: Exactly one '2' in the ID.** - We need to choose four more characters from C, A, T, 0, 1, each used at most once. - So, the number of ways to choose these four characters is C(5,4) = 5. - Then, we have five positions in the ID, and we need to place the '2' in one of these positions. So, 5 choices for where the '2' goes. - Then, the remaining four positions are filled by the four chosen characters, which can be arranged in 4! ways. - So, total permutations for this case: 5 (choices for the four characters) * 5 (positions for '2') * 4! (arrangements of the other characters) = 5 * 5 * 24 = 600.Wait, hold on, that seems high. Let me double-check.Wait, actually, when we choose four characters from five, it's C(5,4) = 5. Then, for each such selection, we have five positions where '2' can be placed, and the remaining four characters can be arranged in 4! ways. So, 5 * 5 * 24 = 600. Hmm, that seems correct.3. **Case 3: Exactly two '2's in the ID.** - Here, we need to choose three more characters from C, A, T, 0, 1, each used at most once. - So, the number of ways to choose these three characters is C(5,3) = 10. - Now, we have five positions in the ID, and we need to place the two '2's. The number of ways to choose positions for the two '2's is C(5,2) = 10. - Then, the remaining three positions are filled by the three chosen characters, which can be arranged in 3! ways. - So, total permutations for this case: 10 (choices for the three characters) * 10 (positions for the two '2's) * 6 (arrangements of the other characters) = 10 * 10 * 6 = 600.Wait, that also seems high. Let me check again.Yes, choosing three characters from five is 10. Choosing two positions out of five is 10. Arranging the three characters is 6. So, 10*10*6=600. That seems correct.Now, adding up all the cases:- Case 1: 120- Case 2: 600- Case 3: 600Total M = 120 + 600 + 600 = 1320.Therefore, M / 10 = 1320 / 10 = 132.Wait, but hold on, I think I might have made a mistake in Case 2. Let me re-examine it.In Case 2, we have exactly one '2'. So, we choose four characters from C, A, T, 0, 1. That's C(5,4)=5. Then, for each such set, we have five positions, and we need to place the '2' in one of them, which is 5 choices. Then, the remaining four characters can be arranged in 4! ways, which is 24. So, 5*5*24=600. Hmm, that seems correct.Similarly, in Case 3, exactly two '2's. Choose three characters from five, which is 10. Choose two positions out of five for the '2's, which is 10. Then, arrange the three characters in the remaining three positions, which is 6. So, 10*10*6=600. That also seems correct.So, adding them up: 120 + 600 + 600 = 1320. So, M=1320, and M/10=132.Wait, but I just realized something. In the problem statement, it says "each ID consists of five characters chosen from the three letters in CAT and the digits in 2021." So, the digits are 2, 0, 1, and the letters are C, A, T. So, total characters available are C, A, T, 2, 0, 1. So, six characters, but with '2' appearing twice in the digits, so we can use '2' up to two times.But in my earlier calculation, I considered all six characters, but in reality, the digits are 2, 0, 1, and the letters are C, A, T, so that's six distinct characters, but '2' can be used twice, while the others can be used once.Wait, but in my cases, I considered that when using '2's, the other characters are chosen from C, A, T, 0, 1, each used once. So, that seems correct.But let me think again: in Case 1, no '2's, so all five characters are from C, A, T, 0, 1, each used once. So, 5P5=120.In Case 2, one '2' and four others from C, A, T, 0, 1, each used once. So, 5C4=5 choices for the four characters, then 5 positions for the '2', and 4! arrangements for the others. So, 5*5*24=600.In Case 3, two '2's and three others from C, A, T, 0, 1, each used once. So, 5C3=10 choices for the three characters, then 5C2=10 positions for the '2's, and 3! arrangements for the others. So, 10*10*6=600.So, total M=120+600+600=1320.Therefore, M/10=132.Wait, but I just thought of another way to approach this problem, using generating functions or something, but maybe that's overcomplicating.Alternatively, think of it as permutations with limited repetitions.The formula for permutations of a multiset is n! divided by the product of (k_i!), where k_i is the number of times each element is repeated.But in this case, since we have constraints on the maximum number of times each element can be used, it's a bit different.But since we're dealing with sequences of length 5, and the maximum repetitions are limited, we can indeed break it down into cases based on the number of '2's, as I did earlier.So, I think my initial approach is correct.Therefore, the final answer is 132.

answer：Alright, so I have this problem here where I need to find the value of ( k ) such that the given parametric line is parallel to the line ( kx + y + 1 = 0 ). Hmm, okay, let me break this down step by step.First, I remember that for two lines to be parallel, their slopes must be equal. So, I need to find the slope of both lines and set them equal to each other. That makes sense.Starting with the parametric equations:[begin{cases}x = sqrt{3} - frac{1}{2}t y = 2 + frac{sqrt{3}}{2}tend{cases}]I need to convert this into a standard linear equation so I can find its slope. I recall that parametric equations can be converted by eliminating the parameter ( t ).From the first equation, I can solve for ( t ):[x = sqrt{3} - frac{1}{2}t Rightarrow frac{1}{2}t = sqrt{3} - x Rightarrow t = 2(sqrt{3} - x)]Okay, so ( t = 2(sqrt{3} - x) ). Now, I can substitute this expression for ( t ) into the second equation to find ( y ) in terms of ( x ).Substituting into the second equation:[y = 2 + frac{sqrt{3}}{2}t Rightarrow y = 2 + frac{sqrt{3}}{2} times 2(sqrt{3} - x) Rightarrow y = 2 + sqrt{3}(sqrt{3} - x)]Let me simplify that:[y = 2 + sqrt{3} times sqrt{3} - sqrt{3}x Rightarrow y = 2 + 3 - sqrt{3}x Rightarrow y = 5 - sqrt{3}x]So, the equation of the line in slope-intercept form is ( y = -sqrt{3}x + 5 ). Therefore, the slope of this line is ( -sqrt{3} ).Now, let's look at the other line: ( kx + y + 1 = 0 ). I can rewrite this in slope-intercept form to find its slope.Starting with:[kx + y + 1 = 0 Rightarrow y = -kx - 1]So, the slope of this line is ( -k ).Since the two lines are parallel, their slopes must be equal. Therefore:[-sqrt{3} = -k Rightarrow k = sqrt{3}]Wait, but let me double-check that. If the slope of the first line is ( -sqrt{3} ) and the slope of the second line is ( -k ), setting them equal gives ( -sqrt{3} = -k ), which simplifies to ( k = sqrt{3} ). That seems correct.But hold on, let me make sure I didn't make a mistake when converting the parametric equations. Let me go through that again.Starting with:[x = sqrt{3} - frac{1}{2}t Rightarrow t = 2(sqrt{3} - x)]Substituting into ( y ):[y = 2 + frac{sqrt{3}}{2} times 2(sqrt{3} - x) Rightarrow y = 2 + sqrt{3}(sqrt{3} - x) Rightarrow y = 2 + 3 - sqrt{3}x Rightarrow y = 5 - sqrt{3}x]Yes, that still gives me a slope of ( -sqrt{3} ). So, the slope of the first line is definitely ( -sqrt{3} ).For the second line, ( kx + y + 1 = 0 ), solving for ( y ) gives:[y = -kx - 1]So, the slope is ( -k ). Setting ( -k = -sqrt{3} ) gives ( k = sqrt{3} ).Hmm, but looking at the answer choices, option D is ( sqrt{3} ), which matches my result. However, I recall that sometimes when dealing with parametric equations, the direction vector is important. Let me think about that.The parametric equations can also be interpreted in terms of direction vectors. The coefficients of ( t ) in the parametric equations give the direction vector of the line. So, for ( x ), the coefficient is ( -frac{1}{2} ) and for ( y ), it's ( frac{sqrt{3}}{2} ). Therefore, the direction vector is ( left( -frac{1}{2}, frac{sqrt{3}}{2} right) ).The slope of the line can also be found by dividing the change in ( y ) by the change in ( x ), which is:[text{slope} = frac{Delta y}{Delta x} = frac{frac{sqrt{3}}{2}}{-frac{1}{2}} = -sqrt{3}]So, that confirms the slope is indeed ( -sqrt{3} ).Now, the line ( kx + y + 1 = 0 ) has a slope of ( -k ), as I found earlier. Setting them equal for parallelism:[-k = -sqrt{3} Rightarrow k = sqrt{3}]So, ( k = sqrt{3} ), which is option D.Wait a second, but I just noticed that in the answer choices, option B is ( -frac{sqrt{3}}{3} ). Is there a chance I might have messed up the signs somewhere? Let me check.Starting from the parametric equations:[x = sqrt{3} - frac{1}{2}t y = 2 + frac{sqrt{3}}{2}t]Solving for ( t ) in the first equation:[t = 2(sqrt{3} - x)]Substituting into ( y ):[y = 2 + frac{sqrt{3}}{2} times 2(sqrt{3} - x) Rightarrow y = 2 + sqrt{3}(sqrt{3} - x) Rightarrow y = 2 + 3 - sqrt{3}x Rightarrow y = 5 - sqrt{3}x]So, the slope is ( -sqrt{3} ), which is correct. The other line has a slope of ( -k ). Setting them equal:[-sqrt{3} = -k Rightarrow k = sqrt{3}]Yes, that seems consistent. So, ( k = sqrt{3} ), which is option D.But just to be thorough, let me consider the direction vector approach again. The direction vector is ( left( -frac{1}{2}, frac{sqrt{3}}{2} right) ). The slope is ( frac{Delta y}{Delta x} = frac{frac{sqrt{3}}{2}}{-frac{1}{2}} = -sqrt{3} ), which matches.Alternatively, if I think about the normal vector of the line ( kx + y + 1 = 0 ), it is ( (k, 1) ). For two lines to be parallel, their normal vectors must be scalar multiples of each other. The normal vector of the first line, which we found to be ( y + sqrt{3}x - 5 = 0 ), is ( (sqrt{3}, 1) ).So, setting ( (k, 1) ) proportional to ( (sqrt{3}, 1) ), we get:[k = sqrt{3} times text{scalar}]But since the scalar can be any real number, the simplest case is ( k = sqrt{3} ).Therefore, after multiple approaches and checks, I'm confident that the correct value of ( k ) is ( sqrt{3} ), which is option D.

answer：Okay, so I have this problem about cups arranged in a circle, and a ball is placed in every nth cup. The process starts at cup 1 and ends when cup 1 gets two balls. There are four parts to this problem, each with different numbers of cups and different steps. I need to figure out each part step by step.Starting with part (a): There are 12 cups, and a ball is placed in every 5th cup. I need to list the order of cups where the balls are placed, starting and ending with cup 1.Alright, so I know that when you place a ball in every nth cup in a circle, it's similar to modular arithmetic. Each time, you add n to the current cup number and take modulo the total number of cups. Since we start at cup 1, the sequence will be 1, 1+5, 1+5+5, and so on, each time taking modulo 12 because there are 12 cups.Let me write this out:1. Start at cup 1.2. Next cup: 1 + 5 = 6. Since 6 is less than 12, no need for modulo.3. Next cup: 6 + 5 = 11. Still less than 12.4. Next cup: 11 + 5 = 16. Now, 16 modulo 12 is 4 (because 12 goes into 16 once with a remainder of 4).5. Next cup: 4 + 5 = 9.6. Next cup: 9 + 5 = 14. 14 modulo 12 is 2.7. Next cup: 2 + 5 = 7.8. Next cup: 7 + 5 = 12. 12 modulo 12 is 0, but since we're numbering cups from 1, 0 corresponds to cup 12.9. Next cup: 12 + 5 = 17. 17 modulo 12 is 5.10. Next cup: 5 + 5 = 10.11. Next cup: 10 + 5 = 15. 15 modulo 12 is 3.12. Next cup: 3 + 5 = 8.13. Next cup: 8 + 5 = 13. 13 modulo 12 is 1.So, the sequence is: 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1. That makes sense because after 12 steps, we end up back at cup 1, which is when the process stops.Moving on to part (b): There are 9 cups, and a ball is placed in every 6th cup. I need to list the numbers of the cups that do not receive a ball.Again, starting at cup 1, and moving every 6th cup. Let's list the cups:1. Start at cup 1.2. Next cup: 1 + 6 = 7.3. Next cup: 7 + 6 = 13. 13 modulo 9 is 4 (since 9*1=9, 13-9=4).4. Next cup: 4 + 6 = 10. 10 modulo 9 is 1.So, the sequence is: 1, 7, 4, 1. The process stops here because we're back at cup 1.Now, the cups that received balls are 1, 4, 7. So, the cups that did not receive a ball are 2, 3, 5, 6, 8, 9.Wait, let me double-check. There are 9 cups, and the ones that got balls are 1, 4, 7. So, the others are 2, 3, 5, 6, 8, 9. That seems correct.Part (c): There are 120 cups, and a ball is placed in every 3rd cup. I need to find how many cups do not contain at least one ball when the process is complete.Hmm, so starting at cup 1, then every 3rd cup. Since 3 and 120 have a common divisor (3), the sequence will cycle through a subset of the cups before returning to cup 1.Wait, actually, 3 and 120 are not coprime because their greatest common divisor is 3. So, the number of unique cups that will receive a ball is equal to the number of steps before returning to cup 1, which is related to the least common multiple.The number of unique cups is given by the formula: LCM(3, 120) / 3 = 120 / 3 = 40. So, 40 unique cups will receive a ball. Therefore, the number of cups without a ball is 120 - 40 = 80.Wait, let me think again. If the step size and the number of cups are not coprime, the number of unique positions is equal to the number of cups divided by the greatest common divisor (GCD). So, GCD(3, 120) = 3. Therefore, the number of unique cups is 120 / 3 = 40. Hence, 40 cups get a ball, and 120 - 40 = 80 cups do not.Yes, that makes sense. So, the answer is 80 cups without a ball.Finally, part (d): There are 1000 cups, and a ball is placed in every 7th cup. I need to determine the number of the cup into which the 338th ball is placed.Starting at cup 1, each subsequent ball is placed 7 cups ahead. So, the position of the k-th ball is given by:Position = (1 + 7*(k-1)) mod 1000But since we're dealing with a circle, if the result is 0, it corresponds to cup 1000.So, for the 338th ball:Position = (1 + 7*(338 - 1)) mod 1000= (1 + 7*337) mod 1000= (1 + 2359) mod 1000= 2360 mod 1000= 360So, the 338th ball is placed in cup 360.Wait, let me verify:7*337 = 23592359 + 1 = 23602360 divided by 1000 is 2 with a remainder of 360, so yes, 2360 mod 1000 is 360.Therefore, the answer is cup 360.I think that covers all parts. Let me just recap:(a) The sequence is 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8, 1.(b) The cups without balls are 2, 3, 5, 6, 8, 9.(c) 80 cups do not contain a ball.(d) The 338th ball is in cup 360.

answer：Alright, so I have this problem about vectors on a plane. Let me try to understand it step by step.We have two vectors, OA and OB, with magnitudes a and b respectively. It's given that a² + b² = 4, which is interesting because it reminds me of the Pythagorean theorem. Also, OA and OB are perpendicular since their dot product is zero. So, OA and OB form a right angle at the origin O.Now, there's another vector OC, which is a linear combination of OA and OB: OC = λOA + μOB, where λ and μ are real numbers. There's also a condition given: (λ - 1/2)²a² + (μ - 1/2)²b² = 1. Hmm, that looks like the equation of a circle, but in terms of λ and μ. Maybe it's parameterizing some kind of ellipse or circle in the λ-μ plane?The question is asking for the maximum value of |OC|, which is the magnitude of vector OC. So, I need to find the maximum possible length of OC given the constraints.Let me write down what I know:1. |OA| = a, |OB| = b, and a² + b² = 4.2. OA · OB = 0, so OA and OB are perpendicular.3. OC = λOA + μOB.4. (λ - 1/2)²a² + (μ - 1/2)²b² = 1.Since OA and OB are perpendicular, the magnitude of OC can be found using the Pythagorean theorem as well. So, |OC|² = (λa)² + (μb)². That simplifies to |OC|² = λ²a² + μ²b².But we have this condition involving (λ - 1/2)²a² + (μ - 1/2)²b² = 1. Maybe I can rewrite this condition to relate it to |OC|.Let me expand the given condition:(λ - 1/2)²a² + (μ - 1/2)²b² = 1Expanding each term:(λ² - λ + 1/4)a² + (μ² - μ + 1/4)b² = 1So, distributing:λ²a² - λa² + (1/4)a² + μ²b² - μb² + (1/4)b² = 1Combine like terms:λ²a² + μ²b² - λa² - μb² + (1/4)(a² + b²) = 1We know that a² + b² = 4, so (1/4)(a² + b²) = 1. Therefore, the equation simplifies to:λ²a² + μ²b² - λa² - μb² + 1 = 1Subtract 1 from both sides:λ²a² + μ²b² - λa² - μb² = 0Hmm, interesting. So, λ²a² + μ²b² = λa² + μb².But from earlier, |OC|² = λ²a² + μ²b². So, substituting, we have |OC|² = λa² + μb².So, |OC|² = λa² + μb².But I need to find the maximum of |OC|. So, I need to maximize |OC|², which is equal to λa² + μb².Wait, but how? I have the condition that (λ - 1/2)²a² + (μ - 1/2)²b² = 1. Maybe I can use this to express λ and μ in terms of some parameter and then maximize |OC|².Alternatively, maybe I can interpret the condition geometrically. Since OA and OB are perpendicular, they form a coordinate system. The condition (λ - 1/2)²a² + (μ - 1/2)²b² = 1 is an ellipse centered at (1/2, 1/2) in the λ-μ plane, scaled by a and b.But since OA and OB are perpendicular, the coefficients a² and b² are just scaling the axes. So, maybe it's an ellipse with semi-axes lengths 1/a and 1/b?Wait, no. Let me think. The standard form of an ellipse is (x - h)²/A² + (y - k)²/B² = 1. So, in this case, it's (λ - 1/2)²/(1/a²) + (μ - 1/2)²/(1/b²) = 1. So, the ellipse has semi-major axis 1/a and semi-minor axis 1/b, or vice versa, depending on which is larger.But since a² + b² = 4, both a and b are less than or equal to 2.But I'm not sure if this is the right direction. Maybe I should think of OC in terms of vectors.Since OC = λOA + μOB, and OA and OB are perpendicular, the magnitude squared is |OC|² = (λa)² + (μb)².We need to maximize this expression given the constraint (λ - 1/2)²a² + (μ - 1/2)²b² = 1.This seems like a problem of maximizing a quadratic function subject to another quadratic constraint. Maybe I can use Lagrange multipliers.Let me set up the function to maximize: f(λ, μ) = λ²a² + μ²b².Subject to the constraint g(λ, μ) = (λ - 1/2)²a² + (μ - 1/2)²b² - 1 = 0.The method of Lagrange multipliers says that at the maximum, the gradient of f is proportional to the gradient of g.So, compute the gradients:∇f = (2λa², 2μb²)∇g = (2(λ - 1/2)a², 2(μ - 1/2)b²)So, setting ∇f = λ∇g, where λ is the Lagrange multiplier:2λa² = λ * 2(λ - 1/2)a²2μb² = λ * 2(μ - 1/2)b²Wait, that seems a bit confusing. Let me write it properly.We have:2λa² = λ * 2(λ - 1/2)a²2μb² = λ * 2(μ - 1/2)b²Wait, no. The Lagrange multiplier is a scalar, let's call it k, to avoid confusion with λ.So, ∇f = k∇g.Thus,2λa² = k * 2(λ - 1/2)a²2μb² = k * 2(μ - 1/2)b²Simplify these equations:Divide both sides by 2a²:λ = k(λ - 1/2)Similarly, divide both sides by 2b²:μ = k(μ - 1/2)So, we have two equations:1. λ = kλ - (k/2)2. μ = kμ - (k/2)Let me rearrange them:1. λ - kλ = -k/2 => λ(1 - k) = -k/22. μ - kμ = -k/2 => μ(1 - k) = -k/2So, from both equations, we have:λ = (-k/2)/(1 - k)μ = (-k/2)/(1 - k)So, λ = μ.Interesting, so λ and μ are equal.Let me denote λ = μ = t.So, t = (-k/2)/(1 - k)Now, let's plug this back into the constraint equation:(λ - 1/2)²a² + (μ - 1/2)²b² = 1Since λ = μ = t,2(t - 1/2)²a² + (t - 1/2)²b² = 1? Wait, no, it's (t - 1/2)²a² + (t - 1/2)²b² = 1.Wait, no, actually, since both terms are (t - 1/2)² multiplied by a² and b² respectively.So, (t - 1/2)²(a² + b²) = 1But a² + b² = 4, so:(t - 1/2)² * 4 = 1 => (t - 1/2)² = 1/4 => t - 1/2 = ±1/2So, t = 1/2 ± 1/2Thus, t = 1 or t = 0.So, λ = μ = 1 or λ = μ = 0.Wait, but let's check if these satisfy the earlier equations.From λ = (-k/2)/(1 - k) = t.So, if t = 1:1 = (-k/2)/(1 - k)Multiply both sides by (1 - k):1 - k = -k/2Multiply both sides by 2:2 - 2k = -k2 = kSo, k = 2.Similarly, if t = 0:0 = (-k/2)/(1 - k)Multiply both sides by (1 - k):0 = -k/2So, k = 0.But if k = 0, from the gradient equations:2λa² = 0 => λ = 0Similarly, μ = 0.So, that gives us t = 0.So, we have two critical points: (λ, μ) = (1, 1) and (0, 0).Now, let's compute |OC|² for these points.First, for (λ, μ) = (1, 1):|OC|² = (1)a² + (1)b² = a² + b² = 4.So, |OC| = 2.For (λ, μ) = (0, 0):|OC|² = 0 + 0 = 0.So, |OC| = 0.Therefore, the maximum value of |OC| is 2.Wait, but let me think again. Is this the only possible maximum? Because sometimes, when dealing with constrained optimization, the maximum could be on the boundary or something else.But in this case, since the constraint is an ellipse, and the function to maximize is quadratic, the maximum should occur at one of the critical points found via Lagrange multipliers. So, 2 is indeed the maximum.Alternatively, maybe I can think geometrically. Since OA and OB are perpendicular, and OC is a linear combination, the condition (λ - 1/2)²a² + (μ - 1/2)²b² = 1 defines an ellipse in the λ-μ plane. The maximum |OC| would correspond to the farthest point from the origin on this ellipse.But since we found via Lagrange multipliers that the maximum occurs at (1,1), which gives |OC| = 2, that should be the answer.Wait, another way to think about it: since a² + b² = 4, and OA and OB are perpendicular, the vectors OA and OB can be considered as axes of a coordinate system. The condition (λ - 1/2)²a² + (μ - 1/2)²b² = 1 is an ellipse centered at (1/2, 1/2) in the λ-μ plane, scaled by a and b.The maximum |OC| would be the maximum distance from the origin to any point on this ellipse. Since the ellipse is centered at (1/2, 1/2), the farthest point from the origin would be in the direction away from the origin, which would be along the line connecting the origin to the center of the ellipse.The distance from the origin to the center is sqrt( (1/2 a)^2 + (1/2 b)^2 ) = (1/2)sqrt(a² + b²) = (1/2)*2 = 1.Since the ellipse has a radius of 1, the maximum distance from the origin would be the distance to the center plus the radius, which is 1 + 1 = 2.Yes, that makes sense. So, the maximum |OC| is 2.**Final Answer**The maximum value of ( |overrightarrow{OC}| ) is (boxed{2}).