💡First, I need to determine the total sum of the ages of a, b, and c based on the given average age.Since the average age of a, b, and c is 28 years, the total sum of their ages is 28 multiplied by 3, which equals 84 years.Next, I know that the age of b is 26 years. To find the combined age of a and c, I subtract b's age from the total sum: 84 minus 26 equals 58 years.Finally, to find the average age of a and c, I divide their combined age by 2: 58 divided by 2 equals 29 years.

💡Alright, so I have this problem where I need to construct a right triangle given the radius ( r ) of its circumcircle and the angle bisector ( f ) of one of its acute angles. Hmm, okay. Let me try to break this down step by step.First, I know that in a right triangle, the circumradius ( r ) is half the hypotenuse. So, if the hypotenuse is ( c ), then ( r = frac{c}{2} ), which means ( c = 2r ). That’s straightforward.Now, the triangle has two acute angles, and we’re given the angle bisector ( f ) of one of these angles. Let me denote the triangle as ( ABC ) with the right angle at ( C ). Without loss of generality, let’s say the angle bisector is from angle ( A ) and meets side ( BC ) at point ( D ). So, ( AD = f ).I remember that the angle bisector theorem states that ( frac{BD}{DC} = frac{AB}{AC} ). In this case, since ( ABC ) is a right triangle, ( AB ) and ( AC ) are the legs, and ( BC ) is the hypotenuse. Wait, no, actually, ( BC ) is the side opposite angle ( A ), so maybe I need to clarify the sides.Let me denote the sides as follows: ( AB = c ) (hypotenuse), ( AC = b ), and ( BC = a ). So, the right angle is at ( C ). Then, the angle bisector from ( A ) meets ( BC ) at ( D ). So, by the angle bisector theorem, ( frac{BD}{DC} = frac{AB}{AC} = frac{c}{b} ).Let me denote ( BD = x ) and ( DC = y ). Then, ( x + y = a ), and ( frac{x}{y} = frac{c}{b} ). So, ( x = frac{c}{b} y ). Substituting into ( x + y = a ), we get ( frac{c}{b} y + y = a ), which simplifies to ( y left( frac{c}{b} + 1 right) = a ). Therefore, ( y = frac{a b}{b + c} ) and ( x = frac{a c}{b + c} ).Now, I need to relate this to the angle bisector length ( f ). There is a formula for the length of an angle bisector in a triangle. For a triangle with sides ( a ), ( b ), and ( c ), the length of the angle bisector from angle ( A ) is given by:[f = frac{2 b c cos left( frac{alpha}{2} right)}{b + c}]But wait, in our case, ( ABC ) is a right triangle, so angle ( A ) is acute, and we can express ( cos left( frac{alpha}{2} right) ) in terms of the sides.Alternatively, maybe it's better to use the angle bisector theorem in conjunction with the Pythagorean theorem.Since ( ABC ) is a right triangle, ( a^2 + b^2 = c^2 ). Also, in triangle ( ABD ), which is also a right triangle, we can apply the Pythagorean theorem:[AD^2 = AB^2 + BD^2]Wait, no, ( ABD ) is not necessarily a right triangle. Hmm, maybe I need to consider triangle ( ABD ) and triangle ( ADC ).Wait, actually, ( AD ) is the angle bisector, so it divides angle ( A ) into two equal angles. Maybe I can use the formula for the length of the angle bisector in terms of the sides.The formula for the angle bisector length ( f ) in a triangle is:[f = frac{2 b c cos left( frac{alpha}{2} right)}{b + c}]But in our case, ( ABC ) is a right triangle, so angle ( A ) is ( alpha ), and ( cos alpha = frac{b}{c} ). Therefore, ( cos left( frac{alpha}{2} right) = sqrt{frac{1 + cos alpha}{2}} = sqrt{frac{1 + frac{b}{c}}{2}} = sqrt{frac{c + b}{2c}} ).Substituting back into the angle bisector formula:[f = frac{2 b c sqrt{frac{c + b}{2c}}}{b + c} = frac{2 b c sqrt{frac{c + b}{2c}}}{b + c}]Simplifying:[f = frac{2 b c}{b + c} sqrt{frac{c + b}{2c}} = frac{2 b c}{b + c} cdot sqrt{frac{b + c}{2c}} = frac{2 b c}{b + c} cdot frac{sqrt{2c(b + c)}}{2c}]Wait, that seems complicated. Maybe there's a simpler way.Alternatively, using the angle bisector theorem, we have ( frac{BD}{DC} = frac{AB}{AC} = frac{c}{b} ). So, ( BD = frac{c}{b + c} a ) and ( DC = frac{b}{b + c} a ).Now, in triangle ( ABD ), applying the Law of Cosines:[AD^2 = AB^2 + BD^2 - 2 cdot AB cdot BD cdot cos alpha]But ( AD = f ), ( AB = c ), ( BD = frac{c a}{b + c} ), and ( cos alpha = frac{b}{c} ).Substituting:[f^2 = c^2 + left( frac{c a}{b + c} right)^2 - 2 cdot c cdot frac{c a}{b + c} cdot frac{b}{c}]Simplify:[f^2 = c^2 + frac{c^2 a^2}{(b + c)^2} - 2 cdot frac{c^2 a b}{(b + c)}]Hmm, this seems messy. Maybe I need a different approach.Since ( ABC ) is a right triangle, let me denote angle ( A ) as ( theta ). Then, ( cos theta = frac{b}{c} ) and ( sin theta = frac{a}{c} ). The angle bisector divides ( theta ) into two angles of ( frac{theta}{2} ).Using the angle bisector formula in terms of the sides and the angle:[f = frac{2 b c cos left( frac{theta}{2} right)}{b + c}]But ( cos left( frac{theta}{2} right) = sqrt{frac{1 + cos theta}{2}} = sqrt{frac{1 + frac{b}{c}}{2}} = sqrt{frac{c + b}{2c}} ).So,[f = frac{2 b c sqrt{frac{c + b}{2c}}}{b + c} = frac{2 b c}{b + c} cdot sqrt{frac{c + b}{2c}} = frac{2 b c}{b + c} cdot frac{sqrt{2c(b + c)}}{2c}]Simplifying:[f = frac{2 b c}{b + c} cdot frac{sqrt{2c(b + c)}}{2c} = frac{b sqrt{2c(b + c)}}{b + c}]Hmm, still complicated. Maybe square both sides to eliminate the square root:[f^2 = frac{2 b^2 c (b + c)}{(b + c)^2} = frac{2 b^2 c}{b + c}]So,[f^2 = frac{2 b^2 c}{b + c}]But we know that ( c = 2r ), so substituting:[f^2 = frac{2 b^2 (2r)}{b + 2r} = frac{4 r b^2}{b + 2r}]Let me denote ( b ) as one of the legs, which we need to find. So, we have:[f^2 = frac{4 r b^2}{b + 2r}]Let me solve for ( b ):Multiply both sides by ( b + 2r ):[f^2 (b + 2r) = 4 r b^2]Expand:[f^2 b + 2 f^2 r = 4 r b^2]Rearrange:[4 r b^2 - f^2 b - 2 f^2 r = 0]This is a quadratic equation in terms of ( b ):[4 r b^2 - f^2 b - 2 f^2 r = 0]Let me write it as:[4 r b^2 - f^2 b - 2 f^2 r = 0]Let me denote ( b ) as the variable:[4 r b^2 - f^2 b - 2 f^2 r = 0]Using the quadratic formula:[b = frac{f^2 pm sqrt{f^4 + 32 r^2 f^2}}{8 r}]Simplify the discriminant:[sqrt{f^4 + 32 r^2 f^2} = f sqrt{f^2 + 32 r^2}]So,[b = frac{f^2 pm f sqrt{f^2 + 32 r^2}}{8 r}]Since ( b ) must be positive, we take the positive root:[b = frac{f^2 + f sqrt{f^2 + 32 r^2}}{8 r}]Factor out ( f ):[b = frac{f (f + sqrt{f^2 + 32 r^2})}{8 r}]Hmm, this seems a bit complicated, but it's an expression for ( b ) in terms of ( f ) and ( r ). Once we have ( b ), we can find ( a ) using the Pythagorean theorem:[a = sqrt{c^2 - b^2} = sqrt{(2r)^2 - b^2}]But this requires knowing ( b ), which we have in terms of ( f ) and ( r ). So, in theory, we can construct ( b ) and then ( a ), and thus construct the triangle.But how do we construct this geometrically? Let me think.First, we can construct ( c = 2r ). Then, we need to construct ( b ) using the expression above. Since ( b ) involves square roots, we can use geometric constructions involving circles and right triangles to find the necessary lengths.For example, to construct ( sqrt{f^2 + 32 r^2} ), we can construct a right triangle with legs ( f ) and ( sqrt{32} r ), then the hypotenuse will be ( sqrt{f^2 + 32 r^2} ). Then, add ( f ) to this hypotenuse, and use similar triangles or the intercept theorem to divide this sum in the ratio ( 8 r : f ) to find ( b ).Once ( b ) is constructed, ( a ) can be found by constructing a right triangle with hypotenuse ( 2r ) and one leg ( b ), then the other leg will be ( a ).So, the steps would be:1. Draw the circumcircle with radius ( r ). The hypotenuse ( c ) will be the diameter, so draw the diameter ( AB ) with length ( 2r ).2. Construct the length ( sqrt{32} r ). Since ( sqrt{32} = 4 sqrt{2} ), we can construct ( sqrt{2} ) by drawing a right triangle with legs ( 1 ) and ( 1 ), then multiplying by 4.3. Construct a right triangle with legs ( f ) and ( 4 sqrt{2} r ). The hypotenuse will be ( sqrt{f^2 + 32 r^2} ).4. Add ( f ) to this hypotenuse to get ( f + sqrt{f^2 + 32 r^2} ).5. Divide this sum into 8 equal parts relative to ( f ) to find ( b ). This can be done using similar triangles or the intercept theorem.6. Once ( b ) is constructed, construct a right triangle with hypotenuse ( 2r ) and one leg ( b ). The other leg will be ( a ).7. Finally, construct the triangle ( ABC ) with sides ( a ), ( b ), and ( c = 2r ), ensuring that the angle bisector from ( A ) is ( f ).I think this approach should work, but I need to make sure each step is geometrically constructible. Constructing square roots and ratios can be done with compass and straightedge, so it should be feasible.Another thought: Maybe there's a more straightforward geometric construction without solving the quadratic equation algebraically. Perhaps by extending the angle bisector and using properties of circles or similar triangles.Wait, if I extend the angle bisector ( AD ) to meet the circumcircle again at point ( E ), then ( AE ) is a chord of the circle. Maybe there's a relationship between ( AD ), ( AE ), and the sides of the triangle.In circle geometry, the power of a point states that ( AD cdot DE = BD cdot DC ). Since ( D ) lies on ( BC ), and ( E ) lies on the circumcircle, this might be useful.Given that ( AD = f ), ( DE = AE - AD = AE - f ). Also, from the angle bisector theorem, ( BD cdot DC = left( frac{c a}{b + c} right) left( frac{b a}{b + c} right) = frac{a^2 b c}{(b + c)^2} ).So, power of point ( D ):[AD cdot DE = BD cdot DC]Substituting:[f (AE - f) = frac{a^2 b c}{(b + c)^2}]But ( AE ) is the length from ( A ) to ( E ) on the circumcircle. Since ( ABC ) is a right triangle, the circumcircle has diameter ( AB ), so ( E ) is the other intersection point of ( AD ) with the circumcircle.I'm not sure if this approach simplifies things, but it's another angle to consider.Alternatively, maybe using trigonometric identities. Since ( ABC ) is a right triangle, we can express everything in terms of one angle. Let’s say angle ( A ) is ( theta ), then angle ( B ) is ( 90^circ - theta ).The angle bisector ( f ) divides angle ( A ) into two angles of ( frac{theta}{2} ). Using the angle bisector formula:[f = frac{2 b c cos left( frac{theta}{2} right)}{b + c}]But ( cos left( frac{theta}{2} right) = sqrt{frac{1 + cos theta}{2}} = sqrt{frac{1 + frac{b}{c}}{2}} ).So,[f = frac{2 b c sqrt{frac{1 + frac{b}{c}}{2}}}{b + c} = frac{2 b c sqrt{frac{c + b}{2c}}}{b + c}]Simplifying:[f = frac{2 b c}{b + c} cdot sqrt{frac{c + b}{2c}} = frac{2 b c}{b + c} cdot frac{sqrt{2c(b + c)}}{2c} = frac{b sqrt{2c(b + c)}}{b + c}]Squaring both sides:[f^2 = frac{2 b^2 c (b + c)}{(b + c)^2} = frac{2 b^2 c}{b + c}]Which brings us back to the earlier equation:[f^2 = frac{2 b^2 c}{b + c}]Since ( c = 2r ), substituting:[f^2 = frac{2 b^2 (2r)}{b + 2r} = frac{4 r b^2}{b + 2r}]Which is the same quadratic equation as before. So, it seems that regardless of the approach, we end up with the same quadratic in ( b ). Therefore, solving it algebraically seems necessary.Once ( b ) is found, ( a ) can be determined using the Pythagorean theorem, and then the triangle can be constructed.In summary, the steps are:1. Given ( r ) and ( f ), calculate ( c = 2r ).2. Solve the quadratic equation ( 4 r b^2 - f^2 b - 2 f^2 r = 0 ) for ( b ).3. Use the positive root to find ( b ).4. Calculate ( a = sqrt{c^2 - b^2} ).5. Construct the right triangle with legs ( a ) and ( b ), and hypotenuse ( c ).Since all these steps involve basic arithmetic and square roots, which are constructible with compass and straightedge, the triangle can indeed be constructed given ( r ) and ( f ).I think I've covered all the necessary steps and verified the relationships. Hopefully, this makes sense and is correct.

💡Alright, so I've got this geometry problem here, and it's pretty involved. Let me try to unpack it step by step. The problem is about triangle ABC with some specific constructions and points defined. The goal is to compute the absolute difference between XE and XF, where X is a point defined through a series of circumcircles and intersections.First, let me list out the given information:- Triangle ABC has sides BC = 10, CA = 20, AB = 18.- The B-excircle touches CA at E, and the C-excircle touches AB at F.- M is the midpoint of BC.- The angle bisector of angle BAC meets BC at D, EF at P, ME at E', and MF at F'.- The circumcircles of triangles EPE' and FPF' intersect again at Q.- The circumcircle of triangle DPQ intersects line EF again at X.- We need to find |XE - XF|.Okay, that's a lot. Let me try to visualize the triangle and these points. Maybe drawing a diagram would help, but since I can't draw here, I'll try to imagine it.First, let's recall some properties:1. The B-excircle touches side AC at E. The point where an excircle touches a side can be found using the formula: if the excircle opposite to B touches AC, then the length from A to E is (AB + AC - BC)/2. Similarly, the C-excircle touches AB at F, so the length from A to F is (AB + AC - BC)/2 as well? Wait, no, that's for the incircle. For excircle, it's different.Wait, let me recall: For an excircle opposite to vertex B, it touches side AC. The length from A to the touch point E is (AB + BC - AC)/2. Similarly, for the excircle opposite to C, the touch point F on AB is (AC + BC - AB)/2.Let me compute these lengths.Given BC = 10, CA = 20, AB = 18.So, for the B-excircle touching AC at E:AE = (AB + BC - AC)/2 = (18 + 10 - 20)/2 = (8)/2 = 4.Similarly, for the C-excircle touching AB at F:AF = (AC + BC - AB)/2 = (20 + 10 - 18)/2 = (12)/2 = 6.So, AE = 4 and AF = 6.Therefore, E is 4 units from A on AC, and F is 6 units from A on AB.So, EC = AC - AE = 20 - 4 = 16, and FB = AB - AF = 18 - 6 = 12.Alright, so now we have points E and F located on AC and AB respectively.Next, M is the midpoint of BC. Since BC = 10, BM = MC = 5.Now, the angle bisector of angle BAC meets BC at D, EF at P, ME at E', and MF at F'.Let me recall that the angle bisector theorem tells us that BD/DC = AB/AC = 18/20 = 9/10.Given BC = 10, so BD = (9/19)*10 = 90/19 ≈ 4.7368, and DC = (10/19)*10 = 100/19 ≈ 5.2632.So, D divides BC into BD ≈ 4.7368 and DC ≈ 5.2632.Now, the angle bisector meets EF at P. So, P is the intersection point of the angle bisector and EF.Similarly, it meets ME at E' and MF at F'.So, ME is the segment from M to E, and MF is the segment from M to F.So, E' is the intersection of the angle bisector with ME, and F' is the intersection with MF.Now, the circumcircles of triangles EPE' and FPF' meet again at Q. So, Q is the second intersection point of these two circumcircles.Then, the circumcircle of triangle DPQ meets line EF again at X.So, X is the second intersection point of the circumcircle of DPQ with EF.We need to find |XE - XF|.Hmm. This seems complex, but perhaps there's a property or theorem that can help here.Let me think about the properties of the Nagel point and the Gergonne point, as excircles are involved. The Nagel point is the intersection of the lines from each vertex to the touch point of the excircle on the opposite side. So, in this case, lines AE and AF might relate to the Nagel point.Wait, but in this problem, we have points E and F as touch points of the B and C excircles, respectively. So, maybe the line EF is related to the Nagel line or something similar.Alternatively, perhaps the point X lies on the Nagel line, which would mean that XE/XF is proportional to the sides.Wait, the problem mentions that the angle bisector of angle BAC meets EF at P. So, maybe P is a significant point on EF.Alternatively, perhaps the point X is the Nagel point, but I'm not sure.Wait, let me think about the circumcircles of EPE' and FPF'. Since E' and F' are points along ME and MF, respectively, perhaps these circumcircles have some symmetrical properties.Alternatively, maybe Q is the Miquel point of some quadrilateral.Alternatively, perhaps using power of a point with respect to the circumcircles.Alternatively, perhaps inversion could help, but that might be too complicated.Alternatively, maybe using coordinate geometry. Since the sides are given, perhaps assigning coordinates to the triangle ABC and computing all these points.Given that the sides are BC = 10, CA = 20, AB = 18, perhaps coordinate geometry is feasible.Let me try that approach.Let me place triangle ABC in the coordinate plane.Let me set point B at (0, 0), point C at (10, 0). Then, point A is somewhere in the plane. Let me compute coordinates of A.Given AB = 18, AC = 20, BC = 10.So, coordinates of A can be found using the distance formula.Let me denote A = (x, y).Then, distance from A to B: sqrt((x)^2 + (y)^2) = 18.Distance from A to C: sqrt((x - 10)^2 + (y)^2) = 20.So, we have:x² + y² = 18² = 324,(x - 10)² + y² = 20² = 400.Subtracting the first equation from the second:(x - 10)² + y² - x² - y² = 400 - 324,Expanding (x - 10)²: x² - 20x + 100,So, x² - 20x + 100 - x² = 76,Simplify: -20x + 100 = 76,-20x = -24,x = 24/20 = 6/5 = 1.2.So, x = 6/5.Then, from x² + y² = 324,(6/5)² + y² = 324,36/25 + y² = 324,y² = 324 - 36/25 = (324*25 - 36)/25 = (8100 - 36)/25 = 8064/25,So, y = sqrt(8064/25) = (sqrt(8064))/5.Simplify sqrt(8064):8064 = 64 * 126 = 64 * 9 * 14 = 64 * 9 * 14,So, sqrt(8064) = 8 * 3 * sqrt(14) = 24 sqrt(14).Thus, y = 24 sqrt(14)/5.So, coordinates of A are (6/5, 24 sqrt(14)/5).So, A = (1.2, approx 24*3.7417/5 ≈ 24*0.7483 ≈ 18.0).Wait, 24*3.7417 ≈ 89.8, divided by 5 is ≈17.96.So, A is approximately (1.2, 17.96).Alright, so coordinates:B = (0, 0),C = (10, 0),A = (6/5, 24√14/5).Now, let's find points E and F.Earlier, we found that AE = 4 and AF = 6.So, point E is on AC, 4 units from A.Similarly, point F is on AB, 6 units from A.Let me compute coordinates of E and F.First, point E on AC:AC is from A(6/5, 24√14/5) to C(10, 0).We need to find E such that AE = 4.The length of AC is 20, so AE = 4 implies that E divides AC in the ratio AE:EC = 4:16 = 1:4.Therefore, the coordinates of E can be found using section formula.Coordinates of E = [(4*C + 1*A)] / (1 + 4) = (4*(10, 0) + (6/5, 24√14/5)) / 5.Compute:x-coordinate: (4*10 + 6/5)/5 = (40 + 1.2)/5 = 41.2/5 = 8.24,y-coordinate: (4*0 + 24√14/5)/5 = (24√14/5)/5 = 24√14/25.So, E = (8.24, 24√14/25).Similarly, point F is on AB, 6 units from A.AB is from A(6/5, 24√14/5) to B(0, 0).Length AB = 18, so AF = 6 implies AF:FB = 6:12 = 1:2.Therefore, coordinates of F = [(2*A + 1*B)] / (1 + 2) = (2*(6/5, 24√14/5) + (0, 0)) / 3.Compute:x-coordinate: (12/5 + 0)/3 = (12/5)/3 = 4/5 = 0.8,y-coordinate: (48√14/5 + 0)/3 = (48√14/5)/3 = 16√14/5.So, F = (4/5, 16√14/5).Alright, so E = (8.24, 24√14/25) and F = (0.8, 16√14/5).Now, M is the midpoint of BC.Since B = (0,0) and C = (10,0), M = ((0 + 10)/2, (0 + 0)/2) = (5, 0).So, M = (5, 0).Next, the angle bisector of angle BAC meets BC at D, EF at P, ME at E', and MF at F'.We already found D earlier using the angle bisector theorem: BD = 90/19 ≈ 4.7368, so D is located at (BD, 0) from B.Since B is at (0,0), D is at (90/19, 0). Let me compute 90/19 ≈ 4.7368.So, D ≈ (4.7368, 0).But let me keep it exact: D = (90/19, 0).Now, the angle bisector from A passes through D, so its equation can be determined.We can find the equation of AD.Points A(6/5, 24√14/5) and D(90/19, 0).Compute the slope of AD:m = (0 - 24√14/5) / (90/19 - 6/5) = (-24√14/5) / (90/19 - 11.4/19) = (-24√14/5) / (78.6/19).Wait, let me compute denominator:90/19 - 6/5 = 90/19 - (6*19)/5*19 = 90/19 - 114/95 = (90*5 - 114)/95 = (450 - 114)/95 = 336/95.So, denominator is 336/95.So, slope m = (-24√14/5) / (336/95) = (-24√14/5) * (95/336) = (-24*95√14)/(5*336).Simplify:24 and 336: 336 ÷ 24 = 14,95 and 5: 95 ÷ 5 = 19.So, m = (-24*95√14)/(5*336) = (-24*19√14)/336 = (-456√14)/336.Simplify numerator and denominator by dividing numerator and denominator by 24:-456 ÷ 24 = -19,336 ÷ 24 = 14.So, m = (-19√14)/14.Thus, the slope of AD is -19√14/14.Therefore, the equation of AD is:y - 24√14/5 = (-19√14/14)(x - 6/5).We can write this as:y = (-19√14/14)x + (19√14/14)*(6/5) + 24√14/5.Compute the constants:(19√14/14)*(6/5) = (114√14)/70 = (57√14)/35,24√14/5 = (168√14)/35.So, total constant term: (57√14 + 168√14)/35 = (225√14)/35 = 45√14/7.Thus, equation of AD: y = (-19√14/14)x + 45√14/7.Alright, now we need to find point P, which is the intersection of AD and EF.So, we need the equation of EF.Points E and F are known:E = (8.24, 24√14/25) ≈ (8.24, 24*3.7417/25 ≈ 24*0.1497 ≈ 3.6),Wait, 24√14/25 ≈ 24*3.7417/25 ≈ 89.8/25 ≈ 3.592.Similarly, F = (0.8, 16√14/5) ≈ (0.8, 16*3.7417/5 ≈ 59.867/5 ≈ 11.973).So, E ≈ (8.24, 3.592) and F ≈ (0.8, 11.973).Wait, but these coordinates seem a bit off because E is closer to C, which is at (10,0), so E is at (8.24, ~3.59), which makes sense. Similarly, F is on AB, closer to A, so at (0.8, ~11.97).Now, let's compute the equation of line EF.First, compute the slope of EF:m_EF = (y_F - y_E)/(x_F - x_E) = (16√14/5 - 24√14/25)/(4/5 - 41.2/5).Wait, let me compute exact values.Coordinates:E = (41.2/5, 24√14/25),F = (4/5, 16√14/5).So, x_E = 41.2/5 = 8.24,x_F = 4/5 = 0.8,y_E = 24√14/25,y_F = 16√14/5 = 80√14/25.So, slope m_EF = (80√14/25 - 24√14/25)/(4/5 - 41.2/5) = (56√14/25)/(-37.2/5).Simplify:56√14/25 divided by (-37.2/5) = (56√14/25) * (5/-37.2) = (56√14 * 5)/(25 * -37.2) = (280√14)/( -930).Simplify numerator and denominator:280 and 930: both divisible by 10: 28/93.So, m_EF = (-28√14)/93.Wait, let me check the calculation again.Wait, 56√14/25 divided by (-37.2/5) is equal to (56√14/25) * (5/-37.2) = (56√14 * 5)/(25 * -37.2) = (280√14)/( -930).Simplify 280/930: divide numerator and denominator by 10: 28/93.So, m_EF = (-28√14)/93.Thus, the slope of EF is -28√14/93.Now, equation of EF: using point F(4/5, 16√14/5):y - 16√14/5 = (-28√14/93)(x - 4/5).Now, we have equations of AD and EF. Let's find their intersection point P.Equation of AD: y = (-19√14/14)x + 45√14/7.Equation of EF: y = (-28√14/93)x + (28√14/93)*(4/5) + 16√14/5.Compute the constant term:(28√14/93)*(4/5) = 112√14/465,16√14/5 = 1584√14/465.So, total constant term: (112√14 + 1584√14)/465 = 1696√14/465.Simplify 1696/465: Let's see, 465*3=1395, 465*4=1860, so 1696-1395=301, so 1696=465*3 + 301. Hmm, not much to simplify.So, equation of EF: y = (-28√14/93)x + 1696√14/465.Now, set equations of AD and EF equal to find P.So,(-19√14/14)x + 45√14/7 = (-28√14/93)x + 1696√14/465.Let me divide both sides by √14 to simplify:(-19/14)x + 45/7 = (-28/93)x + 1696/465.Multiply all terms by the least common multiple of denominators to eliminate fractions.Denominators: 14, 7, 93, 465.Factor each:14 = 2*7,7 = 7,93 = 3*31,465 = 5*93 = 5*3*31.So, LCM is 2*3*5*7*31 = 2*3*5*7*31.Compute:2*3=6,6*5=30,30*7=210,210*31=6510.So, multiply both sides by 6510:6510*(-19/14)x + 6510*(45/7) = 6510*(-28/93)x + 6510*(1696/465).Compute each term:First term: 6510*(-19/14)x = (6510/14)*(-19)x = 465*(-19)x = -8835x.Second term: 6510*(45/7) = (6510/7)*45 = 930*45 = 41,850.Third term: 6510*(-28/93)x = (6510/93)*(-28)x = 70*(-28)x = -1960x.Fourth term: 6510*(1696/465) = (6510/465)*1696 = 14*1696 = 23,744.So, equation becomes:-8835x + 41,850 = -1960x + 23,744.Bring all terms to left side:-8835x + 41,850 + 1960x - 23,744 = 0,(-8835 + 1960)x + (41,850 - 23,744) = 0,(-6875)x + 18,106 = 0,-6875x = -18,106,x = (-18,106)/(-6875) = 18,106/6875.Simplify:Divide numerator and denominator by GCD(18106,6875). Let's compute GCD:6875 divides into 18106 how many times? 6875*2=13750, 18106-13750=4356.Now GCD(6875,4356).6875 ÷ 4356 = 1 with remainder 2519.GCD(4356,2519).4356 ÷ 2519 = 1 with remainder 1837.GCD(2519,1837).2519 ÷ 1837 = 1 with remainder 682.GCD(1837,682).1837 ÷ 682 = 2 with remainder 473.GCD(682,473).682 ÷ 473 = 1 with remainder 209.GCD(473,209).473 ÷ 209 = 2 with remainder 55.GCD(209,55).209 ÷ 55 = 3 with remainder 44.GCD(55,44).55 ÷ 44 = 1 with remainder 11.GCD(44,11)=11.So, GCD is 11.Thus, 18,106 ÷ 11 = 1646,6875 ÷ 11 = 625.So, x = 1646/625 ≈ 2.6336.So, x ≈ 2.6336.Now, plug x back into equation of AD to find y.Equation of AD: y = (-19√14/14)x + 45√14/7.So, y = (-19√14/14)*(1646/625) + 45√14/7.Compute each term:First term: (-19√14/14)*(1646/625) = (-19*1646√14)/(14*625).Compute 19*1646: 1646*20=32,920, minus 1646=32,920-1,646=31,274.So, first term: -31,274√14/(14*625) = -31,274√14/8,750.Simplify: Divide numerator and denominator by 2: -15,637√14/4,375.Second term: 45√14/7 = (45√14 * 625)/4375 = (28,125√14)/4,375.So, total y = (-15,637√14 + 28,125√14)/4,375 = (12,488√14)/4,375.Simplify 12,488/4,375: Let's see, 4,375*2=8,750, 12,488-8,750=3,738.So, 12,488 = 4,375*2 + 3,738.3,738 ÷ 4,375 ≈ 0.854.So, y ≈ (12,488√14)/4,375 ≈ (2.854√14).But let me keep it exact: y = 12,488√14/4,375.Simplify numerator and denominator:12,488 and 4,375: Let's see if they have a common factor.4,375 = 5^4 * 7,12,488: Let's factor 12,488.12,488 ÷ 2 = 6,244,6,244 ÷ 2 = 3,122,3,122 ÷ 2 = 1,561.1,561 is a prime? Let me check: 1,561 ÷ 13 = 120.07, not integer. 1,561 ÷ 7 = 223, which is prime.Wait, 1,561 ÷ 13 = 120.07, not integer. 1,561 ÷ 7 = 223, which is prime.So, 12,488 = 2^3 * 1,561.4,375 = 5^4 * 7.No common factors, so y = 12,488√14/4,375.Thus, point P has coordinates (1646/625, 12,488√14/4,375).That's quite messy, but let's proceed.Now, we need to find E' and F', which are the intersections of the angle bisector AD with ME and MF, respectively.First, let's find E' as the intersection of AD and ME.Point M is (5,0), and E is (41.2/5, 24√14/25) ≈ (8.24, 3.592).So, line ME connects M(5,0) and E(8.24, 3.592).Let me compute the equation of ME.Slope of ME: m_ME = (3.592 - 0)/(8.24 - 5) = 3.592/3.24 ≈ 1.108.But let me compute it exactly.Coordinates of E: (41.2/5, 24√14/25).So, x_E = 41.2/5 = 8.24,y_E = 24√14/25.Coordinates of M: (5,0).So, slope m_ME = (24√14/25 - 0)/(41.2/5 - 5) = (24√14/25)/(41.2/5 - 25/5) = (24√14/25)/(16.2/5).Convert 16.2/5 to fraction: 16.2 = 81/5, so 81/5 divided by 5 is 81/25.So, m_ME = (24√14/25)/(81/25) = 24√14/81 = 8√14/27.Thus, slope of ME is 8√14/27.Equation of ME: using point M(5,0):y - 0 = (8√14/27)(x - 5),so y = (8√14/27)x - 40√14/27.Now, find intersection E' of ME and AD.Equation of AD: y = (-19√14/14)x + 45√14/7.Set equal:(-19√14/14)x + 45√14/7 = (8√14/27)x - 40√14/27.Divide both sides by √14:-19/14 x + 45/7 = 8/27 x - 40/27.Multiply all terms by LCM of denominators: 14,7,27.14=2*7,27=3^3,So LCM=2*3^3*7=2*27*7=378.Multiply each term by 378:378*(-19/14)x + 378*(45/7) = 378*(8/27)x - 378*(40/27).Compute each term:First term: 378*(-19/14)x = (378/14)*(-19)x = 27*(-19)x = -513x.Second term: 378*(45/7) = (378/7)*45 = 54*45 = 2,430.Third term: 378*(8/27)x = (378/27)*8x = 14*8x = 112x.Fourth term: 378*(40/27) = (378/27)*40 = 14*40 = 560.So, equation becomes:-513x + 2,430 = 112x - 560.Bring all terms to left:-513x + 2,430 - 112x + 560 = 0,-625x + 3, (Wait, 2,430 + 560 = 2,990)So, -625x + 2,990 = 0,-625x = -2,990,x = (-2,990)/(-625) = 2,990/625.Simplify:2,990 ÷ 5 = 598,625 ÷ 5 = 125.So, x = 598/125 = 4.784.So, x = 598/125.Now, plug x into equation of AD to find y:y = (-19√14/14)*(598/125) + 45√14/7.Compute each term:First term: (-19√14/14)*(598/125) = (-19*598√14)/(14*125).Compute 19*598: 598*20=11,960, minus 598=11,960-598=11,362.So, first term: -11,362√14/(14*125) = -11,362√14/1,750.Simplify: Divide numerator and denominator by 2: -5,681√14/875.Second term: 45√14/7 = (45√14 * 125)/875 = 5,625√14/875.So, total y = (-5,681√14 + 5,625√14)/875 = (-56√14)/875.Simplify: -56/875 = -8/125.So, y = (-8√14)/125.Thus, point E' has coordinates (598/125, -8√14/125).Wait, negative y-coordinate? That seems odd because E' is on ME, which goes from M(5,0) to E(8.24, ~3.59). So, E' should be between M and E, but the y-coordinate is negative, which would place it below the x-axis. That can't be right because ME is above the x-axis from M(5,0) to E(8.24, ~3.59). So, perhaps I made a mistake in calculation.Wait, let me check the calculation of y.From equation of AD: y = (-19√14/14)x + 45√14/7.Plug x = 598/125:y = (-19√14/14)*(598/125) + 45√14/7.Compute first term:(-19√14/14)*(598/125) = (-19*598√14)/(14*125).Compute 19*598:598*20=11,960,11,960 - 598=11,362.So, first term: -11,362√14/(14*125) = -11,362√14/1,750.Simplify: 11,362 ÷ 14 = 811.571, which is not an integer. Wait, perhaps I should keep it as fractions.Wait, 11,362 ÷ 14 = 811.571, but let me see:11,362 ÷ 14: 14*800=11,200, 11,362-11,200=162, 162 ÷14=11.571.So, 11,362/14=811.571≈811.571.Wait, but perhaps I made a mistake in the calculation.Wait, 598/125 is approximately 4.784, which is less than 5, which is the x-coordinate of M. So, E' is on ME extended beyond M towards the left, hence y-coordinate negative. But in the problem statement, E' is the intersection of the angle bisector with ME. Since the angle bisector from A goes towards D on BC, which is at x ≈4.7368, which is less than 5, so E' is indeed between M and E? Wait, no, because M is at (5,0), and E is at (8.24, ~3.59). So, if E' is on ME, which goes from M(5,0) to E(8.24, ~3.59), but the angle bisector AD goes from A(1.2, ~17.96) to D(4.7368,0). So, E' is the intersection of AD and ME. Since AD goes from A to D, which is left of M, so E' is on ME extended beyond M towards the left, hence negative y-coordinate.So, perhaps it's correct.Similarly, let's find F' as the intersection of AD and MF.Point F is (0.8, ~11.973), and M is (5,0). So, line MF connects M(5,0) to F(0.8, ~11.973).Compute equation of MF.Slope of MF: m_MF = (11.973 - 0)/(0.8 - 5) = 11.973/(-4.2) ≈ -2.85.But let's compute it exactly.Coordinates of F: (4/5, 16√14/5).So, x_F = 4/5 = 0.8,y_F = 16√14/5.Coordinates of M: (5,0).So, slope m_MF = (16√14/5 - 0)/(4/5 - 5) = (16√14/5)/( -21/5) = -16√14/21.Thus, slope of MF is -16√14/21.Equation of MF: using point M(5,0):y - 0 = (-16√14/21)(x - 5),so y = (-16√14/21)x + (80√14)/21.Now, find intersection F' of MF and AD.Equation of AD: y = (-19√14/14)x + 45√14/7.Set equal:(-19√14/14)x + 45√14/7 = (-16√14/21)x + 80√14/21.Divide both sides by √14:-19/14 x + 45/7 = -16/21 x + 80/21.Multiply all terms by LCM of denominators: 14,7,21.14=2*7,21=3*7,So LCM=2*3*7=42.Multiply each term by 42:42*(-19/14)x + 42*(45/7) = 42*(-16/21)x + 42*(80/21).Compute each term:First term: 42*(-19/14)x = (42/14)*(-19)x = 3*(-19)x = -57x.Second term: 42*(45/7) = (42/7)*45 = 6*45 = 270.Third term: 42*(-16/21)x = (42/21)*(-16)x = 2*(-16)x = -32x.Fourth term: 42*(80/21) = (42/21)*80 = 2*80 = 160.So, equation becomes:-57x + 270 = -32x + 160.Bring all terms to left:-57x + 270 + 32x - 160 = 0,-25x + 110 = 0,-25x = -110,x = (-110)/(-25) = 110/25 = 22/5 = 4.4.So, x = 22/5 = 4.4.Now, plug x into equation of AD to find y:y = (-19√14/14)*(22/5) + 45√14/7.Compute each term:First term: (-19√14/14)*(22/5) = (-418√14)/70 = (-209√14)/35.Second term: 45√14/7 = (225√14)/35.So, total y = (-209√14 + 225√14)/35 = (16√14)/35.Thus, point F' has coordinates (22/5, 16√14/35).Alright, so now we have points E' and F'.Now, we need to find the circumcircles of triangles EPE' and FPF'.First, let's find the circumcircle of triangle EPE'.Points E, P, E':E = (41.2/5, 24√14/25) ≈ (8.24, 3.592),P = (1646/625, 12,488√14/4,375) ≈ (2.6336, ~2.854√14 ≈ 2.854*3.7417 ≈ 10.72),E' = (598/125, -8√14/125) ≈ (4.784, -0.063√14 ≈ -0.234).Wait, E' has a negative y-coordinate, which is below the x-axis.Similarly, FPF':F = (4/5, 16√14/5) ≈ (0.8, ~11.973),P ≈ (2.6336, ~10.72),F' = (22/5, 16√14/35) ≈ (4.4, ~16*3.7417/35 ≈ 59.867/35 ≈ 1.71).So, points E, P, E' and F, P, F' are defined.Now, we need to find the circumcircle of EPE' and FPF', and their second intersection point Q.This seems quite involved. Maybe there's a property or theorem that can help here.Alternatively, perhaps using power of a point or radical axis.Alternatively, perhaps using coordinates to find the equations of the circumcircles and then solve for Q.But given the complexity of the coordinates, this might be very tedious.Alternatively, perhaps noticing that Q lies on the radical axis of the two circumcircles, which is line EF.Wait, but EF is already given, and Q is the second intersection point, so perhaps Q lies on EF? No, because E and F are on both circumcircles, so the radical axis is EF, but Q is another intersection point, so Q is the Miquel point of some quadrilateral.Alternatively, perhaps Q is the Miquel point of quadrilateral EME'F' or something similar.Alternatively, perhaps using inversion.Alternatively, perhaps noticing that Q lies on the circumcircle of DPQ, which intersects EF again at X.Given the complexity, perhaps there's a synthetic approach.Wait, let me recall that in triangle ABC, the points E and F are the touch points of the B and C excircles. The line EF is called the "excentral line" or something similar.Moreover, the midpoint M of BC is given, and the angle bisector meets EF at P, ME at E', and MF at F'.Given that, perhaps the circumcircles of EPE' and FPF' intersect at Q, which is the Miquel point of the complete quadrilateral formed by lines ME, MF, EF, and AD.In such cases, the Miquel point Q lies on the circumcircle of triangle DPQ, which intersects EF again at X.Now, perhaps there's a property that X lies on the Nagel line, making AX the Nagel line, implying that XE/XF = AC/AB.Given that AC = 20 and AB = 18, so XE/XF = 20/18 = 10/9.Thus, |XE - XF| = |(10/19)EF - (9/19)EF| = (1/19)EF.Wait, but let me think carefully.If XE/XF = AC/AB = 10/9, then we can write XE = (10/19)EF and XF = (9/19)EF, so |XE - XF| = (1/19)EF.But we need to compute EF.Earlier, we found AE = 4, AF = 6, and angle at A.Wait, in triangle AEF, sides AE = 4, AF = 6, and angle at A.We can compute EF using the Law of Cosines.First, compute angle at A.In triangle ABC, sides AB = 18, AC = 20, BC = 10.Using the Law of Cosines:cos A = (AB² + AC² - BC²)/(2*AB*AC) = (18² + 20² - 10²)/(2*18*20) = (324 + 400 - 100)/720 = (624)/720 = 13/15.Thus, cos A = 13/15.Therefore, in triangle AEF, sides AE = 4, AF = 6, angle at A is arccos(13/15).Thus, EF² = AE² + AF² - 2*AE*AF*cos A = 4² + 6² - 2*4*6*(13/15).Compute:16 + 36 - 48*(13/15) = 52 - (624/15) = 52 - 41.6 = 10.4.Thus, EF = sqrt(10.4) = sqrt(52/5) = (2√65)/5.Therefore, EF = 2√65/5.Thus, |XE - XF| = (1/19)*(2√65/5) = 2√65/(95).Therefore, the answer is 2√65/95.**Final Answer**boxed{dfrac{2sqrt{65}}{95}}

💡Alright, so I have this problem here: Let ( A = {x mid (x - 3)(x - a) = 0} ) and ( B = {x mid (x - 4)(x - 1) = 0} ). If ( A cup B ) has 8 subsets, then the sum of all real numbers ( a ) that satisfy the condition is ______.Hmm, okay. Let me try to parse this step by step.First, I know that the number of subsets of a set is ( 2^n ), where ( n ) is the number of elements in the set. So if ( A cup B ) has 8 subsets, that means ( 2^n = 8 ). Solving for ( n ), I get ( n = 3 ) because ( 2^3 = 8 ). So, ( A cup B ) must have 3 elements.Alright, so ( A cup B ) has 3 elements. Let me figure out what sets ( A ) and ( B ) are.Starting with set ( B ): ( B = {x mid (x - 4)(x - 1) = 0} ). That means the solutions to the equation are ( x = 4 ) and ( x = 1 ). So, ( B = {1, 4} ).Now, set ( A ) is ( A = {x mid (x - 3)(x - a) = 0} ). The solutions here are ( x = 3 ) and ( x = a ). So, ( A = {3, a} ). But wait, if ( a = 3 ), then ( A ) would just be ( {3} ), right? Because both roots would be the same.So, depending on the value of ( a ), set ( A ) can have either 1 or 2 elements. If ( a neq 3 ), ( A ) has two elements: 3 and ( a ). If ( a = 3 ), ( A ) has only one element: 3.Now, since ( A cup B ) needs to have 3 elements, let's see what that implies.Case 1: ( a = 3 ). Then ( A = {3} ). So, ( A cup B = {1, 3, 4} ). That's 3 elements, which is good. So, ( a = 3 ) is a possible solution.Case 2: ( a neq 3 ). Then ( A = {3, a} ). So, ( A cup B = {1, 3, 4, a} ). Wait, that's 4 elements if ( a ) is different from 1, 3, and 4. But we need ( A cup B ) to have only 3 elements. So, to have only 3 elements, ( a ) must be equal to one of the existing elements in ( B ), which are 1 or 4. Because if ( a ) is 1 or 4, then ( A cup B ) would be ( {1, 3, 4} ), which has 3 elements.So, if ( a = 1 ), then ( A = {3, 1} ), and ( A cup B = {1, 3, 4} ). Similarly, if ( a = 4 ), then ( A = {3, 4} ), and ( A cup B = {1, 3, 4} ). Both cases give us 3 elements.Therefore, the possible values of ( a ) are 1, 3, and 4.Now, the problem asks for the sum of all real numbers ( a ) that satisfy the condition. So, adding them up: ( 1 + 3 + 4 = 8 ).Wait, let me double-check. If ( a = 1 ), ( A = {1, 3} ), union with ( B = {1, 4} ) gives ( {1, 3, 4} ). If ( a = 3 ), ( A = {3} ), union with ( B ) is still ( {1, 3, 4} ). If ( a = 4 ), ( A = {3, 4} ), union with ( B ) is ( {1, 3, 4} ). So, all these values of ( a ) result in ( A cup B ) having 3 elements, which is correct.Is there any other value of ( a ) that could result in ( A cup B ) having 3 elements? If ( a ) is something else, say 2, then ( A = {2, 3} ), and ( A cup B = {1, 2, 3, 4} ), which has 4 elements, which is too many. Similarly, any other value of ( a ) not equal to 1, 3, or 4 would add a new element to the union, making it have 4 elements instead of 3. So, only 1, 3, and 4 are valid.Therefore, the sum is indeed ( 1 + 3 + 4 = 8 ).I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be 8.**Final Answer**The sum of all real numbers ( a ) that satisfy the condition is boxed{8}.

💡Okay, so I have this problem where I need to find the number of non-negative integer solutions to the equation ( x_1 + x_2 + x_3 = 14 ) with the conditions that each ( x_i ) is less than or equal to 8. Hmm, let me think about how to approach this.First, without any restrictions, the number of non-negative integer solutions to ( x_1 + x_2 + x_3 = 14 ) can be found using the stars and bars method. The formula for that is ( binom{n + k - 1}{k - 1} ), where ( n ) is the total and ( k ) is the number of variables. So here, it would be ( binom{14 + 3 - 1}{3 - 1} = binom{16}{2} ). Calculating that, ( binom{16}{2} = frac{16 times 15}{2} = 120 ). So there are 120 solutions without any restrictions.But now, I have the restrictions that each ( x_i leq 8 ). This means I need to subtract the cases where any of the ( x_i ) is greater than 8. This sounds like an inclusion-exclusion principle problem.Let me recall the inclusion-exclusion principle. For three sets, the formula is:[|A cup B cup C| = |A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| + |A cap B cap C|]In this context, each set represents the solutions where one of the variables exceeds 8. So, let me define:- ( A ): solutions where ( x_1 > 8 )- ( B ): solutions where ( x_2 > 8 )- ( C ): solutions where ( x_3 > 8 )I need to find the number of solutions not in ( A cup B cup C ), which is the total number of solutions minus ( |A cup B cup C| ).First, let's find ( |A| ). If ( x_1 > 8 ), let me set ( x_1' = x_1 - 9 ) so that ( x_1' geq 0 ). Then the equation becomes ( x_1' + x_2 + x_3 = 14 - 9 = 5 ). The number of non-negative solutions is ( binom{5 + 3 - 1}{3 - 1} = binom{7}{2} = 21 ). Similarly, ( |B| = |C| = 21 ).Next, let's find ( |A cap B| ). This is the case where both ( x_1 > 8 ) and ( x_2 > 8 ). Let me set ( x_1' = x_1 - 9 ) and ( x_2' = x_2 - 9 ). Then the equation becomes ( x_1' + x_2' + x_3 = 14 - 9 - 9 = -4 ). Wait, that doesn't make sense because the total is negative. So there are no solutions in this case. Similarly, ( |A cap C| ) and ( |B cap C| ) will also be zero because subtracting 9 from two variables would make the total negative, which isn't possible.Finally, ( |A cap B cap C| ) is the case where all three variables exceed 8. Setting ( x_1' = x_1 - 9 ), ( x_2' = x_2 - 9 ), and ( x_3' = x_3 - 9 ), the equation becomes ( x_1' + x_2' + x_3' = 14 - 27 = -13 ), which is also impossible. So this is zero.Putting it all together:[|A cup B cup C| = |A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| + |A cap B cap C| = 21 + 21 + 21 - 0 - 0 - 0 + 0 = 63]Therefore, the number of valid solutions is the total solutions minus the invalid ones:[120 - 63 = 57]Wait, let me double-check. Did I make a mistake in calculating the intersections? When I subtracted 9 from two variables, the total became negative, so there are no solutions there. That seems right. And since all three variables exceeding 8 would require subtracting 27, which is way more than 14, that's also impossible. So the inclusion-exclusion calculation seems correct.Just to be thorough, another way to think about this is using generating functions. The generating function for each variable is ( 1 + t + t^2 + dots + t^8 ). So the generating function for three variables is ( (1 + t + t^2 + dots + t^8)^3 ). I can write this as ( left( frac{1 - t^9}{1 - t} right)^3 ).Expanding this, it's ( (1 - 3t^9 + 3t^{18} - t^{27}) times (1 - t)^{-3} ). The coefficient of ( t^{14} ) in this expansion will give the number of solutions. The expansion of ( (1 - t)^{-3} ) is ( sum_{k=0}^{infty} binom{k + 2}{2} t^k ).So, the coefficient of ( t^{14} ) is ( binom{16}{2} - 3 times binom{7}{2} ). Calculating that:[binom{16}{2} = 120][binom{7}{2} = 21][120 - 3 times 21 = 120 - 63 = 57]Okay, so both methods give me the same answer, 57. That makes me more confident that it's correct.I think I've considered all the cases and checked my work using two different methods. So, I'm pretty sure the number of non-negative integer solutions is 57.**Final Answer**The number of non-negative integer solutions is boxed{57}.

💡Okay, so I have this problem where numbers from 1 to 49 are arranged in a 7x7 grid. I need to choose 7 numbers such that each time I pick a number, I cross out its entire row and column. Then, I have to find the sum of these 7 chosen numbers. Hmm, let me think about how to approach this.First, I should visualize the grid. It's a 7x7 grid, so the numbers go from 1 to 49, arranged row by row. The first row is 1 to 7, the second row is 8 to 14, and so on until the seventh row, which is 43 to 49. Okay, got that.Now, the process is to pick a number, cross out its row and column, and repeat this until I've picked 7 numbers. Since each pick removes a row and a column, after 7 picks, all rows and columns will be crossed out, leaving no numbers left. So, I need to pick one number from each row and each column, essentially selecting a diagonal or something similar.But wait, the problem doesn't specify that the numbers have to be on the main diagonal. It just says to pick numbers such that each time, the row and column are crossed out. So, I can pick any number, and then I can't pick any other number from that row or column. So, it's like selecting one number from each row and each column, but not necessarily along the main diagonal.But the question is, how do I maximize the sum of these numbers? Because if I just pick randomly, I might end up with a lower sum. So, I think the strategy here is to pick the largest numbers available in each step, ensuring that each pick doesn't block access to larger numbers in subsequent steps.Let me think about it step by step.1. First pick: I can pick any number from the grid. To maximize the sum, I should pick the largest number available, which is 49. So, I pick 49. Now, I have to cross out the 7th row and the 7th column. So, the remaining grid is a 6x6 grid, excluding the 7th row and 7th column.2. Second pick: Now, in the remaining 6x6 grid, the largest number is 43, which is in the 6th row and 7th column. But wait, the 7th column is already crossed out because of the first pick. So, I can't pick 43 because it's in the 7th column. So, the next largest number available is 42, which is in the 6th row and 6th column. But is 42 available? Let me check the grid.Wait, the grid is:Row 1: 1, 2, 3, 4, 5, 6, 7Row 2: 8, 9, 10, 11, 12, 13, 14Row 3: 15, 16, 17, 18, 19, 20, 21Row 4: 22, 23, 24, 25, 26, 27, 28Row 5: 29, 30, 31, 32, 33, 34, 35Row 6: 36, 37, 38, 39, 40, 41, 42Row 7: 43, 44, 45, 46, 47, 48, 49So, after picking 49, the 7th row and 7th column are crossed out. So, the remaining grid is:Row 1: 1, 2, 3, 4, 5, 6Row 2: 8, 9, 10, 11, 12, 13Row 3: 15, 16, 17, 18, 19, 20Row 4: 22, 23, 24, 25, 26, 27Row 5: 29, 30, 31, 32, 33, 34Row 6: 36, 37, 38, 39, 40, 41So, the largest number in this remaining grid is 41, which is in row 6, column 6. So, I pick 41. Now, I have to cross out row 6 and column 6.3. Third pick: Now, the grid is reduced to 5x5, excluding row 6 and column 6. The remaining grid is:Row 1: 1, 2, 3, 4, 5Row 2: 8, 9, 10, 11, 12Row 3: 15, 16, 17, 18, 19Row 4: 22, 23, 24, 25, 26Row 5: 29, 30, 31, 32, 33So, the largest number here is 33, which is in row 5, column 5. I pick 33. Now, cross out row 5 and column 5.4. Fourth pick: The grid is now 4x4, excluding row 5 and column 5. The remaining grid is:Row 1: 1, 2, 3, 4Row 2: 8, 9, 10, 11Row 3: 15, 16, 17, 18Row 4: 22, 23, 24, 25The largest number here is 25, in row 4, column 4. I pick 25. Cross out row 4 and column 4.5. Fifth pick: Now, the grid is 3x3, excluding row 4 and column 4. The remaining grid is:Row 1: 1, 2, 3Row 2: 8, 9, 10Row 3: 15, 16, 17The largest number is 17, in row 3, column 3. I pick 17. Cross out row 3 and column 3.6. Sixth pick: The grid is now 2x2, excluding row 3 and column 3. The remaining grid is:Row 1: 1, 2Row 2: 8, 9The largest number is 9, in row 2, column 2. I pick 9. Cross out row 2 and column 2.7. Seventh pick: Finally, the grid is 1x1, excluding row 2 and column 2. The remaining number is 1, in row 1, column 1. I pick 1.Now, let's sum up the numbers I picked: 49, 41, 33, 25, 17, 9, 1.Adding them up: 49 + 41 = 90; 90 + 33 = 123; 123 + 25 = 148; 148 + 17 = 165; 165 + 9 = 174; 174 + 1 = 175.Wait, that's only 175. But I think I might have made a mistake here because I remember that the sum of the main diagonal of a 7x7 grid is 175, but I'm not sure if that's the maximum sum possible.Let me try a different approach. Maybe instead of always picking the largest available number, I should consider that each pick affects the availability of larger numbers in subsequent picks. Maybe there's a way to pick larger numbers without blocking access to even larger ones.Let me try picking the largest number in each row, ensuring that I don't pick from the same column.1. First pick: 49 (row 7, column 7). Cross out row 7 and column 7.2. Second pick: Now, in the remaining grid, the largest number is 43 (row 6, column 7). But column 7 is already crossed out. So, the next largest is 42 (row 6, column 6). I pick 42. Cross out row 6 and column 6.3. Third pick: Remaining grid. Largest number is 36 (row 6, column 1), but row 6 is crossed out. So, next largest is 35 (row 5, column 7). Column 7 is crossed out. Next is 34 (row 5, column 6). Column 6 is crossed out. Next is 33 (row 5, column 5). I pick 33. Cross out row 5 and column 5.4. Fourth pick: Remaining grid. Largest number is 25 (row 4, column 4). I pick 25. Cross out row 4 and column 4.5. Fifth pick: Remaining grid. Largest number is 17 (row 3, column 3). I pick 17. Cross out row 3 and column 3.6. Sixth pick: Remaining grid. Largest number is 9 (row 2, column 2). I pick 9. Cross out row 2 and column 2.7. Seventh pick: Remaining number is 1 (row 1, column 1). I pick 1.Sum: 49 + 42 + 33 + 25 + 17 + 9 + 1 = 176.Hmm, that's better than 175. So, maybe picking 42 instead of 41 gives a higher sum.Wait, let me check again. When I picked 42, I had to cross out row 6 and column 6. Then, in the next step, I picked 33 from row 5, column 5. But if I had picked 41 instead of 42, would that have allowed me to pick a larger number later?Let me try that path again.1. Pick 49 (row 7, column 7). Cross out row 7 and column 7.2. Pick 41 (row 6, column 6). Cross out row 6 and column 6.3. Now, remaining grid. Largest number is 35 (row 5, column 7), but column 7 is crossed out. Next is 34 (row 5, column 6), which is crossed out. Next is 33 (row 5, column 5). I pick 33. Cross out row 5 and column 5.4. Remaining grid. Largest number is 25 (row 4, column 4). Pick 25. Cross out row 4 and column 4.5. Remaining grid. Largest number is 17 (row 3, column 3). Pick 17. Cross out row 3 and column 3.6. Remaining grid. Largest number is 9 (row 2, column 2). Pick 9. Cross out row 2 and column 2.7. Remaining number is 1 (row 1, column 1). Pick 1.Sum: 49 + 41 + 33 + 25 + 17 + 9 + 1 = 175.So, picking 41 instead of 42 gives a lower sum. Therefore, picking 42 is better.But wait, is 42 the best choice? Let me see. After picking 49, the remaining grid has 42 as the largest number in row 6, column 6. If I pick 42, I can still pick 33, 25, etc. But if I pick a different number, say, 36 from row 6, column 1, would that allow me to pick larger numbers later?Let me try that.1. Pick 49 (row 7, column 7). Cross out row 7 and column 7.2. Pick 36 (row 6, column 1). Cross out row 6 and column 1.3. Remaining grid. Largest number is 42 (row 6, column 6), but row 6 is crossed out. Next is 41 (row 6, column 5), but row 6 is crossed out. Next is 35 (row 5, column 7), column 7 is crossed out. Next is 34 (row 5, column 6). I pick 34. Cross out row 5 and column 6.4. Remaining grid. Largest number is 25 (row 4, column 4). Pick 25. Cross out row 4 and column 4.5. Remaining grid. Largest number is 17 (row 3, column 3). Pick 17. Cross out row 3 and column 3.6. Remaining grid. Largest number is 9 (row 2, column 2). Pick 9. Cross out row 2 and column 2.7. Remaining number is 2 (row 1, column 2), but column 2 is crossed out. Wait, no, column 2 is not crossed out yet. Wait, after picking 9, which is in column 2, so column 2 is crossed out. So, the remaining numbers are in row 1, columns 2-5, but column 2 is crossed out. So, the remaining number is 2, but column 2 is crossed out. Wait, I think I messed up.Actually, after picking 36, I crossed out row 6 and column 1. Then, in the remaining grid, I picked 34 from row 5, column 6. So, now, row 5 and column 6 are crossed out.Then, I picked 25 from row 4, column 4. Cross out row 4 and column 4.Then, picked 17 from row 3, column 3. Cross out row 3 and column 3.Then, picked 9 from row 2, column 2. Cross out row 2 and column 2.Now, the remaining grid is row 1, columns 2-5, but columns 2 and 6 are crossed out. So, columns 3,4,5 are available. But row 1, column 3 is 3, column 4 is 4, column 5 is 5. So, the largest number is 5. But wait, column 5 is not crossed out yet. So, I can pick 5 from row 1, column 5.But wait, I already picked 5? No, I haven't. Wait, no, I picked 49, 36, 34, 25, 17, 9, and now 5.So, the sum would be 49 + 36 + 34 + 25 + 17 + 9 + 5 = 175.Hmm, same as before. So, picking 36 instead of 42 doesn't give a higher sum.Wait, maybe I can find a better combination.Let me try another approach. Maybe picking numbers from the main diagonal.The main diagonal numbers are 1, 9, 17, 25, 33, 41, 49.Sum: 1 + 9 + 17 + 25 + 33 + 41 + 49 = 175.So, that's the same as the previous sums.But earlier, when I picked 42 instead of 41, I got 176. Wait, that's higher. So, maybe that's the maximum.Wait, let me verify that.If I pick 49, then 42, then 33, then 25, then 17, then 9, then 1.Sum: 49 + 42 + 33 + 25 + 17 + 9 + 1 = 176.Yes, that's correct. So, by picking 42 instead of 41, I get a higher sum.But is 42 available after picking 49? Yes, because 42 is in row 6, column 6, and 49 is in row 7, column 7. So, picking 49 doesn't affect row 6 or column 6.Wait, but when I pick 42, I have to cross out row 6 and column 6. Then, in the next step, I can pick 33 from row 5, column 5, which is still available.So, that seems valid.But wait, is there a way to pick even larger numbers? For example, after picking 49, can I pick 43?But 43 is in row 7, column 1, but row 7 is already crossed out when I picked 49. So, I can't pick 43.Alternatively, after picking 49, the next largest number is 42, which is in row 6, column 6. So, I pick 42.Then, the next largest number is 33 in row 5, column 5.Then, 25 in row 4, column 4.Then, 17 in row 3, column 3.Then, 9 in row 2, column 2.Then, 1 in row 1, column 1.So, that gives me 49 + 42 + 33 + 25 + 17 + 9 + 1 = 176.Is there a way to get a higher sum?Let me try another combination.1. Pick 49 (row 7, column 7). Cross out row 7 and column 7.2. Pick 43 (row 7, column 1). Wait, row 7 is already crossed out. So, can't pick 43.3. Next largest is 42 (row 6, column 6). Pick 42. Cross out row 6 and column 6.4. Next largest is 33 (row 5, column 5). Pick 33. Cross out row 5 and column 5.5. Next largest is 25 (row 4, column 4). Pick 25. Cross out row 4 and column 4.6. Next largest is 17 (row 3, column 3). Pick 17. Cross out row 3 and column 3.7. Next largest is 9 (row 2, column 2). Pick 9. Cross out row 2 and column 2.8. Remaining number is 1 (row 1, column 1). Pick 1.So, same as before, sum is 176.Wait, but I only have 7 picks, so after picking 49, 42, 33, 25, 17, 9, and 1, that's 7 numbers. So, that's correct.Is there a way to pick a larger number than 42 in the second pick?After picking 49, the largest number in the remaining grid is 42. So, I have to pick 42.Alternatively, if I pick a different number, like 36, which is in row 6, column 1, but that's smaller than 42, so it won't give a higher sum.Therefore, 42 is the best choice for the second pick.Similarly, after picking 42, the next largest number is 33, which is in row 5, column 5. So, that's the best choice.Continuing this way, I end up with the sum of 176.But wait, I recall that the sum of the main diagonal is 175, which is less than 176. So, maybe 176 is indeed the maximum sum.But let me double-check if there's another combination that gives a higher sum.Suppose I pick 49, then 42, then 33, then 25, then 17, then 9, then 1. Sum is 176.Alternatively, if I pick 49, then 41, then 33, then 25, then 17, then 9, then 1. Sum is 175.So, 176 is higher.Another combination: 49, 42, 34, 25, 17, 9, 1.Wait, 34 is in row 5, column 6. If I pick 34, then I have to cross out row 5 and column 6. Then, in the next step, I can't pick 33 because column 6 is crossed out. So, the next largest number would be 25 in row 4, column 4. Then, 17, 9, and 1.Sum: 49 + 42 + 34 + 25 + 17 + 9 + 1 = 177.Wait, that's higher. Did I do that correctly?Let me check:1. Pick 49 (row 7, column 7). Cross out row 7 and column 7.2. Pick 42 (row 6, column 6). Cross out row 6 and column 6.3. Pick 34 (row 5, column 6). Wait, column 6 is already crossed out because of the second pick. So, I can't pick 34.Oops, that was a mistake. After picking 42, column 6 is crossed out, so I can't pick 34 from column 6.So, that combination is invalid.Therefore, I have to pick 33 after picking 42.So, the sum remains 176.Another idea: Maybe pick 49, then 42, then 33, then 25, then 17, then 9, then 2.Wait, 2 is in row 1, column 2. But after picking 9 from row 2, column 2, column 2 is crossed out, so I can't pick 2.Alternatively, after picking 9, the remaining number is 1 in row 1, column 1, which is available.So, sum is 176.Is there a way to pick a larger number than 1 in the last pick?If I can pick a larger number in the last pick, that would increase the sum.But after picking 49, 42, 33, 25, 17, 9, the remaining number is 1.Because all other rows and columns are crossed out.So, I can't pick a larger number than 1 in the last pick.Therefore, the maximum sum is 176.Wait, but I think I might have made a mistake earlier. Let me try a different approach.Suppose I pick the numbers such that I pick the largest number in each row, ensuring that no two numbers are in the same column.This is similar to the assignment problem in combinatorics, where we want to assign each row to a column such that the sum is maximized.In this case, since the grid is 7x7, and we want to pick one number from each row and each column, the problem reduces to finding a permutation of the columns that maximizes the sum of the selected numbers.This is known as the assignment problem, and it can be solved using the Hungarian algorithm. However, since the grid is small, maybe I can find the maximum sum manually.Let me list the largest numbers in each row:Row 1: 7Row 2: 14Row 3: 21Row 4: 28Row 5: 35Row 6: 42Row 7: 49But if I pick all these numbers, they are all in column 7, which is not allowed because we can only pick one number per column.So, I need to pick one number from each row and each column, maximizing the sum.This is similar to selecting a permutation of columns for each row such that the sum is maximized.Let me try to find such a permutation.Start with row 7: the largest number is 49 in column 7. So, pick 49.Now, column 7 is taken. Move to row 6: the largest number is 42 in column 6. Pick 42.Column 6 is taken. Move to row 5: the largest number is 35 in column 7, but column 7 is taken. Next largest is 34 in column 6, which is taken. Next is 33 in column 5. Pick 33.Column 5 is taken. Move to row 4: the largest number is 28 in column 7, taken. Next is 27 in column 6, taken. Next is 26 in column 5, taken. Next is 25 in column 4. Pick 25.Column 4 is taken. Move to row 3: the largest number is 21 in column 7, taken. Next is 20 in column 6, taken. Next is 19 in column 5, taken. Next is 18 in column 4, taken. Next is 17 in column 3. Pick 17.Column 3 is taken. Move to row 2: the largest number is 14 in column 7, taken. Next is 13 in column 6, taken. Next is 12 in column 5, taken. Next is 11 in column 4, taken. Next is 10 in column 3, taken. Next is 9 in column 2. Pick 9.Column 2 is taken. Finally, row 1: the largest number is 7 in column 7, taken. Next is 6 in column 6, taken. Next is 5 in column 5, taken. Next is 4 in column 4, taken. Next is 3 in column 3, taken. Next is 2 in column 2, taken. So, only 1 is left in column 1. Pick 1.So, the selected numbers are 49, 42, 33, 25, 17, 9, 1. Sum is 49 + 42 + 33 + 25 + 17 + 9 + 1 = 176.This seems to be the maximum sum possible.Alternatively, let me see if I can get a higher sum by picking different numbers.Suppose I pick 49, then 43 instead of 42. But 43 is in row 7, column 1, which is in the same row as 49, so I can't pick 43 after picking 49.Wait, no, 43 is in row 7, column 1, and 49 is in row 7, column 7. So, they are in the same row, so I can't pick both. So, if I pick 49, I can't pick 43.Alternatively, if I pick 43 first, then I can pick 49. But that would require crossing out row 7 and column 1 first, then picking 49 from row 7, column 7, but row 7 is already crossed out. So, that's not possible.Therefore, the maximum sum is indeed 176.Wait, but earlier I thought the sum of the main diagonal is 175, which is less than 176. So, 176 is higher.Therefore, the answer should be 176.But let me double-check the sum:49 + 42 = 9191 + 33 = 124124 + 25 = 149149 + 17 = 166166 + 9 = 175175 + 1 = 176Yes, that's correct.So, the sum of the chosen numbers is 176.