💡Okay, so I have this problem here where I need to find the minimum value of (frac{1}{x} + frac{1}{y}) given some conditions. Let me try to unpack this step by step.First, the problem states that (a > 0), (b > 0), and (sqrt{2}) is the geometric mean of (a) and (b). I remember that the geometric mean of two numbers (a) and (b) is (sqrt{ab}). So, if (sqrt{2}) is the geometric mean, that means:[sqrt{ab} = sqrt{2}]If I square both sides to get rid of the square roots, I should have:[ab = 2]Okay, so that's a useful equation: (ab = 2). I'll keep that in mind.Next, the problem mentions that (log_a x = log_b y = 3). Hmm, logarithms. I need to decode what this means for (x) and (y). Let's take it one at a time.Starting with (log_a x = 3). By definition, (log_a x = 3) means that (a) raised to the power of 3 equals (x). So,[a^3 = x]Similarly, (log_b y = 3) means that (b^3 = y). So,[b^3 = y]Alright, so now I know that (x = a^3) and (y = b^3). I need to find the minimum value of (frac{1}{x} + frac{1}{y}), which in terms of (a) and (b) becomes:[frac{1}{a^3} + frac{1}{b^3}]Hmm, that's the expression I need to minimize. Let me write that down:[frac{1}{a^3} + frac{1}{b^3}]I also know that (ab = 2). Maybe I can express this in terms of a single variable or find a relationship between (a) and (b) that can help me minimize the expression.Let me think about how to relate (a) and (b). Since (ab = 2), I can express (b) in terms of (a):[b = frac{2}{a}]So, substituting this into the expression for (y), we have:[y = b^3 = left(frac{2}{a}right)^3 = frac{8}{a^3}]Therefore, the expression to minimize becomes:[frac{1}{a^3} + frac{8}{a^3} = frac{9}{a^3}]Wait, that seems too straightforward. If I substitute (b) in terms of (a), the expression simplifies to (frac{9}{a^3}). But that would mean the expression is dependent only on (a), and to minimize it, I need to maximize (a^3). But (a) is a positive real number, so as (a) increases, (a^3) increases, making (frac{9}{a^3}) decrease. However, (a) can't be just any number because (b) is also dependent on (a) through (ab = 2). So, if (a) increases, (b) decreases, and vice versa.But wait, I might have made a mistake here. Let me double-check my substitution.I had (x = a^3) and (y = b^3), so (frac{1}{x} + frac{1}{y} = frac{1}{a^3} + frac{1}{b^3}). Then, since (ab = 2), I expressed (b = frac{2}{a}), so (b^3 = frac{8}{a^3}). Therefore, (frac{1}{b^3} = frac{a^3}{8}). So, substituting back, the expression becomes:[frac{1}{a^3} + frac{a^3}{8}]Ah, okay, I see where I went wrong earlier. I incorrectly substituted (frac{1}{b^3}) as (frac{8}{a^3}), but actually, it's the reciprocal. So, (frac{1}{b^3} = frac{a^3}{8}). That makes more sense.So, now the expression to minimize is:[frac{1}{a^3} + frac{a^3}{8}]Let me denote this expression as (E):[E = frac{1}{a^3} + frac{a^3}{8}]Now, I need to find the minimum value of (E) with respect to (a), where (a > 0).This looks like a calculus problem where I can take the derivative of (E) with respect to (a), set it to zero, and solve for (a). Alternatively, maybe I can use the AM-GM inequality since the expression has two positive terms.Let me try the AM-GM approach first because it might be quicker.The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, for two positive numbers (u) and (v):[frac{u + v}{2} geq sqrt{uv}]Equality holds when (u = v).Applying this to (E), where (u = frac{1}{a^3}) and (v = frac{a^3}{8}):[frac{frac{1}{a^3} + frac{a^3}{8}}{2} geq sqrt{frac{1}{a^3} cdot frac{a^3}{8}} = sqrt{frac{1}{8}} = frac{1}{2sqrt{2}}]Multiplying both sides by 2:[frac{1}{a^3} + frac{a^3}{8} geq frac{1}{sqrt{2}}]So, the minimum value of (E) is (frac{1}{sqrt{2}}), which is approximately 0.7071.But wait, let me check if equality holds. For AM-GM, equality holds when (u = v). So,[frac{1}{a^3} = frac{a^3}{8}]Multiplying both sides by (8a^3):[8 = a^6]Taking the sixth root of both sides:[a = 8^{1/6}]Simplify (8^{1/6}):Since (8 = 2^3), so (8^{1/6} = (2^3)^{1/6} = 2^{3/6} = 2^{1/2} = sqrt{2}).So, (a = sqrt{2}). Then, since (ab = 2), (b = frac{2}{a} = frac{2}{sqrt{2}} = sqrt{2}).Therefore, both (a) and (b) are equal to (sqrt{2}). That makes sense because if the geometric mean is (sqrt{2}), then both numbers are equal to the geometric mean when they are equal.So, with (a = sqrt{2}) and (b = sqrt{2}), let's compute (x) and (y):[x = a^3 = (sqrt{2})^3 = 2^{3/2} = 2 cdot sqrt{2}][y = b^3 = (sqrt{2})^3 = 2^{3/2} = 2 cdot sqrt{2}]Therefore, (x = y = 2sqrt{2}). Plugging these back into (frac{1}{x} + frac{1}{y}):[frac{1}{2sqrt{2}} + frac{1}{2sqrt{2}} = frac{2}{2sqrt{2}} = frac{1}{sqrt{2}} = frac{sqrt{2}}{2}]Wait, that's different from what I got earlier. Earlier, I thought the minimum was (frac{1}{sqrt{2}}), but now it's (frac{sqrt{2}}{2}). But actually, (frac{1}{sqrt{2}}) is equal to (frac{sqrt{2}}{2}), since multiplying numerator and denominator by (sqrt{2}):[frac{1}{sqrt{2}} = frac{sqrt{2}}{2}]So, both expressions are equivalent. Therefore, the minimum value is indeed (frac{sqrt{2}}{2}).Just to make sure, let me try using calculus to verify this result.We had:[E = frac{1}{a^3} + frac{a^3}{8}]Let's take the derivative of (E) with respect to (a):[E' = -frac{3}{a^4} + frac{3a^2}{8}]Set the derivative equal to zero to find critical points:[-frac{3}{a^4} + frac{3a^2}{8} = 0]Multiply both sides by (8a^4) to eliminate denominators:[-24 + 3a^6 = 0]Simplify:[3a^6 = 24][a^6 = 8][a = 8^{1/6} = 2^{3/6} = 2^{1/2} = sqrt{2}]So, (a = sqrt{2}), which is the same result as before. Therefore, the critical point occurs at (a = sqrt{2}), and since the function (E) tends to infinity as (a) approaches 0 or infinity, this critical point must be a minimum.Thus, the minimum value of (frac{1}{x} + frac{1}{y}) is indeed (frac{sqrt{2}}{2}).I think I've covered all the steps and verified the result using both the AM-GM inequality and calculus. It seems consistent, so I'm confident that the minimum value is (frac{sqrt{2}}{2}).**Final Answer**The minimum value of (frac{1}{x} + frac{1}{y}) is (boxed{dfrac{sqrt{2}}{2}}).

💡Okay, so I have these two equations to solve. Let me take them one at a time. Starting with the first equation: 2.4y minus 9.8 equals 1.4y minus 9. Hmm, I need to find the value of y that makes this equation true. Alright, first step, I think I should get all the y terms on one side and the constants on the other. So, I can subtract 1.4y from both sides to move it to the left. That would give me 2.4y minus 1.4y minus 9.8 equals minus 9. Let me calculate 2.4y minus 1.4y. That should be 1.0y, right? So now the equation is 1.0y minus 9.8 equals minus 9. Next, I want to get rid of that minus 9.8 on the left side. To do that, I can add 9.8 to both sides. Adding 9.8 to minus 9.8 gives me zero, so I'm left with 1.0y equals minus 9 plus 9.8. What's minus 9 plus 9.8? That should be 0.8. So, 1.0y equals 0.8. Since 1.0y is just y, that means y equals 0.8. Okay, that seems straightforward. Let me double-check by plugging y back into the original equation. Left side: 2.4 times 0.8 minus 9.8. 2.4 times 0.8 is... let's see, 2 times 0.8 is 1.6, and 0.4 times 0.8 is 0.32. So, 1.6 plus 0.32 is 1.92. Then, 1.92 minus 9.8 is minus 7.88. Right side: 1.4 times 0.8 minus 9. 1.4 times 0.8 is... 1 times 0.8 is 0.8, and 0.4 times 0.8 is 0.32. So, 0.8 plus 0.32 is 1.12. Then, 1.12 minus 9 is minus 7.88. Both sides are equal, so y equals 0.8 is correct. Now, moving on to the second equation: x minus 3 equals three-halves x plus 1. I need to find the value of x that satisfies this equation. Again, I'll try to get all the x terms on one side and the constants on the other. So, I can subtract three-halves x from both sides to move it to the left. That gives me x minus three-halves x minus 3 equals 1. Let me compute x minus three-halves x. Since x is the same as two-halves x, subtracting three-halves x would give me negative one-half x. So, the equation becomes negative one-half x minus 3 equals 1. Now, I want to isolate the x term. I'll add 3 to both sides to get rid of the minus 3 on the left. Adding 3 to both sides gives me negative one-half x equals 4. To solve for x, I need to get rid of the negative one-half coefficient. I can do this by multiplying both sides by the reciprocal, which is negative 2. So, multiplying both sides by negative 2 gives me x equals 4 times negative 2, which is negative 8. Let me check this solution by plugging x back into the original equation. Left side: x minus 3 is negative 8 minus 3, which is negative 11. Right side: three-halves x plus 1 is three-halves times negative 8 plus 1. Three-halves times negative 8 is... well, three times negative 8 is negative 24, divided by 2 is negative 12. Then, negative 12 plus 1 is negative 11. Both sides are equal, so x equals negative 8 is correct. I think I've solved both equations correctly. Just to recap, for the first equation, y equals 0.8, and for the second equation, x equals negative 8. I double-checked both solutions by substituting them back into the original equations, and they worked out. So, I'm confident these are the right answers.

💡Alright, so I have this geometry problem here about a trapezium inscribed in a circle. Let me try to visualize it first. A trapezium (ABCD) with (AD) parallel to (BC), (AD = 7), (BC = 3), and angle (angle BCD = 120^circ). There's a chord (BM) that intersects (AD) at point (N), where (ND = 2). I need to find the area of triangle (BOM), where (O) is the center of the circle.First, since (ABCD) is a trapezium inscribed in a circle, it must be an isosceles trapezium. That means the non-parallel sides (AB) and (CD) are equal in length, and the base angles are equal. So, (AB = CD), and the angles at (A) and (D) are equal, as are the angles at (B) and (C).Given (AD = 7) and (BC = 3), the difference in the lengths of the two bases is (7 - 3 = 4). Since it's an isosceles trapezium, the legs (AB) and (CD) extend beyond the shorter base (BC) by equal amounts on both sides. So, each extension is (4/2 = 2). That means if I drop a perpendicular from (C) to (AD), the foot of this perpendicular will be 2 units away from (D). Let's call this foot (P). So, (DP = 2).Given that (angle BCD = 120^circ), and since (ABCD) is cyclic, the opposite angle (angle BAD) must be equal to (180^circ - 120^circ = 60^circ). Wait, no, in a cyclic quadrilateral, opposite angles are supplementary. So, actually, (angle BAD = 180^circ - 120^circ = 60^circ). That makes sense.Now, let's consider triangle (BCD). We know (BC = 3), (CD = AB) (which we don't know yet), and angle (angle BCD = 120^circ). Maybe I can use the Law of Cosines here to find (BD), the diagonal.Wait, but I don't know (CD). Hmm. Maybe I can find the height of the trapezium first. Since (DP = 2) and (angle CDA = 60^circ), the height (h) can be found using trigonometry. In triangle (CDP), which is a right triangle, (h = DP cdot tan(60^circ)). Since (tan(60^circ) = sqrt{3}), (h = 2sqrt{3}).So, the height of the trapezium is (2sqrt{3}). That might be useful later.Now, moving on to the chord (BM) intersecting (AD) at (N), with (ND = 2). Since (AD = 7), then (AN = AD - ND = 7 - 2 = 5).I remember there's a theorem called the Power of a Point which states that for a point (N) inside a circle, the product of the lengths of the segments of any two chords through (N) is equal. So, in this case, (BN cdot NM = AN cdot ND).Let me write that down:[BN cdot NM = AN cdot ND]We know (AN = 5) and (ND = 2), so:[BN cdot NM = 5 times 2 = 10]But I don't know (BN) or (NM). Maybe I can express (BN) in terms of (BC) and the height.Wait, (BC = 3) and the height is (2sqrt{3}), so triangle (BCN) is a right triangle? Wait, no, because (BC) is the base, and (CN) is the height. So, actually, triangle (BCN) is a right triangle with legs (BC = 3) and (CN = 2sqrt{3}). Therefore, the length of (BN) can be found using the Pythagorean theorem:[BN = sqrt{BC^2 + CN^2} = sqrt{3^2 + (2sqrt{3})^2} = sqrt{9 + 12} = sqrt{21}]So, (BN = sqrt{21}). Now, going back to the Power of a Point theorem:[BN cdot NM = 10][sqrt{21} cdot NM = 10][NM = frac{10}{sqrt{21}} = frac{10sqrt{21}}{21}]Therefore, the length of (BM) is (BN + NM = sqrt{21} + frac{10sqrt{21}}{21} = frac{21sqrt{21} + 10sqrt{21}}{21} = frac{31sqrt{21}}{21}).Now, I need to find the area of triangle (BOM). To do this, I might need to find the lengths of (BO), (OM), and (BM), or perhaps find the height from (O) to (BM).Since (O) is the center of the circle, (BO) and (OM) are radii of the circle. So, if I can find the radius (R) of the circle, then (BO = OM = R).To find the radius (R), I can use the fact that the trapezium is cyclic. For a cyclic trapezium, the radius can be found using the formula:[R = frac{sqrt{AB^2 + BC^2 + CD^2 + DA^2}}{4 sin theta}]Wait, I'm not sure about that formula. Maybe I should use another approach. Since I know the height and the lengths of the bases, perhaps I can find the diameter or use some other properties.Alternatively, since (ABCD) is cyclic, the perpendicular bisectors of the sides meet at the center (O). Maybe I can find coordinates for the points and then find (O).Let me try coordinate geometry. Let's place the trapezium on a coordinate system such that the center (O) is at the origin. Let me assign coordinates to the points.Let me set point (D) at ((0, 0)). Since (AD = 7) and (AD) is parallel to the x-axis, point (A) will be at ((7, 0)). Point (C) is somewhere above (D), and point (B) is somewhere above (A).Given that the height is (2sqrt{3}), point (C) will be at ((2, 2sqrt{3})) because (DP = 2). Similarly, point (B) will be at ((5, 2sqrt{3})) because (AN = 5).Wait, let me confirm that. If (AD = 7), and (DP = 2), then (P) is at ((2, 0)). So, point (C) is at ((2, 2sqrt{3})). Similarly, since (BC = 3), and (BC) is parallel to (AD), point (B) should be at ((5, 2sqrt{3})), because from (C) at ((2, 2sqrt{3})), moving 3 units to the right along the x-axis gives (B) at ((5, 2sqrt{3})).So, coordinates:- (D = (0, 0))- (A = (7, 0))- (C = (2, 2sqrt{3}))- (B = (5, 2sqrt{3}))Now, I need to find the center (O) of the circle passing through these four points. Since (O) is the circumcenter, it is equidistant from all four points.Let me write the general equation of a circle:[x^2 + y^2 + 2gx + 2fy + c = 0]Since the circle passes through points (A), (B), (C), and (D), I can plug in their coordinates to find (g), (f), and (c).First, plugging in (D = (0, 0)):[0 + 0 + 0 + 0 + c = 0 implies c = 0]So, the equation simplifies to:[x^2 + y^2 + 2gx + 2fy = 0]Now, plugging in (A = (7, 0)):[7^2 + 0^2 + 2g(7) + 2f(0) = 0][49 + 14g = 0][14g = -49][g = -frac{49}{14} = -frac{7}{2}]Next, plugging in (C = (2, 2sqrt{3})):[2^2 + (2sqrt{3})^2 + 2g(2) + 2f(2sqrt{3}) = 0][4 + 12 + 4g + 4fsqrt{3} = 0][16 + 4g + 4fsqrt{3} = 0]We already know (g = -frac{7}{2}), so:[16 + 4(-frac{7}{2}) + 4fsqrt{3} = 0][16 - 14 + 4fsqrt{3} = 0][2 + 4fsqrt{3} = 0][4fsqrt{3} = -2][f = -frac{2}{4sqrt{3}} = -frac{1}{2sqrt{3}} = -frac{sqrt{3}}{6}]So, the equation of the circle is:[x^2 + y^2 - 7x - frac{sqrt{3}}{3}y = 0]The center (O) is at ((-g, -f)), which is:[O = left( frac{7}{2}, frac{sqrt{3}}{6} right )]Now, I need to find the coordinates of point (M). Since (BM) is a chord intersecting (AD) at (N), and (N) is at ((5, 0)) because (AN = 5) and (AD) is from ((0, 0)) to ((7, 0)).Wait, actually, (N) is on (AD), which is from (A(7, 0)) to (D(0, 0)). Since (AN = 5), (N) is 5 units from (A), so its coordinate is ((7 - 5, 0) = (2, 0)). Wait, that conflicts with point (C) being at ((2, 2sqrt{3})). Hmm, maybe I made a mistake earlier.Wait, no, (AD) is from (A(7, 0)) to (D(0, 0)). So, (AN = 5) means starting from (A(7, 0)) and moving 5 units towards (D(0, 0)), which would be at (7 - 5 = 2) on the x-axis. So, (N) is at ((2, 0)).But point (C) is also at ((2, 2sqrt{3})). So, chord (BM) passes through (N(2, 0)) and (B(5, 2sqrt{3})). So, I can find the equation of line (BM) and then find its other intersection point (M) with the circle.First, let's find the equation of line (BM). Points (B(5, 2sqrt{3})) and (N(2, 0)).The slope (m) of (BM) is:[m = frac{0 - 2sqrt{3}}{2 - 5} = frac{-2sqrt{3}}{-3} = frac{2sqrt{3}}{3}]So, the equation of line (BM) is:[y - 0 = frac{2sqrt{3}}{3}(x - 2)][y = frac{2sqrt{3}}{3}x - frac{4sqrt{3}}{3}]Now, to find point (M), we need to find the other intersection of this line with the circle. We already know that (B(5, 2sqrt{3})) is one intersection point, so solving the system will give us (M).Substitute (y = frac{2sqrt{3}}{3}x - frac{4sqrt{3}}{3}) into the circle equation:[x^2 + y^2 - 7x - frac{sqrt{3}}{3}y = 0]Plugging in (y):[x^2 + left( frac{2sqrt{3}}{3}x - frac{4sqrt{3}}{3} right)^2 - 7x - frac{sqrt{3}}{3}left( frac{2sqrt{3}}{3}x - frac{4sqrt{3}}{3} right) = 0]Let me compute each term step by step.First, compute (y^2):[left( frac{2sqrt{3}}{3}x - frac{4sqrt{3}}{3} right)^2 = left( frac{2sqrt{3}}{3}x right)^2 - 2 cdot frac{2sqrt{3}}{3}x cdot frac{4sqrt{3}}{3} + left( frac{4sqrt{3}}{3} right)^2][= frac{12}{9}x^2 - 2 cdot frac{8 cdot 3}{9}x + frac{48}{9}][= frac{4}{3}x^2 - frac{16}{3}x + frac{16}{3}]Next, compute (- frac{sqrt{3}}{3}y):[- frac{sqrt{3}}{3} left( frac{2sqrt{3}}{3}x - frac{4sqrt{3}}{3} right ) = - frac{sqrt{3}}{3} cdot frac{2sqrt{3}}{3}x + frac{sqrt{3}}{3} cdot frac{4sqrt{3}}{3}][= - frac{6}{9}x + frac{12}{9}][= - frac{2}{3}x + frac{4}{3}]Now, substitute back into the circle equation:[x^2 + left( frac{4}{3}x^2 - frac{16}{3}x + frac{16}{3} right ) - 7x + left( - frac{2}{3}x + frac{4}{3} right ) = 0]Combine like terms:First, expand all terms:[x^2 + frac{4}{3}x^2 - frac{16}{3}x + frac{16}{3} - 7x - frac{2}{3}x + frac{4}{3} = 0]Combine (x^2) terms:[x^2 + frac{4}{3}x^2 = frac{7}{3}x^2]Combine (x) terms:[- frac{16}{3}x - 7x - frac{2}{3}x = - frac{16}{3}x - frac{21}{3}x - frac{2}{3}x = - frac{39}{3}x = -13x]Combine constant terms:[frac{16}{3} + frac{4}{3} = frac{20}{3}]So, the equation becomes:[frac{7}{3}x^2 - 13x + frac{20}{3} = 0]Multiply through by 3 to eliminate denominators:[7x^2 - 39x + 20 = 0]Now, solve this quadratic equation for (x):Using the quadratic formula:[x = frac{39 pm sqrt{(-39)^2 - 4 cdot 7 cdot 20}}{2 cdot 7}][x = frac{39 pm sqrt{1521 - 560}}{14}][x = frac{39 pm sqrt{961}}{14}][x = frac{39 pm 31}{14}]So, two solutions:1. (x = frac{39 + 31}{14} = frac{70}{14} = 5)2. (x = frac{39 - 31}{14} = frac{8}{14} = frac{4}{7})We already know that (x = 5) corresponds to point (B(5, 2sqrt{3})), so the other intersection point (M) must be at (x = frac{4}{7}).Now, find the corresponding (y) coordinate for (M):[y = frac{2sqrt{3}}{3} cdot frac{4}{7} - frac{4sqrt{3}}{3} = frac{8sqrt{3}}{21} - frac{28sqrt{3}}{21} = -frac{20sqrt{3}}{21}]So, point (M) is at (left( frac{4}{7}, -frac{20sqrt{3}}{21} right )).Now, I have coordinates for points (B(5, 2sqrt{3})), (Oleft( frac{7}{2}, frac{sqrt{3}}{6} right )), and (Mleft( frac{4}{7}, -frac{20sqrt{3}}{21} right )).To find the area of triangle (BOM), I can use the shoelace formula.First, list the coordinates:- (B = (5, 2sqrt{3}))- (O = left( frac{7}{2}, frac{sqrt{3}}{6} right ))- (M = left( frac{4}{7}, -frac{20sqrt{3}}{21} right ))Let me write them in order:1. (B = (5, 2sqrt{3}))2. (O = left( frac{7}{2}, frac{sqrt{3}}{6} right ))3. (M = left( frac{4}{7}, -frac{20sqrt{3}}{21} right ))Shoelace formula:[text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|]Plugging in the coordinates:[text{Area} = frac{1}{2} |5 left( frac{sqrt{3}}{6} - left( -frac{20sqrt{3}}{21} right ) right ) + frac{7}{2} left( -frac{20sqrt{3}}{21} - 2sqrt{3} right ) + frac{4}{7} left( 2sqrt{3} - frac{sqrt{3}}{6} right )|]Let's compute each term step by step.First term: (5 left( frac{sqrt{3}}{6} + frac{20sqrt{3}}{21} right ))Find a common denominator for the fractions inside the parentheses, which is 42:[frac{sqrt{3}}{6} = frac{7sqrt{3}}{42}, quad frac{20sqrt{3}}{21} = frac{40sqrt{3}}{42}][frac{7sqrt{3}}{42} + frac{40sqrt{3}}{42} = frac{47sqrt{3}}{42}][5 times frac{47sqrt{3}}{42} = frac{235sqrt{3}}{42}]Second term: (frac{7}{2} left( -frac{20sqrt{3}}{21} - 2sqrt{3} right ))Convert (2sqrt{3}) to a fraction with denominator 21:[2sqrt{3} = frac{42sqrt{3}}{21}][-frac{20sqrt{3}}{21} - frac{42sqrt{3}}{21} = -frac{62sqrt{3}}{21}][frac{7}{2} times -frac{62sqrt{3}}{21} = -frac{434sqrt{3}}{42} = -frac{217sqrt{3}}{21}]Third term: (frac{4}{7} left( 2sqrt{3} - frac{sqrt{3}}{6} right ))Convert (2sqrt{3}) to a fraction with denominator 6:[2sqrt{3} = frac{12sqrt{3}}{6}][frac{12sqrt{3}}{6} - frac{sqrt{3}}{6} = frac{11sqrt{3}}{6}][frac{4}{7} times frac{11sqrt{3}}{6} = frac{44sqrt{3}}{42} = frac{22sqrt{3}}{21}]Now, combine all three terms:[frac{235sqrt{3}}{42} - frac{217sqrt{3}}{21} + frac{22sqrt{3}}{21}]Convert all terms to have a common denominator of 42:[frac{235sqrt{3}}{42} - frac{434sqrt{3}}{42} + frac{44sqrt{3}}{42}][= frac{235sqrt{3} - 434sqrt{3} + 44sqrt{3}}{42}][= frac{(235 - 434 + 44)sqrt{3}}{42}][= frac{(-155)sqrt{3}}{42}][= -frac{155sqrt{3}}{42}]Taking the absolute value:[left| -frac{155sqrt{3}}{42} right| = frac{155sqrt{3}}{42}]Finally, multiply by (frac{1}{2}):[text{Area} = frac{1}{2} times frac{155sqrt{3}}{42} = frac{155sqrt{3}}{84}]So, the area of triangle (BOM) is (frac{155sqrt{3}}{84}).**Final Answer**The area of triangle (BOM) is boxed{dfrac{155 sqrt{3}}{84}}.

💡Okay, so I have this problem about a function ( f(x) ) that satisfies the condition ( f(x) + f(y) = 2 + f(x+y) ) for any real numbers ( x ) and ( y ). Also, it's given that ( f(3) = 5 ). There are two parts to the problem. First, I need to find the value of ( f(1) + f(-1) ). Hmm, functional equations can sometimes be tricky, but maybe I can find a pattern or figure out the form of ( f(x) ). Let me think about what kind of function satisfies ( f(x) + f(y) = 2 + f(x+y) ). It looks a bit like Cauchy's functional equation, which is ( f(x) + f(y) = f(x+y) ), but here there's an extra constant term, 2. Maybe I can rewrite the equation to make it look more familiar. Let me subtract 2 from both sides: ( f(x) + f(y) - 2 = f(x+y) ). Hmm, so if I define a new function ( g(x) = f(x) - c ), where ( c ) is a constant, maybe I can turn this into Cauchy's equation. Let's try that. Let ( g(x) = f(x) - c ). Then, substituting into the equation: ( (g(x) + c) + (g(y) + c) - 2 = g(x+y) + c ). Simplifying, that's ( g(x) + g(y) + 2c - 2 = g(x+y) + c ). So, ( g(x) + g(y) + c - 2 = g(x+y) ). For this to be Cauchy's equation, ( c - 2 ) must be zero, so ( c = 2 ). Therefore, ( g(x) = f(x) - 2 ) satisfies Cauchy's functional equation: ( g(x) + g(y) = g(x+y) ).Assuming ( g ) is linear (since Cauchy's equation solutions are linear if we assume continuity or monotonicity, which we might have in part 2), ( g(x) = kx ) for some constant ( k ). Therefore, ( f(x) = kx + 2 ). But wait, we also know that ( f(3) = 5 ). Plugging that into our expression: ( f(3) = 3k + 2 = 5 ). Solving for ( k ), we get ( 3k = 3 ), so ( k = 1 ). Therefore, ( f(x) = x + 2 ). Let me verify this function satisfies the original condition. ( f(x) + f(y) = (x + 2) + (y + 2) = x + y + 4 ). On the other hand, ( 2 + f(x+y) = 2 + (x + y + 2) = x + y + 4 ). So yes, it works. Therefore, ( f(x) = x + 2 ). So, ( f(1) = 1 + 2 = 3 ) and ( f(-1) = -1 + 2 = 1 ). Therefore, ( f(1) + f(-1) = 3 + 1 = 4 ). Wait, that seems straightforward. Maybe I should check if there's another way to approach it without assuming the form of the function. Let's see. Starting from the given condition: ( f(x) + f(y) = 2 + f(x+y) ). Let me set ( x = y = 0 ). Then, ( f(0) + f(0) = 2 + f(0) ), which simplifies to ( 2f(0) = 2 + f(0) ), so ( f(0) = 2 ). Next, let me set ( y = -x ). Then, ( f(x) + f(-x) = 2 + f(0) = 2 + 2 = 4 ). So, ( f(x) + f(-x) = 4 ). That's interesting. So, for any ( x ), ( f(-x) = 4 - f(x) ). Given that ( f(3) = 5 ), then ( f(-3) = 4 - 5 = -1 ). But wait, if ( f(x) = x + 2 ), then ( f(-3) = -3 + 2 = -1 ), which matches. So, using this property, ( f(-1) = 4 - f(1) ). Since ( f(1) = 3 ), ( f(-1) = 1 ). Therefore, ( f(1) + f(-1) = 4 ). Okay, that's consistent with the previous method. So, part (1) is 4.Moving on to part (2). It says that ( f(x) ) is an increasing function on ( mathbb{R} ). We need to prove that there exists a unique real number ( t ) such that for any ( x in (0, 1) ), ( f(x^2 + 2t^2x) < 3 ) holds.First, since ( f(x) = x + 2 ), let's substitute that into the inequality: ( x^2 + 2t^2x + 2 < 3 ). Simplifying, ( x^2 + 2t^2x < 1 ). So, for all ( x in (0,1) ), ( x^2 + 2t^2x < 1 ).We need to find a unique ( t ) such that this inequality holds for all ( x ) in (0,1). Since ( f ) is increasing, the inequality ( f(x^2 + 2t^2x) < 3 ) is equivalent to ( x^2 + 2t^2x < f^{-1}(3) ). But since ( f(x) = x + 2 ), ( f^{-1}(y) = y - 2 ). So, ( f^{-1}(3) = 1 ). Therefore, the inequality reduces to ( x^2 + 2t^2x < 1 ) for all ( x in (0,1) ).So, we need ( x^2 + 2t^2x < 1 ) for all ( x in (0,1) ). Let's analyze the function ( g(x) = x^2 + 2t^2x ) on the interval ( (0,1) ).First, note that ( g(x) ) is a quadratic function in ( x ). Since the coefficient of ( x^2 ) is positive, it's a parabola opening upwards. Therefore, its minimum is at the vertex, but since we're looking at ( x in (0,1) ), the maximum will occur either at the endpoints or at critical points within the interval.Let's compute the derivative of ( g(x) ) with respect to ( x ): ( g'(x) = 2x + 2t^2 ). Setting this equal to zero to find critical points: ( 2x + 2t^2 = 0 ) => ( x = -t^2 ). But ( x ) is in ( (0,1) ), so the critical point ( x = -t^2 ) is negative, which is outside our interval. Therefore, the function ( g(x) ) is increasing on ( (0,1) ) because the derivative ( g'(x) = 2x + 2t^2 ) is positive for all ( x > 0 ) (since ( t^2 geq 0 ) and ( x > 0 )).Therefore, the maximum of ( g(x) ) on ( (0,1) ) occurs at ( x = 1 ). So, to ensure ( g(x) < 1 ) for all ( x in (0,1) ), it's sufficient to ensure that ( g(1) leq 1 ). Calculating ( g(1) = 1^2 + 2t^2(1) = 1 + 2t^2 ). We need ( 1 + 2t^2 leq 1 ), which simplifies to ( 2t^2 leq 0 ). Since ( t^2 geq 0 ), this implies ( t^2 = 0 ), so ( t = 0 ).Therefore, the only real number ( t ) that satisfies the condition is ( t = 0 ). To confirm uniqueness, suppose there's another ( t neq 0 ). Then ( t^2 > 0 ), so ( g(1) = 1 + 2t^2 > 1 ), which violates the condition ( g(x) < 1 ) for all ( x in (0,1) ). Hence, ( t = 0 ) is the unique solution.So, putting it all together, part (1) is 4, and part (2) is ( t = 0 ).**Final Answer**(1) boxed{4}(2) boxed{0}

💡Okay, so I need to find the least positive integer that's divisible by each of the first ten positive integers. Hmm, that sounds like I need to find the least common multiple, or LCM, of the numbers from 1 to 10. I remember that the LCM of a set of numbers is the smallest number that each of them can divide into without leaving a remainder. Alright, let me start by recalling how to find the LCM of two numbers. I think it involves prime factorization. For example, to find the LCM of 4 and 6, I would break them down into their prime factors: 4 is 2 squared, and 6 is 2 times 3. Then, the LCM would be the product of the highest powers of all primes that appear, so that would be 2 squared times 3, which is 12. So, applying that to the numbers 1 through 10, I guess I need to do the same thing but with all ten numbers. Let me list them out: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Now, I'll write down their prime factorizations:- 1 is just 1.- 2 is a prime number, so it's 2.- 3 is also a prime number, so it's 3.- 4 is 2 squared.- 5 is a prime number, so it's 5.- 6 is 2 times 3.- 7 is a prime number, so it's 7.- 8 is 2 cubed.- 9 is 3 squared.- 10 is 2 times 5.Now, I need to identify the highest powers of each prime number that appear in these factorizations. Let's see:- For the prime number 2, the highest power is 2 cubed (from 8).- For the prime number 3, the highest power is 3 squared (from 9).- For the prime number 5, the highest power is 5 to the first power (from 5 and 10).- For the prime number 7, the highest power is 7 to the first power (from 7).So, to find the LCM, I should multiply these highest powers together. That would be 2 cubed times 3 squared times 5 times 7. Let me calculate that step by step:First, 2 cubed is 8. Then, 3 squared is 9. So, 8 times 9 is 72. Next, 72 times 5 is 360. Finally, 360 times 7 is 2520.Wait, let me double-check that multiplication to make sure I didn't make a mistake. 8 times 9 is indeed 72. 72 times 5 is 360. 360 times 7... let me do that again. 360 times 7 is 2520. Yeah, that seems right.Just to be thorough, let me verify that 2520 is divisible by each of the numbers from 1 to 10:- 2520 divided by 1 is 2520, which is an integer.- 2520 divided by 2 is 1260, which is an integer.- 2520 divided by 3 is 840, which is an integer.- 2520 divided by 4 is 630, which is an integer.- 2520 divided by 5 is 504, which is an integer.- 2520 divided by 6 is 420, which is an integer.- 2520 divided by 7 is 360, which is an integer.- 2520 divided by 8 is 315, which is an integer.- 2520 divided by 9 is 280, which is an integer.- 2520 divided by 10 is 252, which is an integer.Okay, so it checks out. 2520 is divisible by each of the numbers from 1 to 10. Is there a smaller number that also satisfies this condition? Let me think. If I try to use a smaller number, say 1260, which is half of 2520, let's see if that works. 1260 divided by 8 is 157.5, which is not an integer. So, 1260 doesn't work because it's not divisible by 8. What about 1680? 1680 divided by 9 is approximately 186.666..., which isn't an integer either. So, 1680 doesn't work either. It seems like 2520 is indeed the smallest number that all the numbers from 1 to 10 can divide into without leaving a remainder. Therefore, I'm confident that 2520 is the least positive integer divisible by each of the first ten positive integers.

💡First, I need to find the value of (cos(2013pi)).I know that the cosine function has a period of (2pi), which means that (cos(theta) = cos(theta + 2pi n)) for any integer (n).To simplify (2013pi), I can express it as (2013pi = 1006 times 2pi + pi).Using the periodic property of the cosine function:[cos(2013pi) = cos(1006 times 2pi + pi) = cos(pi)]I recall that (cos(pi) = -1).Therefore, the value of (cos(2013pi)) is (-1).