💡First, determine the passing marks by calculating 45% of the maximum marks of 500.Passing Marks = 45% of 500 = (45/100) * 500 = 225 marksThe candidate fails by 45 marks, meaning they scored 45 marks less than the passing marks.Marks Obtained = Passing Marks - 45 = 225 - 45 = 180 marksTherefore, the candidate obtained 180 marks.

💡Okay, so I have this problem where I need to find the range of values for the real number ( m ) such that the union of sets ( A ) and ( B ) is a subset of set ( C ). Let me try to break this down step by step.First, let's understand the given sets:- Set ( A ) is defined as ( {x mid 0 < x < 2} ). So, this is all the real numbers between 0 and 2, not including 0 and 2.- Set ( B ) is defined as ( {x mid -1 < x < 1} ). This is all the real numbers between -1 and 1, not including -1 and 1.- Set ( C ) is defined as ( {x mid mx + 1 > 0} ). This is a bit more complex because it depends on the value of ( m ).The goal is to find all real numbers ( m ) such that ( A cup B subseteq C ). Let me start by finding ( A cup B ). Since ( A ) is from 0 to 2 and ( B ) is from -1 to 1, the union of these two sets would cover from the smallest number in ( B ) to the largest number in ( A ). So, ( A cup B ) should be ( {x mid -1 < x < 2} ). That makes sense because ( B ) covers from -1 to 1, and ( A ) extends that to 2.Now, I need to ensure that every element in ( A cup B ) is also in ( C ). So, for every ( x ) in ( (-1, 2) ), the inequality ( mx + 1 > 0 ) must hold true.Let me analyze set ( C ) more carefully. The inequality ( mx + 1 > 0 ) can be rewritten as ( mx > -1 ). Depending on the value of ( m ), this inequality can change direction or not. I think I should consider different cases for ( m ):1. **Case 1: ( m > 0 )** If ( m ) is positive, then dividing both sides of the inequality ( mx > -1 ) by ( m ) doesn't change the direction of the inequality. So, ( x > -frac{1}{m} ). Therefore, set ( C ) becomes ( {x mid x > -frac{1}{m}} ). For ( A cup B subseteq C ), the entire interval ( (-1, 2) ) must be contained within ( (-frac{1}{m}, infty) ). That means the lower bound of ( C ), which is ( -frac{1}{m} ), must be less than or equal to the lower bound of ( A cup B ), which is -1. So, ( -frac{1}{m} leq -1 ). Let me solve this inequality for ( m ): Multiply both sides by ( m ) (since ( m > 0 ), the inequality direction remains the same): ( -1 leq -m ) Multiply both sides by -1 (which reverses the inequality): ( 1 geq m ) So, ( m leq 1 ). But since we are in the case where ( m > 0 ), combining these, we get ( 0 < m leq 1 ).2. **Case 2: ( m = 0 )** If ( m = 0 ), the inequality ( mx + 1 > 0 ) simplifies to ( 0 + 1 > 0 ), which is always true. Therefore, set ( C ) becomes all real numbers, ( mathbb{R} ). Since ( A cup B ) is a subset of ( mathbb{R} ), this condition is satisfied. So, ( m = 0 ) is acceptable.3. **Case 3: ( m < 0 )** If ( m ) is negative, then dividing both sides of the inequality ( mx > -1 ) by ( m ) reverses the inequality direction. So, ( x < -frac{1}{m} ). Therefore, set ( C ) becomes ( {x mid x < -frac{1}{m}} ). For ( A cup B subseteq C ), the entire interval ( (-1, 2) ) must be contained within ( (-infty, -frac{1}{m}) ). That means the upper bound of ( A cup B ), which is 2, must be less than or equal to the upper bound of ( C ), which is ( -frac{1}{m} ). So, ( 2 leq -frac{1}{m} ). Let me solve this inequality for ( m ): Multiply both sides by ( m ). But since ( m < 0 ), the inequality direction reverses: ( 2m geq -1 ) Divide both sides by 2: ( m geq -frac{1}{2} ) But since we are in the case where ( m < 0 ), combining these, we get ( -frac{1}{2} leq m < 0 ).Now, let me summarize the results from all three cases:- From Case 1: ( 0 < m leq 1 )- From Case 2: ( m = 0 )- From Case 3: ( -frac{1}{2} leq m < 0 )Combining these intervals, we get ( -frac{1}{2} leq m leq 1 ).Wait, let me double-check if this makes sense.For ( m = 1 ), set ( C ) is ( x > -1 ). Since ( A cup B ) is ( (-1, 2) ), which is entirely within ( (-1, infty) ), so that works.For ( m = -frac{1}{2} ), set ( C ) is ( x < 2 ). Since ( A cup B ) is ( (-1, 2) ), which is entirely within ( (-infty, 2) ), so that also works.What about ( m = 0 )? As we saw, ( C ) becomes all real numbers, so it definitely contains ( A cup B ).What if ( m ) is greater than 1? Let's say ( m = 2 ). Then ( C ) is ( x > -frac{1}{2} ). But ( A cup B ) includes numbers less than -1/2, like -0.75, which are not in ( C ). So, ( m > 1 ) doesn't work.What if ( m ) is less than -1/2? Let's say ( m = -1 ). Then ( C ) is ( x < 1 ). But ( A cup B ) includes numbers up to 2, which are not in ( C ). So, ( m < -1/2 ) doesn't work either.Therefore, the range of ( m ) is indeed from -1/2 to 1, inclusive.I think that's thorough. I considered all possible cases for ( m ), checked the boundaries, and verified with specific examples. So, I'm confident that the range is correct.**Final Answer**The range of values for ( m ) is boxed{left[ -frac{1}{2}, 1 right]}.

💡To compute sin 300^circ, I start by locating the angle on the unit circle. 300^circ is in the fourth quadrant, which means the sine value will be negative.Next, I determine the reference angle by subtracting 300^circ from 360^circ, resulting in 60^circ. I know that sin 60^circ = frac{sqrt{3}}{2}.Since sine is negative in the fourth quadrant, I apply the negative sign to the reference angle's sine value. This gives me sin 300^circ = -frac{sqrt{3}}{2}.Finally, I confirm that this result aligns with the properties of the unit circle and the behavior of sine in the fourth quadrant.

💡Alright, so I have this inequality to prove in an acute triangle ABC. The inequality is:sum sqrt{frac{a^{2}}{b^{2}+c^{2}-a^{2}}} geqslant 3 sqrt{frac{R}{2r}}.Hmm, okay. Let me try to understand what each part means. First, the triangle is acute, so all angles are less than 90 degrees. That might be important because in acute triangles, certain relationships between sides and angles hold that might not hold in obtuse triangles.The left side of the inequality is a sum of square roots. Each term is the square root of (a squared divided by (b squared plus c squared minus a squared)). That denominator looks familiar—it's similar to the expression in the Law of Cosines. Specifically, in the Law of Cosines, we have:a^{2} = b^{2} + c^{2} - 2bc cos A.So, if I rearrange that, we get:b^{2} + c^{2} - a^{2} = 2bc cos A.Ah, that's useful! So, the denominator in each square root term is actually equal to (2bc cos A). So, the term inside the square root becomes:frac{a^{2}}{2bc cos A}.So, the entire sum on the left side can be rewritten as:sum sqrt{frac{a^{2}}{2bc cos A}}.Simplifying that, we can take the square root of (a^{2}) as (a), so:sum frac{a}{sqrt{2bc cos A}}.Okay, so now the left side is expressed in terms of the sides and the cosine of the angles. Maybe this will be helpful later.Now, looking at the right side of the inequality: (3 sqrt{frac{R}{2r}}). Here, R is the circumradius and r is the inradius of the triangle. I remember that in any triangle, there are relationships between R, r, and other elements like the sides and the area.I also recall Euler's inequality, which states that in any triangle, (R geq 2r), with equality if and only if the triangle is equilateral. That might come into play here, especially since we're dealing with an acute triangle, which could potentially be equilateral.So, perhaps I can relate the left side to R and r somehow. Let me think about known inequalities or identities that connect these elements.I also remember that in a triangle, the area can be expressed in terms of the inradius and the semiperimeter:text{Area} = r cdot s,where (s = frac{a + b + c}{2}) is the semiperimeter.Additionally, the area can be expressed in terms of the circumradius:text{Area} = frac{abc}{4R}.So, combining these, we have:r cdot s = frac{abc}{4R},which can be rearranged to:frac{R}{r} = frac{abc}{4r s}.Hmm, not sure if that's directly helpful yet.Let me go back to the left side of the inequality. I have:sum frac{a}{sqrt{2bc cos A}}.Maybe I can manipulate this expression further. Let's consider each term individually.For each term, (frac{a}{sqrt{2bc cos A}}), perhaps I can express this in terms of trigonometric functions or other triangle properties.Wait, in the Law of Sines, we have:frac{a}{sin A} = 2R.So, (a = 2R sin A). Maybe substituting this into the term would help.Let me try that. Substitute (a = 2R sin A) into the term:frac{2R sin A}{sqrt{2bc cos A}}.Simplify the denominator:sqrt{2bc cos A} = sqrt{2bc} cdot sqrt{cos A}.So, the term becomes:frac{2R sin A}{sqrt{2bc} cdot sqrt{cos A}} = frac{2R sin A}{sqrt{2bc cos A}}.Hmm, not sure if that's better. Maybe I can express bc in terms of other variables.Wait, from the Law of Cosines, we have:cos A = frac{b^{2} + c^{2} - a^{2}}{2bc}.So, (bc = frac{b^{2} + c^{2} - a^{2}}{2 cos A}).Substituting back into the denominator:sqrt{2 cdot frac{b^{2} + c^{2} - a^{2}}{2 cos A} cdot cos A} = sqrt{(b^{2} + c^{2} - a^{2})}.Wait, that's interesting. So, the denominator simplifies back to (sqrt{b^{2} + c^{2} - a^{2}}), which is where we started. Hmm, maybe that substitution didn't help much.Let me think differently. Maybe instead of trying to manipulate the left side directly, I can look for known inequalities that relate sums of square roots to other triangle quantities.I recall that in inequalities involving triangle sides and angles, sometimes the Cauchy-Schwarz inequality is useful. Maybe I can apply that here.The Cauchy-Schwarz inequality states that for any real numbers (x_i) and (y_i):left( sum x_i y_i right)^2 leq left( sum x_i^2 right) left( sum y_i^2 right).But I'm not sure how to apply that here directly. Alternatively, maybe I can use the AM-GM inequality, which states that the arithmetic mean is greater than or equal to the geometric mean.But again, I'm not sure how to apply it here. Let me think about the structure of the inequality.The left side is a sum of square roots, each involving a ratio of a squared side to a combination of the other two sides. The right side is a multiple of the square root of the ratio of the circumradius to the inradius.Perhaps I can find a relationship between these square roots and R and r.Wait, I also remember that in a triangle, the distances from the incenter to the sides are related to r, and the distances from the circumradius are related to R.But I'm not sure how to connect that to the given expression.Alternatively, maybe I can express the terms in the sum in terms of trigonometric identities.Earlier, I saw that (b^{2} + c^{2} - a^{2} = 2bc cos A). So, the term inside the square root is:frac{a^{2}}{2bc cos A}.So, the entire sum is:sum sqrt{frac{a^{2}}{2bc cos A}} = sum frac{a}{sqrt{2bc cos A}}.Maybe I can express this in terms of sine and cosine.From the Law of Sines, (a = 2R sin A), so:frac{a}{sqrt{2bc cos A}} = frac{2R sin A}{sqrt{2bc cos A}}.Let me see if I can express bc in terms of R and angles.From the Law of Sines, (b = 2R sin B) and (c = 2R sin C). So, (bc = 4R^2 sin B sin C).Substituting back:frac{2R sin A}{sqrt{2 cdot 4R^2 sin B sin C cdot cos A}} = frac{2R sin A}{sqrt{8R^2 sin B sin C cos A}}.Simplify the denominator:sqrt{8R^2 sin B sin C cos A} = R sqrt{8 sin B sin C cos A}.So, the term becomes:frac{2R sin A}{R sqrt{8 sin B sin C cos A}} = frac{2 sin A}{sqrt{8 sin B sin C cos A}} = frac{sin A}{sqrt{2 sin B sin C cos A}}.Hmm, that's a bit simpler. So, each term in the sum is:frac{sin A}{sqrt{2 sin B sin C cos A}}.Now, let's consider the entire sum:sum frac{sin A}{sqrt{2 sin B sin C cos A}}.This still looks complicated, but maybe I can find a way to relate this to R and r.I know that in a triangle, (A + B + C = pi), so maybe there's a way to use that.Alternatively, perhaps I can use the identity that in any triangle:cos A = frac{b^2 + c^2 - a^2}{2bc},but I already used that earlier.Wait, maybe I can express (sin B sin C) in terms of other trigonometric functions.I recall that:sin B sin C = frac{cos(A - B) - cos(A + B)}{2}.But since (A + B + C = pi), (A + B = pi - C), so:sin B sin C = frac{cos(A - B) - cos(pi - C)}{2} = frac{cos(A - B) + cos C}{2}.Hmm, not sure if that helps.Alternatively, maybe I can use the identity:sin B sin C = frac{cos(A - B) - cos(A + B)}{2} = frac{cos(A - B) + cos C}{2}.But again, not directly helpful.Wait, perhaps I can use the fact that in a triangle, (A + B + C = pi), so (B + C = pi - A). Therefore, (cos(B + C) = cos(pi - A) = -cos A).But I'm not sure how that connects.Alternatively, maybe I can express (sin B sin C) in terms of (cos(B - C)) and (cos(B + C)):sin B sin C = frac{cos(B - C) - cos(B + C)}{2} = frac{cos(B - C) + cos A}{2}.So, substituting back into the denominator:sqrt{2 sin B sin C cos A} = sqrt{2 cdot frac{cos(B - C) + cos A}{2} cdot cos A} = sqrt{(cos(B - C) + cos A) cos A}.So, the term becomes:frac{sin A}{sqrt{(cos(B - C) + cos A) cos A}}.Hmm, this seems to be getting more complicated rather than simpler. Maybe this approach isn't the best.Let me think about another strategy. Perhaps instead of trying to manipulate the left side directly, I can look for known inequalities that relate sums of square roots to other triangle quantities.I remember that in some inequalities, especially those involving symmetric sums, it's useful to use homogenization or substitution to reduce the number of variables.Given that we're dealing with an acute triangle, all angles are less than 90 degrees, so all cosines are positive, which is good because it means the denominators in our square roots are positive.Wait, maybe I can use the fact that in an acute triangle, (b^2 + c^2 > a^2), which we already used earlier.Alternatively, perhaps I can consider substituting variables to simplify the expressions. For example, let me set:x = b^2 + c^2 - a^2,y = a^2 + c^2 - b^2,z = a^2 + b^2 - c^2.Since the triangle is acute, x, y, z are all positive.Then, the left side of the inequality becomes:sqrt{frac{a^2}{x}} + sqrt{frac{b^2}{y}} + sqrt{frac{c^2}{z}}.So, the inequality is:sqrt{frac{a^2}{x}} + sqrt{frac{b^2}{y}} + sqrt{frac{c^2}{z}} geq 3 sqrt{frac{R}{2r}}.Now, perhaps I can relate x, y, z to other triangle properties.From the Law of Cosines, we have:x = 2bc cos A,y = 2ac cos B,z = 2ab cos C.So, substituting back, we have:sqrt{frac{a^2}{2bc cos A}} + sqrt{frac{b^2}{2ac cos B}} + sqrt{frac{c^2}{2ab cos C}} geq 3 sqrt{frac{R}{2r}}.Hmm, this seems similar to what I had earlier. Maybe I can factor out the constants:frac{a}{sqrt{2bc cos A}} + frac{b}{sqrt{2ac cos B}} + frac{c}{sqrt{2ab cos C}} geq 3 sqrt{frac{R}{2r}}.Let me factor out (frac{1}{sqrt{2}}):frac{1}{sqrt{2}} left( frac{a}{sqrt{bc cos A}} + frac{b}{sqrt{ac cos B}} + frac{c}{sqrt{ab cos C}} right) geq 3 sqrt{frac{R}{2r}}.Multiplying both sides by (sqrt{2}):left( frac{a}{sqrt{bc cos A}} + frac{b}{sqrt{ac cos B}} + frac{c}{sqrt{ab cos C}} right) geq 3 sqrt{frac{2R}{r}}.Hmm, interesting. Now, I need to relate the left side to (sqrt{frac{2R}{r}}).I wonder if there's a way to express the left side in terms of R and r. Let me recall some relationships.I know that in a triangle:cos A = frac{b^2 + c^2 - a^2}{2bc},and from the Law of Sines:a = 2R sin A,b = 2R sin B,c = 2R sin C.So, substituting these into the expression:frac{a}{sqrt{bc cos A}} = frac{2R sin A}{sqrt{(2R sin B)(2R sin C) cdot frac{b^2 + c^2 - a^2}{2bc}}}.Wait, that seems messy. Let me try to simplify step by step.First, express bc:bc = (2R sin B)(2R sin C) = 4R^2 sin B sin C.And (cos A) is:cos A = frac{b^2 + c^2 - a^2}{2bc} = frac{(4R^2 sin^2 B) + (4R^2 sin^2 C) - (4R^2 sin^2 A)}{2 cdot 4R^2 sin B sin C}.Simplify numerator:4R^2 (sin^2 B + sin^2 C - sin^2 A).Denominator:8R^2 sin B sin C.So,cos A = frac{sin^2 B + sin^2 C - sin^2 A}{2 sin B sin C}.Therefore, the term inside the square root becomes:sqrt{bc cos A} = sqrt{4R^2 sin B sin C cdot frac{sin^2 B + sin^2 C - sin^2 A}{2 sin B sin C}} = sqrt{2R^2 (sin^2 B + sin^2 C - sin^2 A)}.So, the term:frac{a}{sqrt{bc cos A}} = frac{2R sin A}{sqrt{2R^2 (sin^2 B + sin^2 C - sin^2 A)}} = frac{2R sin A}{R sqrt{2 (sin^2 B + sin^2 C - sin^2 A)}} = frac{2 sin A}{sqrt{2 (sin^2 B + sin^2 C - sin^2 A)}}.Simplify:frac{2 sin A}{sqrt{2 (sin^2 B + sin^2 C - sin^2 A)}} = frac{sqrt{2} sin A}{sqrt{sin^2 B + sin^2 C - sin^2 A}}.Hmm, this seems complicated. Maybe there's a trigonometric identity that can help here.I know that in a triangle, (A + B + C = pi), so perhaps I can express (sin^2 B + sin^2 C) in terms of (sin^2 A).Wait, let's recall that:sin^2 B + sin^2 C = 1 - cos^2 B + 1 - cos^2 C = 2 - (cos^2 B + cos^2 C).But I don't see how that helps.Alternatively, perhaps I can use the identity:sin^2 B + sin^2 C - sin^2 A = 2 sin B sin C cos A.Wait, is that true? Let me check.Using the identity:sin^2 B + sin^2 C - sin^2 A = 2 sin B sin C cos A.Let me verify this. Starting from the right side:2 sin B sin C cos A.Using the identity (cos A = -cos(B + C)), since (A + B + C = pi).So,2 sin B sin C cos A = 2 sin B sin C (-cos(B + C)) = -2 sin B sin C cos(B + C).Expanding (cos(B + C)):cos(B + C) = cos B cos C - sin B sin C.So,-2 sin B sin C (cos B cos C - sin B sin C) = -2 sin B sin C cos B cos C + 2 sin^2 B sin^2 C.Hmm, not sure if that equals (sin^2 B + sin^2 C - sin^2 A).Wait, maybe another approach. Let's use the identity:sin^2 B + sin^2 C - sin^2 A = 2 sin B sin C cos A.Let me test with a specific triangle, say, an equilateral triangle where (A = B = C = 60^circ).Then,Left side: (sin^2 60 + sin^2 60 - sin^2 60 = sin^2 60 = frac{3}{4}).Right side: (2 sin 60 sin 60 cos 60 = 2 cdot frac{sqrt{3}}{2} cdot frac{sqrt{3}}{2} cdot frac{1}{2} = 2 cdot frac{3}{4} cdot frac{1}{2} = frac{3}{4}).Okay, so it holds for an equilateral triangle. Let me try another triangle, say, a right-angled triangle with angles (90^circ, 45^circ, 45^circ).But wait, the original triangle is acute, so all angles are less than 90 degrees. Let's take a triangle with angles (60^circ, 60^circ, 60^circ)—wait, that's equilateral again.Maybe a triangle with angles (50^circ, 60^circ, 70^circ).Compute left side:(sin^2 60 + sin^2 70 - sin^2 50).Compute each term:(sin 60 = frac{sqrt{3}}{2} approx 0.8660), so (sin^2 60 approx 0.75).(sin 70 approx 0.9397), so (sin^2 70 approx 0.8830).(sin 50 approx 0.7660), so (sin^2 50 approx 0.5868).So, left side: (0.75 + 0.8830 - 0.5868 approx 1.0462).Right side: (2 sin 60 sin 70 cos 50).Compute each term:(sin 60 approx 0.8660), (sin 70 approx 0.9397), (cos 50 approx 0.6428).So, right side: (2 cdot 0.8660 cdot 0.9397 cdot 0.6428 approx 2 cdot 0.8660 cdot 0.9397 cdot 0.6428).First, multiply 0.8660 and 0.9397: approx 0.8135.Then, multiply by 0.6428: approx 0.5225.Multiply by 2: approx 1.045.So, left side ≈ 1.0462, right side ≈ 1.045. Close enough, considering rounding errors. So, the identity seems to hold.Therefore, we can use:sin^2 B + sin^2 C - sin^2 A = 2 sin B sin C cos A.So, going back to the term:frac{sqrt{2} sin A}{sqrt{sin^2 B + sin^2 C - sin^2 A}} = frac{sqrt{2} sin A}{sqrt{2 sin B sin C cos A}} = frac{sqrt{2} sin A}{sqrt{2} sqrt{sin B sin C cos A}} = frac{sin A}{sqrt{sin B sin C cos A}}.So, each term simplifies to:frac{sin A}{sqrt{sin B sin C cos A}}.Therefore, the entire sum becomes:sum frac{sin A}{sqrt{sin B sin C cos A}}.Hmm, that's still a bit complicated, but maybe I can find a way to relate this to R and r.I know that in a triangle, the inradius r can be expressed as:r = frac{S}{s},where S is the area and s is the semiperimeter.And the area S can also be expressed as:S = frac{abc}{4R}.So, combining these:r = frac{abc}{4R s}.Therefore,frac{R}{r} = frac{4R^2 s}{abc}.Hmm, not sure if that helps directly.Wait, perhaps I can express (sin A), (sin B), (sin C) in terms of sides and R.From the Law of Sines:sin A = frac{a}{2R},sin B = frac{b}{2R},sin C = frac{c}{2R}.So, substituting back into the term:frac{sin A}{sqrt{sin B sin C cos A}} = frac{frac{a}{2R}}{sqrt{frac{b}{2R} cdot frac{c}{2R} cdot cos A}} = frac{frac{a}{2R}}{sqrt{frac{bc cos A}{4R^2}}} = frac{frac{a}{2R}}{frac{sqrt{bc cos A}}{2R}} = frac{a}{sqrt{bc cos A}}.Wait, that's just the original term we started with. So, this substitution didn't help.Maybe I need to think differently. Perhaps I can consider using the Cauchy-Schwarz inequality on the sum.The Cauchy-Schwarz inequality states that:left( sum frac{a}{sqrt{bc cos A}} right)^2 leq left( sum a^2 right) left( sum frac{1}{bc cos A} right).But I'm not sure if that's helpful because it would give an upper bound, but we need a lower bound.Alternatively, maybe I can use the AM-GM inequality on the terms.The AM-GM inequality states that:frac{x + y + z}{3} geq sqrt[3]{xyz}.But again, not sure how to apply it here.Wait, perhaps I can consider the function (f(x) = sqrt{x}), which is concave, so maybe I can apply Jensen's inequality.Jensen's inequality states that for a concave function:frac{f(x) + f(y) + f(z)}{3} leq fleft( frac{x + y + z}{3} right).But since we have a sum of square roots, which are concave, applying Jensen would give an upper bound, but we need a lower bound.Alternatively, maybe I can use the Cauchy-Schwarz inequality in a different form.Let me consider the sum:sum frac{a}{sqrt{bc cos A}}.Let me set (x_i = frac{a}{sqrt{bc cos A}}) and (y_i = 1). Then, by Cauchy-Schwarz:left( sum x_i y_i right)^2 leq left( sum x_i^2 right) left( sum y_i^2 right).So,left( sum frac{a}{sqrt{bc cos A}} right)^2 leq left( sum frac{a^2}{bc cos A} right) cdot 3.But this gives an upper bound, which is not what we need.Alternatively, maybe I can use the reverse Cauchy-Schwarz inequality, but I'm not sure if that applies here.Wait, perhaps I can use the fact that in an acute triangle, certain inequalities hold that can be leveraged here.I also recall that in an acute triangle, the sum of the angles is (pi), and all angles are less than (pi/2), so their cosines are positive.Wait, maybe I can use the following identity:In any triangle,sum frac{a}{b + c - a} = frac{2R}{r} - 1.I think I've seen this before. Let me verify.Yes, in some triangle identities, it's known that:sum frac{a}{b + c - a} = frac{2R}{r} - 1.So, if I can relate the left side of our inequality to this sum, perhaps I can make progress.Looking back at our left side:sum sqrt{frac{a^2}{b^2 + c^2 - a^2}} = sum frac{a}{sqrt{b^2 + c^2 - a^2}}.But from the Law of Cosines, (b^2 + c^2 - a^2 = 2bc cos A), so:sum frac{a}{sqrt{2bc cos A}}.Hmm, not directly the same as (sum frac{a}{b + c - a}), but maybe there's a relationship.Wait, perhaps I can use the AM-GM inequality on the denominator.We know that (b + c - a geq sqrt{2bc}), because:(b + c - a)^2 geq 2bc.Wait, is that true? Let me check.Expanding ((b + c - a)^2):b^2 + c^2 + a^2 + 2bc - 2ab - 2ac.Hmm, not sure if that's greater than or equal to (2bc).Wait, actually, in an acute triangle, we have (b^2 + c^2 > a^2), so (b + c > a), which is true in any triangle.But I'm not sure about the inequality (b + c - a geq sqrt{2bc}).Let me test with an equilateral triangle where (a = b = c).Then, (b + c - a = a + a - a = a), and (sqrt{2bc} = sqrt{2a^2} = asqrt{2}).So, (a geq asqrt{2})? No, that's false because (sqrt{2} > 1). So, the inequality (b + c - a geq sqrt{2bc}) is not generally true.Therefore, that approach doesn't work.Wait, maybe I can use the Cauchy-Schwarz inequality in a different way.Let me consider the sum:sum frac{a}{sqrt{2bc cos A}}.Let me set (u_i = frac{a}{sqrt{2bc cos A}}) and (v_i = 1). Then, by Cauchy-Schwarz:left( sum u_i v_i right)^2 leq left( sum u_i^2 right) left( sum v_i^2 right).So,left( sum frac{a}{sqrt{2bc cos A}} right)^2 leq left( sum frac{a^2}{2bc cos A} right) cdot 3.But again, this gives an upper bound, not a lower bound.Alternatively, maybe I can use Holder's inequality, which is a generalization of Cauchy-Schwarz.Holder's inequality states that for positive real numbers (a_i, b_i, c_i):sum a_i b_i c_i leq left( sum a_i^p right)^{1/p} left( sum b_i^q right)^{1/q} left( sum c_i^r right)^{1/r},where (1/p + 1/q + 1/r = 1).But I'm not sure how to apply that here.Wait, maybe I can consider the terms in the sum and relate them to the inradius and circumradius through some known identities.I know that in a triangle, the distance from the incenter to a vertex is given by:sqrt{r^2 + left( frac{a}{2} right)^2},but I'm not sure if that's relevant here.Alternatively, perhaps I can use the formula for the area in terms of R and r:S = r s = frac{abc}{4R},where (s) is the semiperimeter.So,r = frac{S}{s},R = frac{abc}{4S}.Therefore,frac{R}{r} = frac{abc}{4S} cdot frac{s}{S} = frac{abc s}{4 S^2}.Hmm, not sure if that helps.Wait, maybe I can express the sum in terms of S and s.But I'm not seeing a direct connection.Let me think differently. Maybe I can use substitution to simplify the problem.Let me set (x = frac{a}{sqrt{2bc cos A}}), (y = frac{b}{sqrt{2ac cos B}}), (z = frac{c}{sqrt{2ab cos C}}).Then, the left side of the inequality is (x + y + z), and we need to show that:x + y + z geq 3 sqrt{frac{R}{2r}}.Now, perhaps I can find expressions for x, y, z in terms of R and r.From earlier, we have:x = frac{a}{sqrt{2bc cos A}} = frac{a}{sqrt{b^2 + c^2 - a^2}}.But I don't see how to express this in terms of R and r.Wait, perhaps I can use the identity that in any triangle:sum frac{a}{b + c - a} = frac{2R}{r} - 1.So, if I can relate (x + y + z) to (sum frac{a}{b + c - a}), perhaps I can make progress.But I'm not sure how to connect (x + y + z) to (sum frac{a}{b + c - a}).Alternatively, maybe I can use the fact that in an acute triangle, (b + c - a > 0), and perhaps relate the denominators.Wait, let me consider the following:From the Law of Cosines, (b^2 + c^2 - a^2 = 2bc cos A), so:sqrt{b^2 + c^2 - a^2} = sqrt{2bc cos A}.Therefore,x = frac{a}{sqrt{2bc cos A}} = frac{a}{sqrt{b^2 + c^2 - a^2}}.Similarly for y and z.So, the sum is:sum frac{a}{sqrt{b^2 + c^2 - a^2}}.Now, perhaps I can consider the function (f(a, b, c) = frac{a}{sqrt{b^2 + c^2 - a^2}}), and see if it has some properties that can be used.Alternatively, maybe I can use the Cauchy-Schwarz inequality in the following way:left( sum frac{a}{sqrt{b^2 + c^2 - a^2}} right)^2 leq left( sum a^2 right) left( sum frac{1}{b^2 + c^2 - a^2} right).But again, this gives an upper bound, not a lower bound.Wait, perhaps I can consider the reciprocal of the terms.Let me define (u = sqrt{frac{a^2}{b^2 + c^2 - a^2}}), and similarly for v and w.Then, the left side is (u + v + w), and we need to show that (u + v + w geq 3 sqrt{frac{R}{2r}}).Now, perhaps I can find a relationship between u, v, w and R, r.Alternatively, maybe I can use the AM-GM inequality on u, v, w.The AM-GM inequality states that:frac{u + v + w}{3} geq sqrt[3]{uvw}.So,u + v + w geq 3 sqrt[3]{uvw}.Therefore, if I can show that:sqrt[3]{uvw} geq sqrt{frac{R}{2r}},then it would follow that:u + v + w geq 3 sqrt{frac{R}{2r}}.So, let's compute (uvw):uvw = sqrt{frac{a^2}{b^2 + c^2 - a^2}} cdot sqrt{frac{b^2}{a^2 + c^2 - b^2}} cdot sqrt{frac{c^2}{a^2 + b^2 - c^2}}.Simplify:uvw = frac{abc}{sqrt{(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}}.Hmm, that looks complicated, but maybe I can relate the denominator to the area.I recall that in a triangle, the area S can be expressed as:S = sqrt{s(s - a)(s - b)(s - c)},where (s = frac{a + b + c}{2}) is the semiperimeter.Also, the denominator in (uvw) is:sqrt{(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}.Wait, in an acute triangle, all these terms are positive, so the square root is defined.I also recall that in a triangle, the product ((b^2 + c^2 - a^2)(a^2 + c^2 - b^2)(a^2 + b^2 - c^2)) is related to the area.Specifically, I think it's equal to (16 S^2), but let me verify.Wait, actually, in a triangle, we have:(b^2 + c^2 - a^2) = 2bc cos A,(a^2 + c^2 - b^2) = 2ac cos B,(a^2 + b^2 - c^2) = 2ab cos C.So, the product is:(2bc cos A)(2ac cos B)(2ab cos C) = 8 a b c cdot a b c cos A cos B cos C = 8 a^2 b^2 c^2 cos A cos B cos C.Therefore, the denominator in (uvw) is:sqrt{8 a^2 b^2 c^2 cos A cos B cos C} = sqrt{8} a b c sqrt{cos A cos B cos C}.So, (uvw) becomes:frac{abc}{sqrt{8} a b c sqrt{cos A cos B cos C}} = frac{1}{sqrt{8} sqrt{cos A cos B cos C}}.Simplify:uvw = frac{1}{2 sqrt{2} sqrt{cos A cos B cos C}}.Therefore,sqrt[3]{uvw} = sqrt[3]{frac{1}{2 sqrt{2} sqrt{cos A cos B cos C}}} = frac{1}{(2 sqrt{2})^{1/3} (cos A cos B cos C)^{1/6}}.Hmm, not sure if that helps. Maybe I can relate (cos A cos B cos C) to R and r.I recall that in a triangle,cos A cos B cos C = frac{r}{4R}.Yes, that's a known identity. So,cos A cos B cos C = frac{r}{4R}.Therefore,sqrt[3]{uvw} = frac{1}{(2 sqrt{2})^{1/3} left( frac{r}{4R} right)^{1/6}}.Simplify the exponents:First, note that (2 sqrt{2} = 2 cdot 2^{1/2} = 2^{3/2}).So,(2 sqrt{2})^{1/3} = (2^{3/2})^{1/3} = 2^{(3/2) cdot (1/3)} = 2^{1/2} = sqrt{2}.Similarly,left( frac{r}{4R} right)^{1/6} = left( frac{r}{4R} right)^{1/6}.So,sqrt[3]{uvw} = frac{1}{sqrt{2} cdot left( frac{r}{4R} right)^{1/6}} = frac{1}{sqrt{2}} cdot left( frac{4R}{r} right)^{1/6}.Simplify (frac{4R}{r}):frac{4R}{r} = 4 cdot frac{R}{r}.So,sqrt[3]{uvw} = frac{1}{sqrt{2}} cdot left( 4 cdot frac{R}{r} right)^{1/6} = frac{1}{sqrt{2}} cdot 4^{1/6} cdot left( frac{R}{r} right)^{1/6}.Simplify (4^{1/6}):4^{1/6} = (2^2)^{1/6} = 2^{1/3}.So,sqrt[3]{uvw} = frac{1}{sqrt{2}} cdot 2^{1/3} cdot left( frac{R}{r} right)^{1/6}.Combine the constants:frac{1}{sqrt{2}} cdot 2^{1/3} = 2^{-1/2} cdot 2^{1/3} = 2^{-1/2 + 1/3} = 2^{-1/6} = frac{1}{2^{1/6}}.Therefore,sqrt[3]{uvw} = frac{1}{2^{1/6}} cdot left( frac{R}{r} right)^{1/6} = left( frac{R}{2 r} right)^{1/6}.Wait, let me check that step.We have:sqrt[3]{uvw} = frac{1}{2^{1/6}} cdot left( frac{R}{r} right)^{1/6} = left( frac{1}{2} cdot frac{R}{r} right)^{1/6} = left( frac{R}{2 r} right)^{1/6}.Yes, that's correct.Therefore,sqrt[3]{uvw} = left( frac{R}{2 r} right)^{1/6}.So, going back to the AM-GM inequality:u + v + w geq 3 sqrt[3]{uvw} = 3 left( frac{R}{2 r} right)^{1/6}.But we need to show that:u + v + w geq 3 sqrt{frac{R}{2 r}}.Comparing the two, we have:3 left( frac{R}{2 r} right)^{1/6} geq 3 sqrt{frac{R}{2 r}}.Dividing both sides by 3:left( frac{R}{2 r} right)^{1/6} geq sqrt{frac{R}{2 r}}.Raise both sides to the power of 6:frac{R}{2 r} geq left( frac{R}{2 r} right)^3.Let me denote (k = frac{R}{2 r}). Then, the inequality becomes:k geq k^3.Which simplifies to:k^3 - k leq 0,k(k^2 - 1) leq 0.So, (k leq 0) or (k^2 - 1 leq 0), i.e., (k leq 1).But (k = frac{R}{2 r}), and from Euler's inequality, (R geq 2 r), so (k geq 1).Therefore, (k geq 1), which implies that (k(k^2 - 1) geq 0), contradicting the earlier inequality.This suggests that our approach using AM-GM leads to a contradiction, meaning that the AM-GM approach is not sufficient to prove the desired inequality.Hmm, so maybe I need to abandon the AM-GM approach and try something else.Let me recall that in the beginning, I thought about using Lemma 1 and Lemma 2 as in the initial proof sketch. Maybe I should follow that approach.**Lemma 1:** In an acute-angled triangle ABC,sum sqrt{frac{a^{2}}{b^{2}+c^{2}-a^{2}}} geqslant sum frac{a}{b+c-a}.**Lemma 2:** In triangle ABC,sum frac{a}{b+c-a} = frac{2R}{r} - 1.Then, combining these,sum sqrt{frac{a^{2}}{b^{2}+c^{2}-a^{2}}} geq frac{2R}{r} - 1.Now, to relate (frac{2R}{r} - 1) to (3 sqrt{frac{R}{2r}}), we can use Euler's inequality (R geq 2r), which gives (frac{R}{r} geq 2).So,frac{2R}{r} - 1 geq 3 sqrt{frac{R}{2r}}.Let me verify this.Let me set (k = sqrt{frac{R}{2r}}). Then, (k geq sqrt{frac{2r}{2r}} = 1), since (R geq 2r).We need to show that:2 cdot 2r cdot k - 1 geq 3k.Wait, that seems messy. Let me express (frac{2R}{r} - 1) in terms of k.Since (k = sqrt{frac{R}{2r}}), then (k^2 = frac{R}{2r}), so (R = 2 r k^2).Therefore,frac{2R}{r} - 1 = frac{2 cdot 2 r k^2}{r} - 1 = 4 k^2 - 1.We need to show that:4 k^2 - 1 geq 3k.Rearranging,4 k^2 - 3k - 1 geq 0.Solving the quadratic inequality:Find the roots of (4 k^2 - 3k - 1 = 0).Using the quadratic formula:k = frac{3 pm sqrt{9 + 16}}{8} = frac{3 pm sqrt{25}}{8} = frac{3 pm 5}{8}.So, the roots are:k = frac{3 + 5}{8} = 1,k = frac{3 - 5}{8} = -frac{1}{4}.Since (k geq 1), the quadratic (4 k^2 - 3k - 1) is positive for (k geq 1).Therefore,4 k^2 - 1 geq 3k quad text{for} quad k geq 1.Thus,frac{2R}{r} - 1 geq 3 sqrt{frac{R}{2r}}.Therefore, combining with Lemma 1 and Lemma 2,sum sqrt{frac{a^{2}}{b^{2}+c^{2}-a^{2}}} geq frac{2R}{r} - 1 geq 3 sqrt{frac{R}{2r}}.Hence, the original inequality is proven.

💡First, I need to determine the original number that was used in the incorrect calculation. The incorrect operation was subtracting 3 and then adding 49, resulting in 66. I can set up the equation:x - 3 + 49 = 66Simplifying this equation:x + 46 = 66Subtracting 46 from both sides gives:x = 20Now that I have the original number (20), I can calculate the correct result by multiplying the number by 3 and then adding 49:3 * 20 + 49 = 60 + 49 = 109Finally, to find the difference between the incorrect and correct results:109 - 66 = 43

💡Alright, so I've got this system of equations to solve:left{begin{array}{l}a + c = -1 ac + b + d = -1 ad + bc = -5 bd = 6end{array}right.Hmm, okay. Let's see. I need to find real numbers a, b, c, and d that satisfy all four equations. First, looking at the first equation: a + c = -1. That seems straightforward. Maybe I can express one variable in terms of the other. Let's say c = -1 - a. That might be useful later on.Next, the second equation: ac + b + d = -1. Hmm, okay. So if I can express c in terms of a, as I did before, then maybe I can substitute that into this equation. Let's try that.So, substituting c = -1 - a into the second equation:a*(-1 - a) + b + d = -1Let's expand that:- a - a^2 + b + d = -1Hmm, so rearranged:- a^2 - a + b + d = -1Maybe I can write this as:b + d = a^2 + a - 1Okay, that's an expression for b + d in terms of a.Moving on to the third equation: ad + bc = -5. Again, since I have c in terms of a, maybe I can substitute that here as well.So, substituting c = -1 - a into ad + bc:a*d + b*(-1 - a) = -5Let's expand that:a*d - b - a*b = -5Hmm, that's a bit messy. Maybe I can rearrange it:a*d - a*b - b = -5Factor out a from the first two terms:a*(d - b) - b = -5Not sure if that helps immediately. Maybe I can hold onto that thought.Looking at the fourth equation: b*d = 6. Okay, so b and d are factors of 6. Since we're dealing with real numbers, the possible pairs (b, d) could be (1,6), (2,3), (-1,-6), (-2,-3), and their permutations. Maybe I can list these possibilities and see which ones fit with the other equations.So, let's list the possible pairs for (b, d):1. (1,6)2. (2,3)3. (-1,-6)4. (-2,-3)5. (6,1)6. (3,2)7. (-6,-1)8. (-3,-2)Okay, that's eight possibilities. Maybe I can test these one by one with the other equations.But before that, maybe there's a smarter way. Let's think about the second equation again: b + d = a^2 + a - 1. So, if I can find b + d for each pair, I can set it equal to a^2 + a - 1 and see if that gives me a consistent a.Let's try that.First pair: (1,6). Then b + d = 1 + 6 = 7.So, 7 = a^2 + a - 1Which simplifies to:a^2 + a - 8 = 0Let's solve this quadratic equation:a = [-1 ± sqrt(1 + 32)] / 2 = [-1 ± sqrt(33)] / 2Hmm, real solutions, but let's see if they fit into the third equation.Wait, before I proceed, maybe I should check if these a values satisfy the third equation.So, for the first pair (b=1, d=6):From the third equation: a*d + b*c = -5We have c = -1 - a, so:a*6 + 1*(-1 - a) = -5Simplify:6a -1 - a = -55a -1 = -55a = -4a = -4/5But from the second equation, we had a^2 + a - 8 = 0, which gave a = [-1 ± sqrt(33)] / 2 ≈ (-1 ± 5.7446)/2So, approximately, a ≈ (4.7446)/2 ≈ 2.3723 or a ≈ (-6.7446)/2 ≈ -3.3723But from the third equation, we have a = -4/5 = -0.8These don't match. So, this pair (1,6) doesn't seem to work.Okay, moving on to the second pair: (2,3). Then b + d = 2 + 3 = 5.So, 5 = a^2 + a - 1Which simplifies to:a^2 + a - 6 = 0Solving this:a = [-1 ± sqrt(1 + 24)] / 2 = [-1 ± 5]/2So, a = (4)/2 = 2 or a = (-6)/2 = -3Alright, so a = 2 or a = -3.Now, let's check these a values with the third equation.First, a = 2:From the third equation: a*d + b*c = -5We have b=2, d=3, c = -1 - a = -1 -2 = -3So:2*3 + 2*(-3) = 6 -6 = 0 ≠ -5Not good.Next, a = -3:c = -1 - (-3) = 2So:a*d + b*c = (-3)*3 + 2*2 = -9 + 4 = -5Perfect! That works.So, for the pair (b=2, d=3), we have a = -3, c = 2.So, one solution is (a, b, c, d) = (-3, 2, 2, 3)Now, let's check the other pair (3,2). So, b=3, d=2.Then, b + d = 3 + 2 = 5, same as before.So, a^2 + a - 6 = 0, same equation, a=2 or a=-3.Now, check a=2:c = -1 -2 = -3Third equation: a*d + b*c = 2*2 + 3*(-3) = 4 -9 = -5Perfect.So, another solution is (a, b, c, d) = (2, 3, -3, 2)Now, let's check the negative pairs.Third pair: (-1,-6). Then b + d = -1 + (-6) = -7So, -7 = a^2 + a -1Which simplifies to:a^2 + a +6 = 0Discriminant: 1 -24 = -23 < 0No real solutions. So, discard this pair.Fourth pair: (-2,-3). Then b + d = -2 + (-3) = -5So, -5 = a^2 + a -1Which simplifies to:a^2 + a +4 = 0Discriminant: 1 -16 = -15 < 0No real solutions. Discard.Fifth pair: (6,1). b + d = 6 +1 =7Same as the first pair. We already saw that this leads to a = -4/5, which didn't satisfy the third equation. So, no solution.Sixth pair: (3,2). We already did this, it's the same as the second pair.Seventh pair: (-6,-1). b + d = -6 + (-1) = -7Same as the third pair, no real solutions.Eighth pair: (-3,-2). b + d = -3 + (-2) = -5Same as the fourth pair, no real solutions.So, the only valid pairs are (2,3) and (3,2), leading to the solutions (a, b, c, d) = (-3, 2, 2, 3) and (2, 3, -3, 2).Let me double-check these solutions in all four equations.First solution: a=-3, b=2, c=2, d=31. a + c = -3 + 2 = -1 ✔️2. ac + b + d = (-3)(2) + 2 + 3 = -6 +5 = -1 ✔️3. ad + bc = (-3)(3) + 2*2 = -9 +4 = -5 ✔️4. bd = 2*3 =6 ✔️Second solution: a=2, b=3, c=-3, d=21. a + c = 2 + (-3) = -1 ✔️2. ac + b + d = 2*(-3) +3 +2 = -6 +5 = -1 ✔️3. ad + bc = 2*2 +3*(-3) =4 -9 = -5 ✔️4. bd =3*2=6 ✔️Both solutions satisfy all four equations. So, I think these are the only real solutions.