answer:Given a triangular pyramid SABC with base ABC being an equilateral triangle with side length a and edge SA = b. We are required to find the volume V of the pyramid under different conditions for b. Step 1: Understanding the Geometry From the problem, we know that the triangular pyramid has congruent lateral faces. This implies that the apex S projects onto the base ABC at specific points. There are three possible positions for the projection of S onto the base ABC: 1. The incenter of ABC. 2. The center of an excircle touching side BC and the extensions of sides AB and AC. 3. Similar centers relative to the other sides. To find the volume, we need to take these projected positions into account and apply the volume formula for pyramids: [ V = frac{1}{3} times text{base area} times text{height} ] Step 2: Calculating the Basic Elements 1. **Area of the Base ABC**: Since ABC is an equilateral triangle with side a, the area A_{ABC} can be found as: [ A_{ABC} = frac{sqrt{3}}{4} a^2 ] 2. **Height of the Equilateral Triangle ABC**: The height h from a vertex to the opposite side in an equilateral triangle is: [ h = frac{sqrt{3}}{2} a ] Step 3: Position of Point S **If ( frac{a}{sqrt{3}} < b leq a )**: - Point S is the incenter of the triangle ABC. - The inradius r of ABC is given by: [ r = frac{a}{2sqrt{3}} ] - The height H of the pyramid SABC is determined by the length from S to the base, using the relation: [ H = sqrt{b^2 - left(frac{a}{2sqrt{3}}right)^2} = sqrt{b^2 - frac{a^2}{12}} ] Thus, the volume V becomes: [ V = frac{1}{3} times A_{ABC} times H = frac{1}{3} times frac{sqrt{3}}{4} a^2 times sqrt{b^2 - frac{a^2}{12}} = frac{a^2}{12} sqrt{3b^2 - a^2} ] **If ( a < b leq a sqrt{3} )**: - There are two possible positions for S: the incenter or one of the excenters. 1. **Incenter result** (same calculation as above): [ V_1 = frac{a^2}{12} sqrt{3b^2 - a^2} ] 2. **Excenter results**: - The distance to this point results in different heights: - If the apex S projects to an excenter, the height H': [ H' = sqrt{b^2 - left(frac{asqrt{3}}{2}right)^2} = sqrt{b^2 - frac{3a^2}{4}} ] - Volume using this height: [ V_2 = frac{1}{3} times frac{sqrt{3}}{4} a^2 times sqrt{b^2 - frac{a^2}{4}} = frac{a^2 sqrt{3}}{12} sqrt{b^2 - a^2} ] **If ( b > asqrt{3} )**: - There are three possible results: 1. **Incenter result**: [ V_1 = frac{a^2}{12} sqrt{3b^2 - a^2} ] 2. **First excenter**: [ V_2 = frac{a^2 sqrt{3}}{12} sqrt{b^2 - a^2} ] 3. **Second excenter**: Shift in position of S results in another height: [ H'' = sqrt{b^2 - (3a^2)} ] Thus: [ V_3 = frac{a^2 sqrt{3}}{12} sqrt{b^2 - 3a^2} ] # Conclusion The volumes of the pyramid SABC under the given conditions are: - If ( frac{a}{sqrt{3}} < b leq a ): [ boxed{V = frac{a^2}{12} sqrt{3b^2 - a^2}} ] - If ( a < b leq a sqrt{3} ): [ boxed{V_1 = frac{a^2}{12} sqrt{3b^2 - a^2}, quad V_2 = frac{a^2 sqrt{3}}{12} sqrt{b^2 - a^2}} ] - If ( b > a sqrt{3} ): [ boxed{V_1 = frac{a^2}{12} sqrt{3b^2 - a^2}, quad V_2 = frac{a^2 sqrt{3}}{12} sqrt{b^2 - a^2}, quad V_3 = frac{a^2 sqrt{3}}{12} sqrt{b^2 - 3a^2}} ]

answer:1. **Identify Possible Triangles**: - **On the Circle**: Triangles formed by any three points among A, B, C, and D. Since the points are equidistant on the circle and form a square inside the circle, the area of triangles like ABC, BCD, CDA, and DAB can be calculated. - **Using Diameter Points**: Triangles such as AEF, BEF, CEF, and DEF involve the diameter segment, which will alter the typical area calculation for circular segment triangles. 2. **Calculate Possible Base Lengths and Unique Triangle Areas**: - Triangles fully on the circle like ABC will have all sides equal to 1, forming an equilateral triangle. The formula for the area of an equilateral triangle with side s is frac{sqrt{3}}{4} s^2. Thus, the area for ABC is frac{sqrt{3}}{4}. - Triangles including the diameter will form right triangles with base EF = 1. The height will be the radius of the circle, say r. The area of these triangles like AEF is frac{1}{2} times 1 times r. - Since the configuration forms a regular arrangement (a square inside a circle with a side on the diameter), we calculate r using the square's diagonal (equal to the diameter), which is 2r = sqrt{2}. Thus, r = frac{sqrt{2}}{2}. The area of AEF becomes frac{1}{2} times 1 times frac{sqrt{2}}{2} = frac{sqrt{2}}{4}. 3. **Conclusion**: - We have two different areas: frac{sqrt{3}}{4} for triangles formed by circle points only and frac{sqrt{2}}{4} for triangles including the diameter. - Therefore, there are exactly two distinct areas possible for the triangles. The number of possible values for the area of the triangle is 2. The final answer is boxed{B}.

answer:1. **Express ( y ) from the given equation**: [ 2y + 5x = 10 implies 2y = 10 - 5x ] Divide both sides by 2: [ y = frac{10 - 5x}{2} = 5 - frac{5}{2}x ] Simplified, this gives: [ y = 5 - frac{5}{2}x ] 2. **Substitute ( y ) in the inequality**: We need to replace ( y ) in the inequality ( 3xy - x^2 - y^2 < 7 ). [ 3xleft(5 - frac{5}{2}xright) - x^2 - left(5 - frac{5}{2}xright)^2 < 7 ] 3. **Simplify the expression**: [ 3xleft(5 - frac{5}{2}xright) = 15x - frac{15}{2}x^2 ] [ left(5 - frac{5}{2}xright)^2 = 25 - 2 cdot 5 cdot frac{5}{2}x + left(frac{5}{2}xright)^2 = 25 - 25x + left(frac{25}{4}right)x^2 ] Plug these into the inequality: [ 15x - frac{15}{2}x^2 - x^2 - (25 - 25x + frac{25}{4}x^2) < 7 ] Combine like terms carefully: [ 15x - frac{15}{2}x^2 - x^2 - 25 + 25x - frac{25}{4}x^2 < 7 ] 4. **Combine and simplify the terms**: Combine the ( x^2 ) terms: [ 15x + 25x - 25 - left(frac{15}{2}x^2 + x^2 + frac{25}{4}x^2right) < 7 ] 5. **Simplify the quadratic terms further**: Factor out common terms from the quadratic term: [ 15x + 25x - 25 - left(left(frac{30}{4} + frac{4}{4} + frac{25}{4}right)x^2right) < 7 ] [ frac{15}{2} + 1 + frac{25}{4} = frac{30}{4} + frac{4}{4} + frac{25}{4} = frac{59}{4} ] [ 15x + 25x - 25 - frac{59}{4} x^2 < 7 ] [ 40x - 25 - frac{59}{4} x^2 < 7 ] 6. **Rearrange the final inequality**: [ 40x - 25 - frac{59}{4} x^2 < 7 implies frac{59}{4}x^2 - 40x + 32 > 0 ] 7. **Solve the quadratic inequality**: Check the nature of the quadratic expression. Its discriminant (Delta) gives the insight: [ Delta = B^2 - 4AC = (-40)^2 - 4 cdot frac{59}{4} cdot 32 = 1600 - 1888 = -288 ] Since (Delta) is negative, the quadratic expression (frac{59}{4}x^2 - 40x + 32) is always positive. Thus, the original inequality is valid for all (x): [ 3xy - x^2 - y^2 < 7 ] The solution is hence completely verified and concluded. [ boxed{} ]

answer:**Analysis** This problem mainly examines the basic relationships of trigonometric functions of the same angle and the application of the cosine double-angle formula, belonging to a basic question. By using the basic relationships of trigonometric functions of the same angle given in the problem, we can find the value of tanalpha, and then use the cosine double-angle formula to find the value of cos 2alpha. **Solution** Given frac{sinalpha - cosalpha}{sinalpha + cosalpha} = frac{1}{2} = frac{tanalpha - 1}{tanalpha + 1}, therefore tanalpha = 3, Then cos 2alpha = frac{cos^2alpha - sin^2alpha}{cos^2alpha + sin^2alpha} = frac{1 - tan^2alpha}{1 + tan^2alpha} = frac{1 - 9}{1 + 9} = -frac{4}{5}, Therefore, the correct answer is: boxed{A}.