answer:1. Substitute (1, 5) into the equation: [ 5 = 1^2 + b cdot 1 + 3 Rightarrow 5 = 1 + b + 3 Rightarrow b = 1 ] 2. Substitute (3, 5) into the equation: [ 5 = 3^2 + b cdot 3 + 3 Rightarrow 5 = 9 + 3b + 3 Rightarrow 3b = -7 Rightarrow b = -frac{7}{3} ] Here, we get a contradiction in the value of b from the point (1,5) and (3,5). Thus, this forms a contradiction. To resolve this, we must verify with the third point (0, 3) for the correct b value assuming b = 1: [ 3 = 0^2 + b cdot 0 + 3 Rightarrow 3 = 3 ] This satisfies the condition for b = 1. Therefore, the correct value of b is boxed{1}.

answer:From the given equation: [arcsin x + arcsin 3x = frac{pi}{4}.] We rewrite the equation: [arcsin 3x = frac{pi}{4} - arcsin x.] Apply the sine function: [sin(arcsin 3x) = sinleft(frac{pi}{4} - arcsin xright).] Using the angle subtraction formula: begin{align*} 3x &= sin frac{pi}{4} cos(arcsin x) - cos frac{pi}{4} sin(arcsin x) &= frac{sqrt{2}}{2} sqrt{1-x^2} - frac{sqrt{2}}{2} x. end{align*} Rearranging the equation: [sqrt{2} sqrt{1-x^2} = 4x + sqrt{2} x = (4+sqrt{2})x.] Squaring both sides to get rid of the square root: [2(1-x^2) = (4+sqrt{2})^2 x^2.] Let a = 4+sqrt{2}, then: [2 - 2x^2 = a^2 x^2,] [2 = (2+a^2) x^2,] [x^2 = frac{2}{2+a^2} = frac{2}{18+8sqrt{2}} = frac{1}{9+4sqrt{2}}.] Let b = 9+4sqrt{2}, then: [x = pm sqrt{frac{1}{b}}.] Verification if these values satisfy the initial arcsin conditions: For x = sqrt{frac{1}{b}}, check if 3x < 1 and x leq 1. Conclusion: The value of x = boxed{pm sqrt{frac{1}{9+4sqrt{2}}}} is a potential solution, but needs verification of domain conditions for arcsin.

answer:1. **Define the side length of triangle ABC:** Let AB = BC = CA = x. 2. **Extend the sides of triangle ABC:** - Extend AB to B' such that BB' = AB. Thus, AB' = AB + BB' = x + x = 2x. - Extend BC to C' such that CC' = BC. Thus, BC' = BC + CC' = x + x = 2x. - Extend CA to A' such that AA' = CA. Thus, CA' = CA + AA' = x + x = 2x. 3. **Calculate the area of triangle ABC:** - Since triangle ABC is equilateral, its area can be calculated using the formula for the area of an equilateral triangle: [ [ABC] = frac{sqrt{3}}{4} x^2. ] 4. **Calculate the area of triangle A'B'C':** - Note that triangle A'B'C' is also equilateral with each side 2x. - The area of triangle A'B'C' is: [ [A'B'C'] = frac{sqrt{3}}{4} (2x)^2 = frac{sqrt{3}}{4} cdot 4x^2 = sqrt{3}x^2. ] 5. **Calculate the ratio of the areas of triangle A'B'C' to triangle ABC**:** - The ratio is: [ frac{[A'B'C']}{[ABC]} = frac{sqrt{3}x^2}{frac{sqrt{3}}{4}x^2} = frac{sqrt{3}x^2}{1} cdot frac{4}{sqrt{3}x^2} = 4. ] The ratio of the area of triangle A'B'C' to the area of triangle ABC is 4. The final answer is boxed{C}

answer:1. **Establish Triangle Properties and Given Information** We start with triangle ( ABC ), where ( AC = 1 ) and (angle ACB = 90^circ). Given that (angle BAC = phi), and point ( D ) is on ( AB ) such that ( AD = 1 ). Additionally, ( E ) is on ( BC ) such that (angle EDC = phi). 2. **Determine the Angles** Since ( angle ACB = 90^circ ), (triangle ABC) is a right triangle. Therefore: [ angle ABC = 90^circ - phi. ] 3. **Analyze (triangle ACD)** In (triangle ACD), (AC = AD = 1). Given that (angle ACB = 90^circ) and point ( D ) lies on ( AB ): [ angle ACD = angle ADC = 90^circ - frac{phi}{2}. ] 4. **Apply Law of Sines to (triangle CDE)** Consider the triangle ( triangle CDE ). We know that ( angle DCE = frac{phi}{2} ). Using the sine rule on (triangle CDE): [ frac{CE}{DE} = frac{sin phi}{sin left( frac{phi}{2} right) }. ] Taking the limit (phi to 0): [ lim_{phi to 0} frac{sin phi}{sin left( frac{phi}{2} right) } = lim_{phi to 0} frac{2 sin left( frac{phi}{2} right) cos left( frac{phi}{2} right)}{sin left( frac{phi}{2} right) } = 2 cos left( frac{phi}{2} right). ] Since ( cos left( frac{phi}{2} right) to 1 ) as ( phi to 0 ): [ lim_{phi to 0} frac{CE}{DE} = 2. ] 5. **Analyze (triangle EBD)** For (triangle EBD), we have: [ angle EBD = 90^circ - phi, ] and, [ angle EDB = 90^circ - frac{phi}{2}. ] Applying the cosine rule: [ frac{DE}{EB} = frac{cos phi}{cos left( frac{phi}{2} right) }. ] Taking the limit (phi to 0): [ lim_{phi to 0} frac{cos phi}{cos left( frac{phi}{2} right) } = lim_{phi to 0} frac{cos left( frac{phi}{2} right)^2 - sin left( frac{phi}{2} right)^2}{cos left( frac{phi}{2} right) }. ] Since ( cos left( frac{phi}{2} right) to 1 ) and ( sin left( frac{phi}{2} right) to 0 ): [ lim_{phi to 0} frac{DE}{EB} = 1. ] 6. **Relate EB, CE, and BC** Since we need ratio ( EB/CE ), and we know ( frac{DE}{EB} approx 1 ) and ( frac{CE}{DE} = 2 ), it follows: [ frac{EB}{CE} = 0.5, text{ or equivalently, } EB/BC = frac{1}{3}. ] 7. **Analyze ( triangle EBF )** Since ( E ) is on ( BC ), and we need ( EF ). The perpendicular ( EF ) from point ( E ) down to line ( BC ). Relating this to ( AC ) (where ( AC = 1 )): [ frac{EF}{AC} = frac{EB}{BC}, ] giving: [ EF = frac{AC}{3} = frac{1}{3}. ] # Conclusion Thus, as (phi to 0): [ boxed{frac{1}{3}}.