answer:Given: - Side (a = BC), - Opposite angle (alpha = angle BAC), - Length of the segment from (A) to the incenter of the triangle, (d = AO). **Steps:** 1. **Triangle Notation and Preliminary Setup:** - Let (O) denote the incenter of the triangle (ABC), where (B) and (C) are the endpoints of side (a). - Let (A'), (B'), and (C') be the points of tangency of the incircle with (BC), (CA), and (AB) respectively. - The radius of the incircle is denoted by (varrho). 2. **Calculation of Angles in (triangle COB):** - In (triangle COB), we use the formula for the internal angle bisectors: [ angle COB = 180^circ - angle OBC - angle OCB ] Since (OB) and (OC) are angle bisectors, we have: [ angle OBC = frac{beta}{2} quad text{and} quad angle OCB = frac{gamma}{2} ] Substituting into the first equation: [ angle COB = 180^circ - frac{beta}{2} - frac{gamma}{2} = 180^circ - frac{beta + gamma}{2} ] Given that the sum of the three interior angles of a triangle is (180^circ): [ beta + gamma = 180^circ - alpha ] Therefore: [ angle COB = 180^circ - frac{180^circ - alpha}{2} = 180^circ - 90^circ + frac{alpha}{2} = 90^circ + frac{alpha}{2} ] 3. **Finding the Geometric Locus and Intersection Points:** Let's consider the locus on which (O) must lie. - The locus (i_1) is a circle with arbitrary point from where (BC) is visible under the angle (90^circ + frac{alpha}{2}). - Another locus (e_1) is composed of parallel lines to the line (BC) at a distance of (varrho). 4. **Geometric Constructions:** - Draw segment (BC = a). - Use the angle (90^circ + frac{alpha}{2}) at point (C) to create a pair of lines (g_1) and (g_2). - Construct the perpendicular bisector, (f), of (BC) to locate the point (O') where the incenter (O) crosses. - Draw a circle centered at (O') with radius (O'B = O'C) to get the circle (i_1). - Mark the points where a distance (d) from line (BC) intersects the extensions along (g_1) and (g_2). 5. **Verification of Points and Distances:** - Validate (O) via the intersection points where the circles (i_1) and the parallel lines (e_1) meet. - Since (A' = text{projection of } O text{ onto } BC) with (OA' = varrho), validate the placement of the incenter. 6. **Final Placement of Vertex (A):** - Draw the incircle of radius (varrho) centered at (O) which touches (BC) at (A'). - The (A) vertex of triangle (ABC) is found as the intersection of the tangents drawn from (B) and (C) to this circle. Taking into account all these steps, we can conclude that the construction of triangle (ABC) is valid. Now that we have the correct segment lengths and angles, we ensure that the final triangle fits all given conditions through the described geometric constructions. (boxed{text{III. solution procedure}} ) provides an alternative and easier approach based on known triangle properties: Given (a = BC), (alpha), and (d = AO), construct supplementary 'known' properties and locate the vertex (A) in a simplified manner. Finally: [ boxed{text{The triangle }ABCtext{ is correctly constructed.}} ]

answer:Since overrightarrow{a}=(1,2) and overrightarrow{b}=(-2,m) are parallel, we have that overrightarrow{a} parallel overrightarrow{b} implies the corresponding components of overrightarrow{a} and overrightarrow{b} are proportional. Specifically, since the first component of overrightarrow{a} is positive and the first component of overrightarrow{b} is negative, we infer that the second component of overrightarrow{b}, namely m, must be negative to maintain the parallel condition. Thus, equating the ratios of the components of these vectors, we obtain: frac{1}{-2} = frac{2}{m} Rightarrow m = -4. Now, we compute 2overrightarrow{a} + 3overrightarrow{b} using m = -4: 2overrightarrow{a} + 3overrightarrow{b} = 2(1,2) + 3(-2,-4) = (2 cdot 1 + 3 cdot (-2), 2 cdot 2 + 3 cdot (-4)) = (2 - 6, 4 - 12) = (-4, -8). To find the magnitude of the resulting vector, we use the formula for the magnitude of a vector in two dimensions: |2overrightarrow{a} + 3overrightarrow{b}| = sqrt{(-4)^2 + (-8)^2} = sqrt{16 + 64} = sqrt{80} = 4sqrt{5}. Therefore, the magnitude of 2overrightarrow{a} + 3overrightarrow{b} is boxed{4sqrt{5}}.

answer:1. This problem requires us to compute a definite integral using its geometric interpretation. Step 1: Interpret the integral geometrically. int_{0}^{1} sqrt{1-{x}^{2}}dx represents the area of a sector of a circle with center at the origin and radius 1 in the first quadrant. Step 2: Compute the area. The area of the sector is given by frac{1}{4} times pi times 1^2 = boxed{frac{pi}{4}}. 2. This problem requires us to find the distance traveled by a particle using a definite integral. Step 1: Determine the time intervals when the velocity is negative. The velocity v(t) = t^2 - 4t + 3 is negative when 0 < t < 4. Step 2: Compute the distance traveled. The distance traveled by the particle from t = 0 to t = 4 is given by: begin{align*} &int_{0}^{1}({t}^{2}-4t+3)dt-int _{1}^{3}({t}^{2}-4t+3)dt+int _{3}^{4}({t}^{2}-4t+3)dt =&left. left( frac{1}{3}{t}^{3}-2{t}^{2}+3t right) right|_{0}^{1}-left. left( frac{1}{3}{t}^{3}-2{t}^{2}+3t right) right|_{1}^{3}+left. left( frac{1}{3}{t}^{3}-2{t}^{2}+3t right) right|_{3}^{4} =&frac{4}{3}+frac{4}{3}+frac{4}{3} = boxed{4}. end{align*} 3. This problem requires us to find the range of values for m such that f(x) > m has a solution in [-1, 2]. Step 1: Find the maximum value of f(x) in [-1, 2]. The derivative of f(x) is f'(x) = 3x^2 - x - 2 = (3x + 2)(x - 1). The critical points of f(x) are x = -frac{2}{3} and x = 1. Evaluate f(x) at these critical points and at the endpoints x = -1 and x = 2 to find the maximum value: begin{align*} f(-1) &= frac{11}{2}, &f(1) &= frac{7}{2}, &fleft(-frac{2}{3}right) &= frac{85}{27}, &f(2) &= 7. end{align*} The maximum value of f(x) is boxed{7}. Step 2: Determine the range of values for m. Since the maximum value of f(x) is 7, m must be less than 7 for f(x) > m to have a solution in [-1, 2]. Therefore, the range for m is boxed{m < 7}. 4. This problem requires us to find the solution set for the inequality x^{2} - f(x) > 0. Step 1: Construct the function g(x) = frac{f(x)}{x}. Note that g'(x) = frac{xf'(x) - f(x)}{x^2} < 0 when x > 0, so g(x) is strictly decreasing on (0, infty). Step 2: Determine when f(x) > 0. Since g(x) is strictly decreasing, f(x) > 0 if and only if g(x) > g(2). This implies that 0 < x < 2. Step 3: Consider the case when x < 0. Since f(x) is an odd function, f(x) > 0 if and only if x < -2 when x < 0. Step 4: Combine the results. The solution set for the inequality f(x) > 0 is (-infty, -2) cup (0, 2). Since x^{2} > 0, the solution set for the inequality x^{2} - f(x) > 0 is the same as the solution set for f(x) > 0. Therefore, the solution set is boxed{(-infty, -2) cup (0, 2)}.

answer:(1) If the side with length 8 is the hypotenuse, then the length of the hypotenuse is 8; (2) If the side with length 8 is not the hypotenuse, then AB and AC are the legs, and the length of the hypotenuse BC is calculated as sqrt{AC^2 + AB^2} = sqrt{8^2 + 6^2} = 10. Therefore, the answer is boxed{10 text{ or } 8}.