answer:Given: - ( angle B = 45^circ ) - ( 45^circ leq angle C leq 90^circ ) - Distance between the circumcenter (O) and the incenter (I) is ( frac{AB - AC}{sqrt{2}} ) - We need to find ( cos A ) 1. **Using Euler's Formula for Incenter and Circumcenter**: [ OI^2 = R^2 - 2Rr ] where (R) is the circumradius and (r) is the inradius. 2. **Assigning Side Lengths**: - Let ( a = BC ), ( b = CA ), ( c = AB ). The inradius (r) can also be expressed using the sides of the triangle. 3. **Expressing Inradius (r):** [ r = frac{1}{2} (a + b - c) tan frac{B}{2} ] 4. **Simplifying ( tan frac{B}{2} )**: - Since ( angle B = 45^circ ): [ tan B = 1 ] - Therefore: [ tan frac{B}{2} = tan 22.5^circ = sqrt{2} - 1 ] 5. **Expressing (a, b, c) in terms of (R) and the angles:** - Since ( sin angle B = sin 45^circ = frac{sqrt{2}}{2} ): [ a = 2R sin A ] [ b = R sqrt{2} ] [ c = 2R sin C ] - Note that (sin C = sin (180^circ - (45^circ + A)) = sin (135^circ - A)): [ sin (135^circ - A) = frac{cos A + sin A}{sqrt{2}} ] 6. **Using the given distance between (O) and (I)**: [ OI^2 = left(frac{AB - AC}{sqrt{2}}right)^2 = frac{(c - b)^2}{2} ] 7. **Putting (a, b, c) into Euler's formula**: [ OI^2 = R^2 - 2R left(frac{1}{2}(2R sin A + R sqrt{2} - 2R sin C) (sqrt{2} - 1)right) ] 8. **Solution Using Identity and Substitution**: - Substituting (c = 2R sin C) and using the identity for (sin C): [ sin C = frac{cos A + sin A}{sqrt{2}} ] - Simplifying ((c - b)^2 / 2): [ left(2R frac{cos A + sin A}{sqrt{2}} - R sqrt{2}right)^2 / 2 = ((R sqrt{2} (cos A + sin A) - R sqrt{2})^2) / 2 R^2 ] 9. **Simplifying and Equating Results**: [ left(2R sin C - R sqrt{2}right)^2 / 2 = left(frac{1}{2}R left((2 sin A - sqrt{2})(sqrt{2} - 1)right)right)^2 ] - After detailed reduction: [ 2 sin A cos A - sqrt{2} cos A - (2 - sqrt{2}) sin A + sqrt{2} - 1 = 0 ] - Factorize and solve the equation to obtain: [ (sqrt{2} sin A - 1)(sqrt{2} cos A - sqrt{2} + 1) = 0 ] 10. **Solving Final Factored Form**: [ cos A = frac{1}{sqrt{2}} quad text{or} quad cos A = 1 - frac{1}{sqrt{2}} ] # Conclusion: [ boxed{cos A = frac{1}{sqrt{2}}} ]

answer:Let x = arccos (sin 3). Then: begin{align*} cos x &= sin 3 &= cos left( frac{pi}{2} - 3 right) &= cos left( 3 - frac{pi}{2} right). end{align*} Since arccos function returns values in the range [0, pi], and 3 - frac{pi}{2} lies in this range (as frac{pi}{2} approx 1.57 and 3 - 1.57 approx 1.43 which is between 0 and pi), the value of x is: [ x = boxed{3 - frac{pi}{2}}. ]

answer:1. **Calculate the height of the equilateral triangle:** The formula for the height ( h ) of an equilateral triangle is: [ h = frac{sqrt{3}}{2} times text{side length} ] For a side length of 12: [ h = frac{sqrt{3}}{2} times 12 = 6sqrt{3} text{ units} ] 2. **Calculate the area of the equilateral triangle:** [ text{Area of equilateral triangle} = frac{sqrt{3}}{4} times (text{side length})^2 ] [ text{Area} = frac{sqrt{3}}{4} times 12^2 = 36sqrt{3} text{ square units} ] 3. **Calculate the area of one isosceles triangle removed:** Since the height of the isosceles triangle can be found using the Pythagorean theorem in a 5-5-x triangle where x is half the base: [ 5^2 = (frac{b}{2})^2 + text{height}^2 ] Let’s assume ( b = 6 ) units (making the assumption clear, though detailed calculations may be needed to confirm based on geometry): [ 25 = 3^2 + text{height}^2 Rightarrow text{height}^2 = 16 Rightarrow text{height} = 4 text{ units} ] [ text{Area of isosceles triangle} = frac{1}{2} times b times text{height} = frac{1}{2} times 6 times 4 = 12 text{ square units} ] 4. **Calculate the total area of removed triangles and remaining area:** With three identical triangles removed: [ text{Total area of removed triangles} = 3 times 12 = 36 text{ square units} ] [ text{Area of remaining hexagon} = text{Area of original triangle} - text{Total area of removed triangles} = 36sqrt{3} - 36 ] 5. **Calculate the ratio:** [ text{Ratio} = frac{text{Area of one removed triangle}}{text{Area of remaining hexagon}} = frac{12}{36sqrt{3} - 36} ] Conclusion: The ratio of the area of one removed triangle to the remaining central hexagon is (boxed{frac{12}{36sqrt{3} - 36}}).

answer:A regular tetrahedron has its surfaces labeled with the numbers (1), (2), (3), (4). If the tetrahedron is rolled randomly twice, and the numbers on the faces that land down are denoted as (a) and (b), respectively. There are a total of n=4times 4 = 16 possible outcomes. The events where the equation x^{2}-ax-b=0 has at least one root min{1,2,3,4} are: - (a,b)=(1,2) - (a,b)=(2,3) - (a,b)=(3,4) There are a total of 3 favorable outcomes. Thus, the probability that the equation is a "beautiful equation" is p=frac{3}{16}. The final answer is p=boxed{frac{3}{16}}. To solve this problem, we first find the total number of possible outcomes. Then, we list all the events where the equation has at least one root in the set {1,2,3,4}. Finally, we calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. This problem tests our understanding of probability and requires careful application of the fundamental counting principle.