answer:Given a right-angled isosceles triangle with a perimeter of (2s), we need to find its area. First Solution: 1. Let's denote the common length of the legs of the triangle by (a). The hypotenuse of such a triangle is (asqrt{2}). 2. Since the perimeter is given by (a + a + asqrt{2}), it follows that: [ a + a + asqrt{2} = 2a + asqrt{2} = a(2 + sqrt{2}) ] Thus, [ a(2 + sqrt{2}) = 2s ] 3. Solving for (a), we get: [ a = frac{2s}{2 + sqrt{2}} ] To simplify this, multiply the numerator and the denominator by the conjugate of the denominator, (2 - sqrt{2}): [ a = frac{2s (2 - sqrt{2})}{(2 + sqrt{2})(2 - sqrt{2})} = frac{2s (2 - sqrt{2})}{4 - 2} = (2 - sqrt{2})s ] 4. The area (t) of the triangle is given by: [ t = frac{1}{2} a^2 = frac{(asqrt{2})^2}{2} ] Using (a = (sqrt{2} - 1)s), we get: [ t = left( frac{a}{sqrt{2}} right)^2 = left( (sqrt{2} - 1)s right)^2 ] 5. Finally, expanding ( (sqrt{2} - 1)^2 ), we find: [ (sqrt{2} - 1)^2 = 2 - 2sqrt{2} + 1 = 3 - 2sqrt{2} ] Hence, [ t = (3 - 2sqrt{2}) s^2 ] So, the area of the triangle is ( boxed{ (3 - 2sqrt{2}) s^2 } ). Second Solution: 1. Consider a right-angled isosceles triangle with leg length (2s). The hypotenuse would be (2ssqrt{2}) and the perimeter is: [ 2s + 2s + 2ssqrt{2} = 2s(2 + sqrt{2}) ] The corresponding area is: [ t' = frac{1}{2} (2s)^2 = 2s^2 ] 2. Given that a right-angled isosceles triangle with a leg length of (a) has perimeter (2s), and comparing: [ frac{text{perimeter of our triangle}}{text{perimeter of this _derived_ triangle}} = 1 : (2 + sqrt{2}) ] [ = (sqrt{2}-1) : sqrt{2} ] 3. By the theorem which states that the ratio of the areas of two similar figures is the square of the ratio of their corresponding lengths: [ frac{t}{2s^2} = left(frac{sqrt{2} - 1}{sqrt{2}}right)^2 ] Simplify: [ = (sqrt{2}-1)^2 : (sqrt{2})^2 ] 4. And so, [ t = (sqrt{2} - 1)^2 s^2 = (3 - 2sqrt{2}) s^2 ] Thus, the area of the triangle is ( boxed{(3 - 2sqrt{2})s^2} ).

answer:For proposition (1), having equal magnitudes |vec{a}| = |vec{b}| does not imply that vectors vec{a} and vec{b} are equal. For example, let vec{a} = (1, 0) and vec{b} = (-1, 0); the vectors have the same magnitude but are not equal. Therefore, proposition (1) is incorrect. For proposition (2), if cos x = -frac{2}{3} where x in [0, pi], then the value of x is obtained by taking the arccosine of -frac{2}{3}, which is arccos(-frac{2}{3}). Since the cos function is decreasing on the interval [0, pi], x would equal pi - arccosleft(frac{2}{3}right). Hence, proposition (2) is correct. For proposition (3), if vec{a} = vec{b} and vec{b} = vec{c}, then by the transitive property of vector equality, it follows that vec{a} = vec{c}. Therefore, proposition (3) is correct. Finally, for proposition (4), if vec{a} = vec{b}, by the definition of vector equality, it necessarily follows that |vec{a}| = |vec{b}|, and furthermore, the vectors are parallel, denoted by vec{a} | vec{b}. This makes proposition (4) also correct. Therefore, the correct answer is: boxed{C}

answer:# Part (a) 1. **Inversion Transformation**: - We perform an inversion with respect to the point ( A ). Let the radii of circles ( c_1 ) and ( c_2 ) be ( r_1 ) and ( r_2 ) respectively, with ( r_1 leq r_2 ). - Under this inversion, the circles ( c_1 ) and ( c_2 ) map to straight lines passing through the points ( B_1' ) and ( C_1' ) for ( c_1 ), and ( B_2' ) and ( C_2' ) for ( c_2 ). 2. **Mapping of Circles**: - The circle ( omega ) circumscribed around triangle ( AB_1B_2 ) maps to a straight line ( overleftrightarrow{B_1'B_2'} ). - The circle ( k ) tangent to ( omega ) at ( A ) maps to a straight line ( overleftrightarrow{D_1'D_2'} ). 3. **Parallelogram Formation**: - Since ( omega ) and ( k ) are tangent at ( A ), the lines ( overleftrightarrow{B_1'B_2'} ) and ( overleftrightarrow{D_1'D_2'} ) are parallel. - This implies that ( B_1'B_2'D_2'D_1' ) forms a parallelogram. 4. **Angle Chasing**: - We have the following angle relationships: [ angle B_1'B_2'C_2' = angle AB_2'C_2' - angle AB_2'B_1' = angle AC_2B_2 - angle AB_1B_2 ] [ = angle C_1B_1B_2 - angle AB_1B_2 = angle AB_1C_1 = angle AC_1'B_1' ] - Therefore, ( B_1', B_2', C_1', C_2' ) are concyclic. 5. **Reim's Theorem**: - Using Reim's theorem, since ( B_1'B_2' parallel D_1'D_2' ), the points ( D_1', D_2', C_2', C_1' ) are concyclic. 6. **Conclusion**: - Since inversion preserves the cyclic nature of points not passing through the center of inversion, ( C_1, C_2, D_1, D_2 ) are concyclic. - If ( D_1', D_2', C_2', C_1' ) are collinear, then ( D_1, D_2, C_1, C_2 ) will be collinear. Thus, part (a) is proved. (blacksquare) # Part (b) 1. **Concyclic Condition**: - If ( B_1, B_2, D_1, D_2 ) are concyclic, their images under inversion form a parallelogram. - This implies that the angles in the parallelogram must be ( frac{pi}{2} ). 2. **Angle Verification**: - We have: [ angle AB_1C_1 = frac{pi}{2} = angle AB_2C_2 ] - This implies that ( AC_1 ) and ( AC_2 ) are diameters of ( c_1 ) and ( c_2 ) respectively. 3. **Conclusion**: - Therefore, ( B_1, B_2, D_1, D_2 ) are concyclic if and only if ( AC_1 ) and ( AC_2 ) are diameters of ( c_1 ) and ( c_2 ). Thus, part (b) is proved. (blacksquare)

answer:To find the total height of all four sandcastles, we need to add the heights of each sandcastle together. However, before we can do that, we need to make sure all the fractions have a common denominator so that they can be added together. Janet's sandcastle: 3 5/6 feet Her sister's sandcastle: 2 7/12 feet Tom's sandcastle: 1 11/20 feet Lucy's sandcastle: 2 13/24 feet First, let's find a common denominator for the fractions. The denominators are 6, 12, 20, and 24. The least common multiple (LCM) of these numbers is 120. Now, let's convert each fraction to have a denominator of 120: Janet's sandcastle: 5/6 = (5×20)/(6×20) = 100/120 Her sister's sandcastle: 7/12 = (7×10)/(12×10) = 70/120 Tom's sandcastle: 11/20 = (11×6)/(20×6) = 66/120 Lucy's sandcastle: 13/24 = (13×5)/(24×5) = 65/120 Now we can add the fractions together: 100/120 + 70/120 + 66/120 + 65/120 = 301/120 Now let's add the whole numbers: 3 (Janet's whole number) + 2 (her sister's whole number) + 1 (Tom's whole number) + 2 (Lucy's whole number) = 8 Now we combine the whole number sum with the fraction sum: 8 + 301/120 = 8 feet + 2 feet 61/120 (since 301/120 = 2 61/120) So the total height of all four sandcastles combined is: 8 feet + 2 feet 61/120 = 10 feet 61/120 Therefore, the total height of all four sandcastles is boxed{10} feet 61/120 feet.