answer:1. **Identify the Key Elements:** - We are given a triangle ( triangle ABC ) with (angle C = 135^circ). - A square is constructed on the side ( AB ), outside of the triangle, with its center at point ( O ). We need to find the length ( OC ), given that ( AB = 6 ). 2. **Construct the Square:** - Let the vertices of the square be ( A, B, D, E ), where ( D ) and ( E ) are outside the triangle and ( ABDE ) forms the square. - Given that ( AB ) is a side of the square and ( AB = 6 ), all sides of the square are equal to 6. 3. **Diagonal of the Square:** - The diagonal ( AD ) of the square forms a right angle of ( 45^circ ) with any of its sides (since all angles in a square are ( 90^circ )). - The length of the diagonal ( AD ) can be calculated using the Pythagorean theorem: [ AD = sqrt{AB^2 + AD^2} = sqrt{6^2 + 6^2} = sqrt{36 + 36} = sqrt{72} = 6sqrt{2} ] 4. **Understanding the Circumscribed Circle:** - Note that the quadrilateral ( ACBD ) can be inscribed in a circle. This means ( angle ACB + angle ADB = 180^circ ). - Given ( angle ACB = 135^circ ), the angles ( angle ADB = 45^circ ). As the diagonals of a square intersect at ( 45^circ ), ( O ), the center of the square, is the midpoint of the diagonal. 5. **Radius of the Circumscribed Circle:** - The radius ( OC ) of the circle circumscribing ( ACBD ) is half of diagonal ( AD ) of the square, since ( O ) is at the midpoint. [ OC = frac{1}{2} times 6sqrt{2} = 3sqrt{2} ] Conclusion Thus, the length ( OC ) is ( 3sqrt{2} ). [ boxed{3sqrt{2}} ]

answer:1. **Initialization and Assumptions:** Let k = 100001 cdot 10^r + x, where x < 10^r. 2. **Representing Factorial in Terms of Variables:** Consider the representation of (k-1)! as: [ (k-1)! = y cdot 10^s + z, quad text{where } z < 10^s text{ and } y text{ is a 6-digit number.} ] 3. **Upper Bound Calculation:** We calculate the upper bound of k! using the following inequality: [ begin{aligned} k! &= (100001 cdot 10^r + x) cdot (y cdot 10^s + z) &< (100001 cdot 10^r + 10^r) cdot (y cdot 10^s + 10^s) &= (100001 + 1) cdot 10^r cdot (y cdot 10^s + 10^s) &< 100002 cdot 10^r cdot y cdot 10^s + 100002 cdot 10^r cdot 10^s &= 100002 cdot y cdot 10^{r+s} + 100002 cdot 10^{r+s+1} &< (y + frac{100}) cdot 10^{r+s+5} + 10^{r+s+5}, end{aligned} ] assuming 100002 leq 10^6 + 10^5 leq y + 100 for the simplification. 4. **Lower Bound Calculation:** On the other hand, we can lower-bound k! as follows: [ begin{aligned} k! &= (100001 cdot 10^r + x) cdot (y cdot 10^s + z) &geq (100001 cdot 10^r) cdot (y cdot 10^s) &= 100001 cdot y cdot 10^{r+s} &geq y cdot 10^{r+s+5}. end{aligned} ] 5. **Combined Inequality:** From the upper and lower bounds, we obtain: [ (y + 1) cdot 10^{r+s+5} leq k! leq (y + 100) cdot 10^{r+s+5} + 10^{r+s+5}. ] 6. **Conclusion on Number of Digits:** Therefore, the factorial k! can either have r+s+11 digits or r+s+12 digits. 7. **Starting Digits of Factorial:** If k! has r+s+12 digits, then the first four digits must be 1000 since it starts with 1000.... If k! has r+s+11 digits, the first four digits must lie between y and y+1 or from y+100. In this way, the first four digits could be derived from y. 8. **Ensuring All Combinations:** For r geq 6 and considering x = 0, 1, 2, ldots, 10^6 - 1, the first four digits of 10^6 possible k! will cover all possible 4-digit numbers. Therefore, there exists some k such that the first four digits of k! can be precisely 1966. # Conclusion: Therefore, there exists a natural number k such that k! starts with 1966. blacksquare

answer:**Answer** The y-intercept of a line is found where x = 0. Substituting x = 0 into the equation y = x - 2, we get y = 0 - 2 = -2. Therefore, the y-intercept of the line is boxed{-2}.

answer:**Step 1:** Use the cofunction identities sin(3π-θ)=sin θ and sin(dfrac{π}{2}+θ)=cos θ. The expression becomes: sqrt{1-2sin θcos θ} **Step 2:** Use the double angle identity cos 2θ = 1 - 2sin θcos θ and solve for 1-2sin θcos θ 1 - 2sin θcos θ = cos 2θ So, the expression becomes: sqrt{cos 2θ} **Step 3:** As θ∈(0, dfrac{π}{4}), then 2θ∈(0, dfrac{π}{2}). In this interval, cos 2θ is positive. So, we can write: sqrt{cos 2θ} = sqrt{(cos θ - sin θ)^2} **Step 4:** Since θ∈(0, dfrac{π}{4}), cos θ > sin θ. Therefore, cos θ - sin θ is positive. This implies: sqrt{(cos θ - sin θ)^2} = cos θ - sin θ So the final answer is: boxed{cos θ - sin θ}