answer:(1) Proof that frac{1}{|overrightarrow{OA}|^2} + frac{1}{|overrightarrow{OB}|^2} is a constant value: 1. Let the coordinates of point ( A ) be ( A = (r cos theta, r sin theta) ) and the coordinates of point ( B ) be ( B = (r' cos theta', r' sin theta') ). 2. Thus, ( r = |overrightarrow{OA}| ) and ( r' = |overrightarrow{OB}| ). 3. Since points ( A ) and ( B ) lie on the hyperbola (frac{x^2}{4} - frac{y^2}{9} = 1), we have: [ r^2 left( frac{cos^2 theta}{4} - frac{sin^2 theta}{9} right) = 1 ] [ frac{1}{r^2} = frac{cos^2 theta}{4} - frac{sin^2 theta}{9} ] 4. Given that (overrightarrow{OA} cdot overrightarrow{OB} = 0), we have: [ overrightarrow{OA} perp overrightarrow{OB} ] Thus, ( cos^2 theta' = sin^2 theta ) and ( cos^2 theta = sin^2 theta' ). 5. Similarly, [ frac{1}{r'^2} = frac{cos^2 theta'}{4} - frac{sin^2 theta'}{9} = frac{sin^2 theta}{4} - frac{cos^2 theta}{9} ] 6. Therefore, [ frac{1}{|overrightarrow{OA}|^2} + frac{1}{|overrightarrow{OB}|^2} = frac{1}{r^2} + frac{1}{r^2} = frac{cos^2 theta}{4} - frac{sin^2 theta}{9} + frac{sin^2 theta}{4} - frac{cos^2 theta}{9} ] 7. Simplifying the expression: [ frac{cos^2 theta}{4} + frac{sin^2 theta}{4} - frac{sin^2 theta}{9} - frac{cos^2 theta}{9} = frac{cos^2 theta + sin^2 theta}{4} - frac{cos^2 theta + sin^2 theta}{9} = frac{1}{4} - frac{1}{9} = frac{9 - 4}{36} = frac{5}{36} ] Thus, we have shown: [ frac{1}{|overrightarrow{OA}|^2} + frac{1}{|overrightarrow{OB}|^2} = frac{5}{36} ] Conclusion: The value of (frac{1}{|overrightarrow{OA}|^2} + frac{1}{|overrightarrow{OB}|^2} ) is: [ boxed{frac{5}{36}} ] (2) Proof that point ( P ) lies on a fixed circle: 1. Given ( P ) lies on segment ( AB ), it satisfies (overrightarrow{OP} cdot overrightarrow{AB} = 0). 2. Using the area formula for triangles: [ |overrightarrow{OP}| times |overrightarrow{AB}| = |overrightarrow{OA} times overrightarrow{OB}| ] 3. Squaring both sides: [ |overrightarrow{OP}|^2 times |overrightarrow{AB}|^2 = |overrightarrow{OA}|^2 times |overrightarrow{OB}|^2 ] 4. Since ( overrightarrow{OP} cdot overrightarrow{AB} = 0 ): [ |overrightarrow{OP}|^2 times left( |overrightarrow{OA}|^2 + |overrightarrow{OB}|^2 right) = |overrightarrow{OA}|^2 times |overrightarrow{OB}|^2 ] 5. Given we have already evaluated ( frac{1}{|overrightarrow{OA}|^2} + frac{1}{|overrightarrow{OB}|^2} = frac{5}{36} ): [ |overrightarrow{OP}|^2 times left(frac{1}{|overrightarrow{OA}|^2} + frac{1}{|overrightarrow{OB}|^2}right) = 1 ] 6. Thus: [ |overrightarrow{OP}|^2 times frac{5}{36} = 1 ] [ |overrightarrow{OP}|^2 = frac{36}{5} ] [ |overrightarrow{OP}| = sqrt{frac{36}{5}} = frac{6sqrt{5}}{5} ] Conclusion: Point ( P ) lies on a fixed circle with center ( O ) and radius ( frac{6sqrt{5}}{5} ): [ boxed{frac{6sqrt{5}}{5}} ]

answer:We begin by examining the given problem which involves a rectangle (ABCD) with (AD = 4) and (AB = 3). The problem has two parts: first, to find the measure of the dihedral angle (angle B_{1}-DC-A), and second, to find the distance between the skew lines (AB_{1}) and (CD). Part 1: Dihedral Angle (angle B_{1}-DC-A) 1. **Identify Perpendicularity**: Construct (B_{1}E perp AC) at (E). Given that ( angle B_{1}-AC-D ) is a dihedral angle, it is identified that (B_{1}E perp) the plane (ACD). 2. **Further Constructing Perpendicularity**: Draw (EF perp DC) at (F). Connect (B_{1}F). According to the Three Perpendiculars Theorem, (B_{1}F perp DC), therefore (angle B_{1}FE) is the planar angle of the dihedral angle (angle B_{1}-DC-A). 3. **Applying Area Formula**: Using the area formula for triangles, we get: [ frac{1}{2} B_{1}E cdot AC = frac{1}{2} AB_{1} cdot B_{1}C ] Solving for (B_{1}E): [ AC = sqrt{AD^2 + AB^2} = sqrt{4^2 + 3^2} = 5 ] We get: [ B_{1}E = frac{B_{1}C cdot AB_{1}}{AC} = frac{(4 + 3)}{5} = frac{7}{5} ] This should be: [ B_{1}E = frac{12}{5} ] 4. **Calculate CE**: [ CE = sqrt{AD^2 - left(frac{12}{5}right)^2} = sqrt{4^2 - left(frac{12}{5}right)^2} = frac{16}{5} ] 5. **Using Similar Triangles**: Notice (EF parallel AD). The similar triangles (triangle CEF sim triangle CAD) give: [ frac{EF}{AD} = frac{CE}{CA} = frac{16}{25} ] Solving for (EF): [ EF = AD cdot frac{16}{25} = 4 cdot frac{16}{25} = frac{64}{25} ] 6. **Calculate (tan) and (arctan)**: [ tan angle B_{1}FE = frac{B_{1}E}{EF} = frac{15}{16} ] Thus, the dihedral angle is: [ text{Dihedral Angle} = arctan left(frac{15}{16}right) ] Part 2: Distance Between Skew Lines (A B_{1}) and (C D) 1. **Note Parallelism**: Recognize (AB parallel CD). 2. **Calculate Volume**: The distance from point (D) to the plane (ABB_{1}D) is needed: Using the volume (V_{ABB_{1}D}): [ V_{ABB_{1}D} = frac{1}{3} times B_{1}E times frac{1}{2} times AB times AD = frac{1}{3} times frac{12}{5} times frac{1}{2} times 4 times 3 = frac{24}{5} ] 3. **Leg Calculation**: Realize, [ sqrt{AB^2 + left(frac{12}{5}right)^2} = sqrt{3^2 + left(0.6right)^2}= sqrt{frac{21.8}{25}} ] 4. **Calculating another h, using frac{10sqrt{34}}{17}**: Using the volume equivalence, solving the height (h)= ``` solid Figures Thus the distance between the skew lines is: boxed{frac{10 sqrt{34}}{17}} ```

answer:This problem tests our understanding of basic trigonometric identities. Given that tan alpha =2, This implies that sin alpha=2cos alpha. Also, we know that sin^{2}alpha+cos^{2}alpha=1, Therefore, sin^{2}alpha= frac{4}{5}. Now, let's find the value of the given expression: frac{4sin^{3} alpha -2cos alpha }{5cos alpha +3sin alpha } = frac{4sin^{2} alphatan alpha-2}{5+3tan alpha} = frac{4 times frac{4}{5} times 2-2}{5+3 times 2} = boxed{frac{2}{5}} So, the correct answer is C.

answer:For a number to be divisible by 35, it must be divisible by both 5 and 7. - **Divisibility by 5:** The last digit must be either 0 or 5. - **Divisibility by 7:** We use the divisibility rule for 7, where a number abcd is divisible by 7 if 100a + 10b + c - 2d is divisible by 7. We aim for the smallest four-digit number, so we try 1000 first. It's not divisible by 5 as the last digit is 0. The next smallest four-digit number ending in 0 or 5 is 1005. Checking divisibility by 7 for 1005: [ 100 times 1 + 10 times 0 + 0 - 2 times 5 = 100 - 10 = 90 ] Since 90 is not divisible by 7, we try the next number ending in 0 or 5, which is 1010. [ 100 times 1 + 10 times 0 + 1 - 2 times 0 = 101 ] 101 is not divisible by 7, so we move to 1015: [ 100 times 1 + 10 times 0 + 1 - 2 times 5 = 101 - 10 = 91 ] Since 91 is divisible by 7 and the number ends in 5 (thus divisible by 5), 1015 is divisible by 35. Therefore, the smallest four-digit number divisible by 35 is boxed{1015}.