answer:1. We start with the given equations: [ a + b - c = 1 ] [ a^2 + b^2 - c^2 = -1 ] 2. From the first equation, we can express (c) in terms of (a) and (b): [ c = a + b - 1 ] 3. Substitute (c = a + b - 1) into the second equation: [ a^2 + b^2 - (a + b - 1)^2 = -1 ] 4. Expand and simplify the equation: [ a^2 + b^2 - (a^2 + b^2 + 1 + 2ab - 2a - 2b) = -1 ] [ a^2 + b^2 - a^2 - b^2 - 1 - 2ab + 2a + 2b = -1 ] [ -1 - 2ab + 2a + 2b = -1 ] 5. Simplify further: [ -2ab + 2a + 2b = 0 ] [ 2(a + b - ab) = 0 ] [ a + b - ab = 0 ] [ ab = a + b ] 6. Rewrite the equation (ab = a + b) as: [ ab - a - b + 1 = 1 ] [ (a-1)(b-1) = 1 ] 7. Solve for (a) and (b) by considering the factor pairs of 1: - If (a-1 = 1) and (b-1 = 1), then (a = 2) and (b = 2). - If (a-1 = -1) and (b-1 = -1), then (a = 0) and (b = 0). 8. Substitute these values back into the expression for (c): - For (a = 2) and (b = 2): [ c = 2 + 2 - 1 = 3 ] - For (a = 0) and (b = 0): [ c = 0 + 0 - 1 = -1 ] 9. Calculate (a^2 + b^2 + c^2) for each set of values: - For ((a, b, c) = (2, 2, 3)): [ a^2 + b^2 + c^2 = 2^2 + 2^2 + 3^2 = 4 + 4 + 9 = 17 ] - For ((a, b, c) = (0, 0, -1)): [ a^2 + b^2 + c^2 = 0^2 + 0^2 + (-1)^2 = 0 + 0 + 1 = 1 ] 10. Sum the possible values of (a^2 + b^2 + c^2): [ 17 + 1 = 18 ] The final answer is (boxed{18}).

answer:1. **Diagram**: Set up the circles such that O at origin, circle O has radius 12, and circle P has radius 4, centered at (12-4, 0) = (8, 0) since they are internally tangent at (12, 0). 2. **External tangent**: Draw the tangent TS, and since TS touches both circles externally, it is perpendicular to the radii at points T and S. Let S' be the reflection of S across O, making S'P = SP = 4 units and S' on the circumference of circle O. The total distance from S' to T through O is twice the radius of O, so S'T = 24 units. 3. **Calculating OP**: The triangles to consider are triangle OS'P and triangle OTS. Since OT = OS' = 12 (radii of larger circle) and S'T = 24 which folds over at O, each half TS' and TO is 12 units. Triangle triangle OS'P is a right triangle with OS' = 12 units, and S'P = 4 units, the distance OP is the hypotenuse, given by the Pythagorean theorem: [ OP = sqrt{OS'^2 + S'P^2} = sqrt{12^2 + 4^2} = sqrt{144 + 16} = sqrt{160} = 4sqrt{10} ] Therefore, the length of OP is boxed{4sqrt{10}}.

answer:Let's denote the width of the splash made by a pebble as P meters. According to the information given: - 3 rocks make a splash that is 3 * (1/2) = 1.5 meters wide in total. - 2 boulders make a splash that is 2 * 2 = 4 meters wide in total. The total width of the splashes made by rocks and boulders is 1.5 + 4 = 5.5 meters. Since the total width of all the splashes is 7 meters, the width of the splashes made by the pebbles is the remaining width after subtracting the width of the splashes made by rocks and boulders. So, the total width of the splashes made by pebbles is 7 - 5.5 = 1.5 meters. TreQuan tossed 6 pebbles, so the width of the splash made by each pebble is 1.5 meters divided by 6. Therefore, the width of the splash made by a pebble is 1.5 / 6 = boxed{0.25} meters.

answer:1. **Understanding the Geometry**: We are given that the base of the tetrahedron (S-ABC) is an equilateral triangle. Point (A) is on the projection line of the face (SBC) (the vertex (H)) which is the orthocenter of (triangle SBC). Let's add that: - (SA = a) - (B) and (C) are vertices such that (S) projects orthogonally to the plane (ABC). 2. **Projection and Orthocenter Relationships**: We can leverage perpendicularity and projections: - (BH perp SC) (where (H) is a projection line meeting at point (E)). - According to the Perpendicular Line Theorem (Triangle Orthocenter), (SC perp AE) and (SC perp AB), implying (SC perp) plane (ABE). 3. **Intersection Analysis**: Let's determine the orthocenter (O) of triangle (ABC): - Construct (SO perp) plane (ABC), intersecting at (O). - Since (ABC) is an equilateral triangle, (O) is both the orthocenter and centroid. Specifically: - (CO) is the altitude of the equilateral triangle inside (ABC), making (CO perp AB). - Similarly, (BO perp AC). Thus, (O) is indeed the centroid (center of gravity) of (triangle ABC). 4. **Geometric Relations and Volumes**: Given the properties of an equilateral triangle: - Let (AB = x) implying area (S_{triangle ABC} = frac{sqrt{3}}{4} x^2). - For centroid to vertex distances, (OC = frac{2}{3} times frac{xsqrt{3}}{2} = frac{sqrt{3}}{3} x). - The height (SO) of the tetrahedron is derived from (SO = sqrt{a^2 - left(frac{xsqrt{3}}{3}right)^2} = sqrt{a^2 - frac{x^2}{3}}). 5. **Calculating the Tetrahedron Volume**: Let's express the volume (V_{S-ABC}) as: [ V_{S-ABC} = frac{1}{3} times S_{triangle ABC} times SO = frac{1}{3} times frac{sqrt{3}}{4} x^2 times sqrt{a^2 - frac{x^2}{3}} ] Simplifying further: [ V_{S-ABC} = frac{sqrt{3}}{12} x^2 sqrt{a^2 - frac{x^2}{3}} ] Using algebraic simplification, we deduce: [ frac{sqrt{3}}{12} x^2 sqrt{a^2 - frac{x^2}{3}} = frac{1}{6} sqrt{frac{x^2}{2} times frac{x^2}{2} times (3a^2 - x^2)} ] 6. **Applying AM-GM Inequality**: Applying the Arithmetic Mean - Geometric Mean Inequality (AM-GM Inequality), we conclude: [ frac{1}{6} sqrt{left(frac{frac{1}{2} x^2 + frac{1}{2} x^2 + 3a^2 - x^2}{3}right)^3} = frac{1}{6} a^3 ] The equality holds when (x = sqrt{2} a). # Conclusion: [ boxed{frac{1}{6} a^3} ]