answer:1. Let A_{1}, B_{1}, and C_{1} be the midpoints of the sides BC, CA, and AB of the acute triangle triangle ABC respectively. 2. Consider the circles circumscribing the triangles triangle A_{1}BC_{1} and triangle A C_{1} B_{1}. Let M be one of the intersection points of these two circles, different from C_{1}. 3. We need to show that M lies on the circumcircle of the triangle triangle A_{1}B_{1}C. To do this, we will utilize the fact of cyclic quadrilaterals and their properties. 4. First, observe the angles at point M: [ angle A_{1} M B_{1} = 360^circ - angle A_{1} M C_{1} - angle B_{1} M C ] [ angle A_{1} M C_{1} = 180^circ - angle A quad text{(Since M lies on the circumcircle of } triangle A_{1}BC_{1}) ] [ angle B_{1} M C = 180^circ - angle B quad text{(Since M lies on the circumcircle of } triangle AC_{1}B_{1}) ] 5. Substituting these back in: [ angle A_{1} M B_{1} = 360^circ - (180^circ - angle A) - (180^circ - angle B) ] [ = 360^circ - 180^circ + angle A - 180^circ + angle B ] [ = angle A + angle B ] 6. Since in triangle ABC, the sum of the interior angles is 180^circ: [ angle A + angle B + angle C = 180^circ ] Therefore, [ angle A + angle B = 180^circ - angle C ] 7. Thus, we have: [ angle A_{1} M B_{1} = 180^circ - angle C ] 8. This implies that points C, A_{1}, M, and B_{1} lie on the same circle (since the opposite angles of a cyclic quadrilateral sum to 180^circ). 9. We have hence shown that M, which was an intersection point of the circumcircles of two selected triangles, indeed lies on the circumcircle of the third triangle formed by the midpoints and one original vertex. 10. By repeating this method for the other pairs of circles, we show that this common intersection point M lies on all three circumcircles. # Conclusion: [ boxed{text{The circles circumscribing the three triangles have a common point.}} ]

answer:1. **Determine the radii**: - Radius of the inner circle, ( R_i ), is ( r ). - Radius of the outer circle, ( R_o ), is ( 3r ). 2. **Calculate the area of the outer circle** using the formula ( A = pi r^2 ): [ A_o = pi (3r)^2 = 9pi r^2 ] 3. **Calculate the area of the inner circle**: [ A_i = pi (r)^2 = pi r^2 ] 4. **Calculate the area of the gray region** by subtracting the area of the inner circle from the area of the outer circle: [ A_{text{gray}} = A_o - A_i = 9pi r^2 - pi r^2 = 8pi r^2 ] Conclusion: The area of the gray region is (boxed{8pi r^2}) square feet.

answer:To solve this problem, we need to visualize or understand the sequence and folding mechanism of the squares to form a cube. The additional challenge here is the yellow lighting, which might modify the perceived colors. Configuration: 1. Assume the sequence starts with red (R) and ends with black (Bl). In a logical configuration, adjacent squares in sequence can be folded to form sides of the cube. 2. Since the squares are consecutive, let's place black (Bl) as the bottom face of the cube. 3. Red (R), being the first in sequence, can be folded to be on top. 4. The remaining colors will fold into the other faces. Given their sequence, purple (P) can be on one side, and directly opposite it would typically be white (W). However, under yellow lighting, white may appear as light yellow. In this configuration, the face opposite black (Bl) is white (W), but under yellow lighting, it appears as light yellow. Conclusion: The face opposite the black face, under yellow lighting, appears as light yellow. Therefore, the answer is text{light yellow}. The final answer is boxed{text{(C)} text{Light Yellow}}

answer:Given the vertical symmetry and vertex at (6, -2), the vertex form of the parabola is: [ y = a(x - 6)^2 - 2 ] Using the point (3, 0), substitute into the equation: [ 0 = a(3 - 6)^2 - 2 ] [ 0 = a(9) - 2 ] [ a cdot 9 = 2 ] [ a = frac{2}{9} ] Now, insert a back into the vertex form and expand: [ y = frac{2}{9}(x - 6)^2 - 2 ] [ y = frac{2}{9}(x^2 - 12x + 36) - 2 ] [ y = frac{2}{9}x^2 - frac{24}{9}x + frac{72}{9} - 2 ] [ y = frac{2}{9}x^2 - frac{24}{9}x + 6 - 2 ] [ y = frac{2}{9}x^2 - frac{24}{9}x + 4 ] Comparing this with y = ax^2 + bx + c, we get: [ a = frac{2}{9}, quad b = -frac{24}{9}, quad c = 4 ] Sum a + b + c: [ a + b + c = frac{2}{9} - frac{24}{9} + 4 ] [ a + b + c = frac{2 - 24}{9} + 4 ] [ a + b + c = frac{-22}{9} + 4 ] [ a + b + c = frac{-22+36}{9} ] [ a + b + c = boxed{frac{14}{9}} ]