answer:1. **Setup:** Let us consider an arbitrary triangle (ABC) on a plane. To find the geometric locus of the centers of equilateral triangles circumscribed around (ABC), we start by describing the setup more precisely. Let the points (E, F,) and (G) be such that (EF, FG,) and (GE) are the sides of an equilateral triangle (EFG), while they also pass through the vertices (A, B,) and (C) respectively. 2. **Angles Property:** Due to the symmetry of the equilateral triangle (EFG), the angles between the lines (FG, GE,) and (EF) are ( pm 60^circ ). Therefore, each vertex of triangle (E, F,) and (G) must satisfy the angle criterion for intersecting lines: [ angle (GE, EF) = angle (EF, FG) = angle (FG, GE) = pm 60^circ ] Consequently, we also have: [ angle(BE, EC) = angle(CF, FA) = angle(AG, GB) = pm 60^circ ] 3. **Determining Centers:** By choosing one of the angle signs (let's say (+60^circ) for this part), we find that there are three circles (S_E, S_F,) and (S_G) that must include points (E, F,) and (G) respectively. Any chosen point (E) on (S_E) uniquely determines the equilateral triangle (EFG). Let the centers of these triangles be denoted as (O). Then other notable points are: [ P, Q, R ] Where (P, Q,) and (R) are points of intersection of a line through (O): [ OE, OF, OG quad ] With their respective circles (S_E, S_F,) and (S_G). 4. **Angle Considerations:** Since (P, Q,) and (R) are the centers of equilateral triangles constructed on sides (AB, BC,) and (CA), we have: [ angle (XB, BP) = angle (CF, FP) = mp 30^circ ] Where (X) is an appropriate point (due to symmetry). Also: [ angle (BR, CR) = angle (GE, EF) = pm 60^circ ] Hence, combining these: [ angle (CB, CP) = pm 30^circ + pm 60^circ = pm 90^circ ] Therefore, (P, Q,) and (R) are indeed the centers of equilateral triangles built on the sides of (ABC). 5. **Equilateral Triangle Property:** The theorem of Napoleon suggests that if three equilateral triangles are built on the three sides of any triangle (either all outward or all inward), then the centers of these equilateral triangles also form an equilateral triangle. Specifically: [ triangle PQR text{ is equilateral} ] Moreover, its center coincides with the intersection point of the medians of triangle (ABC). This geometric property can be verified by checking that ( angle (PR, RQ) = angle (OE, OG) = angle (OP, OQ) = pm 60^circ ). Thus, the center (O) of triangle (EGF) lies on the circumcircle of (triangle PQR). 6. **Conclusion:** The geometric locus of the centers of all equilateral triangles circumscribed about a given triangle (ABC) is the described circumcircle of the equilateral triangle (triangle PQR), with points (P, Q,) and (R) being the centers of the equilateral triangles on the sides of (ABC). Therefore, the center (O) indeed lies on the circle inscribed in (triangle PQR), showing that this is the required geometric locus. [ boxed{text{Circumcircle of the equilateral triangle formed by the centers of equilateral triangles on the sides of } triangle ABC} ]

answer:To solve this problem, we first need the prime factorization of 48. Then, we will determine which factors of 48 are also multiples of 6. 1. **Prime Factorization of 48**: [ 48 = 2^4 cdot 3^1 ] This factorization shows that 48 is composed of 4 twos and one three. 2. **Factors of 48**: The total number of factors of a number given its prime factorization ( p^a cdot q^b ) is ( (a+1)(b+1) ). For 48, this calculation is: [ (4+1)(1+1) = 5 cdot 2 = 10 ] These factors are: ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 ). 3. **Multiples of 6 among the Factors**: A multiple of 6 must have at least one 2 and one 3 in its prime factorization. Therefore, we need to find factors of 48 that include at least ( 2^1 cdot 3^1 ). From the list of factors, we identify those that are multiples of 6: - ( 6 = 2^1 cdot 3^1 ) - ( 12 = 2^2 cdot 3^1 ) - ( 24 = 2^3 cdot 3^1 ) - ( 48 = 2^4 cdot 3^1 ) 4. **Counting the Multiples of 6**: We have identified 4 factors of 48 that are multiples of 6: 6, 12, 24, and 48. Therefore, there are 4 positive factors of 48 that are also multiples of 6. The correct answer is 4. The final answer is boxed{(C) 4}

answer:1. **Identify given information and relationships:** - ( O ) is the center of the circle. - ( AB perp BC ), creating a right triangle ( ABO ). - ( ADOE ) is a straight line, ( AD = AP ). - ( AB = 4r ). 2. **Calculate ( AO ):** Since ( AB perp BC ) and ( AB = 4r ), triangle ( ABO ) is a right triangle with ( AB ) as hypotenuse. Applying the Pythagorean theorem: [ AO = sqrt{AB^2 - OB^2} = sqrt{(4r)^2 - r^2} = sqrt{16r^2 - r^2} = sqrt{15r^2} = rsqrt{15} ] 3. **Determine ( AD ) and ( PB ):** - ( AP = AD = AO - DO = rsqrt{15} - r ) (since ( DO = r ), the radius). - ( PB = AB - AP = 4r - rsqrt{15} ). 4. **Evaluate ( AP^2 = PB times AB ):** [ AP^2 = (rsqrt{15} - r)^2 = r^2(15 - 2sqrt{15} + 1) = r^2(16 - 2sqrt{15}) ] [ PB times AB = (4r - rsqrt{15}) times 4r = (4r^2 - r^2sqrt{15}) times 4 = 16r^2 - 4r^2sqrt{15} ] Simplifying and factoring back in ( r^2 ): [ PB times AB = 16r^2 - 4r^2sqrt{15} ] Both expressions are equal, making the statement true. 5. **Conclusion:** The given statement ( AP^2 = PB times AB ) holds true. Therefore, the final boxed answer is: [ text{True} ] boxed{- The correct answer is (textbf{A. True})}

answer:Given the problem of determining the lengths of the midlines of a triangle when its side lengths are provided, let's proceed with the solution step-by-step. 1. **Define the midpoints and midlines**: Let A, B, and C be the vertices of the triangle. The midpoints of the sides BC, CA and AB are A_{1}, B_{1} and C_{1} respectively. 2. **Concept of midline in geometry**: A midline in a triangle is a line segment connecting the midpoints of two sides of the triangle. By definition, a midline is parallel to the third side and is half its length. 3. **Identifying the midlines**: - ( A_{1}B_{1} ) is the midline parallel to AC. - ( B_{1}C_{1} ) is the midline parallel to AB. - ( C_{1}A_{1} ) is the midline parallel to BC. 4. **Using midline property**: Since midlines are half the length of their respective parallel sides, their lengths can be determined as follows: Let the sides of the triangle be represented as: [ BC = a, quad CA = b, quad AB = c ] 5. **Calculation of midline lengths**: [ text{Length of midline } A_{1}B_{1} = frac{1}{2} b ] [ text{Length of midline } B_{1}C_{1} = frac{1}{2} c ] [ text{Length of midline } C_{1}A_{1} = frac{1}{2} a ] 6. **Combining results**: Therefore, the lengths of the midlines of the triangle are: [ s_{a} = frac{1}{2} b, quad s_{b} = frac{1}{2} c, quad s_{c} = frac{1}{2} a ] # Conclusion: Thus, we conclude that the midlines of the triangle are half the lengths of the sides: [ boxed{ s_{a} = frac{1}{2} b, quad s_{b} = frac{1}{2} c, quad s_{c} = frac{1}{2} a } ]