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question:Two circles X and Y in the plane intersect at two distinct points A and B such that the centre of Y lies on X. Let points C and D be on X and Y respectively, so that C, B and D are collinear. Let point E on Y be such that DE is parallel to AC. Show that AE = AB.
answer:1. **Identify the given elements and their relationships:** - Let the centers of circles (X) and (Y) be (O_X) and (O_Y) respectively. - Since (O_Y) lies on circle (X), (O_XO_Y = R_X), where (R_X) is the radius of circle (X). - Circles (X) and (Y) intersect at points (A) and (B). - Points (C) and (D) lie on circles (X) and (Y) respectively, such that (C), (B), and (D) are collinear. - Point (E) on circle (Y) is such that (DE parallel AC). 2. **Establish the parallelism condition:** - Since (DE parallel AC), the corresponding angles are equal: (angle EDB = angle BCA). 3. **Use the cyclic nature of the circles:** - Since (A) and (B) are points of intersection, quadrilateral (ABCD) is cyclic. - Therefore, (angle BCA = angle BDA). 4. **Analyze the angles:** - Since (DE parallel AC), (angle EDB = angle BCA). - From the cyclic nature, (angle BCA = angle BDA). - Thus, (angle EDB = angle BDA). 5. **Use the properties of cyclic quadrilaterals:** - In cyclic quadrilateral (ABED), (angle EBA = angle EDA) (opposite angles in a cyclic quadrilateral are supplementary). 6. **Conclude the equality of segments:** - Since (angle EBA = angle EDA) and (angle EDB = angle BDA), it follows that (angle EBA = angle BEA). - Therefore, (AB = AE). (blacksquare)
question:triangle DEF, DE = DF = 120, and EF = 80. Circle R has a radius of 20 and is tangent to overline{DF} and overline{EF}. Circle S is externally tangent to R and is tangent to overline{DE} and overline{EF}. No point of circle S lies outside of triangle DEF. The radius of circle S can be expressed in the form p qsqrt {l}, where p, q, and l are positive integers, and l is the product of distinct primes. Find p + ql.
answer:Let X and Y be the feet of the perpendiculars from R and S to EF, respectively. Let the radius of odot S be s. Given that RS = s + 20. Construct overline{SM} parallel overline{EF} with M on RX. The right triangle RSM has sides s + 20, s - 20, and XY. To compute XF, triangle RFX is considered. Since odot R is tangent to overline{DF} and overline{EF}, RF bisects angle DFE. Set angle DFE = 2phi. Construct the altitude from D to EF, noting a scaled 3-4-5 right triangle (9-12-15 by multiplying by 3). Then, tan(phi) = frac{4}{3}. We find XF = 80 - frac{5s + 80}{3} = frac{160 - 5s}{3}. In right triangle RSM, using the Pythagorean Theorem: [ (s + 20)^2 + (s - 20)^2 = left(frac{160 - 5s}{3}right)^2 ] Solving this equation, you may find s in terms of radicals. Calculating actual values: [ s = 54 - 8sqrt{41} ] Thus, the answer is boxed{338}.
question:Given that a is a digit from 1 to 9, if the repeating decimal 0.1overset{cdot }{a} equals frac {1}{a}, then a = ______.
answer:Since 0.1overset{cdot }{a} equals frac {1}{a}, and 0.1overset{cdot }{6} equals frac {1}{6}, thus a = 6. Therefore, the answer is: boxed{6} . According to the definition of repeating decimals, from 0.1overset{cdot }{a} equals frac {1}{a}, substituting digits 1 to 9 into a, it is found that only when a = 6 does it satisfy the condition. This method can be used to solve the problem. This problem examines the representation of numbers with letters, repeating decimals and their classification, and the method of taking special values for solving.
question:What is the greatest product obtainable from two integers whose sum is 300?
answer:Let the two integers be ( x ) and ( 300 - x ). The product of these integers, which we seek to maximize, is: [ P = x(300 - x) = 300x - x^2 ] To find the maximum product, we complete the square: [ P = -x^2 + 300x = -(x^2 - 300x) ] Now, add and subtract (left(frac{300}{2}right)^2 = 22500) inside the expression: [ P = -(x^2 - 300x + 22500 - 22500) = -(x - 150)^2 + 22500 ] Since the square of a real number is always non-negative, (-(x - 150)^2 leq 0). The expression (22500 - (x - 150)^2) is maximized when (x - 150 = 0), i.e., (x = 150). Therefore, the maximum product is: [ P_{text{max}} = 22500 - 0^2 = 22500 ] Thus, the greatest product obtainable is (boxed{22500}).