answer:We start by analyzing the problem where we are given a closed broken line on a plane intersected by a straight line ( l ) exactly in 1985 points, and we need to show that there exists another line intersecting this broken line in more than 1985 points. 1. **Division by Line ( l ):** The line ( l ) divides the plane into two half-planes. Let's call one of them the upper half-plane and the other the lower half-plane. 2. **Counting Vertices:** Let ( n_1 ) be the number of vertices of the broken line lying on ( l ) where both segments extending from these vertices lie in the upper half-plane. Similarly, let ( n_2 ) be the number of vertices where both segments lie in the lower half-plane. Let ( m ) be the number of other intersection points of ( l ) with the broken line. 3. **Closed Path Traversal:** If we traverse the entire broken line starting from a point not on ( l ) and return to the same point, we cross the line ( l ) only at the m points where the broken line intersects ( l ) and not exactly at the vertices on ( l ). Since we return to the starting point, the total number of times we cross ( l ) must be even: [ m quad text{is even.} ] 4. **Equation Setup:** Given ( n_1 + n_2 + m = 1985 ), since ( m ) is even, the sum ( n_1 + n_2 ) must be odd to make the left side odd (to match the right side which is 1985). Therefore, [ n_1 + n_2 quad text{is odd} rightarrow n_1 neq n_2. ] 5. **Assumption for Further Proof:** Without loss of generality, assume ( n_1 > n_2 ). 6. **Introducing New Line ( l_1 ):** Draw a new line ( l_1 ) in the upper half-plane, parallel to ( l ) and sufficiently close to it such that ( l_1 ) doesn't intersect any vertices of the broken line not lying on ( l ). The number of intersection points of the broken line with ( l_1 ) will differ from ( l ). Specifically, we count each intersection by considering the vertices and the crossing points beside vertices: [ text{Number of intersection points} = 2n_1 + m, ] where the factor of 2 accounts for crossings through segments extending through vertices in the upper plane. 7. **Final Calculation:** We now show this new count exceeds 1985: [ 2n_1 + m > n_1 + n_2 + m = 1985. ] 8. **Conclusion:** Thus, ( l_1 ) intersects the broken line in more than 1985 points, fulfilling the required condition. [ boxed{} ]

answer:To solve for the sum of the lengths of all diagonals of pentagon ABCDE, we first denote the lengths of the diagonals as follows: - Let a be the length of a diagonal opposite adjacent sides of length 14 and 3. - Let b be the length of a diagonal opposite adjacent sides of length 14 and 10. - Let c be the length of a diagonal opposite adjacent sides of length 3 and 10. Applying Ptolemy's Theorem to the five possible quadrilaterals formed by the sides and diagonals of the pentagon, we obtain the following equations: 1. c^2 = 3a + 100 2. c^2 = 10b + 9 3. ab = 30 + 14c 4. ac = 3c + 140 5. bc = 10c + 42 From equations (1) and (2), we can express a and b in terms of c: [a = frac{c^2 - 100}{3}] [b = frac{c^2 - 9}{10}] Substituting the expression for a into equation (4), we get: [frac{c^2 - 100}{3}c = 3c + 140] Simplifying this equation, we find: [c^3 - 100c = 9c + 420] [c^3 - 109c - 420 = 0] Factoring this cubic equation, we have: [(c - 12)(c + 7)(c + 5) = 0] Similarly, substituting the expression for b into equation (5) and simplifying, we confirm: [c^3 - 109c - 420 = 0] [(c - 12)(c + 7)(c + 5) = 0] Since c represents a length and must be positive, we conclude that c = 12. Plugging c = 12 back into the expressions for a and b, we find: [a = frac{12^2 - 100}{3} = frac{44}{3}] [b = frac{12^2 - 9}{10} = frac{135}{10} = frac{27}{2}] The sum of the lengths of all diagonals of ABCDE is given by 3c + a + b, which equals: [3 cdot 12 + frac{44}{3} + frac{27}{2} = frac{216 + 88 + 81}{6} = frac{385}{6}] Therefore, the sum of the lengths of all diagonals of ABCDE is frac{385}{6}, and the sum of m + n where frac{m}{n} = frac{385}{6} is 385 + 6 = boxed{391}.

answer:1. **Initialization and Key Concepts:** - Let ( T ) be the concurrency point of lines ( AP, BQ, CR ). - Let ( H ) be the orthocenter of ( triangle ABC ). 2. **Key Claim:** - ( T ) is the midpoint of ( overline{AP}, overline{BQ}, overline{CR}, overline{DH} ). - ( D ) is the orthocenter of ( triangle PQR ). 3. **Proof of the Key Claim:** - Note that ( overline{AQ} parallel overline{BP} ) since both are perpendicular to ( overline{CD} ). This implies that ( overline{AQ} ) and ( overline{BP} ) are distinct lines. - By symmetry and similar reasoning for the other vertices, the hexagon ( AQCPBR ) has opposite sides that are not only parallel but also pairs ( overline{AP}, overline{BQ}, overline{CR} ) intersect at point ( T ). - Hence, the hexagon is centrally symmetric about ( T ). 4. **Principal Ratios:** - We have the ratios: [ frac{AT}{TP} = frac{TQ}{BT} = frac{CT}{TR} = frac{TP}{AT} quad text{and since these are all equal to } +1 ] - Therefore, ( T ) is the midpoint of each segment connecting ( A ) to ( P ), ( B ) to ( Q ), and ( C ) to ( R ). 5. **Orthocentric Configuration:** - ( overline{PD} perp overline{BC} parallel overline{QR} ), indicating symmetry. - Consequently, ( D ) qualifies as the orthocenter of ( triangle PQR ), implying ( T ) is the midpoint of ( overline{DH} ). 6. **Symmetry and Reflections:** - We have symmetry involving four points ( A, B, C, D ) and their reflections at ( T ) corresponding to the orthocenters ( P, Q, R, H ). 7. **Geometric Configuration:** - Let ( S ) denote the centroid of the points ( {A, B, C, D } ). - Define ( O ) as the reflection of ( T ) concerning ( S ). 8. **Final claim and Proof:** - ( A, B, C, D ) are equidistant from ( O ). - Let ( A', O', S', T', D' ) be the projections of ( A, O, S, T, D ) onto the line ( BC ). - Since ( T' ) serves as the midpoint of ( overline{A'D'} ): [ S' = frac{1}{4} left(A' + D' + B + Cright) ] - Providing ( O' ) as midpoint of ( overline{BC} ): - Consequently, ( OB = OC ). - Therefore, ( A, B, C, D ) are equidistant from ( O ). # Conclusion: [ boxed{ text{The points } {A, B, C, D} text{ lie on a circle.} } ]

answer:First, expand and simplify the left-hand side of the equation: [ (3x+8)(x-6) = 3x^2 - 18x + 8x - 48 = 3x^2 - 10x - 48 ] Next, set up the equation with the right-hand side: [ 3x^2 - 10x - 48 = -50 + kx ] Rearranging all terms to one side gives: [ 3x^2 - (10+k)x + 2 = 0 ] For the quadratic equation to have exactly one real solution, the discriminant must be zero: [ b^2 - 4ac = (-10-k)^2 - 4 cdot 3 cdot 2 = 0 ] [ (-10-k)^2 - 24 = 0 ] [ (-10-k)^2 = 24 ] [ -10-k = pm sqrt{24} ] [ k = -10 pm 2sqrt{6} ] Thus, the values of k are: [ boxed{-10 + 2sqrt{6}, -10 - 2sqrt{6}} ]