answer:First, convert 50 to cents: 50 = 5000 text{ cents}. Let n be the number of stamps. Each stamp costs 45 cents, so n stamps cost 45n cents. We seek the largest n such that 45n le 5000. Divide both sides by 45: n le frac{5000}{45} = 111.1111... The largest integer n satisfying this inequality is n = 111. Thus, the maximum number of stamps that can be purchased is boxed{111}.

answer:1. Define ( k ) as the number of liars in the room. 2. Analyze the statements made by the 10 people. The first person said, "In this room, there is at least 1 liar"; the second person said, "In this room, there are at least 2 liars"; and so on, until the tenth person said, "In this room, all 10 people are liars." 3. If there are ( k ) liars, then the first ( k ) statements must be true, because they claim there are at least 1, 2, ..., ( k ) liars, respectively. The people making these statements must be knights (since knights always tell the truth). 4. Therefore, the last ( 10 - k ) statements must be false. The people making these statements must be liars (since liars always lie). To determine ( k ): 5. Since the first ( k ) people are telling the truth, their statements accurately reflect that there are at least ( 1, 2, ..., k ) liars in the room. 6. Since the remaining ( 10 - k ) people are lying, their statements about the number of liars must be false. Paying particular attention to the statement by the ( k+1 )th person, which is a liar, this person claims there are at least ( k + 1 ) liars, but since this is false, there must actually be at most ( k ) liars. 7. Consequently, for the ( left( k+1 right) )th liar to lie about there being at least ( k+1 ) liars, it follows that ( k ) represents exactly the number of liars. If ( k ) were higher, the ( left( k+1 right) )th person would no longer be lying, which contradicts our initial classification. 8. Therefore, we need to solve for ( k ) such that the first ( k ) people are knights and the remaining ( 10 - k ) are liars. [ k = 10 - k ] 9. Simplify the equation: [ 2k = 10 implies k = 5 ] 10. Hence, there are exactly 5 liars in the room. # Conclusion: (boxed{5})

answer:Solution: (1) The polar equation of the circle rho=6costheta can be transformed into rho^2=6rhocostheta, Using the polar coordinate formulas, this can be converted into a standard equation x^2+y^2=6x, which simplifies to (x-3)^2+y^2=9. (2) The equation of circle C is (x-3)^2+y^2=9, with the center of the circle C at (3,0) and radius r=3, The equation of line l is y+3= sqrt {3}(x-3), which simplifies to sqrt {3}x-y-3sqrt {3}-3=0, The distance d from the center of the circle to the line is d= frac {|3sqrt {3}-3sqrt {3}-3|}{sqrt {1+3}}= frac {3}{2}, Therefore, the central angle corresponding to the chord cut by line l on the circle is 120°, and line l divides circle C into two arcs with a length ratio of boxed{1:2}.

answer:1. **Identify the given elements and the goal:** - We have an acute triangle ( triangle ABC ). - From the foot of one altitude, perpendiculars are drawn to the other two sides, meeting them at points ( P ) and ( Q ). - We need to show that the length of ( PQ ) does not depend on which of the three altitudes is selected. 2. **Consider the altitude from vertex ( A ) to side ( BC ):** - Let ( D ) be the foot of the altitude from ( A ) to ( BC ). - Draw perpendiculars from ( D ) to ( AB ) and ( AC ), meeting them at points ( P ) and ( Q ) respectively. 3. **Prove that ( APDQ ) is cyclic:** - Since ( D ) is the foot of the altitude from ( A ), ( angle ADB = angle ADC = 90^circ ). - ( angle APD = angle AQD = 90^circ ) because ( P ) and ( Q ) are points where perpendiculars from ( D ) meet ( AB ) and ( AC ). - Therefore, quadrilateral ( APDQ ) is cyclic (opposite angles sum to ( 180^circ )). 4. **Use the sine rule in cyclic quadrilateral ( APDQ ):** - In a cyclic quadrilateral, the ratio of the length of a side to the sine of the opposite angle is constant. - Applying the sine rule in ( APDQ ): [ frac{PQ}{sin angle PAD} = frac{AD}{sin angle APQ} ] - Since ( angle PAD = angle A ) and ( angle APQ = 90^circ - angle A ): [ frac{PQ}{sin angle A} = frac{AD}{cos angle A} ] - Therefore: [ PQ = AD sin angle A ] 5. **Generalize for other altitudes:** - Let ( BE ) be the altitude from ( B ) to ( AC ), and let ( F ) be the foot of the altitude. - Draw perpendiculars from ( F ) to ( AB ) and ( BC ), meeting them at points ( P' ) and ( Q' ) respectively. - Similarly, quadrilateral ( BP'FQ' ) is cyclic, and by the sine rule: [ P'Q' = BE sin angle B ] 6. **Show that ( AD sin angle A = BE sin angle B ):** - In any triangle, the product of the altitude and the sine of the angle at the vertex is constant: [ AD sin angle A = BE sin angle B = CF sin angle C ] - This is because the area of the triangle can be expressed in terms of any altitude and the corresponding base: [ text{Area} = frac{1}{2} times BC times AD = frac{1}{2} times AC times BE = frac{1}{2} times AB times CF ] - Therefore, ( AD sin angle A = BE sin angle B = CF sin angle C ). 7. **Conclusion:** - Since ( AD sin angle A = BE sin angle B = CF sin angle C ), the length ( PQ ) does not depend on which altitude is selected. (blacksquare)