Aurora

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question:Round 6**p16.** Given that x and y are positive real numbers such that x^3+y = 20 , the maximum possible value of x + y can be written as frac{asqrt{b}}{c} +d where a , b , c , and d are positive integers such that gcd(a,c) = 1 and b is squarefree. Find a +b +c +d .**p17.** In vartriangle DRK , DR = 13 , DK = 14 , and RK = 15 . Let E be the intersection of the altitudes of vartriangle DRK . Find the value of lfloor DE +RE +KE rfloor .**p18.** Subaru the frog lives on lily pad 1 . There is a line of lily pads, numbered 2 , 3 , 4 , 5 , 6 , and 7 . Every minute, Subaru jumps from his current lily pad to a lily pad whose number is either 1 or 2 greater, chosen at random from valid possibilities. There are alligators on lily pads 2 and 5 . If Subaru lands on an alligator, he dies and time rewinds back to when he was on lily pad number 1 . Find the expected number of jumps it takes Subaru to reach pad 7 . Round 7 This set has problems whose answers depend on one another.**p19.** Let B be the answer to Problem 20 and let C be the answer to Problem 21 . Given that f (x) = x^3BxC = (xr )(xs)(xt ) where r , s , and t are complex numbers, find the value of r^2+s^2+t^2 .**p20.** Let A be the answer to Problem 19 and let C be the answer to Problem 21 . Circles omega_1 and omega_2 meet at points X and Y . Let point P ne Y be the point on omega_1 such that PY is tangent to omega_2 , and let point Q ne Y be the point on omega_2 such that QY is tangent to omega_1 . Given that PX = A and QX =C , find XY .**p21.** Let A be the answer to Problem 19 and let B be the answer to Problem 20 . Given that the positive difference between the number of positive integer factors of A^B and the number of positive integer factors of B^A is D , and given that the answer to this problem is an odd prime, find frac{D}{B}40 . Round 8**p22.** Let v_p (n) for a prime p and positive integer n output the greatest nonnegative integer x such that p^x divides n . Find sum^{50}_{i=1}sum^{i}_{p=1} { v_p (i )+1 choose 2}, where the inner summation only sums over primes p between 1 and i .**p23.** Let a , b , and c be positive real solutions to the following equations. frac{2b^2 +2c^2 a^2}{4}= 25 frac{2c^2 +2a^2 b^2}{4}= 49 frac{2a^2 +2b^2 c^2}{4}= 64 The area of a triangle with side lengths a , b , and c can be written as frac{xsqrt{y}}{z} where x and z are relatively prime positive integers and y is squarefree. Find x + y +z .**p24.** Alan, Jiji, Ina, Ryan, and Gavin want to meet up. However, none of them know when to go, so they each pick a random 1 hour period from 5 AM to 11 AM to meet up at Alan’s house. Find the probability that there exists a time when all of them are at the house at one time.**Round 9** **p25.** Let n be the number of registered participantsin this LMT . Estimate the number of digits of left[ {n choose 2} right] in base 10 . If your answer is A and the correct answer is C , then your score will be left lfloor max left( 0,20 left| ln left( frac{A}{C}right) cdot 5 right|right| right rfloor. **p26.** Let gamma be theminimum value of x^x over all real numbers x . Estimate lfloor 10000gamma rfloor . If your answer is A and the correct answer is C , then your score will be left lfloor max left( 0,20 left| ln left( frac{A}{C}right) cdot 5 right|right| right rfloor. **p27.** Let E = log_{13} 1+log_{13}2+log_{13}3+...+log_{13}513513. Estimate lfloor E rfloor . If your answer is A and the correct answer is C , your score will be left lfloor max left( 0,20 left| ln left( frac{A}{C}right) cdot 5 right|right| right rfloor. PS. You should use hide for answers. Rounds 15 have been posted [here](https://artofproblemsolving.com/community/c3h3167127p28823220). Collected [here](https://artofproblemsolving.com/community/c5h2760506p24143309).

answer:To solve this problem, we need to find the orthocenter ( E ) of triangle ( triangle DRK ) and then calculate the sum of the distances from ( E ) to the vertices ( D ), ( R ), and ( K ). Finally, we will take the floor of this sum. 1. **Determine the coordinates of the vertices:** - Let ( D = (0, 0) ) - Let ( K = (14, 0) ) - Let ( R = (9, 12) ) 2. **Calculate the semi-perimeter ( s ) and the area ( Delta ) of the triangle:** - The semi-perimeter ( s ) is given by: [ s = frac{DR + DK + RK}{2} = frac{13 + 14 + 15}{2} = 21 ] - Using Heron's formula, the area ( Delta ) is: [ Delta = sqrt{s(s - DR)(s - DK)(s - RK)} = sqrt{21 cdot 8 cdot 7 cdot 6} = sqrt{7056} = 84 ] 3. **Find the feet of the perpendiculars from each vertex to the opposite side:** - The altitude from ( D ) to ( RK ) is: [ text{Altitude} = frac{2 Delta}{RK} = frac{2 cdot 84}{15} = 11.2 ] - The altitude from ( R ) to ( DK ) is: [ text{Altitude} = frac{2 Delta}{DK} = frac{2 cdot 84}{14} = 12 ] - The altitude from ( K ) to ( DR ) is: [ text{Altitude} = frac{2 Delta}{DR} = frac{2 cdot 84}{13} approx 12.923 ] 4. **Find the equations of the altitudes:** - The equation of the altitude from ( D ) to ( RK ) (line ( RK )) is: [ y = frac{4}{3} x ] - The equation of the altitude from ( R ) to ( DK ) (line ( DK )) is: [ y = 0 ] - The equation of the altitude from ( K ) to ( DR ) (line ( DR )) is: [ y = frac{-12}{5} x + frac{168}{5} ] 5. **Find the orthocenter ( E ):** - The orthocenter ( E ) is the intersection of the altitudes. Solving the equations of the altitudes, we find: [ E = left(9, frac{15}{4}right) ] 6. **Calculate the distances ( DE ), ( RE ), and ( KE ):** - Using the distance formula: [ DE = sqrt{(9 - 0)^2 + left(frac{15}{4} - 0right)^2} = sqrt{81 + left(frac{225}{16}right)} = frac{25}{4} ] [ RE = sqrt{(9 - 9)^2 + left(frac{15}{4} - 12right)^2} = sqrt{0 + left(frac{-33}{4}right)^2} = frac{33}{4} ] [ KE = sqrt{(14 - 9)^2 + left(0 - frac{15}{4}right)^2} = sqrt{25 + left(frac{-15}{4}right)^2} = frac{39}{4} ] 7. **Sum the distances and take the floor:** [ DE + RE + KE = frac{25}{4} + frac{33}{4} + frac{39}{4} = frac{97}{4} = 24.25 ] [ lfloor 24.25 rfloor = 24 ] The final answer is ( boxed{ 24 } )

question:Determine the product of all possible values for c if the length of the segment between the points (3c, c+5) and (1, 4) is 5 units.

answer:Using the distance formula, the distance from (3c, c+5) to (1, 4) is sqrt{(3c-1)^2 + ((c+5)-4)^2}. Setting this equal to 5 units, we solve: [ sqrt{(3c-1)^2 + (c+1)^2} = 5 ] Squaring both sides: [ (3c-1)^2 + (c+1)^2 = 25 ] Expanding and combining like terms: [ 9c^2 - 6c + 1 + c^2 + 2c + 1 = 25 ] [ 10c^2 - 4c - 23 = 0 ] Using the quadratic formula, c = frac{-b pm sqrt{b^2 - 4ac}}{2a}: [ c = frac{-(-4) pm sqrt{(-4)^2 - 4 cdot 10 cdot (-23)}}{2 cdot 10} ] [ c = frac{4 pm sqrt{16 + 920}}{20} ] [ c = frac{4 pm sqrt{936}}{20} ] [ c = frac{4 pm 2sqrt{234}}{20} ] [ c = frac{2 pm sqrt{234}}{10} ] Thus, the possible values for c are frac{2 + sqrt{234}}{10} and frac{2 - sqrt{234}}{10}. The product of these values is: [ frac{(2 + sqrt{234})}{10} times frac{(2 - sqrt{234})}{10} = frac{4 - 234}{100} = frac{-230}{100} = boxed{-2.3} ]

question:Robert now reads at a speed of 75 pages per hour. How many 300page books can he read in 9 hours?

answer:First, calculate the time it takes Robert to read one 300-page book: [ text{Time per book} = frac{300 text{ pages}}{75 text{ pages per hour}} = 4 text{ hours per book} ] Now, determine how many complete books Robert can read in 9 hours: [ text{Number of books} = frac{9 text{ hours}}{4 text{ hours per book}} = 2.25 ] Since Robert cannot read a fraction of a book in this context, he can read boxed{2} complete 300-page books in 9 hours.

question:Given that y=fleft(xright) is an odd function defined on R, when x gt 0, fleft(xright)=x^{2}+x+1, then when x lt 0, fleft(xright)=____.

answer:Given that y=f(x) is an odd function defined on mathbb{R}, and when x > 0, f(x)=x^2+x+1, we need to find the expression for f(x) when x < 0. Step 1: Recognize that for an odd function, f(-x) = -f(x) for all x in the domain. Step 2: Given f(x) = x^2 + x + 1 for x > 0, we substitute -x for x to find f(-x) because -x > 0 when x < 0. Step 3: Calculate f(-x): [f(-x) = (-x)^2 + (-x) + 1 = x^2 - x + 1] Step 4: Use the property of odd functions, f(-x) = -f(x), to find f(x) when x < 0: [f(x) = -f(-x) = -(x^2 - x + 1) = -x^2 + x - 1] Therefore, when x < 0, f(x) = -x^2 + x - 1. Hence, the answer is boxed{-x^2 + x - 1}.