Home » 国际竞赛 » 数学国际竞赛 » AMC美国数学竞赛 » Details

1985AIME 真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月12日下午12:01

1985AIME 真题及答案解析

答案解析请参考文末

Problem 1

Let $x_1=97$ , and for $n>1$ let $x_n=frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2 ldots x_8$ .

Problem 2

When a right triangle is rotated about one leg, the volume of the cone produced is $800pi ;textrm{cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920pi ;textrm{cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle?

Problem 3

Find $c$ if $a$ , $b$ , and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$ , where $i^2 = -1$ .

Problem 4

A small square is constructed inside a square of area $1$ by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices, as shown in the figure. Find the value of $n$ if the the area of the small square is exactly $frac1{1985}$ . AIME 1985 Problem 4.png

Problem 5

A sequence of integers $a_1, a_2, a_3, ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n ge 3$ . What is the sum of the first $2001$ terms of this sequence if the sum of the first $1492$ terms is $1985$ , and the sum of the first $1985$ terms is $1492$ ?

Problem 6

As shown in the figure, $triangle ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of $triangle ABC$ . AIME 1985 Problem 6.png

Problem 7

Assume that $a$ , $b$ , $c$ and $d$ are positive integers such that $a^5 = b^4$ , $c^3 = d^2$ and $c - a = 19$ . Determine $d - b$ .

Problem 8

The sum of the following seven numbers is exactly 19: $a_1 = 2.56$ , $a_2 = 2.61$ , $a_3 = 2.65$ , $a_4 = 2.71$ , $a_5 = 2.79$ , $a_6 = 2.82$ , $a_7 = 2.86$ . It is desired to replace each $a_i$ by an integer approximation $A_i$ , $1le i le 7$ , so that the sum of the $A_i$ 's is also $19$ and so that $M$ , the maximum of the "errors" $lvert A_i-a_i rvert$ , is as small as possible. For this minimum $M$ , what is $100M$ ?

Problem 9

In a circle, parallel chords of lengths $2$ , $3$ , and $4$ determine central angles of $alpha$ , $beta$ and $alpha + beta$ radians, respectively, where $alpha + beta < pi$ . If $cos alpha$ , which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

Problem 10

How many of the first $1000$ positive integers can be expressed in the form $lfloor 2x rfloor + lfloor 4x rfloor + lfloor 6x rfloor + lfloor 8x rfloor$ ,
where $x$ is a real number, and $lfloor z rfloor$ denotes the greatest integer less than or equal to $z$ ?

Problem 11

An ellipse has foci at $(9, 20)$ and $(49, 55)$ in the $xy$ -plane and is tangent to the $x$ -axis. What is the length of its major axis?

Problem 12

Let $A$ , $B$ , $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ .

Problem 13

The numbers in the sequence $101$ , $104$ , $109$ , $116$ , $ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,ldots$ . For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers.

Problem 14

In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned $frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned in games against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Problem 15

Three $12$ cm $times 12$ cm squares are each cut into two pieces $A$ and $B$ , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon, as shown in the second figure, so as to fold into a polyhedron. What is the volume (in $mathrm{cm}^3$ ) of this polyhedron? AIME 1985 Problem 15.png

1985AIME 详细解析

Since $x_n=frac{n}{x_{n-1}}$ , $x_n cdot x_{n - 1} = n$ . Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$ , $x_3x_4 = 4$ , $x_5x_6 = 6$ and $x_7x_8 = 8$ so $[x_1x_2x_3x_4x_5x_6x_7x_8 = 2cdot4cdot6cdot8 = boxed{384}.]$ Notice that the value of $x_1$ was completely unneeded!
Let one leg of the triangle have length $a$ and let the other leg have length $b$ . When we rotate around the leg of length $a$ , the result is a cone of height $a$ and radius $b$ , and so of volume $frac 13 pi ab^2 = 800pi$ . Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and radius $a$ and so of volume $frac13 pi b a^2 = 1920 pi$ . If we divide this equation by the previous one, we get $frac ab = frac{frac13 pi b a^2}{frac 13 pi ab^2} = frac{1920}{800} = frac{12}{5}$ , so $a = frac{12}{5}b$ . Then $frac{1}{3} pi (frac{12}{5}b)b^2 = 800pi$ so $b^3 = 1000$ and $b = 10$ so $a = 24$ . Then by the Pythagorean Theorem, the hypotenuse has length $sqrt{a^2 + b^2} = boxed{026}$ .
Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$ . Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$ . Since $a, b$ are integers, this means $b$ is a divisor of 107, which is a prime number. Thus either $b = 1$ or $b = 107$ . If $b = 107$ , $3a^2 - 107^2 = 1$ so $3a^2 = 107^2 + 1$ , but $107^2 + 1$ is not divisible by 3, a contradiction. Thus we must have $b = 1$ , $3a^2 = 108$ so $a^2 = 36$ and $a = 6$ (since we know $a$ is positive). Thus $c = 6^3 - 3cdot 6 = boxed{198}$ .
The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram. Using the smaller side of the parallelogram, $1/n$ , as the base, where the height is 1, we find that the area of the parallelogram is $A = frac{1}{n}$ . By the Pythagorean Theorem, the longer base of the parallelogram has length $l = sqrt{1^2 + left(frac{n - 1}{n}right)^2} = frac{1}{n}sqrt{2n^2 - 2n + 1}$ , so the parallelogram has height $h = frac{A}{l} = frac{1}{sqrt{2n^2 - 2n + 1}}$ . But the height of the parallelogram is the side of the little square, so $2n^2 - 2n + 1 = 1985$ . Solving this quadratic equation gives $n = boxed{32}$ .Surrounding the square with area $frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + frac{1}{1985} = 1$ , where $X$ is the area of the of the 4 triangles. We can thus use proportions to solve this problem. $begin{eqnarray*} frac{GF}{BE}=frac{CG}{CB}implies frac{frac{1}{sqrt{1985}}}{BE}=frac{frac{1}{n}}{1}implies BE=frac{nsqrt{1985}}{1985} end{eqnarray*}$ Also, $begin{eqnarray*} frac{BE}{1}=frac{EC}{frac{n-1}{n}}implies EC=frac{sqrt{1985}}{1985}(n-1) end{eqnarray*}$ Thus, $begin{eqnarray*} 2(BE)(EC)+frac{1}{1985}=1\ 2n^{2}-2n+1=1985\ n(n-1)=992 end{eqnarray*}$ Simple factorization and guess and check gives us $boxed{32}$ .
The problem gives us a sequence defined by a recursion, so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$ . Then $a_3 = b - a$ , $a_4 = (b - a) - b = -a$ , $a_5 = -a - (b - a) = -b$ , $a_6 = -b - (-a) = a - b$ , $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$ . Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular $a_{j + 6} = a_j$ for all $j$ , and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$ .Because of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 cdot 248$ is the largest multiple of 6 less than 1492, so $sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + sum_{n = 0}^{247}sum_{j = 1}^6 a_{6n + j} = (a + b + (b - a) + (-a)) + sum_{n = 0}^{247}sum_{j = 1}^6 a_j = 2b - a$ , where we can make this last step because $sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero.Similarly, since $1980 = 6 cdot 360$ , $sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$ .Finally, $sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$ .Then by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2cdot 493 = boxed{986}$ .
Let the interior point be $P$ , let the points on $overline{BC}$ , $overline{CA}$ and $overline{AB}$ be $D$ , $E$ and $F$ , respectively. Let $x$ be the area of $triangle APE$ and $y$ be the area of $triangle CPD$ . Note that $triangle APF$ and $triangle BPF$ share the same altitude from $P$ , so the ratio of their areas is the same as the ratio of their bases. Similarly, $triangle ACF$ and $triangle BCF$ share the same altitude from $C$ , so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have: $frac{40}{30} = frac{124 + x}{65 + y}$ or equivalently $372 + 3x = 260 + 4y$ and so $4y = 3x+ 112$ .Applying identical reasoning to the triangles with bases $overline{CD}$ and $overline{BD}$ , we get $frac{y}{35} = frac{x+y+84}{105}$ so that $3y = x + y + 84$ and $2y = x + 84$ . Substituting from this equation into the previous one gives $x = 56$ , from which we get $y = 70$ and so the area of $triangle ABC$ is $56 + 40 + 30 + 35 + 70 + 84 = Rightarrow boxed{315}$ .This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight $times$ side) = (Other weight) $times$ (The other side), the problem yields the answer $boxed{315}$ Let the interior point be $P$ and let the points on $overline{BC}$ , $overline{CA}$ and $overline{AB}$ be $D$ , $E$ and $F$ , respectively. Also, let $[APE]=x,[CPD]=y.$ Then notice that by Ceva's, $frac{FBcdot DCcdot EA}{DBcdot CEcdot AF}=1.$ However, we can deduce $frac{FB}{AF}=frac{3}{4}$ from the fact that $[AFP]$ and $[BPF]$ share the same height. Similarly, $x=frac{84CE}{EA}$ and $y=frac{35DC}{BD}.$ Plug and chug and you get $xy=84cdot 35cdot frac{3}{4}=2205.$ Then notice by the same height reasoning, $frac{84}{x}=frac{119+y}{x+70}.$ Clear the fractions and combine like terms to get $35x=5880-xy.$ We know $xy=2205$ so subtraction yields $35x=3675,$ or $x=105.$ Plugging this in to our previous ratio statement yields $frac{84}{105}=frac{4}{5}=frac{119+y}{175},$ so $y=21.$ Basic addition gives us $105+84+21+35+30+40=boxed{315}.$
It follows from the givens that $a$ is a perfect fourth power, $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube. Thus, there exist integers $s$ and $t$ such that $a = t^4$ , $b = t^5$ , $c = s^2$ and $d = s^3$ . So $s^2 - t^4 = 19$ . We can factor the left-hand side of this equation as a difference of two squares, $(s - t^2)(s + t^2) = 19$ . 19 is a prime number and $s + t^2 > s - t^2$ so we must have $s + t^2 = 19$ and $s - t^2 = 1$ . Then $s = 10, t = 3$ and so $d = s^3 = 1000$ , $b = t^5 = 243$ and $d-b=boxed{757}$ .
If any of the approximations $A_i$ is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the $A_i$ are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3. Then there must be five 3s and two 2s. It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the $a_i$ , so our approximations are $A_1 = A_2 = 2$ and $A_3 = A_4 = A_5 = A_6 = A_7 = 3$ and the largest error is $|A_2 - a_2| = 0.61$ , so the answer is $061$ .
$[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3); D(CR(O,r)); D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle); D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle); D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle); MP("2",(rotate(a/2)*A+rotate(-a/2)*A)/2,NE); MP("3",(rotate(b/2)*A+rotate(-b/2)*A)/2,NE); MP("4",(rotate((a+b)/2)*A+rotate(-(a+b)/2)*A)/2,NE); D(anglemark(rotate(-(a+b)/2)*A,O,rotate((a+b)/2)*A,5)); label("(alpha+beta)",(0.08,0.08),NE,fontsize(8)); [/asy]$ All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle. $[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A; D(CR(O,r)); D(O--A1--A2--cycle); D(O--A2--A3--cycle); D(O--A1--A3--cycle); MP("2",(A1+A2)/2,NE); MP("3",(A2+A3)/2,E); MP("4",(A1+A3)/2,E); D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); label("(alpha)",(0.07,0.16),NE,fontsize(8)); label("(beta)",(0.12,-0.16),NE,fontsize(8)); [/asy]$ This triangle has semiperimeter $frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = sqrt{frac92 cdot frac52 cdot frac32 cdot frac12} = frac{3}{4}sqrt{15}$ . The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = frac{abc}{4R}$ , so $frac6R = frac{3}{4}sqrt{15}$ and $R = frac8{sqrt{15}}$ .Now, consider the triangle formed by two radii and the chord of length 2. This isosceles triangle has vertex angle $alpha$ , so by the Law of Cosines, $[2^2 = R^2 + R^2 - 2R^2cos alpha Longrightarrow cos alpha = frac{2R^2 - 4}{2R^2} = frac{17}{32}]$ and the answer is $17 + 32 = boxed{049}$ . $[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A; D(CR(O,r)); D(O--A1--A2--cycle); D(O--A2--A3--cycle); D(O--A1--A3--cycle); MP("2",(A1+A2)/2,NE); MP("3",(A2+A3)/2,E); MP("4",(A1+A3)/2,E); D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); D(anglemark(A2,A3,A1,18)); label("(alpha)",(0.07,0.16),NE,fontsize(8)); label("(beta)",(0.12,-0.16),NE,fontsize(8)); label("(alpha)/2",(0.82,-1.25),NE,fontsize(8)); [/asy]$ It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is $frac{alpha}{2}$ , and using the Law of Cosines, we get: $[2^2 = 3^2 + 4^2 - 2cdot3cdot4cosfrac{alpha}{2}]$ Which, rearranges to: $[21 = 24cosfrac{alpha}{2}]$ And, that gets us: $[cosfrac{alpha}{2} = 7/8]$ Using $cos 2theta = 2cos^2 theta - 1$ , we get that: $[cosalpha = 17/32]$ Which gives an answer of $boxed{049}$
We will be able to reach the same number of integers while $x$ ranges from 0 to 1 as we will when $x$ ranges from $n$ to $n + 1$ for any integer $n$ (Quick proof: $lfloor 2(n+x)rfloor + ldots = lfloor 2n + 2xrfloor + ldots = 2n + lfloor 2xrfloor ldots$ ). Since $lfloor 2cdot50 rfloor + lfloor 4cdot50 rfloor + lfloor 6cdot50 rfloor + lfloor 8cdot50 rfloor = 100 + 200 + 300 + 400$ , the answer must be exactly 50 times the number of integers we will be able to reach as $x$ ranges from 0 to 1, including 1 but excluding 0.
An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant. Let $F_1 = (9, 20)$ , $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$ -axis. Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is minimal. (The last claim begs justification: Let $F'_2$ be the reflection of $F_2$ across the $y$ -axis. Let $Y$ be where the line through $F_1$ and $F’_2$ intersects the ellipse. We will show that $X=Y$ . Note that $X F_2 = X F’_2$ since $X$ is on the $x$ -axis. Also, since the entire ellipse is on or above the $x$ -axis and the line through $F_2$ and $F’_2$ is perpendicular to the $x$ -axis, we must have $F_2 Y leq F’_2 Y$ with equality if and only if $Y$ is on the $x$ -axis. Now, we have $[F_1 X + F'_2 X = F_1 X + F_2 X = F_1 Y + F_2 Y leq F_1 Y + F’_2 Y]$ But the right most sum is the straight-line distance from $F_1$ to $F’_2$ and the left is the distance of some path from $F_1$ to $F_2$ ., so this is only possible if we have equality and thus $X = Y$ ). Finding the optimal location for $X$ is a classic problem: for any path from $F_1$ to $X$ and then back to $F_2$ , we can reflect (as above) the second leg of this path (from $X$ to $F_2$ ) across the $x$ -axis. Then our path connects $F_1$ to the reflection $F_2'$ of $F_2$ via some point on the $x$ -axis, and this path will have shortest length exactly when our original path has shortest length. This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the $x-$ axis. $[asy] size(200); pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(10); pair F1=(9,20),F2=(49,55); D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle); D((-20,0)--(80,0)--(0,0)--(0,80)--(0,-60)); path p = F1--(49,-55); pair X = IP(p,(0,0)--(80,0)); D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((49,-55)); MP("X",X,SW,f); MP("F_1",F1,NW,f); MP("F_2",F2,NW,f); MP("F_2'",(49,-55),NE,f); [/asy]$ The sum of the two distances $F_1 X$ and $F_2X$ is therefore equal to the length of the segment $F_1F_2'$ , which by the distance formula is just $d = sqrt{(9 - 49)^2 + (20 + 55)^2} = sqrt{40^2 + 75^2} = 5sqrt{8^2 + 15^2} = 5cdot 17 = 85$ .Finally, let $A$ and $B$ be the two endpoints of the major axis of the ellipse. Then by symmetry $AF_1 = F_2B$ so $AB = AF_1 + F_1B = F_2B + F_1B = d$ (because $B$ is on the ellipse), so the answer is $boxed{085}$ .
Let $P(n)$ denote the probability that the bug is at $A$ after it has crawled $n$ meters. Since the bug can only be at vertex $A$ if it just left a vertex which is not $A$ , we have $P(n + 1) = frac13 (1 - P(n))$ . We also know $P(0) = 1$ , so we can quickly compute $P(1)=0$ , $P(2) = frac 13$ , $P(3) = frac29$ , $P(4) = frac7{27}$ , $P(5) = frac{20}{81}$ , $P(6) = frac{61}{243}$ and $P(7) = frac{182}{729}$ , so the answer is $boxed{182}$ . One can solve this recursion fairly easily to determine a closed-form expression for $P(n)$ .
If $(x,y)$ denotes the greatest common divisor of $x$ and $y$ , then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$ . Now assuming that $d_n$ divides $100+n^2$ , it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$ .Thus the equation turns into $d_n=(100+n^2,2n+1)$ . Now note that since $2n+1$ is odd for integral $n$ , we can multiply the left integer, $100+n^2$ , by a power of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multiply by $4$ so that we can get a $(2n+1)^2$ in there.So $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$ . It simplified the way we wanted it to! Now using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$ . Thus $d_n$ must divide $boxed{401}$ for every single $n$ . This means the largest possible value for $d_n$ is $401$ , and we see that it can be achieved when $n = 200$ .We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$ . Since we want to find the GCD of $a_n$ and $a_{n+1}$ , we can use the Euclidean algorithm: $a_{n+1}-a_n = 2n+1$ Now, the question is to find the GCD of $2n+1$ and $100+n^2$ . We subtract $2n+1$ 100 times from $100+n^2$ . This leaves us with $n^2-200n$ . We want this to equal 0, so solving for $n$ gives us $n=200$ . The last remainder is 0, thus $200*2+1 = boxed{401}$ is our GCD.
If Solution 2 is not entirely obvious, our answer is the max possible range of $frac{x(x-200)}{2x+1}$ . Using the Euclidean Algorithm on $x$ and $2x+1$ yields that they are relatively prime. Thus, the only way the GCD will not be 1 is if the $x-200$ term share factors with the $2x+1$ . Using the Euclidean Algorithm, $gcd(x-200,2x+1)=gcd(x-22,2x+1-2(x-200))=gcd(x-200,401)$ . Thus, the max GCD is 401.
We can just plug in Euclidean algorithm, to go from $gcd(n^2 + 100, n^2 + 2n + 101)$ to $gcd(n^2 + 100, 2n + 1)$ to $gcd(n^2 + 100 - 100(2n + 1), 2n + 1)$ to get $gcd(n^2 - 200n, 2n + 1)$ . Now we know that no matter what, $n$ is relatively prime to $2n + 1$ . Therefore the equation can be simplified to: $gcd(n - 200, 2n + 1)$ . Subtracting $2n - 400$ from $2n + 1$ results in $gcd(n - 200,401)$ . The greatest possible value of this is $boxed{401}$ , an happens when $n equiv 200 mod{401}$ .
Let us suppose for convenience that there were $n + 10$ players overall. Among the $n$ players not in the weakest 10 there were $n choose 2$ games played and thus $n choose 2$ points earned. By the givens, this means that these $n$ players also earned $n choose 2$ points against our weakest 10. Now, the 10 weakest players playing amongst themselves played ${10 choose 2} = 45$ games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger $n$ players. Since every point earned falls into one of these categories, It follows that the total number of points earned was $2{n choose 2} + 90 = n^2 - n + 90$ . However, there was one point earned per game, and there were a total of ${n + 10 choose 2} = frac{(n + 10)(n + 9)}{2}$ games played and thus $frac{(n + 10)(n + 9)}{2}$ points earned. So we have $n^2 -n + 90 = frac{(n + 10)(n + 9)}{2}$ so $2n^2 - 2n + 180 = n^2 + 19n + 90$ and $n^2 -21n + 90 = 0$ and $n = 6$ or $n = 15$ . Now, note that the top $n$ players got $n(n - 1)$ points in total (by our previous calculation) for an average of $n - 1$ , while the bottom 10 got 90 points total, for an average of 9. Thus we must have $n > 10$ , so $n = 15$ and the answer is $15 + 10 = boxed{25}$ .Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people. Thus, the $n-10$ people earned a total of $10(n-10)-45 = 10n-145$ points playing vs. this group of 10 people, and also earned a total of $10n-145$ playing against themselves. Since each match gives a total of one point, we must have that $dbinom{n-10}{2}=10n-145$ . Expanding and simplifying gives us $n^2-41+400=0$ . Thus, $n=16$ or $n=25$ . Note however that if $n=16$ , then the strongest $16$ people get a total of $16*10-145=15$ playing against the weakest $10$ who gained $45$ points vs them, which is a contradiction since it must be larger. Thus, $n=boxed{25}$ .Solution by GameMaster402
Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^circ$ from each other) yields a cube, so the volume is $frac12 cdot 12^3 = 864$ .Image: