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

1991AIME 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日上午11:28

1991AIME 真题及答案解析

答案解析请参考文末

Problem 1

Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that

$xy_{}^{}+x+y = 71$ $x^2y+xy^2 = 880^{}_{}.$

Problem 2

Rectangle $ABCD_{}^{}$ has sides $overline {AB}$ of length 4 and $overline {CB}$ of length 3. Divide $overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, ldots, P_{168}=B$ , and divide $overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, ldots, Q_{168}=B$ . For $1_{}^{} le k le 167$ , draw the segments $overline {P_kQ_k}$ . Repeat this construction on the sides $overline {AD}$ and $overline {CD}$ , and then draw the diagonal $overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn.

Problem 3

Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives

${1000 choose 0}(0.2)^0+{1000 choose 1}(0.2)^1+{1000 choose 2}(0.2)^2+cdots+{1000 choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + cdots + A_{1000},$ where $A_k = {1000 choose k}(0.2)^k$ for $k = 0,1,2,ldots,1000$ . For which $k_{}^{}$ is $A_k^{}$ the largest?

Problem 4

How many real numbers $x^{}_{}$ satisfy the equation $frac{1}{5}log_2 x = sin (5pi x)$ ?

Problem 5

Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $20_{}^{}!$ be the resulting product?

Problem 6

Suppose $r^{}_{}$ is a real number for which

$leftlfloor r + frac{19}{100} rightrfloor + leftlfloor r + frac{20}{100} rightrfloor + leftlfloor r + frac{21}{100} rightrfloor + cdots + leftlfloor r + frac{91}{100} rightrfloor = 546.$ Find $lfloor 100r rfloor$ . (For real $x^{}_{}$ , $lfloor x rfloor$ is the greatest integer less than or equal to $x^{}_{}$ .)

Problem 7

Find $A^2_{}$ , where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation:

$x = sqrt{19} + frac{91}{{sqrt{19}+frac{91}{{sqrt{19}+frac{91}{{sqrt{19}+frac{91}{{sqrt{19}+frac{91}{x}}}}}}}}}$

Problem 8

For how many real numbers $a^{}_{}$ does the quadratic equation $x^2 + ax^{}_{} + 6a=0$ have only integer roots for $x^{}_{}$ ?

Problem 9

Suppose that $sec x+tan x=frac{22}7$ and that $csc x+cot x=frac mn,$ where $frac mn$ is in lowest terms. Find $m+n^{}_{}.$

Problem 10

Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ is written as a fraction in lowest terms, what is its numerator?

Problem 11

Twelve congruent disks are placed on a circle $C^{}_{}$ of radius 1 in such a way that the twelve disks cover $C^{}_{}$ , no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be written in the form $pi(a-bsqrt{c})$ , where $a,b,c^{}_{}$ are positive integers and $c^{}_{}$ is not divisible by the square of any prime. Find $a+b+c^{}_{}$ .

$[asy] unitsize(100); draw(Circle((0,0),1)); dot((0,0)); draw((0,0)--(1,0)); label("$1$", (0.5,0), S); for (int i=0; i<12; ++i) { dot((cos(i*pi/6), sin(i*pi/6))); } for (int a=1; a<24; a+=2) { dot(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))); draw(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))--((1/cos(pi/12))*cos((a+2)*pi/12), (1/cos(pi/12))*sin((a+2)*pi/12))); draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); } [/asy]$

Problem 12

Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ , $Q^{}_{}$ , $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $overline{AB}$ , $overline{BC}$ , $overline{CD}$ , and $overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ , $BQ^{}_{}=20$ , $PR^{}_{}=30$ , and $QS^{}_{}=40$ . Let $m/n^{}_{}$ , in lowest terms, denote the perimeter of $ABCD^{}_{}$ . Find $m+n^{}_{}$ .

Problem 13

A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?

Problem 14

A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by $overline{AB}$ , has length 31. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$ .

Problem 15

For positive integer $n_{}^{}$ , define $S_n^{}$ to be the minimum value of the sum

$sum_{k=1}^n sqrt{(2k-1)^2+a_k^2},$ where $a_1,a_2,ldots,a_n^{}$ are positive real numbers whose sum is 17. There is a unique positive integer $n^{}_{}$ for which $S_n^{}$ is also an integer. Find this $n^{}_{}$ .

1991AIME 详细解析

Define $a = x + y$ and $b = xy$ . Then $a + b = 71$ and $ab = 880$ . Solving these two equations yields a quadratic: $a^2 - 71a + 880 = 0$ , which factors to $(a - 16)(a - 55) = 0$ . Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$ . For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $le 16$ , no two factors of $16$ can sum greater than $32$ , and so there are no integral solutions for $(x,y)$ . The solution is $5^2 + 11^2 = boxed{146}$ .
$[asy] real r = 0.35; size(220); pointpen=black;pathpen=black+linewidth(0.65);pen f = fontsize(8); pair A=(0,0),B=(4,0),C=(4,3),D=(0,3); D(A--B--C--D--cycle); pair P1=A+(r,0),P2=A+(2r,0),P3=B-(r,0),P4=B-(2r,0); pair Q1=C-(0,r),Q2=C-(0,2r),Q3=B+(0,r),Q4=B+(0,2r); D(A--C);D(P1--Q1);D(P2--Q2);D(P3--Q3);D(P4--Q4); MP("A",A,f);MP("B",B,SE,f);MP("C",C,NE,f);MP("D",D,W,f); MP("P_1",P1,f);MP("P_2",P2,f);MP("P_{167}",P3,f);MP("P_{166}",P4,f);MP("Q_1",Q1,E,f);MP("Q_2",Q2,E,f);MP("Q_{167}",Q3,E,f);MP("Q_{166}",Q4,E,f); MP("4",(A+B)/2,N,f);MP("cdots",(A+B)/2,f); MP("3",(B+C)/2,W,f);MP("vdots",(C+B)/2,E,f); [/asy]$ The length of the diagonal is $sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle). For each $k$ , $overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 cdot frac{168-k}{168}, 4 cdot frac{168-k}{168}$ . Thus, its length is $5 cdot frac{168-k}{168}$ . Let $a_k=frac{5(168-k)}{168}$ . We want to find $2sumlimits_{k=1}^{168} a_k-5$ since we are over counting the diagonal. $2sumlimits_{k=1}^{168} frac{5(168-k)}{168}-5 =2frac{(0+5)cdot169}{2}-5 =168cdot5 =boxed{840}$
Let $0<x_{}^{}<1$ . Then we may write $A_{k}^{}={Nchoose k}x^{k}=frac{N!}{k!(N-k)!}x^{k}=frac{(N-k+1)!}{k!}x^{k}$ . Taking logarithms in both sides of this last equation and using the well-known fact $log(a_{}^{}b)=log a + log b$ (valid if $a_{}^{},b_{}^{}>0$ ), we have $log(A_{k})=logleft[frac{(N-k+1)!}{k!}x^{k}right]=logleft[prod_{j=1}^{k}frac{(N-j+1)x}{j}right]=sum_{j=1}^{k}logleft[frac{(N-j+1)x}{j}right], .$ Now, $log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $frac{(N-j+1)x}{j}>1$ in each of the $logbig[;big]$ terms (recall that $log y_{}^{} <0$ if $0<y_{}^{}<1$ ). Therefore, the integer $k_{}^{}$ that we are looking for must satisfy $k=Biglfloorfrac{(N+1)x}{1+x}Bigrfloor$ , where $lfloor z_{}^{}rfloor$ denotes the greatest integer less than or equal to $z_{}^{}$ .In summary, substituting $N_{}^{}=1000$ and $x_{}^{}=0.2$ we finally find that $k_{}^{}=166$ .
The range of the sine function is $-1 le y le 1$ . It is periodic (in this problem) with a period of $frac{2}{5}$ .Thus, $-1 le frac{1}{5} log_2 x le 1$ , and $-5 le log_2 x le 5$ . The solutions for $x$ occur in the domain of $frac{1}{32} le x le 32$ . When $x > 1$ the logarithm function returns a positive value; up to $x = 32$ it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of $x$ ) of the sine curve and another curve that is $< 1$ , so there are $frac{32}{2} cdot 10 - 6 = 160 - 6 = 154$ values (the subtraction of 6 since all the “intersections” when $x < 1$ must be disregarded). When $y = 0$ , there is exactly $1$ touching point between the two functions: $left(frac{1}{5},0right)$ . When $y < 0$ or $x < 1$ , we can count $4$ more solutions. The solution is $154 + 1 + 4 = boxed{159}$ .
If the fraction is in the form $frac{a}{b}$ , then $a < b$ and $gcd(a,b) = 1$ . There are 8 prime numbers less than 20 ( $2, 3, 5, 7, 11, 13, 17, 19$ ), and each can only be a factor of one of $a$ or $b$ . There are $2^8$ ways of selecting some combination of numbers for $a$ ; however, since $a<b$ , only half of them will be between $0 < frac{a}{b} < 1$ . Therefore, the solution is $frac{2^8}{2} = 128$ .
There are $91 - 19 + 1 = 73$ numbers in the sequence. Since the terms of the sequence can be at most $1$ apart, all of the numbers in the sequence can take one of two possible values. Since $frac{546}{73} = 7 R 35$ , the values of each of the terms of the sequence must be either $7$ or $8$ . As the remainder is $35$ , $8$ must take on $35$ of the values, with $7$ being the value of the remaining $73 - 35 = 38$ numbers. The 39th number is $lfloor r+frac{19 + 39 - 1}{100}rfloor= lfloor r+frac{57}{100}rfloor$ , which is also the first term of this sequence with a value of $8$ , so $8 le r + frac{57}{100} < 8.01$ . Solving shows that $frac{743}{100} le r < frac{744}{100}$ , so $743le 100r < 744$ , and $lfloor 100r rfloor = 743$ .
Let $f(x) = sqrt{19} + frac{91}{x}$ . Then $x = f(f(f(f(f(x)))))$ , from which we realize that $f(x) = x$ . This is because if we expand the entire expression, we will get a fraction of the form $frac{ax + b}{cx + d}$ on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic $f(x)=x$ .The given finite expansion can then be easily seen to reduce to the quadratic equation $x_{}^{2}-sqrt{19}x-91=0$ . The solutions are $x_{pm}^{}=$ $frac{sqrt{19}pmsqrt{383}}{2}$ . Therefore, $A_{}^{}=vert x_{+}vert+vert x_{-}vert=sqrt{383}$ . We conclude that $A_{}^{2}=boxed{383}$ .
By Vieta's formulas, $x_1 + x_2 = -a$ where $x_1, x_2$ are the roots of the quadratic, and since $x_1,x_2$ are integers, $a$ must be an integer. Applying the quadratic formula, $[x = frac{-a pm sqrt{a^2 - 24a}}{2}]$ Since $-a$ is an integer, we need $sqrt{a^2-24a}$ to be an integer (let this be $b$ ): $b^2 = a^2 - 24a$ . Completing the square, we get $[(a - 12)^2 = b^2 + 144]$ Which implies that $b^2 + 144$ is a perfect square also (let this be $c^2$ ). Then
$[c^2 - b^2 = 144 Longrightarrow (c+b)(c-b) = 144]$

The pairs of factors of $144$ are $(pm1,pm144),( pm 2, pm 72),( pm 3, pm 48),( pm 4, pm 36),( pm 6, pm 24),( pm 8, pm 18),( pm 9, pm 16),( pm 12, pm 12)$ ; since $c$ is the average of each respective pair and is also an integer, the pairs that work must have the same parity. Thus we get $boxed{10}$ pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work.
Use the two trigonometric Pythagorean identities $1 + tan^2 x = sec^2 x$ and $1 + cot^2 x = csc^2 x$ .If we square the given $sec x = frac{22}{7} - tan x$ , we find that $begin{align*} sec^2 x &= left(frac{22}7right)^2 - 2left(frac{22}7right)tan x + tan^2 x \ 1 &= left(frac{22}7right)^2 - frac{44}7 tan x end{align*}$ This yields $tan x = frac{435}{308}$ .Let $y = frac mn$ . Then squaring,
$[csc^2 x = (y - cot x)^2 Longrightarrow 1 = y^2 - 2ycot x.]$

Substituting $cot x = frac{1}{tan x} = frac{308}{435}$ yields a quadratic equation: $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$ . It turns out that only the positive root will work, so the value of $y = frac{29}{15}$ and $m + n = boxed{044}$ .

Note: The problem is much easier computed if we consider what sec x is, then find the relationship between sin x and cos x (using tan x = $frac{435}{308}$ , and then computing csc x + cot x using 1/sin x and then the reciprocal of tan x.
Let us make a chart of values in alphabetical order, where $P_a, P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$ , and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = sum_{n=1}^{b} P_b$ ): $[begin{array}{|r||r|r|r|} hline text{String}&P_a&P_b&S_b\ hline aaa & 8 & 1 & 1 \ aab & 4 & 2 & 3 \ aba & 4 & 2 & 5 \ abb & 2 & 4 & 9 \ baa & 4 & 2 & 11 \ bab & 2 & 4 & 15 \ bba & 2 & 4 & 19 \ bbb & 1 & 8 & 27 \ hline end{array}]$ The probability is $p=sum P_a cdot (27 - S_b)$ , so the answer turns out to be $frac{8cdot 26 + 4 cdot 24 ldots 2 cdot 8 + 1 cdot 0}{27^2} = frac{532}{729}$ , and the solution is $boxed{532}$ .
We wish to find the radius of one circle, so that we can find the total area.Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius $1$ .We thus know that the apothem of the dodecagon is equal to $1$ . To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote $A, M,$ and $O$ respectively. Notice that $OM=1$ , and that $triangle OMA$ is a right triangle with hypotenuse $OA$ and $m angle MOA = 15^circ$ . Thus $AM = (1) tan{15^circ} = 2 - sqrt {3}$ , which is the radius of one of the circles. The area of one circle is thus $pi(2 - sqrt {3})^{2} = pi (7 - 4 sqrt {3})$ , so the area of all $12$ circles is $pi (84 - 48 sqrt {3})$ , giving an answer of $84 + 48 + 3 = boxed{135}$ .
Note that it is very useful to understand the side lengths of a 15-75-90 triangle, as these triangles often appear on higher level math contests. The side lengths are in the ratio $sqrt {3} - 1$ , $sqrt {3} + 1$ , and $2sqrt {2}$
Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent ( $triangle BPQ cong triangle DRS$ , $triangle APS cong triangle CRQ$ ). Quickly we realize that $O$ is also the center of the rectangle.By the Pythagorean Theorem, we can solve for a side of the rhombus; $PQ = sqrt{15^2 + 20^2} = 25$ . Since the diagonals of a rhombus are perpendicular bisectors, we have that $OP = 15, OQ = 20$ . Also, $angle POQ = 90^{circ}$ , so quadrilateral $BPOQ$ is cyclic. By Ptolemy's Theorem, $25 cdot OB = 20 cdot 15 + 15 cdot 20 = 600$ .By similar logic, we have $APOS$ is a cyclic quadrilateral. Let $AP = x$ , $AS = y$ . The Pythagorean Theorem gives us $x^2 + y^2 = 625quad mathrm{(1)}$ . Ptolemy’s Theorem gives us $25 cdot OA = 20x + 15y$ . Since the diagonals of a rectangle are equal, $OA = frac{1}{2}d = OB$ , and $20x + 15y = 600quad mathrm{(2)}$ . Solving for $y$ , we get $y = 40 - frac 43x$ . Substituting into $mathrm{(1)}$ , $begin{eqnarray*}x^2 + left(40-frac 43xright)^2 &=& 625\ 5x^2 - 192x + 1755 &=& 0\ x = frac{192 pm sqrt{192^2 - 4 cdot 5 cdot 1755}}{10} &=& 15, frac{117}{5}end{eqnarray*}$ We reject $15$ because then everything degenerates into squares, but the condition that $PR neq QS$ gives us a contradiction. Thus $x = frac{117}{5}$ , and backwards solving gives $y = frac{44}5$ . The perimeter of $ABCD$ is $2left(20 + 15 + frac{117}{5} + frac{44}5right) = frac{672}{5}$ , and $m + n = boxed{677}$ .
Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$ . The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by $[frac{r(r-1)}{(r+b)(r+b-1)}+frac{b(b-1)}{(r+b)(r+b-1)}=frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=frac{1}{2}.]$ Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$ , for $r$ in terms of $t$ , one obtains that $[r=frac{tpmsqrt{t}}{2}, .]$ Now, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$ , with $ninmathbb{N}$ . Hence, $r=n(npm 1)/2$ would correspond to the general solution. For the present case $tleq 1991$ , and so one easily finds that $n=44$ is the largest possible integer satisfying the problem conditions.
In summary, the solution is that the maximum number of red socks is $r=boxed{990}$ .
$[asy]defaultpen(fontsize(9)); pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18)); draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D..A--E); dot(A^^B^^C^^D^^E^^F); label("(A)",A,(-1,-1));label("(B)",B,(1,-1));label("(C)",C,(1,0)); label("(D)",D,(1,1));label("(E)",E,(-1,1));label("(F)",F,(-1,0)); label("31",A/2+B/2,(0.7,1));label("81",B/2+C/2,(0.45,-0.2)); label("81",C/2+D/2,(-1,-1));label("81",D/2+E/2,(0,-1)); label("81",E/2+F/2,(1,-1));label("81",F/2+A/2,(1,1)); label("(x)",A/2+C/2,(-1,1));label("(y)",A/2+D/2,(1,-1.5)); label("(z)",A/2+E/2,(1,0)); [/asy]$ Let $x=AC=BF$ , $y=AD=BE$ , and $z=AE=BD$ .Ptolemy's Theorem on $ABCD$ gives $81y+31cdot 81=xz$ , and Ptolemy on $ACDF$ gives $xcdot z+81^2=y^2$ . Subtracting these equations give $y^2-81y-112cdot 81=0$ , and from this $y=144$ . Ptolemy on $ADEF$ gives $81y+81^2=z^2$ , and from this $z=135$ . Finally, plugging back into the first equation gives $x=105$ , so $x+y+z=105+144+135=boxed{384}$ .
Interpret the problem geometrically. Consider $n$ right triangles joined at their vertices, with bases $a_1,a_2,ldots,a_n$ and heights $1,3,ldots, 2n - 1$ . The sum of their hypotenuses is the value of $S_n$ . The minimum value of $S_n$ , then, is the length of the straight line connecting the bottom vertex of the first right triangle and the top vertex of the last right triangle, so $[S_n ge sqrt {left(sum_{k = 1}^n (2k - 1)right)^2 + left(sum_{k = 1}^n a_kright)^2}.]$ Since the sum of the first $n$ odd integers is $n^2$ and the sum of $a_1,a_2,ldots,a_n$ is 17, we get $[S_n ge sqrt {17^2 + n^4}.]$ If this is an integer, we can write $17^2 + n^4 = m^2$ , for an integer $m$ . Thus, $(m - n^2)(m + n^2) = 289cdot 1 = 17cdot 17 = 1cdot 289.$ The only possible value, then, for $m$ is $145$ , in which case $n^2 = 144$ , and $n = boxed {012}$ .