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1986AIME 真題及答案解析

Category: AMC美國數(shù)學(xué)競(jìng)賽, 國際競(jìng)賽 Date: 2019年12月12日上午11:55

1986AIME 真題及答案解析

Problem 1

What is the sum of the solutions to the equation? $sqrt[4]{x} = frac{12}{7 - sqrt[4]{x}}$ ?

Problem 2

Evaluate the product? $(sqrt 5+sqrt6+sqrt7)(-sqrt 5+sqrt6+sqrt7)(sqrt 5-sqrt6+sqrt7)(sqrt 5+sqrt6-sqrt7)$ .

Problem 3

If? $tan x+tan y=25$ ?and? $cot x + cot y=30$ , what is? $tan(x+y)$ ?

Problem 4

Determine? $3x_4+2x_5$ ?if? $x_1$ ,? $x_2$ ,? $x_3$ ,? $x_4$ , and? $x_5$ ?satisfy the system of equations below.

$2x_1+x_2+x_3+x_4+x_5=6$ $x_1+2x_2+x_3+x_4+x_5=12$ $x_1+x_2+2x_3+x_4+x_5=24$ $x_1+x_2+x_3+2x_4+x_5=48$ $x_1+x_2+x_3+x_4+2x_5=96$

Problem 5

What is that largest?positive integer? $n$ ?for which? $n^3+100$ ?is?divisible?by? $n+10$ ?

Problem 6

The pages of a book are numbered? $1_{}^{}$ ?through? $n_{}^{}$ . When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of? $1986_{}^{}$ . What was the number of the page that was added twice?

Problem 7

The increasing sequence? $1,3,4,9,10,12,13cdots$ ?consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the? $100^{mbox{th}}$ ?term of this sequence.

Problem 8

Let? $S$ ?be the sum of the base? $10$ ?logarithms of all the proper divisors of? $1000000$ . What is the integer nearest to? $S$ ?

Problem 9

In? $triangle ABC$ ,? $AB= 425$ ,? $BC=450$ , and? $AC=510$ . An interior point? $P$ ?is then drawn, and segments are drawn through? $P$ ?parallel to the sides of the triangle. If these three segments are of an equal length? $d$ , find? $d$ .

Problem 10

In a parlor game, the magician asks one of the participants to think of a three digit number? $(abc)$ ?where? $a$ ,? $b$ , and? $c$ ?represent digits in base? $10$ ?in the order indicated. The magician then asks this person to form the numbers? $(acb)$ ,? $(bca)$ ,? $(bac)$ ,? $(cab)$ , and? $(cba)$ , to add these five numbers, and to reveal their sum,? $N$ . If told the value of? $N$ , the magician can identify the original number,? $(abc)$ . Play the role of the magician and determine the? $(abc)$ ?if? $N= 3194$ .

Problem 11

The polynomial? $1-x+x^2-x^3+cdots+x^{16}-x^{17}$ ?may be written in the form? $a_0+a_1y+a_2y^2+cdots +a_{16}y^{16}+a_{17}y^{17}$ , where? $y=x+1$ ?and the? $a_i$ 's are constants. Find the value of? $a_2$ .

Problem 12

Let the sum of a set of numbers be the sum of its elements. Let? $S$ ?be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of? $S$ ?have the same sum. What is the largest sum a set? $S$ ?with these properties can have?

Problem 13

In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH, HH, and etc. For example, in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH, three HT, four TH, and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT subsequences?

Problem 14

The shortest distances between an interior diagonal of a rectangular parallelepiped,? $P$ , and the edges it does not meet are? $2sqrt{5}$ ,? $frac{30}{sqrt{13}}$ , and? $frac{15}{sqrt{10}}$ . Determine the volume of? $P$ .

Problem 15

Let triangle? $ABC$ ?be a right triangle in the? $xy$ -plane with a right angle at? $C_{}$ . Given that the length of the hypotenuse? $AB$ ?is? $60$ , and that the medians through? $A$ ?and? $B$ ?lie along the lines? $y=x+3$ ?and? $y=2x+4$ ?respectively, find the area of triangle? $ABC$ .

1986AIME 詳細(xì)解析

Let? $y = sqrt[4]{x}$ . Then we have? $y(7 - y) = 12$ , or, by simplifying, $[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.]$ This means that? $sqrt[4]{x} = y = 3$ ?or? $4$ .Thus the sum of the possible solutions for? $x$ ?is? $4^4 + 3^4 = 337$ .
Simplify by repeated application of the?difference of squares.

$left((sqrt{6} + sqrt{7})^2 - sqrt{5}^2right)left(sqrt{5}^2 - (sqrt{6} - sqrt{7})^2right)$

$= (13 + 2sqrt{42} - 5)(5 - (13 - 2sqrt{42}))$

$= (2sqrt{42} + 8)(2sqrt{42} - 8)$

$= (2sqrt{42})^2 - 8^2 =$ ? $boxed{104}$
Since? $cot$ ?is the reciprocal function of? $tan$ : $cot x + cot y = frac{1}{tan x} + frac{1}{tan y} = frac{tan x + tan y}{tan x cdot tan y} = 30$ Thus,? $tan x cdot tan y = frac{tan x + tan y}{30} = frac{25}{30} = frac{5}{6}$ Using the tangent addition formula: $tan(x+y) = frac{tan x + tan y}{1-tan x cdot tan y} = frac{25}{1-frac{5}{6}} = boxed{150}$ .
Adding all five?equations?gives us? $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ ?so? $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ . Subtracting this from the fourth given equation gives? $x_4 = 17$ ?and subtracting it from the fifth given equation gives? $x_5 = 65$ , so our answer is? $3cdot17 + 2cdot65 = boxed{181}$ .
If? $n+10 mid n^3+100$ ,? $gcd(n^3+100,n+10)=n+10$ . Using the?Euclidean algorithm, we have? $gcd(n^3+100,n+10)= gcd(-10n^2+100,n+10)$ ? $= gcd(100n+100,n+10)$ ? $= gcd(-900,n+10)$ , so? $n+10$ ?must divide? $900$ . The greatest?integer? $n$ ?for which? $n+10$ ?divides? $900$ ?is? $boxed{890}$ ; we can double-check manually and we find that indeed? $900 mid 890^3+100$ .In a similar manner, we can apply synthetic division. We are looking for? $frac{n^3 + 100}{n + 10} = n^2 - 10n - 100 - frac{900}{n + 10}$ . Again,? $n + 10$ ?must be a factor of? $900 Longrightarrow n = boxed{890}$ .The key to this problem is to realize that? $n+10 mid n^3 +1000$ ?for all? $n$ . Since we are asked to find the maximum possible? $n$ ?such that? $n+10 mid n^3 +100$ , we have:? $n+10 mid ((n^3 +1000) - (n^3 +100) longrightarrow n+10 mid 900$ . This is because of the property that states that if? $a mid b$ ?and? $a mid c$ , then? $a mid b pm c$ . Since, the largest factor of 900 is itself we have:? $n+10=900 Longrightarrow boxed{n = 890}$
Denote the page number as? $x$ , with? $x < n$ . The sum formula shows that? $frac{n(n + 1)}{2} + x = 1986$ . Since? $x$ ?cannot be very large, disregard it for now and solve? $frac{n(n+1)}{2} = 1986$ . The positive root for? $n approx sqrt{3972} approx 63$ . Quickly testing, we find that? $63$ ?is too large, but if we plug in? $62$ ?we find that our answer is? $frac{62(63)}{2} + x = 1986 Longrightarrow x = boxed{033}$ .
Rewrite all of the terms in base 3. Since the numbers are sums of?distinct?powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number.? $100$ ?is equal to? $64 + 32 + 4$ , so in binary form we get? $1100100$ . However, we must change it back to base 10 for the answer, which is? $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = boxed {981}$ .Notice that the first term of the sequence is? $1$ , the second is? $3$ , the fourth is? $9$ , and so on. Thus the? $64th$ ?term of the sequence is? $729$ . Now out of? $64$ ?terms which are of the form? $729$ ?+? $'''S'''$ ,? $32$ ?of them include? $243$ ?and? $32$ ?do not. The smallest term that includes? $243$ , i.e.? $972$ , is greater than the largest term which does not, or? $854$ . So the? $95$ th term will be? $972$ , then? $973$ , then? $975$ , then? $976$ , and finally? $boxed{981}$ After the? $n$ th power of 3 in the sequence, the number of terms after that power but before the? $(n+1)$ th power of 3 is equal to the number of terms before the? $n$ th power, because those terms after the? $n$ th power are just the? $n$ th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of? $3$ ?and the terms that come after them, we see that the? $100$ th term is after? $729$ , which is the? $64$ th term. Also, note that the? $k$ th term after the? $n$ th power of 3 is equal to the power plus the? $k$ th term in the entire sequence. Thus, the? $100$ th term is? $729$ ?plus the? $36$ th term. Using the same logic, the? $36$ th term is? $243$ ?plus the? $4$ th term,? $9$ . We now have? $729+243+9=boxed{981}$
The?prime factorization?of? $1000000 = 2^65^6$ , so there are? $(6 + 1)(6 + 1) = 49$ ?divisors, of which? $48$ ?are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers.Writing out the first few terms, we see that the answer is equal to $[log 1 + log 2 + log 4 + ldots + log 1000000 = log (2^05^0)(2^15^0)(2^25^0)cdots (2^65^6).]$ Each power of? $2$ ?appears? $7$ ?times; and the same goes for? $5$ . So the overall power of? $2$ ?and? $5$ ?is? $7(1+2+3+4+5+6) = 7 cdot 21 = 147$ . However, since the question asks for proper divisors, we exclude? $2^65^6$ , so each power is actually? $141$ ?times. The answer is thus? $S = log 2^{141}5^{141} = log 10^{141} = boxed{141}$ .Since the prime factorization of? $10^6$ ?is? $2^6 cdot 5^6$ , the number of factors in? $10^6$ ?is? $7 cdot 7=49$ . You can pair them up into groups of two so each group multiplies to? $10^6$ . Note that? $log n+log{(10^6/n)}=log{n}+log{10^6}-log{n}=6$ . Thus, the sum of the logs of the divisors is half the number of divisors of? $10^6 cdot 6 -6$ ?(since they are asking only for proper divisors), and the answer is? $(49/2)cdot 6-6=141$ .
$[asy] size(200); pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425)); /* construct remaining points */ pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); dot(MP("D",D,NE,s));dot(MP("E",E,NW,s));dot(MP("F",F,s));dot(MP("D'",Da,NE,s));dot(MP("E'",Ea,NW,s));dot(MP("F'",Fa,s)); D(D--Ea);D(Da--F);D(Fa--E); MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); /*P copied from above solution*/ pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); [/asy]$ Let the points at which the segments hit the triangle be called? $D, D', E, E', F, F'$ ?as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are?similar?( $triangle ABC sim triangle DPD' sim triangle PEE' sim triangle F'PF$ ). The remaining three sections are?parallelograms.By similar triangles,? $BE'=fraccqle8lm{510}cdot450=frac{15}{17}d$ ?and? $EC=fraccqle8lm{425}cdot450=frac{18}{17}d$ . Since? $FD'=BC-EE'$ , we have? $900-frac{33}{17}d=d$ , so? $d=boxed{306}$ .
Let? $m$ ?be the number? $100a+10b+c$ . Observe that? $3194+m=222(a+b+c)$ ?so $[mequiv -3194equiv -86equiv 136pmod{222}]$ This reduces? $m$ ?to one of? $136, 358, 580, 802$ . But also? $a+b+c=frac{3194+m}{222}>frac{3194}{222}>14$ ?so? $a+b+cgeq 15$ . Of the four options, only? $m = boxed{358}$ ?satisfies this inequality.
Using the?geometric series?formula,? $1 - x + x^2 + cdots - x^{17} = frac {1 - x^{18}}{1 + x} = frac {1-x^{18}}{y}$ . Since? $x = y - 1$ , this becomes? $frac {1-(y - 1)^{18}}{y}$ . We want? $a_2$ , which is the coefficient of the? $y^3$ ?term in? $-(y - 1)^{18}$ ?(because the? $y$ ?in the denominator reduces the degrees in the numerator by? $1$ ). By the?Binomial Theorem, this is? $(-1) cdot (-1)^{15}{18 choose 3} = boxed{816}$ .Again, notice? $x = y - 1$ . So $begin{align*}1 - x + x^2 + cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + cdots - (y - 1)^{17} \ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 cdots + (1 - y)^{17}end{align*}.$ We want the coefficient of the? $y^2$ ?term of each power of each binomial, which by the binomial theorem is? ${2choose 2} + {3choose 2} + cdots + {17choose 2}$ . The?Hockey Stick Identity?tells us that this quantity is equal to? ${18choose 3} = boxed{816}$ .
By using the greedy algorithm, we obtain? $boxed{061}$ , with? $S={ 15,14,13,11,8}$ . We must now prove that no such set has sum greater than 61. Suppose such a set? $S$ ?existed. Then? $S$ ?must have more than 4 elements, otherwise its sum would be at most? $15+14+13+12=54$ . $S$ ?can't have more than 5 elements. To see why, note that at least? $dbinom{6}{0} + dbinom{6}{1} + dbinom{6}{2} + dbinom{6}{3} + dbinom{6}{4}=57$ ?of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element and so on), and each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum.Thus,? $S$ ?would have to have 5 elements.? $S$ ?contains both 15 and 14 (otherwise its sum is at most? $10+11+12+13+15=60$ ). It follows that? $S$ ?cannot contain both? $a$ ?and? $a-1$ ?for any? $aleq 13$ , or the subsets? ${a,14}$ ?and? ${a-1,15}$ ?would have the same sum. So now? $S$ ?must contain 13 (otherwise its sum is at most? $15+14+12+10+8=59$ ), and? $S$ ?cannot contain 12, or the subsets? ${12,15}$ ?and? ${13,14}$ ?would have the same sum.Now the only way? $S$ ?could have sum at least? $62=15+14+13+11+9$ ?would be if? $S={ 15,14,13,11,9}$ . But the subsets? ${9,15}$ ?and? ${11,13}$ ?have the same sum, so this set does not work. Therefore no? $S$ ?with sum greater than 61 is possible and 61 is indeed the maximum.
Let's consider each of the sequences of two coin tosses as an?operation?instead; this operation takes a string and adds the next coin toss on (eg,?THHTH?+?HT?=?THHTHT). We examine what happens to the last coin toss. Adding?HH?or?TT?is simply an?identity?for the last coin toss, so we will ignore them for now. However, adding?HT?or?TH?switches the last coin.?H?switches to?T?three times, but?T?switches to?H?four times; hence it follows that our string will have a structure of?THTHTHTH.Now we have to count all of the different ways we can add the identities back in. There are 5?TT?subsequences, which means that we have to add 5?T?into the strings, as long as the new?Ts are adjacent to existing?Ts. There are already 4?Ts in the sequence, and since order doesn’t matter between different tail flips this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives? ${{5+3}choose3} = 56$ ?combinations. We do the same with 2?Hs to get? ${{2+3}choose3} = 10$ ?combinations; thus there are? $56 cdot 10 = boxed {560}$ ?possible sequences.SLIGHT VARIATION ON FINAL ARGUMENTThe structure of the final order is?T_H_T_H_T_H_T_H_, and there are 4 spots to put the 2 heads in, and 4 spots to put the 5 tails in. By using the formula for distributing r identical objects into n distinct boxes? $dbinom{r+n-1}{r}$ ?and multiplication, the answer is? ${{2+4-1}choose2}$ ?*? ${{5+4-1}choose5}$ ?=560
In the above diagram, we focus on the line that appears closest and is parallel to? $BC$ . All the blue lines are perpendicular lines to? $BC$ ?and their other points are on? $AB$ , the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and reach out to? $AC$ , and have the same lengths as their corresponding blue lines. So we want to find the shortest distance between? $AC$ ?and that corner, which is? $frac {wl}{sqrt {w^2 + l^2}}$ .So we have: $[frac {lw}{sqrt {l^2 + w^2}} = frac {10}{sqrt {5}}]$ $[frac {hw}{sqrt {h^2 + w^2}} = frac {30}{sqrt {13}}]$ $[frac {hl}{sqrt {h^2 + l^2}} = frac {15}{sqrt {10}}]$ Notice the familiar roots:? $sqrt {5}$ ,? $sqrt {13}$ ,? $sqrt {10}$ , which are? $sqrt {1^2 + 2^2}$ ,? $sqrt {2^2 + 3^2}$ ,? $sqrt {1^2 + 3^2}$ , respectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.) $[frac {l^2w^2}{l^2 + w^2} = frac {1}{frac {1}{l^2} + frac {1}{w^2}} = 20]$ $[frac {h^2w^2}{h^2 + w^2} = frac {1}{frac {1}{h^2} + frac {1}{w^2}} = frac {900}{13}]$ $[frac {h^2l^2}{h^2 + l^2} = frac {1}{frac {1}{h^2} + frac {1}{l^2}} = frac {45}{2}]$ We invert the above equations to get a system of linear equations in? $frac {1}{h^2}$ ,? $frac {1}{l^2}$ , and? $frac {1}{w^2}$ : $[frac {1}{l^2} + frac {1}{w^2} = frac {45}{900}]$ $[frac {1}{h^2} + frac {1}{w^2} = frac {13}{900}]$ $[frac {1}{h^2} + frac {1}{l^2} = frac {40}{900}]$
We see that? $h = 15$ ,? $l = 5$ ,? $w = 10$ . Therefore? $V = 5 cdot 10 cdot 15 = 750$ .
Translate so the?medians?are? $y = x$ , and? $y = 2x$ , then model the?points? $A: (a,a)$ ?and? $B: (b,2b)$ .? $(0,0)$ ?is the?centroid, and is the average of the vertices, so? $C: (- a - b, - a - 2b)$
$AB = 60$ ?so

$3600 = (a - b)^2 + (2b - a)^2$
$3600 = 2a^2 + 5b^2 - 6ab (1)$

$AC$ ?and? $BC$ ?are?perpendicular, so the product of their?slopes?is? $-1$ , giving

$left(frac {2a + 2b}{2a + b}right)left(frac {a + 4b}{a + 2b}right) = - 1$
$2a^2 + 5b^2 = - frac {15}{2}ab (2)$

Combining? $(1)$ ?and? $(2)$ , we get? $ab = - frac {800}{3}$

Using the?determinant?product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is? $left|frac {3}{2}abright|$ , so we get the answer to be? $400$ .