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2018 USAMO Problems真题及答案

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年8月13日上午11:08

2018 USAMO Problems真题及答案

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Day 1

Note: For any geometry problem whose statement begins with an asterisk ( $*$ ), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction.

Problem 1

Let $a,b,c$ be positive real numbers such that $a+b+c=4sqrt[3]{abc}$ . Prove that $[2(ab+bc+ca)+4min(a^2,b^2,c^2)ge a^2+b^2+c^2.]$

Problem 2

Find all functions $f:(0,infty) rightarrow (0,infty)$ such that

$[fleft(x+frac{1}{y}right)+fleft(y+frac{1}{z}right) + fleft(z+frac{1}{x}right) = 1]$ for all $x,y,z >0$ with $xyz =1.$

Problem 3

For a given integer $nge 2,$ let ${a_1,a_2,…,a_m}$ be the set of positive integers less than $n$ that are relatively prime to $n.$ Prove that if every prime that divides $m$ also divides $n,$ then $a_1^k+a_2^k + dots + a_m^k$ is divisible by $m$ for every positive integer $k.$

Day 2

Problem 4

Let $p$ be a prime, and let $a_1, dots, a_p$ be integers. Show that there exists an integer $k$ such that the numbers $[a_1 + k, a_2 + 2k, dots, a_p + pk]$ produce at least $tfrac{1}{2} p$ distinct remainders upon division by $p$ .

Problem 5

In convex cyclic quadrilateral $ABCD,$ we know that lines $AC$ and $BD$ intersect at $E,$ lines $AB$ and $CD$ intersect at $F,$ and lines $BC$ and $DA$ intersect at $G.$ Suppose that the circumcircle of $triangle ABE$ intersects line $CB$ at $B$ and $P$ , and the circumcircle of $triangle ADE$ intersects line $CD$ at $D$ and $Q$ , where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$ , then $angle MAC = 90^{circ}.$

Problem 6

Let $a_n$ be the number of permutations $(x_1, x_2, dots, x_n)$ of the numbers $(1,2,dots, n)$ such that the $n$ ratios $frac{x_k}{k}$ for $1le kle n$ are all distinct. Prove that $a_n$ is odd for all $nge 1.$

2018USAMO真题参考答案及详解

Problem 1

Solution 1

WLOG let $a leq b leq c$ . Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: $[4(a(a+b+c)+bc) geq (a+b+c)^2]$ $[frac{4asqrt[3]{abc}+bc}{2} geq 2sqrt[3]{a^2b^2c^2}]$

The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.

Problem 2&3

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Problem 4

Solution

$textbf{Lemma: }$ For fixed $ineq j,$ where $i, jin{1, 2, ..., p},$ the statement $a_i + ikequiv a_j + jktext{ (mod } ptext{)}$ holds for exactly one $kin {1, 2, ..., p}.$

$textbf{Proof: }$ Notice that the left side minus the right side is congruent to $(a_i - a_j) + (i - j)k$ modulo $p.$ For this difference to equal $0,$ there is a unique solution for $k$ modulo $p$ given by $kequiv (a_j - a_i)(i - j)^{-1}text{ (mod } ptext{)},$ where we have used the fact that every nonzero residue modulo $p$ has a unique multiplicative inverse. Therefore, there is exactly one $kin {1, 2, ..., p}$ that satisfies $a_i + ikequiv a_j + jktext{ (mod } ptext{)}$ for any fixed $ineq j. textbf{ End Lemma}$

Suppose that you have $p$ graphs $G_1, G_2, ..., G_p,$ and graph $G_k$ consists of the vertices $(i, k)$ for all $1le ile p.$ Within any graph $G_k,$ vertices $(i_1, k)$ and $(i_2, k)$ are connected by an edge if and only if $a_{i_1} + i_1kequiv a_{i_2} + i_2ktext{ (mod } ptext{)}.$ Notice that the number of disconnected components of any graph $G_k$ equals the number of distinct remainders when divided by $p$ given by the numbers $a_1 + k, a_2 + 2k, ..., a_p + pk.$

These $p$ graphs together have exactly one edge for every unordered pair of elements of ${1, 2, ..., p},$ so they have a total of exactly $frac{p(p-1)}{2}$ edges. Therefore, there exists at least one graph $G_k$ that has strictly fewer than $frac{p}{2}$ edges, meaning that it has more than $frac{p}{2}$ disconnected components. Therefore, the collection of numbers ${a_i + ik: 1le ile p}$ for this particular value of $k$ has at least $frac{p}{2}$ distinct remainders modulo $p.$ This completes the proof.

$textbf{Note: This is the same problem as 2018 USAJMO Problem 5.}$

Problem 5

Solution

We invert about $C$ swapping $E$ and $A$ .

Under this inversion, we must have $A mapsto E$ , $B mapsto P$ , and $Dmapsto Q$ ; therefore as $ABCD$ is a cyclic quadrilateral, $P, E,$ and $Q$ are collinear.

Next, let $G'$ be the image of $G$ under this inversion. I claim that $EB = EG'$ . Note that $angle{EBC} = angle{EAC} = 180^{circ} - angle{EAG} = angle{EG'G}$ , so the claim follows. Similarly if $F'$ is the image of $F$ , then $ED = EF'$ .

Then the image of line $GQ$ is the circumcircle of $triangle{G'DC}$ , and similarly the image of line $FP$ is the circumcircle of $triangle{F'BC}$ .Suppose these circles meet at $X$ , which is the image of $M$ under the inversion. Then we want to show that $angle{CXE} = 90^{circ}$ .

The good thing about this inversion is that it completely eliminates the necessity of points $A, P, Q, G$ and $F$ ; we may restate the problem as follows:

Inverted USAMO 5 wrote:

Suppose that $CDB$ is a triangle, and let $E$ be some point on side $BD$ . Let $U$ and $V$ be the feet of the perpendiculars from $E$ to $CD$ and $CB$ , respectively, and let $F'$ and $G'$ be the reflections of $D$ and $B$ over $U$ and $V$ , respectively. Let the circumcircles of $triangle{G'DC}$ and $triangle{F'BC}$ meet at $X neq C$ . Show that $angle{CXE} = 90^{circ}$ .

Equivalently, it suffices to demonstrate that $CUXEV$ is a cyclic quadrilateral; we know that $CUEV$ is cyclic with diameter $CE$ , so we want to show that $angle{UXV} = 180^{circ} - angle{C}$ . We claim that $X$ is the center of the spiral similarity mapping $DG'$ to $F'B$ . Observe that $angle{XDG'} = angle{XCB} = angle{XF'B}$ ; similarly, $angle{XG'D} = angle{XCD} = angle{XBF'}$ . Thus this claim is proven.

Therefore, by the Gliding Principle and the principle that spiral similarities come in pairs, we observe that $X$ maps $U$ , the midpoint of $DF'$ , to $V$ , the midpoint of $G'B$ . Therefore $angle{UXV} = angle{F'XB} = 180^{circ} - angle{C}$ , and we may conclude.

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Problem 6

Solution

Write out the mapping of each $k$ to $x_k$ as follows: $[1 2 3 dots n]$ $[x_1 x_2 x_3 dots x_n]$

Now, consider what happens when the two rows are swapped (and the top-bottom pairs are reordered so that the top reads (1,2,3,...,n)). This will result in a new permutation if and only if $x_k = j$ does not imply $x_j = k$ for all $k,j$ in $(1,2,3,dots,n)$ . I will denote this new permutation as $(y_1,y_2,y_3,dots,y_n)$ . Notice that $(y_1,y_2,y_3,dots,y_n)$ is a valid permutation if and only if $(x_1,x_2,x_3,dots,x_n)$ is valid, because each fraction $frac{y_k}{k}$ is the reciprocal of $frac{x_k}{k}$ . This means that we need only consider the parity of number of cases in which $(y_1,y_2,y_3,dots, y_n) = (x_1,x_2,x_3,dots, x_n)$ , as there will be an even number of cases where $(y_1,y_2,y_3,dots, y_n) ne (x_1,x_2,x_3,dots, x_n)$ . Let the number of valid permutations where $(y_1,y_2,y_3,dots, y_n) = (x_1,x_2,x_3,dots, x_n)$ be $b_n$ .

Notice that the only permutations that have the property $x_{x_k}=k$ (which is an equivalent statement to the one given above) are those that are formed by taking pairs of elements and swapping their positions having the maximum number of pairs possible and having no two pairs both contain the same element. Some more necessary conditions will be outlined below after we split $n$ into two cases.

Case 1: $n$ is odd. If $n$ is odd then there must be exactly one $k$ such that $x_k = k$ . This yields $b_n=ncdot b_{n-1}$ * which has the same parity as $b_{n-1}$ , so we need only consider the parity of $b_n$ for odd $n$ . (*This is because we must select one $k$ so that $x_k=k$ and the other $n-1$ numbers can be determined in $b_{n-1}$ ways, because there can be no $k$ such that $x_k=k$ if $k$ is even.)

Case 2: $n$ is even. If $n$ is even then can be no $k$ so that $x_k=k$ , or else there must be at least two distinct numbers $k$ and $j$ so that $x_k=k$ and $x_j=j$ which violates the given conditions. Denote a pair of numbers that are swapped to arrive at the final permutation as the pair $(a,b)$ . Then if a permutation yields an invalid arrangement there must be two pairs $(a,b)$ and $(c,d)$ such that $frac{a^2}{b^2} = frac{c^2}{d^2}$ . But notice that the two pairs $(a,c)$ and $(b,d)$ will also result in an invalid arrangement. So, there must be an even number of invalid arrangements, meaning the parity we desire is just the number of ways to separate $n$ objects into $frac n2$ pairs! However, this is quite simply just $(n-1)(n-3)(n-5)cdots(3)(1)$ , which is clearly the product of odd numbers. So we conclude that there are an odd number of valid permutations $(x_1,x_2,x_3,dots,x_n)$ .