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2014 USAJMO 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11:48

2014 USAJMO Problems真题及答案

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Day 1

Problem 1

Let $a$ , $b$ , $c$ be real numbers greater than or equal to $1$ . Prove that $[min{left (frac{10a^2-5a+1}{b^2-5b+10},frac{10b^2-5b+1}{c^2-5c+10},frac{10c^2-5c+1}{a^2-5a+10}right )}leq abc.]$

Problem 2

Let $triangle{ABC}$ be a non-equilateral, acute triangle with $angle A=60^circ$ , and let $O$ and $H$ denote the circumcenter and orthocenter of $triangle{ABC}$ , respectively.

(a) Prove that line $OH$ intersects both segments $AB$ and $AC$ .

(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$ , respectively. Denote by $s$ and $t$ the respective areas of triangle $APQ$ and quadrilateral $BPQC$ . Determine the range of possible values for $s/t$ .

Problem 3

Let $mathbb{Z}$ be the set of integers. Find all functions $f : mathbb{Z} rightarrow mathbb{Z}$ such that $[xf(2f(y)-x)+y^2f(2x-f(y))=frac{f(x)^2}{x}+f(yf(y))]$ for all $x, y in mathbb{Z}$ with $x neq 0$ .

Day 2

Problem 4

Let $bgeq 2$ be an integer, and let $s_b(n)$ denote the sum of the digits of $n$ when it is written in base $b$ . Show that there are infinitely many positive integers that cannot be represented in the form $n+s_b(n)$ , where $n$ is a positive integer.

Problem 5

Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.

Problem 6

Let $ABC$ be a triangle with incenter $I$ , incircle $gamma$ and circumcircle $Gamma$ . Let $M,N,P$ be the midpoints of sides $overline{BC}$ , $overline{CA}$ , $overline{AB}$ and let $E,F$ be the tangency points of $gamma$ with $overline{CA}$ and $overline{AB}$ , respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$ , respectively, and let $X$ be the midpoint of arc $BAC$ of $Gamma$ .

(a) Prove that $I$ lies on ray $CV$ .

(b) Prove that line $XI$ bisects $overline{UV}$ .

Problem 1

Solution

Since $(a-1)^5ge 0$ , $[a^5-5a^4+10a^3-10a^2+5a-1ge 0]$ or $[10a^2-5a+1le a^3(a^2-5a+10)]$ Since $a^2-5a+10=left( a-dfrac{5}{2}right)^2 +dfrac{15}{4}>0$ , $[frac{10a^2-5a+1}{a^2-5a+10}le a^3]$ Also note that $10a^2-5a+1=10left( a-dfrac{1}{4}right)^2+dfrac{3}{8}> 0$ , We conclude $[0le frac{10a^2-5a+1}{a^2-5a+10}le a^3]$ Similarly, $[0le frac{10b^2-5b+1}{b^2-5b+10}le b^3]$ $[0le frac{10c^2-5c+1}{c^2-5c+10}le c^3]$ So $[left(frac{10a^2-5a+1}{a^2-5a+10}right)left(frac{10b^2-5b+1}{b^2-5b+10}right)left(frac{10c^2-5c+1}{c^2-5c+10}right)le a^3b^3c^3]$ or $[left(frac{10a^2-5a+1}{b^2-5b+10}right)left(frac{10b^2-5b+1}{c^2-5c+10}right)left(frac{10c^2-5c+1}{a^2-5a+10}right) le(abc)^3]$ Therefore, $[minleft(frac{10a^2-5a+1}{b^2-5b+10},frac{10b^2-5b+1}{c^2-5c+10},frac{10c^2-5c+1}{a^2-5a+10}right )le abc.]$

Problem 2

Solution

$[asy] import olympiad; unitsize(1inch); pair A,B,C,O,H,P,Q,i1,i2,i3,i4; //define dots A=3*dir(50); B=(0,0); C=right*2.76481496; O=circumcenter(A,B,C); H=orthocenter(A,B,C); i1=2*O-H; i2=2*i1-O; i3=2*H-O; i4=2*i3-H; //These points are for extending PQ. DO NOT DELETE! P=intersectionpoint(i2--i4,A--B); Q=intersectionpoint(i2--i4,A--C); //draw dot(P); dot(Q); draw(P--Q); dot(A); dot(B); dot(C); dot(O); dot(H); draw(A--B--C--cycle); //label label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,NW); label("$Q$",Q,NE); label("$O$",O,N); label("$H$",H,N); //change O and H label positions if interfering with other lines to be added //further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistance and are not to be used [/asy]$

Lemma: $H$ is the reflection of $O$ over the angle bisector of $angle BAC$ (henceforth 'the' reflection)

Proof: Let $H'$ be the reflection of $O$ , and let $B'$ be the reflection of $B$ .

Then reflection takes $angle ABH'$ to $angle AB'O$ .

$Delta ABB'$ is equilateral, and $O$ lies on the perpendicular bisector of $overline{AB}$

It's well known that $O$ lies strictly inside $Delta ABC$ (since it's acute), meaning that $angle ABH' = angle AB'O = 30^{circ},$ from which it follows that $overline{BH'} perp overline{AC}$ . Similarly, $overline{CH'} perp overline{AB}$ . Since $H'$ lies on two altitudes, $H$ is the orthocenter, as desired.

So $overline{OH}$ is perpendicular to the angle bisector of $angle OAH$ , which is the same line as the angle bisector of $angle BAC$ , meaning that $Delta APQ$ is equilateral.

Let its side length be $s$ , and let $PH=t$ , where $0 < t < s, t neq s/2$ because $O$ lies strictly within $angle BAC$ , as must $H$ , the reflection of $O$ . Also, it's easy to show that if $O=H$ in a general triangle, it's equilateral, and we know $Delta ABC$ is not equilateral. Hence H is not on the bisector of $angle BAC implies t neq s/2$ . Let $overrightarrow{BH}$ intersect $overline{AC}$ at $P_B$ .

Since $Delta HP_BQ$ and $BP_BA$ are 30-60-90 triangles, $AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t$

Similarly, $AC=2s-t$

The ratio $frac{[APQ]}{[ABC]-[APQ]}$ is $frac{AP cdot AQ}{AB cdot AC - AP cdot AQ} = frac{s^2}{(s+t)(2s-t)-s^2}$ The denominator equals $(1.5s)^2-(.5s-t)^2-s^2$ where $.5s-t$ can equal any value in $(-.5s, .5s)$ except $0$ . Therefore, the denominator can equal any value in $(s^2, 5s^2/4)$ , and the ratio is any value in $boxed{left(frac{4}{5},1right)}.$

Note: It's easy to show that for any point $H$ on $overline{PQ}$ except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.

Solution 2

Let $J$ be the farthest point on the circumcircle of $triangle{ABC}$ from line $BC$ . Lemma: Line $AJ$ ||Line $OH$ Proof: Set $b=-dfrac{1}{2}+idfrac{sqrt{3}}{2}$ and $c=-dfrac{1}{2}-idfrac{sqrt{3}}{2}$ , and $a$ on the unit circle. It is well known that $o=0$ and $h=a+b+c$ , so we have $h-o=a+b+c=a-1=a-j$ , so $dfrac{a-j}{h-o}$ is real and thus the 2 lines are parallel.

WLOG let $A$ be in the first quadrant. Clearly by the above lemma $OH$ must intersect line $BC$ closer to $B$ than to $C$ . Intersect $AJ$ and $BC$ at $D$ and $OH$ and $BC$ at $E$ . We clearly have $0 < dfrac{overarc{JC}-overarc{AB}}{2} = dfrac{120-overarc{AC}}{2} = angle{ADB} = angle{OEC} < 30 = angle{OBC}$ , $OH$ must intersect $AB$ . We also have, letting the intersection of line $OH$ and line $AC$ be $Q$ , and letting intersection of $OH$ and $AB$ be $P$ , $angle{OQA} = angle{JAB} = dfrac{overarc{JB}}{2} = 60$ . Since $angle{OCA}=angle{BCA}-angle{BCO} < 90-30 =angle{OQA}$ , and $angle{OQC} = 120>angle{OAC}$ , $OH$ also intersects $AC$ . We have $angle{OPA}=angle{PAJ}=60$ , so $triangle{APQ}$ is equilateral. Letting $AJ=2x$ , and letting the foot of the perpendicular from $O$ to $AJ$ be $L$ , we have $OL=sqrt{1-x^{2}}$ , and since $OL$ is an altitude of $triangle{APQ}$ , we have $[APQ]=dfrac{OL^{2}sqrt{3}}{3} = dfrac{sqrt{3}(1-x^{2})}{3}$ . Letting the foot of the perpendicular from $A$ to $OJ$ be $K$ , we have $triangle{JKA}sim triangle{JLO}$ by AA with ratio $dfrac{JA}{JO} = 2x$ . Therefore, $JK = 2x(JL) = 2x^{2}$ . Letting $D$ be the foot of the altitude from $J$ to $BC$ , we have $KD=JD-JK=dfrac{3}{2}-2x^{2}$ , since $re(B)=re(C) implies JD=re(j)-(-dfrac{1}{2})=dfrac{3}{2}$ . Thus, since $BC=sqrt{3}$ we have $[ABC]=dfrac{(dfrac{3}{2}-2x^{2})(sqrt{3})}{2} = dfrac{(3-4x^{2})(sqrt{3})}{2}$ , so $[PQCB]=[ABC]-[APQ]=dfrac{(5-8x^{2})(sqrt{3})}{12}$ , so $dfrac{[APQ]}{[PQCB]} = dfrac{4-4x^{2}}{5-8x^{2}} = dfrac{1}{2}+dfrac{3}{10-16x^{2}}$ . We have $x=sin(dfrac{JOA}{2})$ , with $0<120-2angle{ACB}=angle{JOA}<60$ , so $x$ can be anything in the interval $(0, dfrac{1}{2})$ . Therefore, the desired range is $(dfrac{4}{5}, 1)$ .

Problem 3

Solution

Let's assume $f(0)neq 0.$ Substitute $(x,y)=(2f(0),0)$ to get $[2f(0)^2=f(2f(0))^2/2f(0)+f(0)]$ $[2f(0)^2(2f(0)-1)=f(2f(0))^2]$

This means that $2(2f(0)-1)$ is a perfect square. However, this is impossible, as it is equivalent to $2pmod{4}.$ Therefore, $f(0)=0.$ Now substitute $xneq 0, y=0$ to get $[xf(-x)=frac{f(x)^2}{x} implies x^2f(-x)=f(x)^2.]$ Similarly, $[x^2f(x)=f(-x)^2.]$ From these two equations, we can find either $f(x)=f(-x)=0,$ or $f(x)=f(-x)=x^2.$ Both of these are valid solutions on their own, so let's see if there are any solutions combining the two.

Let's say we can find $f(x)=x^2, f(y)=0,$ and $x,yneq 0.$ Then $[xf(-x)+y^2f(2x)=f(x)^2/x.]$ $[y^2f(2x)=x-x^3.]$ (NEEDS FIXING: $f(x)^2/x= x^4/x = x^3$ , so the RHS is $0$ instead of $x-x^3$ .)

If $f(2x)=4x^2,$ then $y^2=frac{x-x^3}{4x^2}=frac{1-x^2}{4x},$ which is only possible when $y=0.$ This contradicts our assumption. Therefore, $f(2x)=0.$ This forces $x=pm 1$ due to the right side of the equation. Let's consider the possibility $f(2)=0, f(1)=1.$ Substituting $(x,y)=(2,1)$ into the original equation yields $[0=2f(0)+1f(2)=0+f(1)=1,]$ which is impossible. So $f(2)=f(-2)=4$ and there are no solutions "combining" $f(x)=x^2$ and $f(x)=0.$

Therefore our only solutions are $boxed{f(x)=0}$ and $boxed{f(x)=x^2.}$

Problem 4

Solution 1

Define $S(n) = n + s_b(n)$ , and call a number unrepresentable if it cannot equal $S(n)$ for a positive integer $n$ . We claim that in the interval $(b^p, b^{p+1}]$ there exists an unrepresentable number, for every positive integer $p$ .

If $b^{p+1}$ is unrepresentable, we're done. Otherwise, time for our lemma:

Lemma: Define the function $f(p)$ to equal the number of integers x less than $b^p$ such that $S(x) ge b^p$ . If $b^{p+1} = S(y)$ for some y, then $f(p+1) > f(p)$ .

Proof: Let $F(p)$ be the set of integers x less than $b^p$ such that $S(x) ge b^p$ . Then for every integer in $F(p)$ , append the digit $(b-1)$ to the front of it to create a valid integer in $F(p+1)$ . Also, notice that $(b-1) cdot b^p le y < b^{p+1}$ . Removing the digit $(b-1)$ from the front of y creates a number that is not in $F(p)$ . Hence, $F(p) rightarrow F(p+1)$ , but there exists an element of $F(p+1)$ not corresponding with $F(p)$ , so $f(p+1) > f(p)$ .

Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.

Operation Intuitive Number Theory (INT)- Solution 2

I hope this solution is quite intuitive, because it is without complicated notation. It didn't take me very long to discover. As with Solution 1, define $S(n)=n+s_b(n)$ . We will use numbers from $b^k$ to $b^{k+1}-1$ for induction. Call this interval Class $k$ .

Start with $b^0=1$ and go up to $b-1$ . There are $b-1$ numbers covered. Easily $S(n)$ ranges from $1+1=2$ to $b-1+b-1=2b-2$ , or a range of $2b-3$ . Thus, there are at LEAST $b-2$ numbers lacking coverage. Now, in order to fill up these gaps, we consult Class $1$ : $b^1=b$ to $b^2-1$ . If we are to fill up all these gaps, then we need at least $b-2$ numbers in the next series with their $S$ values. Unfortunately, there are only at MOST $b-3$ numbers that can satisfy a value, from $b$ to $2b-3$ , otherwise the $S$ value is too big (note that none of them are zero)! Thus, between Class 0 and 1, there is at least one value lacking.

After setting up the base case, consider the Class $k$ numbers from $b^k$ to $b^{k+1}-1$ there are clearly $b^{k+1}-b^k$ integers. Yet the $S(n)$ can range from $b^k+1=b^k+1$ to $(b^{k+1}-1)+(k+1)(b-1)=b^{k+1}+b(k+1)-(k+1)$ . This gives a range of $b^{k+1}-b^k+b(k+1)-(k+1)$ . This leaves an extra $b(k+1)-(k+1)$ numbers. Now, we invoke the next class: $k+1$ . Again, to fill in the gap there are sadly only $b(k+1)-(k+2)$ numbers available: $b^{k+1}$ to $b^{k+1}+b(k+1)-(k+3)$ , because $s_b(n)$ is at least 1.

By induction we are done. Because each Class $k$ for integral $k$ at least $0$ misses at least one $S$ value, we miss an infinite number of numbers. Game over!

Note: If you are unconvinced of the range of $S(n)$ values we go from the smallest value of both the number and the $s_b(n)$ , or $b^k$ and $1$ , to the greatest of both, i.e. $b^{k+1}-1$ and $k(b+1)$ , because in that series the largest possible sum is a bunch of $b-1$ s.

Problem 5

Solution

The answer is $k=6$ . We prove that $A$ can win for $k=5$ (which hence proves it for $k<5$ as well) and show that $B$ can thwart $A$ for $kgeq 6$ .

Arrange the board so that a pair of opposite sides are horizontal. Create a coordinate system on the board by setting the center of some hexagon as the origin and setting the hexagons directly above and above-and-right as $(0, 1)$ and $(1, 0)$ , respectively. Then, for example, the below-and-right hexagon touching the origin is $(1, -1)$ . So two hexagons touch if their coordinate difference is one of these.

Now for $k=5$ , person $A$ places his counters only in ${0, 1, 2, 3, 4}times{0, 1}$ . Note that if, at $A$ 's turn, there are 4 counters in either column, then $A$ can win immediately, so let us assume that in both columns there are at least 2 missing, meaning that at most $6$ counters are on the board. We would like to find when $A$ cannot play under these circumstances. If we look at the disjoint sets

${(0, 0), (0, 1), (1, 0)}, {(1, 1), (2, 0), (2, 1)}, {(3, 0), (3, 1), (4, 0)}, {(4, 1)}$

we see that either some set has at least $2$ hexagons without counters, in which case $A$ can move, or all four sets have exactly $1$ missing counter. Similarly for the sets

${(0, 0)}, {(0, 1), (1, 0), (1, 1)}, {(2, 0), (2, 1), (3, 0)}, {(3, 1), (4, 0), (4, 1)}$

So both $(0, 0), (4, 1)$ have no token on them. This means that $(0, 1), (1, 0), (3, 1), (4, 0)$ do. Thus $(3, 0)$ and $(1, 1)$ do not. So this is the only situation in which $A$ can neither win immediately nor play in only these 10 hexagons.

So $A$ plays only in these 10 hexagons until either he has a win or he can't anymore. If he wins, then we're done. Otherwise, $A$ plays in hexagons $(5, 0)$ and $(5, 1)$ . Then $B$ either removes $(5, 1)$ so that $A$ can win at $(2, 0)$ , or $B$ removes $(5, 0)$ and $A$ plays at $(5, 0)$ and $(4, 1)$ and then at either $(3, 0)$ or $(1, 1)$ the next turn, or $B$ removes one of ${0, 1, 2, 3, 4}times{0, 1}$ in which case $A$ can either win immediately at $(3, 0)$ or can play in both columns, and then win the next turn.

Now if $kgeq 6$ , then if $A$ plays on anything in the lattice generated by $(2, -1)$ and $(1, 1)$ , that is $(2a+b, b-a)$ for $a, b$ integers, then $B$ removes it. Otherwise, $B$ removes any of $A$ 's counters. This works because in order for $A$ to win, there must be at least 2 counters in this lattice, but $A$ can only put a counter on $1$ at any time, so there's at most 1 on the lattice at any time.

So $A$ wins if $kleq 5$ , and $B$ wins if $kgeq 6$ .

Problem 6

Solution

$[asy] unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label("$A$", A, W); label("$B$", B, dir(90)); label("$C$", C, dir(0)); dot(I); label("$I$", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label("$Gamma$", dir(35), dir(35)); label("$gamma$", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label("$M$", M, NE); label("$N$", N, SE); label("$P$", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label("$E$", E, SW); label("$F$", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label("$U$", U, S); label("$V$", V, NE); draw(P--M--U--F); X = dir(235); label("$X$", X, dir(235)); draw(X--I, dashed); draw(C--V, dashed); draw(A--I); label("$x^circ$", A + (0.2,0), dir(90)); label("$y^circ$", C + (-0.4,0), dir(90)); [/asy]$ (a)

Solution 1: We will prove this via contradiction: assume that line $IC$ intersects line $MP$ at $Q$ and line $EF$ and $R$ , with $R$ and $Q$ not equal to $V$ . Let $x = angle A/2 = angle IAE$ and $y = angle C/2 = angle ICA$ . We know that $overline{MP} parallel overline{AC}$ because $MP$ is a midsegment of triangle $ABC$ ; thus, by alternate interior angles (A.I.A) $angle MVE = angle FEA = (180^circ - 2x) / 2 = 90^circ - x$ , because triangle $AFE$ is isosceles. Also by A.I.A, $angle MQC = angle QCA = y$ . Furthermore, because $AI$ is an angle bisector of triangle $AFE$ , it is also an altitude of the triangle; combining this with $angle QIA = x + y$ from the Exterior Angle Theorem gives $angle FRC = 90 - x - y$ . Also, $angle VRQ = angle FRC = 90 - x - y$ because they are vertical angles. This completes part (a).

Solution 2: First we show that the intersection $V'$ of $MP$ with the internal angle bisector of $C$ is the same as the intersection $V''$ of $EF$ with the internal angle bisector of $C.$ Let $D$ denote the intersection of $AB$ with the internal angle bisector of $C,$ and let $a,b,c$ denote the side lengths of $BC, AC, AB.$ By Menelaus on $V'', F, E,$ with respect to $triangle ADC,$ $[frac{AF}{FD}cdot frac{DV''}{V''C}cdot frac{CE}{EA}=-1]$ $[frac{DV''}{V''C}=-frac{frac{bc}{a+b}+frac{a-b-c}{2}}{frac{a+b-c}{2}}=frac{b-a}{b+a}.]$ Similarly, $[frac{BP}{PD}cdot frac{DV'}{V'C}cdot frac{CM}{MB}=-1]$ $[frac{DV'}{V'C}=-frac{frac{ac}{a+b}-frac{c}{2}}{frac{c}{2}}=frac{b-a}{b+a}.]$ Since $V'$ and $V''$ divide $CD$ in the same ratio, they must be the same point. Now, since $frac{b-a}{b+a}>-1,$ $I$ lies on ray $CV.$

(b)

Solution 1: Using a similar argument to part (a), point U lies on line $BI$ . Because $angle MVC = angle VCA = angle MCV$ , triangle $VMC$ is isosceles. Similarly, triangle $BMU$ is isosceles, from which we derive that $VM = MC = MB = MU$ . Hence, triangle $VUM$ is isosceles.

Note that $X$ lies on both the circumcircle and the perpendicular bisector of segment $BC$ . Let $D$ be the midpoint of $UV$ ; our goal is to prove that points $X$ , $D$ , and $I$ are collinear, which equates to proving $X$ lies on ray $ID$ .

Because $MD$ is also an altitude of triangle $MVU$ , and $MD$ and $IA$ are both perpendicular to $EF$ , $overline{MD} parallel overline{IA}$ . Furthermore, we have $angle VMD = angle UMD = x$ because $APMN$ is a parallelogram. (incomplete)

Solution 2: Let $I_A$ , $I_B$ , and $I_C$ be the excenters of $ABC$ . Note that the circumcircle of $ABC$ is the nine-point circle of $I_AI_BI_C$ . Since $AX$ is the external angle bisector of $angle BAC$ , $X$ is the midpoint of $I_BI_C$ . Now $UV$ and $I_BI_C$ are parallel since both are perpendicular to the internal angle bisector of $angle BAC$ . Since $IX$ bisects $I_BI_C$ , it bisects $UV$ as well.

Solution 3: Let $X'$ be the antipode of $X$ with respect to the circumcircle of triangle $ABC$ . Then, by the Incenter-Excenter lemma, $X'$ is the center of a circle containing $B$ , $I$ , and $C$ . Because $XX'$ is a diameter, $XB$ and $XC$ are tangent to the aforementioned circle; thus by a well-known symmedian lemma, $XI$ coincides with the $I$ -symmedian of triangle $IBC$ . From part (a); we know that $BVUC$ is cyclic (we can derive a similar argument for point $U$ ); thus $XI$ coincides with the median of triangle $VIU$ , and we are done.