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2017 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11:25

2017 USAJMO 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

Prove that there are infinitely many distinct pairs $(a,b)$ of relatively prime positive integers $a > 1$ and $b > 1$ such that $a^b + b^a$ is divisible by $a + b.$

Problem 2

Consider the equation $\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.$

(a) Prove that there are infinitely many pairs $(x,y)$ of positive integers satisfying the equation.

(b) Describe all pairs $(x,y)$ of positive integers satisfying the equation.

Problem 3

( $*$ ) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$ ; let lines $PB$ and $CA$ intersect at $E$ ; and let lines $PC$ and $AB$ intersect at $F$ . Prove that the area of triangle $DEF$ is twice the area of triangle $ABC$ .

Day 2

Problem 4

Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2) + 12$ is a prime that properly divides the positive number $a^2 + b^2 + c^2 + abc - 2017$ ?

Problem 5

( $*$ ) Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$ . Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$ . Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$ . Prove that $\angle ADO = \angle HAN$ .

Problem 6

Let $P_1, \ldots, P_{2n}$ be $2n$ distinct points on the unit circle $x^2 + y^2 = 1$ other than $(1,0)$ . Each point is colored either red or blue, with exactly $n$ of them red and exactly $n$ of them blue. Let $R_1, \ldots, R_n$ be any ordering of the red points. Let $B_1$ be the nearest blue point to $R_1$ traveling counterclockwise around the circle starting from $R_1$ . Then let $B_2$ be the nearest of the remaining blue points to $R_2$ traveling counterclockwise around the circle from $R_2$ , and so on, until we have labeled all the blue points $B_1, \ldots, B_n$ . Show that the number of counterclockwise arcs of the form $R_i \rightarrow B_i$ that contain the point $(1,0)$ is independent of the way we chose the ordering $R_1, \ldots, R_n$ of the red points.

Problem 1

Solution 1

Let $a = 2n-1$ and $b = 2n+1$ . We see that $(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}$ . Therefore, we have $(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}$ , as desired.

Solution 2

Let $x$ be odd where $x>1$ . We have $x^2-1=(x-1)(x+1),$ so $x^2-1 \equiv 0 \pmod{2x+2}.$ This means that $x^{x+2}-x^x \equiv 0 \pmod{2x+2},$ and since x is odd, $x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},$ or $x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},$ as desired.

Solution 3

Because problems such as this usually are related to expressions along the lines of $x\pm1$ , it's tempting to try these. After a few cases, we see that $\left(a,b\right)=\left(2x-1,2x+1\right)$ is convenient due to the repeated occurrence of $4x$ when squared and added. We rewrite the given expressions as: $\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.$ After repeatedly factoring the initial equation,we can get: $\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).$ Expanding each of the squares, we can compute each product independently then sum them: $\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},$ $\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.$ Now we place the values back into the expression: $\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.$ Plugging any positive integer value for $x$ into $\left(a,b\right)=\left(2x-1,2x+1\right)$ yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs $\left(a,b\right)$ .

Problem 2

Solution 1

We have $(3x^3+xy^2)(yx^2+3y^3)=(x-y)^7$ , which can be expressed as $xy(3x^2+y^2)(x^2+3y^2)=(x-y)^7$ . At this point, we

think of substitution. A substitution of form $a=x+y, b=x-y$ is slightly derailed by the leftover x and y terms, so

instead, seeing the xy in front, we substitute $x=a+b, y=a-b$ . This leaves us with $(a^2-b^2)(4a^2+4ab+4b^2)(4a^2- 4ab+4b^2)=128b^7$ , so $(a^2-b^2)(a^2+ab+b^2)(a^2-ab+b^2)=8b^7$ . Expanding yields $a^6-b^6=8b^7$ . Rearranging, we have

$b^6(8b+1)=a^6$ . To satisfy this equation in integers, $8b+1$ must obviously be a $6th$ power, and further inspection shows

that it must also be odd. Also, since it is a square and all odd squares are 1 mod 8, every odd sixth power gives a

solution. Since the problem asks for positive integers, the pair $(a,b)=(0,0)$ does not work. We go to the next highest odd

$6th$ power, $3^6$ or $729$ . In this case, $b=91$ , so the LHS is $91^6\cdot3^6=273^6$ , so $a=273$ . Using the original

substitution yields $(x,y)=(364,182)$ as the first solution. We have shown part a by showing that there are an infinite

number of positive integer solutions for $(a,b)$ , which can then be manipulated into solutions for $(x,y)$ . To solve part

b, we look back at the original method of generating solutions. Define $a_n$ and $b_n$ to be the pair representing the nth

solution. Since the nth odd number is $2n+1$ , $b_n=\frac{(2n+1)^6-1}{8}$ . It follows that $a_n=(2n+1)b_n=\frac{(2n+1)^7- (2n+1)}{8}$ . From our original substitution, $(x,y)=\left(\frac{(2n+1)^7+(2n+1)^6-2n-2}{8}, \frac{(2n+1)^7-(2n+1)^6-2n} {8}\right)$ .

Solution 2 (and motivation)

First, we shall prove a lemma:

LEMMA:

$\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = \frac{(x+y)^6-(x-y)^6}{4}$ PROOF: Expanding and simplifying the right side, we find that $\frac{(x+y)^6-(x-y)^6}{4}=\frac{12x^5y+40x^3y^3+12xy^5}{4}$ $=3x^5y+10x^3y^3+3xy^5$ $=\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right)$ which proves our lemma.

Now, we have that $\frac{(x+y)^6-(x-y)^6}{4}=(x-y)^7$ Rearranging and getting rid of the denominator, we have that $(x+y)^6=4(x-y)^7+(x-y)^6$ Factoring, we have $(x+y)^6=(x-y)^6(4(x-y)+1)$ Dividing both sides, we have $\left(\frac{x+y}{x-y}\right)^6=4x-4y+1$ Now, since the LHS is the 6th power of a rational number, and the RHS is an integer, the RHS must be a perfect 6th power. Define $a=\frac{x+y}{x-y}$ . By inspection, $a$ must be a positive odd integer satistisfying $a \geq 3$ . We also have $a^6=4x-4y+1$ Now, we can solve for $x$ and $y$ in terms of $a$ : $x-y=\frac{a^6-1}{4}$ and $x+y=a(x-y)=\frac{a(a^6-1)}{4}$ . Now we have: $(x,y)=\left(\frac{(a+1)(a^6-1)}{8},\frac{(a-1)(a^6-1)}{8}\right)$ and it is trivial to check that this parameterization works for all such $a$ (to keep $x$ and $y$ integral), which implies part (a).

MOTIVATION FOR LEMMA: I expanded the LHS, noticed the coefficients were $(3,10,3)$ , and immediately thought of binomial coefficients. Looking at Pascal's triangle, it was then easy to find and prove the lemma.

-sunfishho

Solution $(1.5, x)$ where $|x|>0$

So named because it is a mix of solutions 1 and 2 but differs in other aspects. After fruitless searching, let $x+y=a$ , $x-y=b$ . Clearly $a, b > 0$ . Then $x=\frac{a+b}{2}, y=\frac{a-b}{2}, x^2+y^2=\frac{a^2+b^2}{2}, x^2=\frac{a^2+2ab+b^2}{4}, y^2=\frac{a^2-2ab+b^2}{4}, xy=\frac{a^2-b^2}{4}$ .

Change the LHS to $xy(3x^2+y^2)(x^2+3y^2)=(a-b)(a+b)(a^2+ab+b^2)(a^2-ab+b^2)*\frac{1}{4}=(a^3-b^3)(a^3+b^3)*\frac{1}{4}=(a^6-b^6)*\frac{1}{4}$ . Change the RHS to $b^7$ . Therefore $(\frac{a}{b})^6=4b+1$ . Let $n=\frac{a}{b}$ , and note that $n$ is an integer. Therefore $a=n(\frac{n^6-1}{4}), b=\frac{n^6-1}{4}$ . Because $n^6 \equiv 1 (\mod{4})$ , $n$ is odd and $>1$ because $b>0$ . Therefore, substituting for $x$ and $y$ we get: $x=\frac{(n^4+n^2+1)(n-1)(n+1)^2}{8}, y=\frac{(n+1)(n-1)^2(n^4+n^2+1)}{8}$ .

Problem 3

WLOG, let $AB = 1$ . Let $[ABD] = X, [ACD] = Y$ , and $\angle BAD = \theta$ . After some angle chasing, we find that $\angle BCF \cong \angle BEC \cong \theta$ and $\angle FBC \cong \angle BCE \cong 120^{\circ}$ . Therefore, $\triangle FBC$ ~ $\triangle BCE$ .

Lemma 1: If $BF = k$ , then $CE = \frac 1k$ . This lemma results directly from the fact that $\triangle FBC$ ~ $\triangle BCE$ ; $\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}$ , or $CE = \frac{1}{BF}$ .

Lemma 2: $[AEF] = (k+\frac 1k + 2)(X+Y)$ . We see that $[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)$ , as desired.

Lemma 3: $\frac{X}{Y} = k$ . We see that $\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.$ However, after some angle chasing and by the Law of Sines in $\triangle BCF$ , we have $\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}$ , or $k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}$ , which implies the result.

By the area lemma, we have $[BDF] = kX$ and $[CDF] = \frac{Y}{k}$ .

We see that $[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk$ . Thus, it suffices to show that $X + Y + \frac Xk + Yk = 2X + 2Y$ , or $\frac Xk + Yk = X + Y$ . Rearranging, we find this to be equivalent to $\frac XY = k$ , which is Lemma 3, so the result has been proven.

Solution 2

We will use barycentric coordinates and vectors. Let $\vec{X}$ be the position vector of a point $X.$ The point $(x, y, z)$ in barycentric coordinates denotes the point $x\vec{A} + y\vec{B} + z\vec{C}.$ For all points in the plane of $\triangle{ABC},$ we have $x + y + z = 1.$ It is clear that $A = (1, 0, 0)$ ; $B = (0, 1, 0)$ ; and $C = (0, 0, 1).$

Define the point $P$ as $P = \left(x_P, y_P, z_P\right).$ The fact that $P$ lies on the circumcircle of $\triangle{ABC}$ gives us $x^2_P + y^2_P + z^2_P = 1.$ This, along with the condition $x_P + y_P + z_P = 1$ inherent to barycentric coordinates, gives us $x_Py_P + y_Pz_P + z_Px_P = 0.$

We can write the equations of the following lines: $BC: x = 0$ $CA: y = 0$ $AB: z = 0$ $PA: \frac{y}{y_P} = \frac{z}{z_P}$ $PB: \frac{x}{x_P} = \frac{z}{z_P}$ $PC: \frac{x}{x_P} = \frac{y}{y_P}.$

We can then solve for the points $D, E, F$ : $D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)$ $E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)$ $F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).$

The area of an arbitrary triangle $XYZ$ is: $[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|$ $[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.$

To calculate $[DEF],$ we wish to compute $(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).$ After a lot of computation, we obtain the following: $(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].$

Evaluating the denominator, $(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)$ $(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.$

Since $x_P + y_P + z_P = 1$ and $x_Py_P + y_Pz_P + z_Px_P = 0,$ it follows that: $(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.$

We thus conclude that: $(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].$

From this, it follows that $[DEF] = 2[ABC],$ and we are done.

Solution 3

$[asy] import cse5; import graph; import olympiad; size(3inch); pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3)); pair P = (-1, -sqrt(11)+sqrt(3)); path circle = Circle(O, 2sqrt(3)); pair D = extension(A,P,B,C); pair E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label("A", A, N); label("B", B, W); label("C", C, E); label("P", P, S); label("D", D, NW); label("E", E1, SE); label("F", F, SW); dot("O", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$ . Let $AB=2$ , and $BF = 2x$ , then $F=(-x-1,-x\sqrt{3})$ . Using the facts $\triangle{CBP} \sim \triangle{CFB}$ and $\triangle{BCP} \sim \triangle{BEC}$ , we have $BF * CE = BC^2$ , so $CE = \frac{1}{2x}$ , and $E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})$ .

The slope of $FE$ is $k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}$ It is well-known that $\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$ , i.e., $OD$ is perpendicular to $FE$ . Therefore, the slope of $OD$ is $-\frac{1}{k}$ . Since $O=(0,\frac{1}{\sqrt{3}})$ , we get the x-coordinate of $D$ , $x_D = \frac{k}{\sqrt{3}}$ , i.e., $D = (\frac{k}{\sqrt{3}},0)$ . Using shoelace, $2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})- (-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}}$ $= 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)$ $= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2} {2+\frac{1}{x}+x})$ $= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})$ $= 4\sqrt{3}$ So $[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]$ . Q.E.D

Problem 4

Solution

Yes. Let $p = (a-2)(b-2)(c-2)+12 = abc - 2(ab+ac+bc)+4(a+b+c)+4$ . Also define $\alpha=a+b+c$ . We want $p$ to divide the positive number $a^2+b^2+c^2+abc-2017=(\alpha-47)(\alpha+43)+p$ . This equality can be verified by expanding the righthand side. Because $a^2+b^2+c^2+abc-2017$ will be trivially positive if $(\alpha-47)(\alpha+43)$ is non-negative, we can just assume that $\alpha=47$ . Analyzing the structure of $p$ , we see that $a-2$ , $b-2$ , and $c-2$ must be $1$ or $5$ mod $6$ , or $p$ will not be prime (divisibility by $2$ and $3$ ). Thus, we can guess any $a$ , $b$ , and $c$ which satisfies those constraints. For example, $a = 21$ , $b=13$ , and $c=13$ works. $p=2311$ is prime, and it divides the positive number $a^2+b^2+c^2+abc-2017=2311$ .

This solution is wrong. No $p$ actually exist.

Problem 5

Solution

$[asy] size(9cm); pair A = dir(130); pair B = dir(220); pair C = dir(320); draw(unitcircle, lightblue); pair P = dir(-90); pair Q = dir(90); pair D = extension(A, P, B, C); pair O = origin; pair M = extension(B, C, O, P); pair N = 2*M-P; draw(A--B--C--cycle, lightblue); draw(A--P--Q, lightblue); draw(A--N--D--O--A, lightblue); draw(A--D--N--O--cycle, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$D$", D, dir(225)); dot("$O$", O, dir(315)); dot("$M$", M, dir(315)); dot("$N$", N, dir(315)); [/asy]$

Suppose ray $OM$ intersects the circumcircle of $BHC$ at $N'$ , and let the foot of the A-altitude of $ABC$ be $E$ . Note that $\angle BHE=90-\angle HBE=90-90+\angle C=\angle C$ . Likewise, $\angle CHE=\angle B$ . So, $\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C$ . $BHCN'$ is cyclic, so $\angle BN'C=180-\angle BHC=180-\angle B-\angle C=\angle A$ . Also, $\angle BAC=\angle A$ . These two angles are on different circles and have the same measure, but they point to the same line $BC$ ! Hence, the two circles must be congruent. (This is also a well-known result)

We know, since $M$ is the midpoint of $BC$ , that $OM$ is perpendicular to $BC$ . $AH$ is also perpendicular to $BC$ , so the two lines are parallel. $AN$ is a transversal, so $\angle HAN=\angle ANO$ . We wish to prove that $\angle ANO=\angle ADO$ , which is equivalent to $AOND$ being cyclic.

Now, assume that ray $OM$ intersects the circumcircle of $ABC$ at a point $P$ . Point $P$ must be the midpoint of $\stackrel{\frown}{BC}$ . Also, since $AD$ is an angle bisector, it must also hit the circle at the point $P$ . The two circles are congruent, which implies $MN=MP\implies ND=DP\implies$ NDP is isosceles. Angle ADN is an exterior angle, so $\angle ADN=\angle DNP+\angle DPO=2\angle DPO$ . Assume WLOG that $\angle B>\angle C$ . So, $\angle DPO=\angle APO=\frac{\angle B+\angle C}{2}-\angle C=\frac{\angle B-\angle C}{2}$ . In addition, $\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A$ . Combining these two equations, $\angle AON+\angle ADN=\angle B-\angle C+2\angle C+\angle A=\angle A+\angle B+\angle C=180$ .

Opposite angles sum to $180$ , so quadrilateral $AOND$ is cyclic, and the condition is proved.

Problem 6

Solution

I define a sequence to be, starting at $(1,0)$ and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include $RB$ , $RBBR$ , $BBRRRB$ , $BRBRRBBR$ , etc. Note that choosing an $R_1$ is equivalent to choosing an $R$ in a sequence, and $B_1$ is defined as the $B$ closest to $R_1$ when moving rightwards. If no $B$ s exist to the right of $R_1$ , start from the far left. For example, if I have the above example $RBBR$ , and I define the 2nd $R$ to be $R_1$ , then the first $B$ will be $B_1$ . Because no $R$ or $B$ can be named twice, I can simply remove $R_1$ and $B_1$ from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of $BBRRRB$ is: $BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3$

Note that, if, in a move, $B_n$ appears to the left of $R_n$ , then $\stackrel{\frown}{R_nB_n}$ intersects $(1,0)$

Now, I define a commencing $B$ to be a $B$ which appears to the left of all $R$ s, and a terminating $R$ to be a $R$ which appears to the right of all $B$ s. Let the amount of commencing $B$ s be $j$ , and the amount of terminating $R$ s be $k$ , I claim that the number of arcs which cross $(1,0)$ is constant, and it is equal to $\text{max}(j,k)$ . I will show this with induction.

Base case is when $n=1$ . In this case, there are only two possible sequences - $RB$ and $BR$ . In the first case, $\stackrel{\frown}{R_1B_1}$ does not cross $(1, 0)$ , but both $j$ and $k$ are $0$ , so $\text{max}(j,k)=0$ . In the second example, $j=1$ , $k=1$ , so $\text{max}(j,k)=1$ . $\stackrel{\frown}{R_1B_1}$ crosses $(1,0)$ since $B_1$ appears to the left of $R_1$ , so there is one arc which intersects. Hence, the base case is proved.

For the inductive step, suppose that for a positive number $n$ , the number of arcs which cross $(1,0)$ is constant, and given by $\text{max}(j, k)$ for any configuration. Now, I will show it for $n+1$ .

Suppose I first choose $R_1$ such that $B_1$ is to the right of $R_1$ in the sequence. This implies that $\stackrel{\frown}{R_1B_1}$ does not cross $(1,0)$ . But, neither $R_1$ nor $B_1$ is a commencing $B$ or terminating $R$ . These numbers remain constant, and now after this move we have a sequence of length $2n$ . Hence, by assumption, the total amount of arcs is $0+\text{max}(j,k)=\text{max}(j,k)$ .

Now suppose that $R_1$ appears to the right of $B_1$ , but $B_1$ is not a commencing $B$ . This implies that there are no commencing $B$ s in the series, because there are no $B$ s to the left of $B_1$ , so $j=0$ . Note that this arc does intersect $(1,0)$ , and $R_1$ must be a terminating $R$ . $R_1$ must be a terminating $R$ because there are no $B$ s to the right of $R_1$ , or else that $B$ would be $B_1$ . The $2n$ length sequence that remains has $0$ commencing $B$ s and $k-1$ terminating $R$ s. Hence, by assumption, the total amount of arcs is $1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)$ .

Finally, suppose that $R_1$ appears to the right of $B_1$ , and $B_1$ is a commencing $B$ . We know that this arc will cross $(1,0)$ . Analogous to the previous case, $R_1$ is a terminating $R$ , so the $2n$ length sequence which remains has $j-1$ commencing $B$ s and $k-1$ terminating $R$ s. Hence, by assumption, the total amount of arcs is $1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)$ .

There are no more possible cases, hence the induction is complete, and the number of arcs which intersect $(1,0)$ is indeed a constant which is given by $\text{max}(j,k)$ .

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2017 USAJMO Problems真题及答案

2017 USAJMO Problems真题及答案

Day 1

Problem 1

Problem 2

Problem 3

Day 2

Problem 4

Problem 5

Problem 6

Problem 1

Solution 1

Solution 2

Solution 3

Problem 2

Solution 1

Solution 2 (and motivation)

Solution where

Problem 3

Solution 2

Solution 3

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

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Solution $(1.5, x)$ where $|x|>0$