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2016 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11:33

2016 USAJMO Problems真题及答案

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Day 1

Problem 1

The isosceles triangle $\triangle ABC$ , with $AB=AC$ , is inscribed in the circle $\omega$ . Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$ , and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$ , respectively.

Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point.

Problem 2

Prove that there exists a positive integer $n < 10^6$ such that $5^n$ has six consecutive zeros in its decimal representation.

Problem 3

Let $X_1, X_2, \ldots, X_{100}$ be a sequence of mutually distinct nonempty subsets of a set $S$ . Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$ , that is, $X_i\cap X_{i+1}=\emptyset$ and $X_i\cup X_{i+1}\neq S$ , for all $i\in\{1, \ldots, 99\}$ . Find the smallest possible number of elements in $S$ .

Day 2

Problem 4

Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,\dots,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ .

Problem 5

Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$ , and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$ , respectively.

Given that $AH^2=2\cdot AO^2,$ prove that the points $O,P,$ and $Q$ are collinear.

Problem 6

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$ , $(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$

Problem 1

Solution 1

Step 1: Set $x = y = 0$ to obtain $f(0) = 0.$

Step 2: Set $x = 0$ to obtain $f(y)f(-y) = f(y)^2.$

$\indent$ In particular, if $f(y) \ne 0$ then $f(y) = f(-y).$

$\indent$ In addition, replacing $y \to -t$ , it follows that $f(t) = 0 \implies f(-t) = 0$ for all $t \in \mathbb{R}.$

Step 3: Set $x = 3y$ to obtain $\left[f(y) + 3y^2\right]f(8y) = f(4y)^2.$

$\indent$ In particular, replacing $y \to t/8$ , it follows that $f(t) = 0 \implies f(t/2) = 0$ for all $t \in \mathbb{R}.$

Step 4: Set $y = -x$ to obtain $f(4x)\left[f(x) + f(-x) - 2x^2\right] = 0.$

$\indent$ In particular, if $f(x) \ne 0$ , then $f(4x) \ne 0$ by the observation from Step 3, because $f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0.$ Hence, the above equation implies that $2x^2 = f(x) + f(-x) = 2f(x)$ , where the last step follows from the first observation from Step 2.

$\indent$ Therefore, either $f(x) = 0$ or $f(x) = x^2$ for each $x.$

$\indent$ Looking back on the equation from Step 3, it follows that $f(y) + 3y^2 \ne 0$ for any nonzero $y.$ Therefore, replacing $y \to t/4$ in this equation, it follows that $f(t) = 0 \implies f(2t) = 0.$

Step 5: If $f(a) = f(b) = 0$ , then $f(b - a) = 0.$

$\indent$ This follows by choosing $x, y$ such that $x - 3y = a$ and $3x - y = b.$ Then $x + y = \tfrac{b - a}{2}$ , so plugging $x, y$ into the given equation, we deduce that $f\left(\tfrac{b - a}{2}\right) = 0.$ Therefore, by the third observation from Step 4, we obtain $f(b - a) = 0$ , as desired.

Step 6: If $f \not\equiv 0$ , then $f(t) = 0 \implies t = 0.$

$\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \ne 0$ and $x + y = t.$ The following three facts are crucial:

$\indent$ 1. $f(y) \ne 0.$ This is because $(x + y) - y = x$ , so by Step 5, $f(y) = 0 \implies f(x) = 0$ , impossible.

$\indent$ 2. $f(x - 3y) \ne 0.$ This is because $(x + y) - (x - 3y) = 4y$ , so by Step 5 and the observation from Step 3, $f(x - 3y) = 0 \implies f(4y) = 0 \implies f(2y) = 0 \implies f(y) = 0$ , impossible.

$\indent$ 3. $f(3x - y) \ne 0.$ This is because by the second observation from Step 2, $f(3x - y) = 0 \implies f(y - 3x) = 0.$ Then because $(x + y) - (y - 3x) = 4x$ , Step 5 together with the observation from Step 3 yield $f(3x - y) = 0 \implies f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0$ , impossible.

$\indent$ By the second observation from Step 4, these three facts imply that $f(y) = y^2$ and $f(x - 3y) = \left(x - 3y\right)^2$ and $f(3x - y) = \left(3x - y\right)^2.$ By plugging into the given equation, it follows that $\begin{align*} \left(x^2 + xy\right)\left(x - 3y\right)^2 + \left(y^2 + xy\right)\left(3x - y\right)^2 = 0. \end{align*}$ But the above expression miraculously factors into $\left(x + y\right)^4$ ! This is clearly a contradiction, since $t = x + y \ne 0$ by assumption. This completes Step 6.

Step 7: By Step 6 and the second observation from Step 4, the only possible solutions are $f \equiv 0$ and $f(x) = x^2$ for all $x \in \mathbb{R}.$ It's easy to check that both of these work, so we're done.

Alternative Solution 1

From steps 1 and 2 of Solution 1 we have that $f(0)=0$ , and $f(-y) \cdot f(y)=(f(y))^2$ . Therefore, if $f(y) \ne 0$ , then $f(y)=f(-y)$ . Furthermore, setting $y=-x$ gives us $(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0$ . The LHS can be factored as $(f(x)+f(-x)-2x^2) \cdot f(4x) = 0$ . In particular, if $f(4x) \ne 0$ , then we have $f(x)+f(-x)=2x^2$ . However, since we have from step 2 that $f(x)=f(-x)$ , assuming $f(x) \ne 0$ , the equation becomes $2f(x)=2x^2$ , so for every $x$ , $f(x)$ is equivalent to either $0$ or $x^2$ . From step 6 of Solution 1, we can prove that $f(x)=0$ , and $f(x)=x^2$ are the only possible solutions.

Solution 2

Step 1: $x=y=0 \implies f(0)=0$

Step 2: $x=0 \implies f(y)f(-y)=f(y)^{2}$ . Now, assume $y \not = 0$ . Then, if $f(y)=0$ , we substitute in $-y$ to get $f(y)f(-y)=f(-y)^{2}$ , or $f(y)=f(-y)=0$ . Otherwise, we divide both sides by $f(y)$ to get $f(y)=f(-y)$ . If $y=0$ , we obviously have $f(0)=f(0)$ . Thus, the function is even. . Step 3: $y=-x \implies 2f(4x)(f(x)-x^{2})=0$ . Thus, $\forall x$ , we have $f(4x)=0$ or $f(x)=x^{2}$ .

Step 4: We now assume $f(x) \not = 0$ , $x\not = 0$ . We have $f(\frac{x}{4})=\frac{x^{2}}{16}$ . Now, setting $x=y=\frac{x}{4}$ , we have $f(\frac{x}{2}=\frac{x^{2}}{4}$ or $f(\frac{x}{2})=0$ . The former implies that $f(x)=0$ or $x^{2}$ . The latter implies that $f(x)=0$ or $f(x)=\frac{x^{2}}{2}$ . Assume the latter. $y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}$ . Clearly, this implies that $f(x)$ is negative for some $m$ . Now, we have $f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0$ , which is a contradiction. Thus, $\forall x$ $f(x)=0$ or $f(x)=x^{2}$ .

Step 5: We now assume $f(x)=0$ , $f(y)=y^{2}$ for some $x,y \not = 0$ . Let $m$ be sufficiently large integer, let $z=|4^{m}x|$ and take the absolute value of $y$ (since the function is even). Choose $c$ such that $3z-c=y$ . Note that we have $\frac{c}{z}$ ~ $3$ and $\frac{y}{z}$ ~ $0$ . Note that $f(z)=0$ . Now, $x=z, y=c \implies$ LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to $(z+c)^{4}$ ~ $256z^{4}$ . Now if $f(z-3c)=0$ , the second term of the LHS/RHS clearly ~0 as $m \to \infty$ . if $f(z-3c)=0$ , then we have LHS/RHS ~ $0$ , otherwise, we have LHS/RHS~ $\frac{8^{2}\cdot 3z^{4}}{256z^{4}}$ ~ $\frac{3}{4}$ , a contradiction, as we're clearly not dividing by $0$ , and we should have LHS/RHS=1.

Problem 2

Solution 1

Define $v_p(N)$ for all rational numbers $N$ and primes $p$ , where if $N=\frac{x}{y}$ , then $v_p(N)=v_p(x)-v_p(y)$ , and $v_p(x)$ is the greatest power of $p$ that divides $x$ for integer $x$ . Note that the expression(that we're trying to prove is an integer) is clearly rational, call it $N$ .

$v_p(N)=\sum_{i=1}^\infty \left\lfloor \frac{k^{2}}{p^{i}} \right\rfloor+\sum_{j=0}^{k-1} \sum_{i=1}^\infty \left\lfloor \frac{j}{p^{i}}\right\rfloor-\sum_{j=k}^{2k-1} \sum_{i=1}^\infty \left\lfloor \frac{j}{p^{i}} \right\rfloor$ , by Legendre. Clearly, $\left\lfloor{\frac{x}{p}}\right\rfloor={\frac{x-r(x,p)}{p}}$ , and $\sum_{i=0}^{k-1} r(i,m)\leq \sum_{i=k}^{2k-1} r(i,m)$ , where $r(i,m)$ is the remainder function(we take out groups of $m$ which are just permutations of numbers $1$ to $m$ until there are less than $m$ left, then we have $m$ distinct values, which the minimum sum is attained at $0$ to $k-1$ ). Thus, $v_p(N)=\sum_{m=p^{i}, i\in \mathbb{N}_{+}}-\frac{k^{2}}{m}+\left\lfloor{\frac{k^{2}}{m}}\right\rfloor-\frac{\sum_{i=0}^{k-1} r(i,m)-\sum_{i=k}^{2k-1} r(i,m)}{m} \geq \sum_{m=p^{i}, i\in \mathbb{N}} \left\lceil -\frac{k^{2}}{m}+\lfloor{\frac{k^{2}}{m}}\rfloor\right\rceil \geq 0$ , as the term in each summand is a sum of floors also and is clearly an integer.

Solution 2 (Controversial)

Consider an $k\times k$ grid, which is to be filled with the integers $1$ through $k^2$ such that the numbers in each row are in increasing order from left to right, and such that the numbers in each column are in increasing order from bottom to top. In other words, we are creating an $k\times k$ standard Young tableaux.

The Hook Length Formula is the source of the controversy, as it is very powerful and trivializes this problem. The Hook Length Formula states that the number of ways to create this standard Young tableaux (call this $N$ for convenience) is: $N = \frac{\left(k^2\right)!}{\prod_{1\le i, j\le k}(i+j-1)}.$ Now, we do some simple rearrangement: $N = \left(k^2\right)!\cdot\prod_{j=1}^{k}\prod_{i=1}^{k}\frac{1}{i+j-1} = \left(k^2\right)!\cdot\prod_{j=1}^{k}\frac{\left(j-1\right)!}{\left(j+k-1\right)!}$ $= \left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}.$ This is exactly the expression given in the problem! Since the expression given in the problem equals the number of distinct $k\times k$ standard Young tableaux, it must be an integer, so we are done.

Problem 3

Solution

This problem can be proved in the following two steps.

1. Let $I_A$ be the $A$ -excenter, then $I_A,O,$ and $P$ are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for $\triangle I_AI_BI_C.$

2. Show that $I_AY^2-I_AZ^2=OY^2-OZ^2,$ which implies $\overline{OI_A}\perp\overline{YZ}.$ This can be proved by multiple applications of the Pythagorean Thm.

Problem 4

Solution 1

Step 1: Set $x = y = 0$ to obtain $f(0) = 0.$

Step 2: Set $x = 0$ to obtain $f(y)f(-y) = f(y)^2.$

$\indent$ In particular, if $f(y) \ne 0$ then $f(y) = f(-y).$

$\indent$ In addition, replacing $y \to -t$ , it follows that $f(t) = 0 \implies f(-t) = 0$ for all $t \in \mathbb{R}.$

Step 3: Set $x = 3y$ to obtain $\left[f(y) + 3y^2\right]f(8y) = f(4y)^2.$

$\indent$ In particular, replacing $y \to t/8$ , it follows that $f(t) = 0 \implies f(t/2) = 0$ for all $t \in \mathbb{R}.$

Step 4: Set $y = -x$ to obtain $f(4x)\left[f(x) + f(-x) - 2x^2\right] = 0.$

$\indent$ Therefore, either $f(x) = 0$ or $f(x) = x^2$ for each $x.$

Step 5: If $f(a) = f(b) = 0$ , then $f(b - a) = 0.$

Step 6: If $f \not\equiv 0$ , then $f(t) = 0 \implies t = 0.$

$\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \ne 0$ and $x + y = t.$ The following three facts are crucial:

$\indent$ 1. $f(y) \ne 0.$ This is because $(x + y) - y = x$ , so by Step 5, $f(y) = 0 \implies f(x) = 0$ , impossible.

Alternative Solution 1

Solution 2

Step 1: $x=y=0 \implies f(0)=0$

Problem 5

Solution 1

$[asy] size(8cm); pair O=(0,0); pair A=dir(110); pair B=dir(-29); pair C=dir(209); pair H=foot(A,B,C); pair P=foot(H,A,B); pair Q=foot(H,A,C); draw(A--B--C--A--H--P); draw(circle(O,1)); draw(Q--H); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, S); dot("$P$", P, NE); dot("$Q$", Q, NW); dot("$O$", O, S); [/asy]$

It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or standard area formulas). Then by Power of a Point, $AP\cdot AB=AH^2=AQ\cdot AC$ Consider the transformation $X\mapsto \Psi(X)$ which dilates $X$ from $A$ by a factor of $\dfrac{AB}{AQ}=\dfrac{AC}{AP}$ and reflects about the $A$ -angle bisector. Then $\Psi(O)$ clearly lies on $AH$ , and its distance from $A$ is $AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH$ so $\Psi(O)=H$ , hence we conclude that $O,P,Q$ are collinear, as desired.

Solution 2

We will use barycentric coordinates with respect to $\triangle ABC.$ The given condition is equivalent to $(\sin B\sin C)^2=\frac{1}{2}.$ Note that $O=(\sin(2A):\sin(2B):\sin(2C)), P=(\cos^2B,\sin^2B,0), Q=(\cos^2C,0,\sin^2C).$ Therefore, we must show that $\begin{vmatrix} \sin(2A) & \sin(2B) & \sin(2C) \\ \cos^2B & \sin^2B & 0 \\ \cos^2C & 0 & \sin^2C \\ \end{vmatrix}=0.$ Expanding, we must prove $\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)$ $\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)$ $\begin{align*} \frac{\sin(2A)+\sin(2B)+\sin(2C)}{2}&=\sin^2B\sin(2C)+\sin^2C\sin(2B)\\ &=2\sin B\sin C(\sin B\cos C+\cos B\sin C) \\ &=2\sin B\sin C\sin A.\end{align*}$

Let $x=e^{iA}, y=e^{iB}, z=e^{iC},$ such that $xyz=-1.$ The left side is equal to $\frac{x^2+y^2+z^2-\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}}{4i}.$ The right side is equal to $\begin{align*} 2\cdot \frac{x-\frac{1}{x}}{2i}\cdot \frac{y-\frac{1}{y}}{2i}\cdot \frac{z-\frac{1}{z}}{2i}&=\frac{xyz-\frac{1}{xyz}-\frac{xy}{z}-\frac{yz}{x}-\frac{xz}{y}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}}{-4i}\\ &=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*}$ which is equivalent to the left hand side. Therefore, the determinant is $0,$ and $O,P,Q$ are collinear. $\blacksquare$

Solution 3

For convenience, let $a, b, c$ denote the lengths of segments $BC, CA, AB,$ respectively, and let $\alpha, \beta, \gamma$ denote the measures of $\angle CAB, \angle ABC, \angle BCA,$ respectively. Let $R$ denote the circumradius of $\triangle ABC.$

Since the central angle $\angle AOB$ subtends the same arc as the inscribed angle $\angle ACB$ on the circumcircle of $\triangle ABC,$ we have $\angle AOB = 2\gamma.$ Note that $OA = OB,$ so $\angle OAB = \angle OBA.$ Thus, $\angle OAB = \frac{\pi}{2} - \gamma.$ Similarly, one can show that $\angle OAC = \frac{\pi}{2} - \beta.$ (One could probably cite this as well-known, but I have proved it here just in case.)

Clearly, $AO = R.$ Since $AH^2 = 2\cdot AO^2,$ we have $AH = \sqrt{2}R.$ Thus, $AH\cdot AO = \sqrt{2}R^2.$

Note that $AH = b\sin\gamma = c\sin\beta.$ The Extended Law of Sines states that: $\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.$ Therefore, $AH = \frac{bc}{2R} = \sqrt{2}R.$ Thus, $bc = \sqrt{2}R^2.$

Since $\angle PHA = \beta$ and $\angle QHA = \gamma,$ we have: $AP = AH\sin\beta = c\sin^2\beta = \frac{b^2 c}{4R^2}$ $AQ = AH\sin\gamma = b\sin^2\gamma = \frac{bc^2}{4R^2}$ It follows that: $AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.$ We see that $AP\cdot AQ = AH\cdot AO.$

Rearranging $AP\cdot AQ = AH\cdot AO,$ we get $\frac{AP}{AH} = \frac{AO}{AQ}.$ We also have $\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,$ so $\triangle PAH\sim\triangle OAQ$ by SAS similarity. Thus, $\angle AOQ = \angle APH,$ so $\angle AOQ$ is a right angle.

Rearranging $AP\cdot AQ = AH\cdot AO,$ we get $\frac{AP}{AO} = \frac{AO}{AH}.$ We also have $\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,$ so $\triangle PAO\sim\triangle HAQ$ by SAS similarity. Thus, $\angle AOP = \angle AQH,$ so $\angle AOP$ is a right angle.

Since $\angle AOP$ and $\angle AOQ$ are both right angles, we get $\angle POQ = \pi,$ so we conclude that $P, O, Q$ are collinear, and we are done. (We also obtain the extra interesting fact that $AO\perp PQ.$ )

Solution 4

Draw the altitude from $O$ to $AB$ , and let the foot of this altitude be $D$ .

Then, by the Right Triangle Altitude Theorem on triangle $AHB$ , we have: $AB\cdot AP=AH^{2}$ .

Since $OD$ is the perpendicular bisector of $AB$ , $2\cdot AD = AB$ .

Substituting this into our previous equation gives $2\cdot AD \cdot AP = AH^{2}$ , which equals $2\cdot AO^{2}$ by the problem condition.

Thus, $2\cdot AD\cdot AP = 2\cdot AO^{2} \implies AD\cdot AP = AO^{2}$ .

Again, by the Right Triangle Altitude Theorem, angle $AOP$ is right.

By dropping an altitude from $O$ to $AC$ and using the same method, we can find that angle $AOQ$ is right. Since $\angle AOP=\angle AOQ=90$ , $P$ , $O$ , $Q$ are collinear and we are done.

~champion999

Solution 5

We use complex numbers. Let lower case letters represent their respective upper case points, with $|a| = |b| = |c| = 1$ . Spamming the foot from point to segment formula, we obtain $h = \dfrac{1}{2}\left(a+b+c-\dfrac{bc}{a}\right),$ $p = \dfrac{1}{2}(h+a+b-ab\overline{h}) = \dfrac{1}{4}\left(2a+2b+c-\dfrac{bc}{a}-\dfrac{ab}{c}+\dfrac{a^2}{c}\right),$ and $q = \dfrac{1}{2}(h+a+c-ac\overline{h}) = \dfrac{1}{4}\left(2a+b+2c-\dfrac{bc}{a}-\dfrac{ac}{b}+\dfrac{a^2}{b}\right).$ We now simplify the given length condition: $\begin{align*} AH^2 &= 2AO^2 \\ \implies (h-a)\overline{(h-a)} &= 2 \\ \implies \dfrac{1}{2}\left(-a+b+c-\dfrac{bc}{a}\right) \cdot \dfrac{1}{2}\left(-\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{a}{bc}\right) &= 2 \\ \implies (ab+ac-bc-a^2)^2 &= 8a^2bc \\ \implies (a-b)^2(a-c)^2 &= 8a^2bc.\end{align*}$ We would like to show that $P$ , $O$ , $Q$ are collinear, or $\begin{vmatrix} p & \overline{p} & 1 \\ 0 & 0 & 1 \\ q & \overline{q} & 1 \end{vmatrix} = \overline{p}q - p\overline{q} = 0.$ After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to $\overline{p}q - p\overline{q} = 0 \iff (a-b)(a-c)(b-c)((a-b)^2(a-c)^2 - 8a^2bc) = 0,$ which is true.

Problem 6

Solution 1

Step 1: Set $x = y = 0$ to obtain $f(0) = 0.$

Step 2: Set $x = 0$ to obtain $f(y)f(-y) = f(y)^2.$

$\indent$ In particular, if $f(y) \ne 0$ then $f(y) = f(-y).$

$\indent$ In addition, replacing $y \to -t$ , it follows that $f(t) = 0 \implies f(-t) = 0$ for all $t \in \mathbb{R}.$

Step 3: Set $x = 3y$ to obtain $\left[f(y) + 3y^2\right]f(8y) = f(4y)^2.$

$\indent$ In particular, replacing $y \to t/8$ , it follows that $f(t) = 0 \implies f(t/2) = 0$ for all $t \in \mathbb{R}.$

Step 4: Set $y = -x$ to obtain $f(4x)\left[f(x) + f(-x) - 2x^2\right] = 0.$

$\indent$ Therefore, either $f(x) = 0$ or $f(x) = x^2$ for each $x.$

Step 5: If $f(a) = f(b) = 0$ , then $f(b - a) = 0.$

Step 6: If $f \not\equiv 0$ , then $f(t) = 0 \implies t = 0.$

$\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \ne 0$ and $x + y = t.$ The following three facts are crucial:

$\indent$ 1. $f(y) \ne 0.$ This is because $(x + y) - y = x$ , so by Step 5, $f(y) = 0 \implies f(x) = 0$ , impossible.

Alternative Solution 1

Solution 2

Step 1: $x=y=0 \implies f(0)=0$

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2016 USAJMO Problems真题及答案

2016 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

Alternative Solution 1

Solution 2

Problem 2

Solution 1

Solution 2 (Controversial)

Problem 3

Solution

Problem 4

Solution 1

Alternative Solution 1

Solution 2

Problem 5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Problem 6

Solution 1

Alternative Solution 1

Solution 2

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