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2012 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日下午12:09

2012 USAJMO Problems真题及答案

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Day 1

Problem 1

Given a triangle $ABC$ , let $P$ and $Q$ be points on segments $overline{AB}$ and $overline{AC}$ , respectively, such that $AP = AQ$ . Let $S$ and $R$ be distinct points on segment $overline{BC}$ such that $S$ lies between $B$ and $R$ , $angle BPS = angle PRS$ , and $angle CQR = angle QSR$ . Prove that $P$ , $Q$ , $R$ , $S$ are concyclic (in other words, these four points lie on a circle).

Problem 2

Find all integers $n ge 3$ such that among any $n$ positive real numbers $a_1$ , $a_2$ , $dots$ , $a_n$ with $[max(a_1, a_2, dots, a_n) le n cdot min(a_1, a_2, dots, a_n),]$ there exist three that are the side lengths of an acute triangle.

Problem 3

Let $a$ , $b$ , $c$ be positive real numbers. Prove that $[frac{a^3 + 3b^3}{5a + b} + frac{b^3 + 3c^3}{5b + c} + frac{c^3 + 3a^3}{5c + a} ge frac{2}{3} (a^2 + b^2 + c^2).]$

Day 2

Problem 4

Let $alpha$ be an irrational number with $0 < alpha < 1$ , and draw a circle in the plane whose circumference has length 1. Given any integer $n ge 3$ , define a sequence of points $P_1$ , $P_2$ , $dots$ , $P_n$ as follows. First select any point $P_1$ on the circle, and for $2 le k le n$ define $P_k$ as the point on the circle for which the length of arc $P_{k - 1} P_k$ is $alpha$ , when travelling counterclockwise around the circle from $P_{k - 1}$ to $P_k$ . Suppose that $P_a$ and $P_b$ are the nearest adjacent points on either side of $P_n$ . Prove that $a + b le n$ .

Problem 5

For distinct positive integers $a$ , $b < 2012$ , define $f(a,b)$ to be the number of integers $k$ with $1 le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$ , where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$ .

Problem 6

Let $P$ be a point in the plane of triangle $ABC$ , and $gamma$ a line passing through $P$ . Let $A'$ , $B'$ , $C'$ be the points where the reflections of lines $PA$ , $PB$ , $PC$ with respect to $gamma$ intersect lines $BC$ , $AC$ , $AB$ , respectively. Prove that $A'$ , $B'$ , $C'$ are collinear.

2012USAJMO真题参考答案及详解

Problem 1

Solution

Since $angle BPS = angle PRS$ , the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$ . Similarly, since $angle CQR = angle QSR$ , the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$ .

$[asy] import markers; unitsize(0.5 cm); pair A, B, C, O, P, Q, R, S; A = (2,12); B = (0,0); C = (14,0); P = intersectionpoint(A--B,Circle(A,8)); Q = intersectionpoint(A--C,Circle(A,8)); O = extension(P, P + rotate(90)*(A - P), Q, Q + rotate(90)*(A - Q)); S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270)); R = intersectionpoint(B--C,arc(O, abs(O - P), 270, 360)); draw(A--B--C--cycle); draw(Circle(O, abs(O - P))); draw(S--P--R); draw(S--Q--R); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$P$", P, W); label("$Q$", Q, NE); label("$R$", R, SE); label("$S$", S, SW); markangle(1, B, P, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, P, R, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, R, Q, C, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); markangle(1, R, S, Q, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); [/asy]$

For the sake of contradiction, suppose that the circumcircles of triangles $PRS$ and $QRS$ are not the same circle. Since $AP = AQ$ , $A$ lies on the radical axis of both circles. However, both circles pass through $2012 USAJMO Problems真题及答案$ and $S$ , so the radical axis of both circles is $RS$ . Hence, $A$ lies on $RS$ , which is a contradiction.

Therefore, the two circumcircles are the same circle. In other words, $P$ , $Q$ , $R$ , and $S$ all lie on the same circle.

Solution 2

Note that (as in the first solution) the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$ . Similarly, since $angle CQR = angle QSR$ , the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$ .

Now, suppose these circumcircles are not the same circle. They already intersect at $R$ and $S$ , so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points $M$ and $N$ , with $M$ on the circumcircle of triangle $PRS$ . By Power of a Point, $AQ^2 = AM cdot AS$ and $AP^2 = AN cdot AS$ . Hence, because $AP = AQ$ , $AM = AN$ , a contradiction because then, as they lie on the same line segment, M and N must be the same point! (Note line segment, not line.) Hence, the two circumcircles are the same circle.

Problem 2

Solution

We claim that any configuration of $0$ 's produces a distinct garden. To verify this claim, we show that, for any cell that is nonzero, the value of that cell is its distance away from the nearest zero, where distance means the shortest chain of adjacent cells connecting two cells. Now, since we know that any cell with a nonzero value must have a cell adjacent to it that is less than its value, there is a path that goes from this cell to the $0$ that is decreasing, which means that the value of the cell must be its distance from the $0 rightarrow$ as the path must end. From this, we realize that, for any configuration of $0$ 's, the value of each of the cells is simply its distance from the nearest $0$ , and therefore one garden is produced for every configuration of $0$ 's.

However, we also note that there must be at least one $0$ in the garden, as otherwise the smallest number in the garden, which is less than or equal to all of its neighbors, is $>0$ , which violates condition $(ii)$ . There are $2^{mn}$ possible configurations of $0$ and not $0$ in the garden, one of which has no $0$ 's, so our total amount of configurations is $boxed{2^{mn} -1}$

Problem 3

Solution

By the Cauchy-Schwarz inequality, $[[a(5a + b) + b(5b + c) + c(5c + a)] left( frac{a^3}{5a + b} + frac{b^3}{5b + c} + frac{c^3}{5c + a} right) ge (a^2 + b^2 + c^2)^2,]$ so $[frac{a^3}{5a + b} + frac{b^3}{5b + c} + frac{c^3}{5c + a} ge frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc}.]$ Since $a^2 + b^2 + c^2 ge ab + ac + bc$ , $[frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc} ge frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = frac{1}{6} (a^2 + b^2 + c^2).]$ Hence, $[frac{a^3}{5a + b} + frac{b^3}{5b + c} + frac{c^3}{5c + a} ge frac{1}{6} (a^2 + b^2 + c^2).]$

Again by the Cauchy-Schwarz inequality, $[[b(5a + b) + c(5b + c) + a(5c + a)] left( frac{b^3}{5a + b} + frac{c^3}{5b + c} + frac{a^3}{5c + a} right) ge (a^2 + b^2 + c^2)^2,]$ so $[frac{b^3}{5a + b} + frac{c^3}{5b + c} + frac{a^3}{5c + a} ge frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc}.]$ Since $a^2 + b^2 + c^2 ge ab + ac + bc$ , $[frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc} ge frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = frac{1}{6} (a^2 + b^2 + c^2).]$ Hence, $[frac{b^3}{5a + b} + frac{c^3}{5b + c} + frac{a^3}{5c + a} ge frac{1}{6} (a^2 + b^2 + c^2).]$

Therefore, $[frac{a^3 + 3b^3}{5a + b} + frac{b^3 + 3c^3}{5b + c} + frac{c^3 + 3a^3}{5c + a} ge frac{1 + 3}{6} (a^2 + b^2 + c^2) = frac{2}{3} (a^2 + b^2 + c^2).]$

Solution 2

Titu's Lemma: The sum of multiple fractions in the form $frac{a_n^2}{b_n}$ where $a_n$ and $b_n$ are sequences of real numbers is greater than of equal to the square of the sum of all $a_i$ divided by the sum of all $b_i$ , where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS.

Consider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS $[sum_{cyc} frac {a^4} {5a^2+ab}+sum_{cyc} frac {9a^4} {15ac+3a^2}]$ (Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.)

Then use Titu's Lemma on all terms: $[sum_{cyc} frac {a^4} {5a^2+ab} +sum_{cyc} frac {9a^4} {15ac+3a^2} ge frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} ge frac {16(a^2 + b^2 + c^2)^2}{24(a^2 + b^2 + c^2)} = frac{2}{3} (a^2 + b^2 + c^2)]$ owing to the fact that $a^2+b^2+c^2 ge ab+bc+ca$ , which is actually equivalent to $(a-b)^2 + (b-c)^2 + (c-a)^2 ge 0$ !

Solution 3

We proceed to prove that $[frac{a^3 + 3b^3}{5a + b} ge -frac{1}{36} a^2 + frac{25}{36} b^2]$

(then the inequality in question is just the cyclic sum of both sides, since $[sum_{cyc} (-frac{1}{36} a^2 + frac{25}{36} b^2) = frac{24}{36}sum_{cyc} a^2 = frac{2}{3} (a^2+b^2+c^2)]$ )

Indeed, by AP-GP, we have

$[41 (a^3 + b^3+b^3) ge 41 cdot 3 ab^2]$

and

$[b^3 + a^2b ge 2 ab^2]$

Summing up, we have

$[41a^3 + 83b^3 + a^2 b ge 125 ab^2]$

which is equivalent to:

$[36(a^3 + 3b^3) ge (5a + b)(-a^2 + 25b^2)]$

Dividing $36(5a+b)$ from both sides, the desired inequality is proved.

Solution 4

By Cauchy-Schwarz,

$[left(sum_{cyc} dfrac{a^3}{5a + b} + dfrac{b^3}{5a + b} + dfrac{b^3}{5a + b} + dfrac{b^3}{5a + b} right) ge dfrac{left( sum_{cyc} a^2 + b^2 + b^2 + b^2 right)^2}{ left( sum_{cyc} a(5a + b) + b(5a + b) + b(5a + b) + b(5a + b) right) }]$

$[= dfrac{left(4a^2 + 4b^2 + 4c^2right)^2}{left(8a^2 + 8b^2 + 8c^2right) + left(16ab + 16bc + 16caright)}]$

$[ge dfrac{16left(a^2 + b^2 + c^2right)^2}{left(8a^2 + 8b^2 + 8c^2right) + left(8a^2 + 8b^2right) + left(8b^2 + 8c^2right) + left(8c^2 + 8a^2right)}]$ (by AM-GM) $[= dfrac{2}{3}left(a^2 + b^2 + c^2right)]$ as desired.

Problem 4

Solution

Use mathematical induction. For $n=3$ it is true because one point can't be closest to $P_3$ in both ways, and that $1+2le 3$ . Suppose that for some $n$ , the nearest adjacent points $P_a$ and $P_b$ on either side of $P_n$ satisfy $a+b le n$ . Then consider the nearest adjacent points $P_c$ and $P_d$ on either side of $P_{n+1}$ . It is by the assumption of the nearness we can see that either $(c,d)=(a+1,b+1)$ still holds, or $P_1$ jumps into the interior of the arc $P_{a+1}P_{n}P_{b+1}$ , so that $c$ or $d$ equals to $1$ . Let's consider the following two cases.

(i) Suppose $a+b=n$ .

Since the length of the arc $P_nP_a$ is ${(a-n)alpha}$ (where ${x}$ equals to $x$ subtracted by the greatest integer not exceeding $x$ ) and length of the arc $P_bP_n$ is ${(n-b)alpha} = {aalpha}$ , we now consider a point $P_0$ which is defined by $P_1$ traveling clockwise on the circle such that the length of arc $P_0P_1$ is $alpha$ . We claim that $P_0$ is in the interior of the arc $P_bP_nP_a$ . Algebraically, it is equivalent to either ${0-nalpha} < {aalpha-nalpha}$ or ${nalpha -0 } < {nalpha - balpha} = {aalpha}$ .

Suppose the latter fails, i.e. ${nalpha} ge {aalpha}$ . Then suppose $nalpha = m_1 + r_1$ and $aalpha = m_2 + r_2$ , where $m_1$ , $m_2$ are integers and $0< r_2le r_1 <1$ ( $r_2$ is not zero because $aalpha$ is irrational). We now have $[{0-nalpha} = {-m_1-1 + (1 -r_1)}=1-r_1]$ and $[{aalpha -nalpha} = {m_2-m_1-1 + (1+r_2-r_1)} = 1+r_2-r_1>1-r_1]$

Therefore $P_0$ is either closer to $P_n$ than $P_a$ on the $P_a$ side, or closer to $P_n$ than $P_b$ on the $P_b$ side. In other words, $P_1$ is the closest adjacent point of $P_{n+1}$ on the $P_{a+1}$ side, or the closest adjacent point of $P_{n+1}$ on the $P_{b+1}$ side. Hence $P_c$ or $P_d$ is $P_1$ , therefore $c+d le n+1$ .

(ii) Suppose $a+ble n-1$ Then either $c+d = (a+1)+(b+1) le n+1$ when $c=a+1$ and $d=b+1$ , or $c+d le 1+n$ when one of $P_c$ or $P_d$ is $P_1$ .

In either case, $c+dle n+1$ is true.

Problem 5

Solution 1

First we'll show that $S geq 502$ , then we'll find an example $(a, b)$ that have $f(a, b)=502$ .

Let $x_k$ be the remainder when $ak$ is divided by 2012, and let $y_k$ be defined similarly for $bk$ . First, we know that, if $x_k > y_k >0$ , then $x_{2012-k} equiv a(2012-k) equiv 2012-ak equiv 2012-x_k pmod 2012$ and $y_{2012-k} equiv 2012-y_k pmod 2012$ . This implies that, since $2012 - x_k neq 0$ and $2012 -y_k neq 0$ , $x_{2012-k} < y_{2012-k}$ . Similarly, if $0< x_k < y_k$ then $x_{2012-k} > y_{2012-k}$ , establishing a one-to-one correspondence between the number of $k$ such that $x_k < y_k$ . Thus, if $n$ is the number of $k$ such that $x_k neq y_k$ and $y_k neq 0 neq x_k$ , then $S geq frac{1}{2}n$ . Now I'll show that $n geq 1004$ .

If $gcd(k, 2012)=1$ , then I'll show you that $x_k neq y_k$ . This is actually pretty clear; assume that's not true and set up a congruence relation: $[ak equiv bk pmod {2012}]$ Since $k$ is relatively prime to 2012, it is invertible mod 2012, so we must have $a equiv b pmod {2012}$ . Since $0 < a, b <2012$ , this means $a=b$ , which the problem doesn't allow, thus contradiction, and $x_k neq y_k$ . Additionally, if $gcd(k, 2012)=1$ , then $x_k neq 0 neq y_k$ , then based on what we know about $n$ from the previous paragraph, $n$ is at least as large as the number of k relatively prime to 2012. Thus, $n geq phi(2012) = phi(503*4) = 1004$ . Thus, $S geq 502$ .

To show 502 works, consider $(a, b)=(1006, 2)$ . For all even $k$ we have $x_k=0$ , so it doesn't count towards $f(1006, 2)$ . Additionally, if $k = 503, 503*3$ then $x_k = y_k = 1006$ , so the only number that count towards $f(1006, 2)$ are the odd numbers not divisible by 503. There are 1004 such numbers. However, for all such odd k not divisible by 503 (so numbers relatively prime to 2012), we have $x_k neq 0 neq y_k$ and $2012-k$ is also relatively prime to 2012. Since under those conditions exactly one of $x_k > y_k$ and $x_{2012-k} > y_{2012-k}$ is true, we have at most 1/2 of the 1004 possible k actually count to $f(1006, 2)$ , so $frac{1004}{2} = 502 geq f(1006, 2) geq S geq 502$ , so $S=502$ .

Solution 2

Let $ak equiv r_{a} pmod{2012}$ and $bk equiv r_{b} pmod{2012}$ . Notice that this means $a(2012 - k) equiv 2012 - r_{a} pmod{2012}$ and $b(2012 - k) equiv 2012 - r_{b} pmod{2012}$ . Thus, for every value of $k$ where $r_{a} > r_{b}$ , there is a value of $k$ where $r_{b} > r_{a}$ . Therefore, we merely have to calculate $frac{1}{2}$ times the number of values of $k$ for which $r_{a} neq r_{b}$ and $r_{a} neq 0$ .

However, the answer is NOT $frac{1}{2}(2012) = 1006$ ! This is because we must count the cases where the value of $k$ makes $r_{a} = r_{b}$ or where $r_{a} = 0$ .

So, let's start counting.

If $k$ is even, we have either $a equiv 0 pmod{1006}$ or $a - b equiv 0 pmod{1006}$ . So, $a = 1006$ or $a = b + 1006$ . We have $1005$ even values of $k$ (which is all the possible even values of $k$ , since the two above requirements don't put any bounds on $k$ at all).

If $k$ is odd, if $k = 503$ or $k = 503 cdot 3$ , then $a equiv 0 pmod{4}$ or $a equiv b pmod{4}$ . Otherwise, $ak equiv 0 pmod{2012}$ or $ak equiv bk pmod{2012}$ , which is impossible to satisfy, given the domain $a, b < 2012$ . So, we have $2$ values of $k$ .

In total, we have $2 + 1005 = 1007$ values of $k$ which makes $r_{a} = r_{b}$ or $r_{a} = 0$ , so there are $2011 - 1007 = 1004$ values of $k$ for which $r_{a} neq r_{b}$ and $r_{a} neq 0$ . Thus, by our reasoning above, our solution is $frac{1}{2} cdot 1004 = boxed{502}$ .

Solution 3

The key insight in this problem is noticing that when $ak$ is higher than $bk$ , $a(2012-k)$ is lower than $b(2012-k)$ , except at $2 pmod{4}$ residues*. Also, they must be equal many times. $2012=2^2*503$ . We should have multiples of $503$ . After trying all three pairs and getting $503$ as our answer, we win. But look at the $2pmod{4}$ idea. What if we just took $2$ and plugged it in with $1006$ ? We get $502$ .

Solution 4

Say that the problem is a race track with $2012$ spots. To intersect the most, we should get next to each other a lot so the negation is high. As $2012=2^2*503$ , we intersect at a lot of multiples of $503$ .

Problem 6

Solution

By the law of sines on triangle $AB'P$ , $[frac{AB'}{sin angle APB'} = frac{AP}{sin angle AB'P},]$ so $[AB' = AP cdot frac{sin angle APB'}{sin angle AB'P}.]$

$[asy] import graph; import geometry; unitsize(0.5 cm); pair[] A, B, C; pair P, R; A[0] = (2,12); B[0] = (0,0); C[0] = (14,0); P = (4,5); R = 5*dir(70); A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0])); B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0])); C[1] = extension(A[0],B[0],P,reflect(P + R,P - R)*(C[0])); draw((P - R)--(P + R),red); draw(A[1]--B[1]--C[1]--cycle,blue); draw(A[0]--B[0]--C[0]--cycle); draw(A[0]--P); draw(B[0]--P); draw(C[0]--P); draw(P--A[1]); draw(P--B[1]); draw(P--C[1]); draw(A[1]--B[0]); draw(A[1]--B[0]); label("$A$", A[0], N); label("$B$", B[0], S); label("$C$", C[0], SE); dot("$A'$", A[1], SW); dot("$B'$", B[1], NE); dot("$C'$", C[1], W); dot("$P$", P, SE); label("$gamma$", P + R, N); [/asy]$

Similarly, $begin{align*} B'C &= CP cdot frac{sin angle CPB'}{sin angle CB'P}, \ CA' &= CP cdot frac{sin angle CPA'}{sin angle CA'P}, \ A'B &= BP cdot frac{sin angle BPA'}{sin angle BA'P}, \ BC' &= BP cdot frac{sin angle BPC'}{sin angle BC'P}, \ C'A &= AP cdot frac{sin angle APC'}{sin angle AC'P}. end{align*}$ Hence, $begin{align*} &frac{AB'}{B'C} cdot frac{CA'}{A'B} cdot frac{BC'}{C'A} \ &= frac{sin angle APB'}{sin angle AB'P} cdot frac{sin angle CB'P}{sin angle CPB'} cdot frac{sin angle CPA'}{sin angle CA'P} cdot frac{sin angle BA'P}{sin angle BPA'} cdot frac{sin angle BPC'}{sin angle BC'P} cdot frac{sin angle AC'P}{sin angle APC'}. end{align*}$

Since angles $angle AB'P$ and $angle CB'P$ are supplementary or equal, depending on the position of $B'$ on $AC$ , $[sin angle AB'P = sin angle CB'P.]$ Similarly, $begin{align*} sin angle CA'P &= sin angle BA'P, \ sin angle BC'P &= sin angle AC'P. end{align*}$

By the reflective property, $angle APB'$ and $angle BPA'$ are supplementary or equal, so $[sin angle APB' = sin angle BPA'.]$ Similarly, $begin{align*} sin angle CPA' &= sin angle APC', \ sin angle BPC' &= sin angle CPB'. end{align*}$ Therefore, $[frac{AB'}{B'C} cdot frac{CA'}{A'B} cdot frac{BC'}{C'A} = 1,]$ so by Menelaus's theorem, $A'$ , $B'$ , and $C'$ are collinear.