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2013 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11:58

2013 USAJMO Problems真题及答案

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Day 1

Problem 1

Are there integers $a$ and $b$ such that $a^5b+3$ and $ab^5+3$ are both perfect cubes of integers?

Problem 2

Each cell of an $mtimes n$ board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:

(i) The difference between any two adjacent numbers is either $0$ or $1$ .

(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to $0$ .

Determine the number of distinct gardens in terms of $m$ and $n$ .

Problem 3

In triangle $ABC$ , points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $omega_A$ , $omega_B$ , $omega_C$ denote the circumcircles of triangles $AQR$ , $BRP$ , $CPQ$ , respectively. Given the fact that segment $AP$ intersects $omega_A$ , $omega_B$ , $omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$ .

Day 2

Problem 4

Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ , $2+2$ , $2+1+1$ , $1+2+1$ , $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd.

Problem 5

Quadrilateral $XABY$ is inscribed in the semicircle $omega$ with diameter $XY$ . Segments $AY$ and $BX$ meet at $P$ . Point $Z$ is the foot of the perpendicular from $P$ to line $XY$ . Point $C$ lies on $omega$ such that line $XC$ is perpendicular to line $AZ$ . Let $Q$ be the intersection of segments $AY$ and $XC$ . Prove that $[dfrac{BY}{XP}+dfrac{CY}{XQ}=dfrac{AY}{AX}.]$ 、

Problem 6

Find all real numbers $x,y,zgeq 1$ satisfying $[min(sqrt{x+xyz},sqrt{y+xyz},sqrt{z+xyz})=sqrt{x-1}+sqrt{y-1}+sqrt{z-1}.]$

2013USAJMO真题参考答案及详解

Problem 1

Solution

No, such integers do not exist. This shall be proven by contradiction, by showing that if $a^5b+3$ is a perfect cube then $ab^5+3$ cannot be.

Remark that perfect cubes are always congruent to $0$ , $1$ , or $-1$ modulo $9$ . Therefore, if $a^5b+3equiv 0,1,text{ or} -1pmod{9}$ , then $a^5bequiv 5,6,text{ or }7pmod{9}$ .

If $a^5bequiv 6pmod 9$ , then note that $3|b$ . (This is because if $3|a$ then $a^5bequiv 0pmod 9$ .) Therefore $ab^5equiv 0pmod 9$ and $ab^5+3equiv 3pmod 9$ , contradiction.

Otherwise, either $a^5bequiv 5pmod 9$ or $a^5bequiv 7pmod 9$ . Note that since $a^6b^6$ is a perfect sixth power, and since neither $a$ nor $b$ contains a factor of $3$ , $a^6b^6equiv 1pmod 9$ . If $a^5bequiv 5pmod 9$ , then $[a^6b^6equiv (a^5b)(ab^5)equiv 5ab^5equiv 1pmod 9implies ab^5equiv 2pmod 9.]$ Similarly, if $a^5bequiv 7pmod 9$ , then $[a^6b^6equiv (a^5b)(ab^5)equiv 7ab^5equiv 1pmod 9implies ab^5equiv 4pmod 9.]$ Therefore $ab^5+3equiv 5,7pmod 9$ , contradiction.

Therefore no such integers exist.

Solution 2

We shall prove that such integers do not exist via contradiction. Suppose that $a^5b + 3 = x^3$ and $ab^5 + 3 = y^3$ for integers x and y. Rearranging terms gives $a^5b = x^3 - 3$ and $ab^5 = y^3 - 3$ . Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = $(x^3 - 3)^frac{5}{24} (y^3 - 3)^frac{-1}{24}$ and b = $(y^3 - 3)^frac{5}{24} (x^3 - 3)^frac{-1}{24}$ . Consider a prime p in the prime factorization of $x^3 - 3$ and $y^3 - 3$ . If it has power $r_1$ in $x^3 - 3$ and power $r_2$ in $y^3 - 3$ , then $5r_1$ - $r_2$ is a multiple of 24 and $5r_2$ - $r_1$ also is a multiple of 24.

Adding and subtracting the divisions gives that $r_1$ - $r_2$ divides 12. (actually, $r_1 - r_2$ is a multiple of 4, as you can verify if ${^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}$ . So the rest of the proof is invalid.) Because $5r_1$ - $r_2$ also divides 12, $4r_1$ divides 12 and thus $r_1$ divides 3. Repeating this trick for all primes in $x^3 - 3$ , we see that $x^3 - 3$ is a perfect cube, say $q^3$ . Then $x^3 - q^3 = 3,$ and $(x-q)(x^2 + xq + q^2) = 3$ , so that $x - q = 1$ and $x^2 + xq + q^2 = 3$ . Clearly, this system of equations has no integer solutions for $x$ or $q$ , a contradiction, hence completing the proof.

Therefore no such integers exist.

Solution 3

Let $a^5b+3=x^3$ and $ab^5+3=y^3$ . Then, $a^5b=x^3-3$ , $ab^5=y^3-3$ , and $[(ab)^6=(x^3-3)(y^3-3)]$ Now take $text{mod }9$ (recall that perfect cubes $equiv -1,0,1pmod{9}$ and perfect sixth powers $equiv 0,1pmod{9}$ ) on both sides. There are $3times 3=9$ cases to consider on what values $text{mod }9$ that $x^3$ and $y^3$ take. Checking these $9$ cases, we see that only $x^3equiv y^3equiv 0pmod{9}$ or $xequiv yequiv 0pmod{3}$ yield a valid residue $text{mod }9$ (specifically, $(x^3-3)(y^3-3)equiv 0pmod{9}$ ). But this means that $3mid ab$ , so $729mid (ab)^6$ so $[729mid (x^3-3)(y^3-3)iff 729mid (27x'^3-3)(27y'^3-3)iff 81mid (9x'^3-1)(9y'^3-1)]$ contradiction.

Solution 4

If $a^5b+3$ is a perfect cube, then $a^5b$ can be one of $5,6,7 pmod 9$ , so $(a^5b)^5 = a^{25} b^5$ can be one of $5^5 equiv 2$ , $6^5 equiv 0$ , or $7^5 equiv 4 pmod 9$ . If $a$ were divisible by $3$ , we'd have $a^5 b equiv 0 pmod 9$ , which we've ruled out. So $gcd(a,9) = 1$ , which means $a^6 equiv 1 pmod 9$ , and therefore $a^{25} b^5 equiv ab^5 pmod 9$ .

We've shown that $a b^5$ can be one of $0, 2, 4 pmod 9$ , so $ab^5 + 3$ can be one of $3, 5, 7 pmod 9$ . None of these are possibilities for a perfect cube, so if $a^5b+3$ is a perfect cube, $ab^5+3$ cannot be.

Solution 5

As in previous solutions, notice $ab^5,a^5b equiv 5,6,7 pmod 9$ . Now multiplying gives $a^6b^6$ , which is only $0,1 pmod 9$ , so after testing all cases we find that $ab^5equiv a^5b equiv 6 mod 9$ . Then since $phi (9) = 6$ , $ab^5equiv frac{a}{b}pmod 9$ and $a^5b equiv frac{b}{a}pmod 9$ (Note that $a,b$ cannot be $0pmod 9$ ). Thus we find that the inverse of $6$ is itself under modulo $9$ , a contradiction.

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 1

$[asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-1.0, 3.0); pair B = (-3.0, -3.0); pair C = (4.0, -3.0); pair P = (-0.6698198198198195, -3.0); pair Q = (1.1406465288818244, 0.43122416534181074); pair R = (-1.6269590345062048, 1.119122896481385); path w_A = circumcircle(A,Q,R); path w_B = circumcircle(B,P,R); path w_C = circumcircle(P,Q,C); pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5)); pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5)); pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5)); pair X = (2)*(foot(O_A,A,P))-A; pair Y = (2)*(foot(O_B,A,P))-P; pair Z = (2)*(foot(O_C,A,P))-P; pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P; pair D = (2)*(foot(O_B,X,M))-M; pair E = (2)*(foot(O_C,X,M))-M; /* Draw objects */ draw(A--B, rgb(0.6,0.6,0.0)); draw(B--C, rgb(0.6,0.6,0.0)); draw(C--A, rgb(0.6,0.6,0.0)); draw(w_A, rgb(0.4,0.4,0.0)); draw(w_B, rgb(0.4,0.4,0.0)); draw(w_C, rgb(0.4,0.4,0.0)); draw(A--P, rgb(0.0,0.2,0.4)); draw(D--E, rgb(0.0,0.2,0.4)); draw(P--D, rgb(0.0,0.2,0.4)); draw(P--E, rgb(0.0,0.2,0.4)); draw(P--M, rgb(0.4,0.2,0.0)); draw(R--M, rgb(0.4,0.2,0.0)); draw(Q--M, rgb(0.4,0.2,0.0)); draw(B--M, rgb(0.0,0.2,0.4)); draw(C--M, rgb(0.0,0.2,0.4)); draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(Z); dot(M); dot(D); dot(E); /* Label points */ label("$A$", A, lsf * dir(110)); label("$B$", B, lsf * unit(B-M)); label("$C$", C, lsf * unit(C-M)); label("$P$", P, lsf * unit(P-M) * 1.8); label("$Q$", Q, lsf * dir(90) * 1.6); label("$R$", R, lsf * unit(R-M) * 2); label("$X$", X, lsf * dir(-60) * 2); label("$Y$", Y, lsf * dir(45)); label("$Z$", Z, lsf * dir(5)); label("$M$", M, lsf * dir(M-P)*2); label("$D$", D, lsf * dir(150)); label("$E$", E, lsf * dir(5));[/asy]$

In this solution, all lengths and angles are directed.

Firstly, it is easy to see by that $omega_A, omega_B, omega_C$ concur at a point $M$ . Let $XM$ meet $omega_B, omega_C$ again at $D$ and $E$ , respectively. Then by Power of a Point, we have $[XM cdot XE = XZ cdot XP quadtext{and}quad XM cdot XD = XY cdot XP]$ Thusly $[frac{XY}{XZ} = frac{XD}{XE}]$ But we claim that $triangle XDP sim triangle PBM$ . Indeed, $[measuredangle XDP = measuredangle MDP = measuredangle MBP = - measuredangle PBM]$ and $[measuredangle DXP = measuredangle MXY = measuredangle MXA = measuredangle MRA = 180^circ - measuredangle MRB = measuredangle MPB = -measuredangle BPM]$ Therefore, $frac{XD}{XP} = frac{PB}{PM}$ . Analogously we find that $frac{XE}{XP} = frac{PC}{PM}$ and we are done.

Solution 2

Diagram Refer to the Diagram link.

By Miquel's Theorem, there exists a point at which $omega_A, omega_B, omega_C$ intersect. We denote this point by $M.$ Now, we angle chase: $[angle YMX = 180^{circ} - angle YXM - angle XYM]$ $[= 180^{circ} - angle AXM - angle PYM]$ $[= left(180^{circ} - angle ARMright) - angle PRM]$ $[= angle BRM - angle PRM]$ $[= angle BRP = angle BMP.]$ In addition, we have $[angle ZMX = 180^{circ} - angle MZY - angle ZYM - angle YMX]$ $[= 180^{circ} - angle MZP - angle PYM - angle BMP]$ $[= 180^{circ} - angle MCP - angle PBM - angle BMP]$ $[= left(180^{circ} - angle PBM - angle BMPright) - angle MCP]$ $[= angle BPM - angle MCP]$ $[= 180^{circ} - angle MPC - angle MCP]$ $[= angle CMP.]$ Now, by the Ratio Lemma, we have $[frac{XY}{XZ} = frac{MY}{MZ} cdot frac{sin angle YMX}{sin angle ZMX}]$ $[= frac{sin angle YZM}{sin angle ZYM} cdot frac{sin angle BMP}{sin angle CMP}]$ (by the Law of Sines in $triangle MZY$ ) $[= frac{sin angle PZM}{sin angle PYM} cdot frac{sin angle BMP}{sin angle CMP}]$ $[= frac{sin angle PCM}{sin angle PBM} cdot frac{sin angle BMP}{sin angle CMP}]$ $[= frac{MB}{MC} cdot frac{sin angle BMP}{sin angle CMP}]$ (by the Law of Sines in $triangle MBC$ ) $[= frac{PB}{PC}]$ by the Ratio Lemma. The proof is complete.

Solution 3

Use directed angles modulo $pi$ .

Lemma. $angle{XRY} equiv angle{XQZ}.$

Proof. $[angle{XRY} equiv angle{XRA} - angle{YRA} equiv angle{XQA} + angle{YRB} equiv angle{XQA} + angle{CPY} = angle{XQA} + angle{AQZ} = angle{XQZ}.]$

Now, it follows that (now not using directed angles) $[frac{XY}{YZ} = frac{frac{XY}{sin angle{XRY}}}{frac{YZ}{sin angle{XQZ}}} = frac{frac{RY}{sin angle{RXY}}}{frac{QZ}{sin angle{QXZ}}} = frac{BP}{PC}]$ using the facts that $ARY$ and $APB$ , $AQZ$ and $APC$ are similar triangles, and that $frac{RA}{sin angle{RXA}} = frac{QA}{sin angle{QXA}}$ equals twice the circumradius of the circumcircle of $AQR$ .

Problem 4

Solution

First of all, note that $f(n)$ = $sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k le n$ . We let $f(0) = 1$ for convenience.

From here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a in mathbb{Z}$

We induct on $a$ . It is trivially true for $a = 0$ and $a = 1$ . From here, we show that, if the only numbers $n le 2^{a-1} - 1$ where $f(n)$ is odd are of the form described above, then the only numbers $n le 2^{a} -1$ that are odd are of that form. We first consider all numbers $b$ , such that $2^{a-1} le b le 2^{a} - 2$ , going from the lower bound to the upper bound (a mini induction, you might say). We know that $f(b) = sum_{i=0}^{a-1} f(b-2^{i})$ . For a number in this summation to be odd, $b - 2^i = 2^m -1 rightarrow b = 2^i + 2^m - 1$ . However, we know that $b > 2^{a-1}$ , so $m$ must be equal to $a-1$ , or else $b$ cannot be in that interval. Now, from this, we know that $i < a-1$ , as $b<2^{a} - 1$ . Therefore, $i$ and $m$ are distinct, and thus $f(b - 2^i)$ and $f(b- 2^{a-1})$ are odd; since there are just two odd numbers, the ending sum for any $b$ is even. Finally, considering $2^{a} - 1$ , the only odd number is $f(2^{a} - 1 - 2^{a-1})$ , so the ending sum is odd. $Box$

The smallest $n$ greater than $2013$ expressible as $2^d - 1, d in mathbb{N}$ is $2^{11} -1 = boxed{2047}$

Solution with Explanations

Of course, as with any number theory problem, use actual numbers to start, not variables! By plotting out the first few sums (do it!) and looking for patterns, we observe that $f(n)=sum_{power=0}^{pow_{larg}} f(n-2^{power})$ , where $pow_{larg}$ represents the largest power of $2$ that is smaller than $n$ . I will call this sum the Divine Sign, or DS.

But wait a minute... we are trying to determine odd/even of $f(n)$ . Why not call all the evens 0 and odds 1, basically using mod 2? Sounds so simple. Draw a small table for the values: as $n$ goes up from $0$ , you get: $1,1,0,1,0,0,0,1,0...$ . We have to set $f(0)=1$ for this to work. Already it looks like $f(n)$ is only odd if $n=2^{power}-1$ .

The only tool here is induction. The base case is clearly established. Then let's assume we successfully made our claim up to $2^n-1$ . We need to visit numbers from $2^n$ to $2^{n+1}-1$ . Realize that $2^n$ has $0$ for $f$ because there will be two numbers in DS that give a $f$ of one: $2^{n-1}$ and $1$ .

But to look at whether a value of $f(number)$ is 1 or 0, we need to revisit our first equation. We can answer this rather natural question: When will a number to be inducted upon, say $2^n+k$ , ever have a 1 as $f(number)$ in the DS equation? Well- because by our assumption of the claim up to $2^n-1$ , we know that the only way for that to happen is if $2^n+k-2^{power}$ in the DS is equal to $2^{somePower} - 1$ . Clearly $1 leq k leq 2^n - 1$ .

Finally, we can simplify. Using our last equation, $2^n+k-2^{power}=2^{somePower}-1$ , regrouping gives $2^n+k=2^{power}+2^{somePower}-1$ .

Most importantly, realize that $power$ can be from $0$ to $n$ , because of the restraints on $k$ mentioned earlier. Same with $somePower$ . Immediately at least one of $power$ and $somePower$ has to be $n$ . If both were smaller, LHS is greater, contradiction. If both were greater, RHS is greater, contradiction.

Therefore, by setting one of $power$ or $somePower$ to $n$ , we realize $k=2^{aCertainPower}-1$ .

The conclusion is clear, right? Each $k$ from $1$ to $2^{n-1}-1$ yields two distinct cases: one of $power$ and $somePower$ is equal to $n$ , while the other is LESS THAN $n$ . But for $k=2^n-1$ , there is ONE CASE: BOTH values have to equal $n$ . Therefore, the only $k$ that has $f(2^n+k)$ as odd must only be $2^n-1$ , because the other ones yield a $f$ of 1+1=0 in our mod. That proves our induction for a new power of 2, namely $n+1$ , meaning that $f(number)$ is only odd if $number = 2^{powerOfTwo} - 1$ , and we are almost done...

Thus, the answer is $2^{11}-1=boxed{2047}$ .

This was pretty intuitive, and realizing quite $elementary$ steps about how powers of 2 work gave us the solution! Estimated time: ~50 minutes. Cheers- expiLnCalc.

Problem 5

Solution 1

Let us use coordinates. Let O, the center of the circle, be (0,0). WLOG the radius of the circle is 1, so set Y (1,0) and X (-1,0). Also, for arbitrary constants $a$ and $b$ set A $(cos a, sin a)$ and B $(cos b, sin b)$ . Now, let's use our coordinate tools. It is easily derived that the equation of $BX$ is $y = frac{sin b}{1 + cos b}(x + 1) = v(x+1)$ and the equation of $AY$ is $y = frac{sin a}{1 - cos a}(x - 1) = u(x-1)$ , where $u$ and $v$ are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, $P$ , is $left(frac{u-v}{u+v}right), frac{2uv}{u+v})$ . Also, $Zleft(frac{u-v}{u+v}right), 0)$ . It shall be left to the reader to find the slope of $AZ$ , the coordinates of Q and C, and use the distance formula to verify that $frac{BY}{XP} + frac{CY}{XQ} = frac{AY}{AX}$ .

Solution 2

First of all

$[angle BXY = angle PAZ =angle AXQ =angle AXC]$ since the quadrilateral $APZX$ is cyclic, and triangle $AXQ$ is rectangle, and $CX$ is orthogonal to $AZ$ . Now

$[angle BXY =angle BAY =angle AXC]$ because $XABY$ is cyclic and we have proved that

$[angle AXC = angle BXY]$ so $BC$ is parallel to $AY$ , and $[AC=BY, CY=AB]$ Now by Ptolomey's theorem on $APZX$ we have $[(AX)(PZ)+(AP)(XZ)=(AZ)(PX)]$ we see that triangles $PXZ$ and $QXA$ are similar since $[angle QAX= angle PZX= 90]$ and $[angle AXC = angle BXY]$ is already proven, so $[(AX)(PZ)=(AQ)(XZ)]$ Substituting yields $[(AQ)(XZ)+(AP)(XZ)=(AZ)(PX)]$ dividing by $(PX)(XZ)$ We get $[frac {AQ+AP}{XP} = frac {AZ}{XZ}]$ Now triangles $AYZ$ , and $XYP$ are similar so $[frac {AY}{AZ}= frac {XY}{XP}]$ but also triangles $XPY$ and $XZB$ are similar and we get $[frac {XY}{XP}= frac {XB}{XZ}]$ Comparing we have, $[frac {AY}{XB}= frac {AZ}{XZ}]$ Substituting, $[frac {AQ+AP}{XP}= frac {AY}{XB}]$ Dividing the new relation by $AX$ and multiplying by $XB$ we get $[frac{XB(AQ+AP)}{(XP)(AX)} = frac {AY}{AX}]$ but $[frac {XB}{AX}= frac {XY}{XQ}]$ since triangles $AXB$ and $QXY$ are similar, because $[angle AYX= angle ABX]$ and $[angle AXB= angle CXY]$ since $CY=AB$ Substituting again we get $[frac {XY(AQ)+XY(AP)}{(XP)(XQ)} =frac {AY}{AX}]$ Now since triangles $ACQ$ and $XYQ$ are similar we have $[XY(AQ)=AC(XQ)]$ and by the similarity of $APB$ and $XPY$ , we get $[AB(CP)=XY(AP)]$ so substituting, and separating terms we get $[frac{AC}{XP} + frac{AB}{XQ} = frac{AY}{AX}]$ In the beginning we prove that $AC=BY$ and $AB=CY$ so $[frac{BY}{XP} + frac{CY}{XQ} = frac{AY}{AX}]$ $blacksquare$

Solution 3

It is obvious that $[angle AXB=angle CXY=alpha]$ for some value $alpha$ . Also, note that $angle BYA=alpha$ . Set $[angle BXC=angle BYC=beta.]$ We have $[frac{XC}{CY}=tan {angle CYZ}=tan (90-alpha)]$ and $[frac{CQ}{CY}=tan {angle CYQ}=tan (alpha+beta).]$ This gives $[frac{CY}{XQ}=frac{1}{tan (90-alpha)-tan (alpha+beta)}.]$ Similarly, we can deduce that $[frac{BY}{XP}=frac{1}{tan (90-alpha-beta)-tan {alpha}}.]$ Adding gives $[frac{tan alpha}{1-tan (alpha+beta)tan {beta}}+frac{tan (alpha+beta)}{1-tan (alpha+beta)tan {beta}}=tan (2alpha +beta)=frac{AY}{AX}.]$

Solution 4

First, since $XY$ is the diameter and $A$ , $B$ , and $C$ lie on the circle, $[angle {XAY} = angle {XBY} = angle{XCY} = 90]$ . Next, because $AZ$ and $CY$ are both perpendicular to $CX$ , we have $AZ$ to be parallel to $CY$ .

Now looking at quadrilateral $APZX$ , we see that this is cyclic because $[angle {PAX} + angle {PZX} = 90+90 = 180.]$ Set $alpha = angle{BXA} = angle{BYA}$ , and $beta = angle{BXC} = angle{BYC}$ . Now, $[angle{AYC} = angle{YAZ}]$ since $AZ$ and $CY$ are parallel. Also, $[angle{PAZ} = angle{PXZ} = alpha + beta.]$ That means $[angle{PXZ} = angle{PXQ} + angle{QXZ} = beta + angle{QXZ} = alpha + beta]$ so $[angle{QXZ} = alpha.]$ This means $angle{QXZ} = angle{YBC} = alpha$ , so $BC$ and $AY$ are parallel. Finally, we can look at the equation. We know $[XPcos{alpha} = AX,]$ so $XP = frac{AX}{cos{alpha}}.$ We also know $[XQcos(alpha+beta) = AX,]$ so $XQ = frac{AX}{cos(alpha+beta)}.$ Plugging this into the LHS of the equation, we get $[frac{BYcosalpha}{AX}+frac{CYcos(alpha+beta)}{AZ}.]$ Now, let $H$ be the point on $AY$ such that $BH$ is perpendicular to $AY$ . Also, since $angle{AYB} = angle{CXY}$ , their arcs have equal length, and $AB=CY$ . Now, the LHS is simplified even more to $[frac{ABcosalpha}{AX}+frac{CYcos(alpha+beta)}{AZ}]$ which is equal to $[frac{AH+YH}{AZ}]$ which is equal to $[frac{AY}{AX}.]$ This completes the proof.

Problem 6

Solution (Cauchy or AM-GM)

The key Lemma is: $[sqrt{a-1}+sqrt{b-1} le sqrt{ab}]$ for all $a,b ge 1$ . Equality holds when $(a-1)(b-1)=1$ .

This is proven easily. $[sqrt{a-1}+sqrt{b-1} = sqrt{a-1}sqrt{1}+sqrt{1}sqrt{b-1} le sqrt{(a-1+1)(b-1+1)} = sqrt{ab}]$ by Cauchy. Equality then holds when $a-1 =frac{1}{b-1} implies (a-1)(b-1) = 1$ .

Now assume that $x = min(x,y,z)$ . Now note that, by the Lemma,

$[sqrt{x-1}+sqrt{y-1}+sqrt{z-1} le sqrt{x-1} + sqrt{yz} le sqrt{x(yz+1)} = sqrt{xyz+x}]$ . So equality must hold. So $(y-1)(z-1) = 1$ and $(x-1)(yz) = 1$ . If we let $z = c$ , then we can easily compute that $y = frac{c}{c-1}, x = frac{c^2+c-1}{c^2}$ . Now it remains to check that $x le y, z$ .

But by easy computations, $x = frac{c^2+c-1}{c^2} le c = z Longleftrightarrow (c^2-1)(c-1) ge 0$ , which is obvious. Also $x = frac{c^2+c-1}{c^2} le frac{c}{c-1} = y Longleftrightarrow 2c ge 1$ , which is obvious, since $c ge 1$ .

So all solutions are of the form $boxed{left(frac{c^2+c-1}{c^2}, frac{c}{c-1}, cright)}$ , and all permutations for $c > 1$ .

Remark: An alternative proof of the key Lemma is the following: By AM-GM, $[(ab-a-b+1)+1 = (a-1)(b-1) + 1 ge 2sqrt{(a-1)(b-1)}]$ $[abge (a-1)+(b-1)+2sqrt{(a-1)(b-1)}]$ . Now taking the square root of both sides gives the desired. Equality holds when $(a-1)(b-1) = 1$ .

Solution with Thought Process

Without loss of generality, let $1 le x le y le z$ . Then $sqrt{x + xyz} = sqrt{x - 1} + sqrt{y - 1} + sqrt{z - 1}$ .

Suppose x = y = z. Then $sqrt{x + x^3} = 3sqrt{x-1}$ , so $x + x^3 = 9x - 9$ . It is easily verified that $x^3 - 8x + 9 = 0$ has no solution in positive numbers greater than 1. Thus, $sqrt{x + xyz} ge sqrt{x - 1} + sqrt{y - 1} + sqrt{z - 1}$ for x = y = z. We suspect if the inequality always holds.

Let x = 1. Then we have $sqrt{1 + yz} ge sqrt{y-1} + sqrt{z-1}$ , which simplifies to $[1 + yz ge y + z - 2 + 2sqrt{(y-1)(z-1)}]$ and hence $[yz - y - z + 3 ge 2sqrt{(y-1)(z-1)}]$ Let us try a few examples: if y = z = 2, we have $3 > 2$ ; if y = z, we have $y^2 - 2y + 3 ge 2(y-1)$ , which reduces to $y^2 - 4y + 5 ge 0$ . The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let $u = sqrt{(y-1)(z-1)}$ ! Thus, $[u^2 - 2u + 2 = (u-1)^2 + 1 > 0]$ and the claim holds for x = 1.

If x > 1, we see the $sqrt{x - 1}$ will provide a huge obstacle when squaring. But, using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz$ : $[x + xyz ge x - 1 + y - 1 + z - 1 + 2sqrt{(x-1)(y-1)} + 2sqrt{(y-1)(z-1)} + 2sqrt{(x-1)(y-1)}]$ which leads to $[xyz ge y + z - 3 + 2sqrt{(x-1)(y-1)} + 2sqrt{(y-1)(z-1)} + 2sqrt{(x-1)(z-1)}]$ Again, we experiment. If x = 2, y = 3, and z = 3, then $18 > 7 + 4sqrt{6}$ .

Now, we see the finish: setting $u = sqrt{x-1}$ gives $x = u^2 + 1$ . We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations: $[u^2(yz) - u(2sqrt{y-1} + 2sqrt{z-1}) + (3 + yz - y - z - 2sqrt{(y-1)(z-1)}) ge 0]$

Because the coefficient of $u^2$ is positive, all we need to do is to verify that the discriminant is nonpositive: $[b^2 - 4ac = 4(y-1) + 4(z-1) - 8sqrt{(y-1)(z-1)} - yz(12 + 4yz - 4y - 4z - 8sqrt{(y-1)(z-1)})]$

Let us try a few examples. If y = z, then the discriminant D = $8(y-1) - 8(y-1) - yz(12 + 4y^2 - 8y - 8(y-1)) = -yz(4y^2 - 16y + 20) = -4yz(y^2 - 4y + 5) < 0$ .

We are almost done, but we need to find the correct argument. (How frustrating!) Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis.

--Thinking Process by suli

Solution 2

WLOG, assume that $x = min(x,y,z)$ . Let $a=sqrt{x-1},$ $b=sqrt{y-1}$ and $c=sqrt{z-1}$ . Then $x=a^2+1$ , $y=b^2+1$ and $z=c^2+1$ . The equation becomes $[(a^2+1)+(a^2+1)(b^2+1)(c^2+1)=(a+b+c)^2.]$ Rearranging the terms, we have $[(1+a^2)(bc-1)^2+[a(b+c)-1]^2=0.]$ Therefore $bc=1$ and $a(b+c)=1.$ Express $a$ and $b$ in terms of $c$ , we have $a=frac{c}{c^2+1}$ and $b=frac{1}{c}.$ Easy to check that $a$ is the smallest among $a$ , $b$ and $c.$ Then $x=frac{c^4+3c^2+1}{(c^2+1)^2}$ , $y=frac{c^2+1}{c^2}$ and $z=c^2+1.$ Let $c^2=t$ , we have the solutions for $(x,y,z)$ as follows: $(frac{t^2+3t+1}{(t+1)^2}, frac{t+1}{t}, t+1)$ and permutations for all $t>0.$