exam: USAMO

problem: In triangle $A B C$, points $P, Q, R$ lie on sides $B C, C A, A B$, respectively. Let $\omega_{A}$, $\omega_{B}, \omega_{C}$ denote the circumcircles of triangles $A Q R, B R P, C P Q$, respectively. Given the fact that segment $A P$ intersects $\omega_{A}, \omega_{B}, \omega_{C}$ again at $X, Y, Z$ respectively, prove that $Y X / X Z=B P / P C$.

solution: Let $M$ be the concurrence point of $\omega_{A}, \omega_{B}, \omega_{C}$ (by Miquel's theorem). ![](https://cdn.mathpix.com/cropped/2024_11_19_a0b18c84d2376739ca3eg-03.jpg?height=707&width=806&top_left_y=960&top_left_x=631)  Then $M$ is the center of a spiral similarity sending $\overline{Y Z}$ to $\overline{B C}$. So it suffices to show that this spiral similarity also sends $X$ to $P$, but  $$ \measuredangle M X Y=\measuredangle M X A=\measuredangle M R A=\measuredangle M R B=\measuredangle M P B $$  so this follows.


problem: For a positive integer $n \geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2 n$ distinct moves available; two from each point. Let $a_{n}$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move. Prove that $a_{n-1}+a_{n}=2^{n}$ for all $n \geq 4$.

solution: 【 First solution. Imagine the counter is moving along the set $S=\{0,1, \ldots, 2 n\}$ instead, starting at 0 and ending at $2 n$, in jumps of length 1 and 2 . We can then record the sequence of moves as a matrix of the form  $$ \left[\begin{array}{cccccc} p_{0} & p_{1} & p_{2} & \ldots & p_{n-1} & p_{n} \\ p_{n} & p_{n+1} & p_{n+2} & \ldots & p_{2 n-1} & p_{2 n} \end{array}\right] $$  where $p_{i}=1$ if the point $i$ was visited by the counter, and $p_{i}=0$ if the point was not visited by the counter. Note that $p_{0}=p_{2 n}=1$ and the upper-right and lower-left entries are equal. Then, the problem amounts to finding the number of such matrices which avoid the contiguous submatrices  $$ \left[\begin{array}{ll} 0 & 0 \end{array}\right] \quad\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \quad\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] $$  which correspond to forbidding jumps of length greater than 2 , repeating a length 2 jump and repeating a length 1 jump.  We give a nice symmetric phrasing suggested by fclvbfm934 at https://aops.com/ community/p27834267. If we focus on just the three possible column vectors that appear, say  $$ \mathbf{u}:=\left[\begin{array}{l} 1 \\ 0 \end{array}\right], \quad \mathbf{v}:=\left[\begin{array}{l} 0 \\ 1 \end{array}\right], \quad \mathbf{w}:=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] $$  then we can instead describe valid matrices as sequences of $n+1$ such column vectors, where no two column vectors are adjacent, and where the boundary condition is that  - either we start with $\mathbf{u}$ and end with $\mathbf{v}$, or - either we start with $\mathbf{w}$ and end with $\mathbf{w}$.  Let $x_{n}$ and $y_{n}$ denote the number of such $2 \times(n+1)$ matrices. (Hence $a_{n}=x_{n}+y_{n}$.) But owing to the symmetry of the setup with $\mathbf{u}, \mathbf{v}, \mathbf{w}$, we could instead view $x_{n}$ and $y_{n}$ as the number of $2 \times(n+1)$ matrices for a fixed starting first column whose final column is the same/different. So we have the recursions  $$ \begin{aligned} x_{n+1} & =x_{n}+y_{n} \\ y_{n+1} & =2 x_{n} . \end{aligned} $$  We also have that  $$ 2 x_{n}+y_{n}=2^{n} $$  which may either be proved directly from the recursions (using $x_{1}=1$ and $y_{1}=0$ ), or by noting the left-hand side counts the total number of sequences of $n+1$ column vectors with no restrictions on the final column at all (in which case there are simply 2 choices for each of the $n$ columns after the first one). Thus,  $$ \begin{aligned} a_{n+1}+a_{n} & =\left(x_{n+1}+y_{n+1}\right)+\left(x_{n}+y_{n}\right) \\ & =\left(\left(x_{n}+y_{n}\right)+2 x_{n}\right)+\left(x_{n}+y_{n}\right) \\ & =2\left(2 x_{n}+y_{n}\right)=2^{n+1} \end{aligned} $$  as needed.


solution: Second (longer) solution. If one does not notice the nice rephrasing with $\mathbf{u}, \mathbf{v}, \mathbf{w}$ above, one may still proceed with the following direct calculation. Retain the notation of  $$ \left[\begin{array}{cccccc} p_{0} & p_{1} & p_{2} & \ldots & p_{n-1} & p_{n} \\ p_{n} & p_{n+1} & p_{n+2} & \cdots & p_{2 n-1} & p_{2 n} \end{array}\right] $$  described earlier. We will for now ignore the boundary conditions. Instead we say a $2 \times m$ matrix is silver ( $m \geq 2$ ) if it avoids the three shapes above. We consider three types of silver matrices (essentially doing casework on the last column):  - type $B$ matrices, of the shape $\left[\begin{array}{lll}1 & \cdots & 1 \\ 0 & \cdots & 0\end{array}\right]$ - type C matrices, of the shape $\left[\begin{array}{lll}1 & \cdots & 0 \\ 0 & \cdots & 1\end{array}\right]$. - type D matrices, of the shape $\left[\begin{array}{lll}1 & \cdots & 1 \\ 0 & \cdots & 1\end{array}\right]$.  We let $b_{m}, c_{m}, d_{m}$ denote matrices of each type, of size $2 \times m$, and claim the following two recursions for $m \geq 4$ :  $$ \begin{aligned} b_{m} & =c_{m-1}+d_{m-1} \\ c_{m} & =b_{m-1}+d_{m-1} \\ d_{m} & =b_{m-1}+c_{m-1} \end{aligned} $$  Indeed, if we delete the last column of a type B matrix and consider what used to be the second-to-last column, we find that it is either type C or type D. This establishes the first recursion and the others are analogous.  Note that $b_{2}=0$ and $c_{2}=d_{2}=1$. So using this recursion, the first few values are  | $m$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | | $b_{m}$ | 0 | 2 | 2 | 6 | 10 | 22 | 42 | | $c_{m}$ | 1 | 1 | 3 | 5 | 11 | 21 | 43 | | $d_{m}$ | 1 | 1 | 3 | 5 | 11 | 21 | 43 |  and a calculation gives $b_{m}=\frac{2^{m-1}+2(-1)^{m-1}}{3}, c_{m}=d_{m}=\frac{2^{m-1}-(-1)^{m-1}}{3}$.  We now relate $a_{n}$ to $b_{m}, c_{m}, d_{m}$. Observe that a matrix as described in the problem is a silver matrix of one of two forms:  $$ \left[\begin{array}{cccccc} 1 & p_{1} & p_{2} & \ldots & p_{n-1} & 0 \\ 0 & p_{n+1} & p_{n+2} & \ldots & p_{2 n-1} & 1 \end{array}\right] \quad \text { or } \quad\left[\begin{array}{cccccc} 1 & p_{1} & p_{2} & \ldots & p_{n-1} & 1 \\ 1 & p_{n+1} & p_{n+2} & \ldots & p_{2 n-1} & 1 \end{array}\right] $$  There are $c_{n+1}$ matrices of the first form. Moreover, there are $2 d_{n}$ matrices of the second form (to see this, delete the first column; we either get a type-D matrix or an upside-down type-D matrix). Thus we get  $$ a_{n}=c_{n+1}+2 d_{n}=\frac{2^{n+1}+(-1)^{n+1}}{3} $$  This implies the result.


problem: Let $n$ be a positive integer. There are $\frac{n(n+1)}{2}$ tokens, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing $n$ tokens. Initially, each token has the white side up. An operation is to choose a line parallel to the sides of the triangle, and flip all the token on that line. A configuration is called admissible if it can be obtained from the initial configuration by performing a finite number of operations. For each admissible configuration $C$, let $f(C)$ denote the smallest number of operations required to obtain $C$ from the initial configuration. Find the maximum value of $f(C)$, where $C$ varies over all admissible configurations.

solution: The answer is  $$ \max _{C} f(C)= \begin{cases}6 k & n=4 k \\ 6 k+1 & n=4 k+1 \\ 6 k+2 & n=4 k+2 \\ 6 k+3 & n=4 k+3\end{cases} $$  The main point of the problem is actually to determine all linear dependencies among the $3 n$ possible moves (since the moves commute and applying a move twice is the same as doing nothing). In what follows, assume $n>1$ for convenience.  To this end, we consider sequences of operations as additive vectors in $v \in \mathbb{F}_{2}^{3 n}$, with the linear map $T: \mathbb{F}_{2}^{3 n} \rightarrow \mathbb{F}_{2}^{\frac{1}{2} n(n+1)}$ denoting the result of applying a vector $v$. We in particular focus on the following four vectors.  - Three vectors $x, y, z$ are defined by choosing all $n$ lines parallel to one axis. Note $T(x)=T(y)=T(z)=1$ (i.e. these vectors flip all tokens). - The vector $\theta$ which toggles all lines with an even number of tokens. One can check that $T(\theta)=\mathbf{0}$. (Easiest to guess from $n=2$ and $n=3$ case.) One amusing proof that this works is to use Vivani's theorem: in an equilateral triangle $A B C$, the sum of distances from an interior point $P$ to the three sides is equal.  The main claim is: Claim - For $n \geq 2$, the kernel of $T$ has exactly eight elements, namely $\{\mathbf{0}, x+$ $y, y+z, z+x, \theta, \theta+x+y, \theta+y+z, \theta+z+x\}$.   - If $v$ uses the $y$-move of length $n$, then we replace $v$ with $v+(x+y)$ to obtain a vector in the kernel not using the $y$-move of length $n$. - If $v$ uses the $z$-move of length $n$, then we replace $v$ with $v+(x+z)$ to obtain a vector in the kernel not using the $z$-move of length $n$. - If $v$ uses the $x$-move of length 2 , then - if $n$ is odd, replace $v$ with $v+\theta$; - if $n$ is even, replace $v$ with $v+(\theta+y+z)$ to obtain a vector in the kernel not using the $x$-move of length 2 . A picture is shown below, with the unused rows being dotted. ![](https://cdn.mathpix.com/cropped/2024_11_19_a0b18c84d2376739ca3eg-08.jpg?height=801&width=809&top_left_y=519&top_left_x=632)  Then, it is easy to check inductively that $v$ must now be the zero vector, after the replacements. The idea is that for each token $t$, if two of the moves involving $t$ are unused, so is the third, and in this way we can show all rows are unused. Thus the original $v$ was in the kernel we described.  Then problem is a coordinate bash, since given any $v$ we now know exactly which vectors $w$ have $T(v)=T(w)$, so given any admissible configuration $C$ one can exactly compute $f(C)$ as the minimum of eight values.  To be explicit, we could represent a vector $v$ as  $$ v \longleftrightarrow\left(a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}\right) $$  where $a_{1}$ is the number of 1's in odd $x$-indices, $a_{2}$ number of 1 's in even $x$-indices. Then for example  $$ \begin{aligned} v & \longleftrightarrow\left(a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}\right) \\ v+x+y & \longleftrightarrow\left(\left\lceil\frac{n}{2}\right\rceil-a_{1},\left\lfloor\frac{n}{2}\right\rfloor-a_{2},\left\lceil\frac{n}{2}\right\rceil-b_{1},\left\lfloor\frac{n}{2}\right\rfloor-b_{2}, c_{1}, c_{2}\right) \\ v+\theta & \longleftrightarrow\left(a_{1},\left\lfloor\frac{n}{2}\right\rfloor-a_{2}, b_{1},\left\lfloor\frac{n}{2}\right\rfloor-b_{2}, c_{1},\left\lfloor\frac{n}{2}\right\rfloor-c_{2}\right) \end{aligned} $$  and $f(T(v))$ is the smallest sum of the six numbers across all eight 6 -tuples. So you expect to answer about $\frac{3}{2} n$ if all things are about $n / 4$. The details are too annoying to reproduce here, so they are omitted.


problem: Find all real numbers $x, y, z \geq 1$ satisfying  $$ \min (\sqrt{x+x y z}, \sqrt{y+x y z}, \sqrt{z+x y z})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} $$

solution: Set $x=1+a, y=1+b, z=1+c$ which eliminates the $x, y, z \geq 1$ condition. Assume without loss of generality that $a \leq b \leq c$. Then the given equation rewrites as  $$ \sqrt{(1+a)(1+(1+b)(1+c))}=\sqrt{a}+\sqrt{b}+\sqrt{c} $$  In fact, we are going to prove the left-hand side always exceeds the right-hand side, and then determine the equality cases. We have:  $$ \begin{aligned} (1+a)(1+(1+b)(1+c)) & =(a+1)(1+(b+1)(1+c)) \\ & \leq(a+1)\left(1+(\sqrt{b}+\sqrt{c})^{2}\right) \\ & \leq(\sqrt{a}+(\sqrt{b}+\sqrt{c}))^{2} \end{aligned} $$  by two applications of Cauchy-Schwarz. Equality holds if $b c=1$ and $1 / a=\sqrt{b}+\sqrt{c}$. Letting $c=t^{2}$ for $t \geq 1$, we recover $b=t^{-2} \leq t^{2}$ and $a=\frac{1}{t+1 / t} \leq t^{2}$.   $$ (x, y, z)=\left(1+\left(\frac{t}{t^{2}+1}\right)^{2}, 1+\frac{1}{t^{2}}, 1+t^{2}\right) $$  and permutations, for any $t>0$.


problem: Let $m$ and $n$ be positive integers. Prove that there exists a positive integer $c$ such that cm and cn have the same nonzero decimal digits.

solution: One-line spoiler: 142857. More verbosely, the idea is to look at the decimal representation of $1 / D, m / D, n / D$ for a suitable denominator $D$, which have a "cyclic shift" property in which the digits of $n / D$ are the digits of $m / D$ shifted by 3 .  Remark (An example to follow along). Here is an example to follow along in the subsequent proof If $m=4$ and $n=23$ then the magic numbers $e=3$ and $D=41$ obey  $$ 10^{3} \cdot \frac{4}{41}=97+\frac{23}{41} $$  The idea is that  $$ \begin{aligned} \frac{1}{41} & =0 . \overline{02439} \\ \frac{4}{41} & =0 . \overline{09756} \\ \frac{23}{41} & =0.56097 \end{aligned} $$  and so $c=2349$ works; we get $4 c=9756$ and $23 c=56097$ which are cyclic shifts of each other by 3 places (with some leading zeros appended).  Here is the one to use: Claim - There exists positive integers $D$ and $e$ such that $\operatorname{gcd}(D, 10)=1, D>$ $\max (m, n)$, and moreover  $$ \frac{10^{e} m-n}{D} \in \mathbb{Z} $$   $$ A=10^{e} n-m . $$  Let $r=\nu_{2}(m)$ and $s=\nu_{5}(m)$. As long as $e>m a x(r, s)$ we have $\nu_{2}(A)=r$ and $\nu_{5}(A)=s$, too. Now choose any $e>\max (r, s)$ big enough that $A>2^{r} 5^{s} \max (m, n)$ also holds. Then the number $D=\frac{A}{2^{r 5^{s}}}$ works; the first two properties hold by construction and  $$ 10^{e} \cdot \frac{n}{D}-\frac{m}{D}=\frac{A}{D}=2^{r} 5^{s} $$  is an integer.  Remark (For people who like obscure theorems). Kobayashi's theorem implies we can actually pick $D$ to be prime.  Now we take $c$ to be the number under the bar of $1 / D$ (leading zeros removed). Then the decimal representation of $\frac{m}{D}$ is the decimal representation of cm repeated (possibly including leading zeros). Similarly, $\frac{n}{D}$ has the decimal representation of cm repeated (possibly including leading zeros). Finally, since  $$ 10^{e} \cdot \frac{m}{D}-\frac{n}{D} \text { is an integer } $$  it follows that these repeating decimal representations are rotations of each other by $e$ places, so in particular they have the same number of nonzero digits.  Remark. Many students tried to find a $D$ satisfying the stronger hypothesis that $1 / D$, $2 / D, \ldots,(D-1) / D$ are cyclic shifts of each other. For example, this holds in the famous $D=7$ case.   One may be tempted to resort to using large primes $D$ rather than powers of 7 to deal with this issue. However it is an open conjecture (a special case of Artin's primitive root conjecture) whether or not $10(\bmod p)$ is primitive infinitely often, which is the necessary conjecture so this is harder than it seems.


problem: Let $A B C$ be a triangle. Find all points $P$ on segment $B C$ satisfying the following property: If $X$ and $Y$ are the intersections of line $P A$ with the common external tangent lines of the circumcircles of triangles $P A B$ and $P A C$, then  $$ \left(\frac{P A}{X Y}\right)^{2}+\frac{P B \cdot P C}{A B \cdot A C}=1 $$

solution: Let $O_{1}$ and $O_{2}$ denote the circumcenters of $P A B$ and $P A C$. The main idea is to notice that $\triangle A B C$ and $\triangle A O_{1} O_{2}$ are spirally similar. ![](https://cdn.mathpix.com/cropped/2024_11_19_a0b18c84d2376739ca3eg-12.jpg?height=712&width=812&top_left_y=1009&top_left_x=628)  Claim (Salmon theorem) - We have $\triangle A B C \stackrel{ \pm}{\sim} \triangle A O_{1} O_{2}$.  $$ \angle A O_{1} B=2 \angle A P B $$  but  $$ \angle A O_{2} C=2(180-\angle A P C)=2 \angle A B P $$  Hence $\angle A O_{1} B=\angle A O_{2} C$. Moreover, both triangles are isosceles, establishing the first similarity. The second part follows from spiral similarities coming in pairs.  Claim - We always have  $$ \left(\frac{P A}{X Y}\right)^{2}=1-\left(\frac{a}{b+c}\right)^{2} . $$  (In particular, this does not depend on $P$.)  Without loss of generality $A$ is closer to $X$ than $Y$, and let the common tangents be $\overline{X_{1} X_{2}}$ and $\overline{Y_{1} Y_{2}}$. We'll perform the main calculation with the convenient scaling $O_{B} O_{C}=a, A O_{C}=b$, and $A O_{B}=c$. Let $B_{1}$ and $C_{1}$ be the tangency points of $X$, and let $h=A M$ be the height of $\triangle A O_{B} O_{C}$. ![](https://cdn.mathpix.com/cropped/2024_11_19_a0b18c84d2376739ca3eg-13.jpg?height=641&width=812&top_left_y=639&top_left_x=625)  Note that by Power of a Point, we have $X X_{1}^{2}=X X_{2}^{2}=X M^{2}-h^{2}$. Also, by Pythagorean theorem we easily obtain $X_{1} X_{2}=a^{2}-(b-c)^{2}$. So putting these together gives  $$ X M^{2}-h^{2}=\frac{a^{2}-(b-c)^{2}}{4}=\frac{(a+b-c)(a-b+c)}{4}=(s-b)(s-c) . $$  Therefore, we have Then  $$ \begin{aligned} \frac{X M^{2}}{h^{2}} & =1+\frac{(s-b)(s-c)}{h^{2}}=1+\frac{a^{2}(s-b)(s-c)}{a^{2} h^{2}} \\ & =1+\frac{a^{2}(s-b)(s-c)}{4 s(s-a)(s-b)(s-c)}=1+\frac{a^{2}}{4 s(s-a)} \\ & =1+\frac{a^{2}}{(b+c)^{2}-a^{2}}=\frac{(b+c)^{2}}{(b+c)^{2}-a^{2}} . \end{aligned} $$  Thus  $$ \left(\frac{P A}{X Y}\right)^{2}=\left(\frac{h}{X M}\right)^{2}=1-\left(\frac{a}{b+c}\right)^{2} $$  To finish, note that when $P$ is the foot of the $\angle A$-bisector, we necessarily have  $$ \frac{P B \cdot P C}{A B \cdot A C}=\frac{\left(\frac{b}{b+c} a\right)\left(\frac{c}{b+c} a\right)}{b c}=\left(\frac{a}{b+c}\right)^{2} $$