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Đề thi IMO 2012 và Lời giải

VnMaTh.CoM 11 tháng 7, 2012 0

VNMATH xin giới thiệu đề thi Olympic Toán quốc tế lần thứ 53 năm 2012 (IMO 2012). Tác giả của các bài toán trong đề thi năm nay là: 1. Evangelos Psychas (Hy Lạp), 2. Angelo di Pasquale (Australia), 3. David Arthur (Canada), 4. Liam Baker (Nam Phi), 5. Josef "Pepa" Tkadlec (Cộng hòa Séc), 6. Dusan Djukic (Serbia).

Ngày thi thứ nhất (10/7/2012)

Bài 1. Cho tam giác $ABC$ và điểm $J$ là tâm đường tròn bàng tiếp trong góc $A$ của tam giác. Đường tròn này tiếp xúc với $AB,AC,BC$ tại $K,L,M$ theo thứ tự. $LM$ cắt $BJ$ tại $F$, $KM$ cắt $CJ$ tại $G$. Gọi $S,T$ lần lượt là giao điểm của $AF,AG$ với $BC$. Chứng minh rằng $M$ là trung điểm của $ST$.
Bài này không khó
Ta có $\triangle FKJ=\triangle FMJ$, do đó $\widehat{KFJ}=\widehat{JFM}$.
Mặt khác $\widehat{JFM}=\widehat{KAJ}$ nên $AFKJ$ nội tiếp. Suy ra $FJ\perp AS$. Từ đó $FS=FA$
Tương tự $AT$ vuông góc với $GJ$. Từ đó $AT//FM$.
Vậy $M$ là trung điểm của $ST$.

Bài 2. Cho số nguyên $n \ge 3$ và các số thực dương $a_2,a_3,\ldots,a_n$ thỏa mãn $a_2 \cdots a_n= 1$. Chứng minh rằng
$$(1+a_2)^2(1+a_3)^3 \cdots (1+a_n)^n > n^n.$$
Câu này khá dễ:Theo AM - GM ta có $(a_{k}+1)=\left(a_{k}+\frac {1}{k-1}+\cdots+\frac {1}{k-1}\right)\geq k\sqrt[k]{\frac{a_{k}}{(k-1)^{k-1}}};$.
Suy ra
$(a_{k}+1)^{k}\geq\frac{k^{k}}{(k-1)^{k-1}}\cdot a_{k}.$
Do đó
$\prod_{k=2}^{n}(a_{k}+1)^{k}\geq n^{n}a_{2}a_{3}\cdots a_{n}=n^{n}.$
Dấu bằng xảy ra nếu và chỉ nếu $a_{k}=\frac {1}{k-1}$, vô lí.

Bài 3. Trò chơi đoán kẻ nói dối là một trò chơi giữa hai người chơi $A$ và $B$. Quy tắc của trò chơi phụ thuộc vào hai số nguyên dương $k$ và $n$ mà cả hai người chơi đều đã biết trước.

Bắt đầu trò chơi, $A$ sẽ chọn các số nguyên $x$ và $N$ với $1 \le x \le N$. $A$ giữ bí mật số $x$ và nói số $N$ cho $B$. $B$ sẽ cố thu nhận thông tin về số $x$ bằng cách hỏi $A$ các câu hỏi như sau : mỗi câu hỏi bao gồm việc $B$ xác định một tập $S$ tùy ý các số nguyên dương (có thể là một tập đã được nhắc đến trong câu hỏi trước đó) và hỏi $A$ xem $x$ có thuộc $S$ hay không. Sau mỗi câu hỏi, $A$ phải trả lời hoặc không, nhưng có thể nói dối bao nhiêu lần tùy thích, chỉ có điều là phải trả lời đúng ít nhất một trong số $k+1$ câu hỏi liên tiếp.

Sau khi $B$ đã hỏi xong, $B$ phải chỉ ra một tập $X$ có tối đa $n$ số nguyên dương. Nếu $x \in X$, $B$ thắng; nếu ngược lại, $B$ thua. Chứng minh rằng :
1. Nếu $n \ge 2^k$, $B$ có thể đảm bảo một chiến thắng.
2. Với mọi $k$ đủ lớn, tồn tại một số nguyên $n \ge 1.99^k$ sao cho $B$ không thể đảm bảo có một chiến thắng.
Đây là bài khó.
 Đề thi IMO 2012 ngày thi thứ nhất

Ngày thi thứ hai (11/7/2012).
Bài 4. Tìm tất cả các hàm số $f : \mathbb{Z} \to \mathbb{Z}$ sao cho với mọi $a+b+c=0$ thì
$$f(a)^2+f(b)^2+f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).$$

Bài 5. Cho tam giác $ABC$ có $\widehat{ACB} = 90^\circ$ và $D$ là chân đường cao tương ứng với đỉnh $C$. Gọi $X$ là một điểm trong của đoạn thẳng $CD$. Gọi $K$ là điểm trên đoạn thẳng $AX$ sao cho $BK=BC$. Tương tự, gọi $L$ là điểm trên đoạn thẳng $BX$ sao cho $AL=AC$. Gọi $M$ là giao điểm của $AL$ và $BK$. Chứng minh rằng $MK=ML$.

Bài 6. Tìm tất cả các số nguyên dương $n$ sao cho tồn tại các số nguyên không âm $a_1,a_2,\ldots,a_n$ thỏa mãn
$$\frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1$$
Cập nhật: lời giải Đề thi IMO 2012 từ IMO Math
IMO 2012 Problems and Solutions

Day 1 (July 10th, 2012)

Problem 1

Given a triangle $ABC$, let $J$ be the center of the excircle opposite to the vertex $A$. This circle is tangent to lines $AB$, $AC$, and $BC$ at $K$, $L$, and $M$, respectively. The lines $BM$ and $JF$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$. Let $S$ be the intersection of $AF$ and $BC$, and let $T$ be the intersection of $AG$ and $BC$. Prove that $M$ is the midpoint of $BC$.
Solution

We have $KM\perp BJ$ hence $BM$ is parallel to the bisector of $\angle ABC$. Therefore $\angle BMK=\frac12\angle ABC$ and $\angle FMB=\frac12\angle ACB$. Denote by $X$ the intersection of $KM$ and $FJ$. From the triangle $FXM$ we derive:

$\angle XFM=90^{\circ}-\angle FMB-\angle BMK=\frac12\angle BAC.$

The points $K$ and $M$ are symmetric with respect to the line $FJ$ hence $\angle KFJ=\angle JFM=\frac12\angle BAC=\angle KAJ$, therefore $K$, $J$, $A$, and $F$ belong to a circle which implies that $\angle JFA=\angle JKA=90^{\circ}$ and $KM\| AS$. The quadrilateral $SKMA$ is a trapezoid and from $\angle SMK=\angle AKM$ we obtain $SM=AK$. Similarly, we get $AL=TM$. Since $AK=AL$ (tangents from $A$ to the excircle) we get $SM=TM$.

Problem 2
Let $a_2$, $a_3$, $\dots$, $a_n$ be positive real numbers that satisfy $a_2\cdot a_3\cdots a_n=1$. Prove that $\left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+n\right)^n\gneq n^n.$
Solution

The inequality between arithmetic and geometric mean implies

$\left(a_k+1\right)^k=\left(a_k+\frac1{k-1}+\frac1{k-1}+\cdots+\frac1{k-1}\right)^k \geq k^k\cdot a_k\cdot \frac1{(k-1)^{k-1}}=\frac{k^k}{(k-1)^{k-1}}\cdot a_k.$

The inequality is strict unless $a_k=\frac1{k-1}$.

Multiplying analogous inequalities for $k=2$, $3$, $\dots$, $n$ yields

$\left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+n\right)^n\gneq \frac{3^3}{2^2}\cdot \frac{4^4}{3^3}\cdots \frac{n^n}{(n-1)^{n-1}}\cdot a_2a_3\cdots a_n=n^n.$

The inequality is strict because at least one of $a_2$, $\dots$, $a_n$ has to be greater than or equal than $1$ and thus $a_k\gneq \frac1{k-1}$ holds for at least one integer $k\in\{2,\dots, n\}$.

Problem 3
The liar’s guessing game is a game played between two players $A$ and $B$. The rules of the game depend on two positive integers $k$ and $n$ which are known to both players.

At the start of the game the player $A$ chooses integers $x$ and $N$ with $1\leq x\leq N$. Player $A$ keeps $x$ secret, and truthfully tells $N$ to the player $B$. The player $B$ now tries to obtain information about $x$ by asking player $A$ questions as follows: each question consists of $B$ specifying an arbitrary set $S$ of positive integers (possibly one specified in some previuos question), and asking $A$ whether $x$ belongs to $S$. Player $B$ may ask as many questions as he wishes. After each question, player $A$ must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any $k+1$ consecutive answers, at least one answer must be truthful.

After $B$ has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers. If $x\in X$, then $B$ wins; otherwise, he loses. Prove that:

(a) If $n\geq 2^k$ then $B$ has a winning strategy.

(b) There exists a positive integer $k_0$ such that for every $k\geq k_0$ there exists an integer $n\geq 1.99^k$ for which $B$ cannot guarantee a victory.
Solution

The game can be re-formulated in an equivalent one: The player $A$ chooses an element $x$ from the set $S$ (with $|S|=N$) and the player $B$ asks the sequence of questions. The $j$-th question consists of $B$ choosing a set $D_j\subseteq S$ and player $A$ selecting a set $P_j\in\left\{ Q_j,Q_j^C\right\}$. For every $j\geq 1$ the following relation holds: $x\in P_j\cup P_{j+1}\cup\cdots \cup P_{j+k}.$

The player $B$ wins if after a finite number of steps he can choose a set $X$ with $|X|\leq n$ such that $x\in X$.

(a) It suffices to prove that if $N\geq 2^k+1$ then the player $B$ can determine a set $S^{\prime}\subseteq S$ with $|S^{\prime}|\leq N-1$ such that $x\in S^{\prime}$.

Assume that $N\geq 2^n+1$.

In the first move $B$ selects any set $D_1\subseteq S$ such that $|D_1|\geq 2^{k-1}$ and $|D_1^C|\geq 2^{k-1}$. After receiving the set $P_1$ from $A$, $B$ makes the second move. The player $B$ selects a set $D_2\subseteq S$ such that $|D_2\cap P_1^C|\geq 2^{k-2}$ and $|D_2^C\cap P_1^C|\geq 2^{k-2}$. The player $B$ continues this way: in the move $j$ he/she chooses a set $D_j$ such that $|D_j\cap P_j^C|\geq 2^{k-j}$ and $|D_j^C\cap P_j^C|\geq 2^{k-j}$.

In this way the player $B$ has obtained the sets $P_1$, $P_2$, $\dots$, $P_k$ such that $\left(P_1\cup \cdots \cup P_k\right)^C\geq 1$. Then $B$ chooses the set $D_{k+1}$ to be a singleton containing any element of $P_1\cup\cdots \cup P_k$. There are two cases now:

• $1^{\circ}$ The player $A$ selects $P_{k+1}=D_{k+1}^C$. Then $B$ can take $S^{\prime}=S\setminus D_{k+1}$ and the statement is proved.

• $2^{\circ}$ The player $A$ selects $P_{k+1}=D_{k+1}$. Now the player $B$ repeats the previous procedure on the set $S_1=S\setminus D_{k+1}$ to obtain the sequence of sets $P_{k+2}$, $P_{k+3}$, $\dots$, $P_{2k+1}$. The following inequality holds: $\left|S_1\setminus \left(P_{k+2}\cdots P_{2k+1}\right)\right|\geq 1,$

since $|S_1|\geq 2^k$. However, now we have $\left|\left(P_{k+1}\cup P_{k+2}\cup\cdots\cup P_{2k+1}\right)^C\right|\geq 1,$ and we may take $S^{\prime}=P_{k+1}\cup \cdots \cup P_{2k+1}$.

(b) Let $p$ and $q$ be two positive integers such that $1.99\lneq p\lneq q\lneq 2$. Let us choose $k_0$ such that $\left(\frac{p}{q}\right)^{k_0}\leq 2\cdot \left(1-\frac{q}2\right)\quad\quad\quad\mbox{and}\quad\quad\quad p^k-1.99^k\gneq 1.$ We will prove that for every $k\geq k_0$ if $|S|\in\left(1.99^k, p^k\right)$ then there is a strategy for the player $A$ to select sets $P_1$, $P_2$, $\dots$ (based on sets $D_1$, $D_2$, $\dots$ provided by $B$) such that for each $j$ the following relation holds: $P_j\cup P_{j+1}\cup\cdots\cup P_{j+k}=S.$
Assuming that $S=\{1,2,\dots, N\}$, the player $A$ will maintain the following sequence of $N$-tuples: $(\mathbf{x})_{j=0}^{\infty}=\left(x_1^j, x_2^j, \dots, x_N^j\right)$. Initially we set $x_1^0=x_2^0=\cdots =x_N^0=1$. After the set $P_j$ is selected then we define $\mathbf x^{j+1}$ based on $\mathbf x^j$ as follows: $x_i^{j+1}=\left\{\begin{array}{rl} 1,&\mbox{ if } i\in S\\ q\cdot x_i^j, &\mbox{ if } i\not\in S. \end{array}\right.$ The player $A$ can keep $B$ from winning if $x_i^j\leq q^k$ for each pair $(i,j)$. For a sequence $\mathbf x$, let us define $T(\mathbf x)=\sum_{i=1}^N x_i$. It suffices for player $A$ to make sure that $T\left(\mathbf x^j\right)\leq q^{k}$ for each $j$.
Notice that $T\left(\mathbf x^0\right)=N\leq p^k \lneq q^k$.
We will now prove that given $\mathbf x^j$ such that $T\left(\mathbf x^j\right)\leq q^k$, and a set $D_{j+1}$ the player $A$ can choose $P_{j+1}\in\left\{D_{j+1},D_{j+1}^C\right\}$ such that $T\left(\mathbf x^{j+1}\right)\leq q^k$. Let $\mathbf y$ be the sequence that would be obtained if $P_{j+1}=D_{j+1}$, and let $\mathbf z$ be the sequence that would be obtained if $P_{j+1}=D_{j+1}^C$. Then we have $T\left(\mathbf y\right)=\sum_{i\in D_{j+1}^C} qx_i^j+\left|D_{j+1}\right|$ $T\left(\mathbf z\right)=\sum_{i\in D_{j+1}} qx_i^j+\left|D_{j+1}^C\right|.$ Summing up the previous two equalities gives: $T\left(\mathbf y\right)+T\left(\mathbf z\right)= q\cdot T\left(\mathbf x^j\right)+ N\leq q^{k+1}+ p^k, \mbox{ hence}$ $\min\left\{T\left(\mathbf y\right),T\left(\mathbf z\right)\right\}\leq \frac{q}2\cdot q^k+\frac{p^k}2\leq q^k,$ because of our choice of $k_0$.

Day 2 (July 11th, 2012)

Problem 4
Find all functions $f:\mathbb Z\to\mathbb Z$ such that, for all integers $a$, $b$, $c$ with $a+b+c=0$ the following inequality holds: $f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$ Solution

Placing $a=b=c=0$ yields $3f(0)^2=6f(0)^2$ which implies $f(0)=0$. Now we can place $b=-a$, $c=0$ to obtain $f(a)^2+f(-a)^2=2f(a)f(-a)$, or, equivalently $\left(f(a)-f(-a)\right)^2=0$ which implies $f(a)=f(-a)$.
Assume now that $f(a)=0$ for some $a\in\mathbb Z$. Then for any $b$ we have $a+b+(-a-b)=0$ hence $f(a)^2+f(b)^2+f(a+b)^2=2f(b)f(a+b)$, which is equivalent to $\left(f(b)-f(a+b)\right)^2=0$, or $f(a+b)=f(b)$. Therefore if $f(a)=0$ for some $a\neq 0$, then $f$ is a periodic function with period $a$.
Placing $b=a$ and $c=2a$ in the original equation yields $f(2a)\cdot \left(f(2a)-4f(a)\right)=0$. Choosing $a=1$ we get $f(2)=0$ or $f(2)=4f(1)$.
If $f(2)=0$, then $f$ is periodic with period 2 and we must have $f(n)=f(1)$ for all odd $n$. It is easy to verify that for each $c\in\mathbb Z$ the function $f(x)=\left\{\begin{array}{rl}0,& 2\mid n,\\ c, & 2\not\mid n\end{array}\right.$ satisfies the conditions of the problem.
Assume now that $f(2)=4f(1)$ and that $f(1)\neq 0$. We will now prove by induction that $f(n)=n^2\cdot f(1)$. The statement holds for $n\in\{0,1,2\}$. Assume that the statement holds for all $i\in\{1,2,\dots, n\}$ and let us prove it for $n+1$. We place $a=1$, $b=n$, $c=-n-1$ in the original equation to obtain: $f(1)^2+n^4f(1)^2+f(n+1)^2=2n^2f(1)^2+2(n^2+1)f(n+1)f(1)$ $\Leftrightarrow \quad\quad\quad \left(f(n+1)-(n+1)^2f(1)\right)\cdot \left(f(n+1)-(n-1)^2f(1)\right)=0.$
If $f(n+1)=(n-1)^2f(1)$ then setting $a=n+1$, $b=1-n$, and $c=-2$ in the original equation yields $2(n-1)^4f(1)^2+16f(1)^2=2\cdot 4\cdot 2(n-1)^2f(1)^2+2\cdot (n-1)^4f(1)$ which implies $(n-1)^2=1$ hence $n=2$. Therefore $f(3)=f(1)$. Placing $a=1$, $b=3$, and $c=4$ into the original equation implies that $f(4)=0$ or $f(4)=4f(1)=f(2)$. If $f(4)\neq 0$ we get $f(2)^2+f(2)^2+f(4)^2=2f(2)^2+4f(2)f(4)$ hence $f(4)=4f(2)$. We already have that $f(4)=f(2)$ and this implies that $f(2)=0$, which is impossible according to our assumption. Therefore $f(4)=0$ and the function $f$ has period $4$. Then $f(4k)=0$, $f(4k+1)=f(4k+3)=c$, and $f(4k+2)=4c$. It is easy to verify that this function satisfies the requirements of the problem.
Thus the solutions are: $f(x)=cx^2$ for some $c\in\mathbb Z$; $f(x)=\left\{\begin{array}{rl}0,& 2\mid n,\\ c, & 2\not\mid n\end{array}\right.$ for some $c\in\mathbb Z$; and $f(x)=\left\{\begin{array}{rl}0,& 4\mid n,\\ c, & 2\nmid n,\\ 4c, & n\equiv 2 \mbox{ (mod } 4)\end{array}\right.$ for some $c\in\mathbb Z$.

Problem 5
Given a triangle $ABC$, assume that $\angle C=90^{\circ}$. Let $D$ be the foot of the perpendicular from $C$ to $AB$, and let $X$ be any point of the segment $CD$. Let $K$ and $L$ be the points on the segments $AX$ and $BX$ such that $BK=BC$ and $AL=AC$, respectively. Let $M$ be the intersection of $AL$ and $BK$.
Prove that $MK=ML$.

Solution

Since $AL^2=AC^2=AD\cdot AB$ the triangles $ALD$ and $ABL$ are similar hence $\angle ALD=\angle XBA$. Let $R$ be the point on the extension of $DC$ over point $C$ such that $DX\cdot DR=BD\cdot AD$. Since $\angle BDX=\angle RDA=90^{\circ}$ we conclude $\triangle RAD\sim\triangle BXD$ hence $\angle XBD=\angle ARD$, therefore $\angle ALD=\angle ARD$ and the points $R$, $A$, $D$, and $L$ belong to a circle. This implies that $\angle RLA=90^{\circ}$ hence $RL^2=AR^2-AL^2=AR^2-AC^2$. Analogously we prove that $RK^2=BR^2-BC^2$ and $\angle RKB=90^{\circ}$. Since $RC\perp AB$ we have $AR^2-AC^2=BR^2-BC^2$, therefore $RL^2=RK^2$ hence $RL=RK$. Together with $\angle RLM=\angle RKM=90^{\circ}$ we conclude $\triangle RLM\cong \triangle LKM$ hence $MK=ML$.

Problem 6
Find all positive integers $n$ for which there exist non-negative integers $a_1$, $a_2$, $\dots$, $a_n$ such that $\frac1{2^{a_1}}+\frac1{2^{a_2}}+\cdots+\frac1{2^{a_n}}=\frac1{3^{a_1}}+\frac2{3^{a_2}}+\cdots+\frac{n}{3^{a_n}}=1.$
Solution

Let $M=\max\{a_1,\dots, a_n\}$. Then we have $3^M=1\cdot 3^{M-a_1}+ 2\cdot 3^{M-a_2}+\cdots + n\cdot 3^{M-a_n}\equiv 1+2+\cdots+n=\frac{n(n+1)}2$ (mod $n$). Therefore, the number $\frac{n(n+1)}2$ must be odd and hence $n\equiv 1$ (mod 4) or $n\equiv 2$ (mod 4).
We will now prove that each $n\in\mathbb N$ of the form $4k+1$ or $4k+2$ (for some $k\in\mathbb N$) there exist integers $a_1$, $\dots$, $a_n$ with the described property.
For a sequence $\mathbf a=\left(a_1, a_2, \dots, a_n\right)$ let us introduce the following notation: $L(\mathbf a)=\frac1{2^{a_1}}+\frac1{2^{a_2}}+\cdots+\frac1{2^{a_n}}\quad\quad\quad\mbox{ and }\quad\quad\quad R(\mathbf a)=\frac1{3^{a_1}}+\frac2{3^{a_2}}+\cdots+\frac{n}{3^{a_n}}.$ Assume that for $n=2m+1$ there exists a sequence $\mathbf a=(a_1, \dots, a_n)$ of non-negative integers with $L(\mathbf a)=R(\mathbf a)=1$. Consider the sequence $\mathbf a^{\prime}=(a_1^{\prime},\dots, a_{n+1}^{\prime})$ defined in the following way: $a_j^{\prime}=\left\{\begin{array}{rl} a_j,& \mbox{ if } j\not \in \{m+1,2m+2\}\\ a_{m+1}+1,& \mbox{ if } j\in \{m+1,2m+2\}.\end{array}\right.$ Then we have $L\left(\mathbf a^{\prime}\right)=L(\mathbf a)-\frac1{2^{a_{m+1}}}+2\cdot \frac1{2^{a_{m+1}+1}}=1\;\;\;\mbox{ and }\;\;\; R\left(\mathbf a^{\prime}\right)=R(\mathbf a)- \frac{m+1}{3^{a_{m+1}}}+\frac{m+1}{3^{a_{m+1}+1}}+\frac{2m+2}{3^{a_{m+1}+1}}=1.$ This implies that if the statement holds for $2m+1$, then it holds for $2m+2$.
Assume now that the statement holds for $n=4m+2$ for some $m\geq 2$, and assume that $\mathbf a=\left(a_1, \dots, a_{4m+2}\right)$ is the corresponding sequence of $n$ non-negative integers. We will construct a following sequence $\mathbf a^{\prime}=\left(a_1^{\prime}, a_2^{\prime}, \dots, a^{\prime}_{4m+13}\right)$ that satisfies $L\left(\mathbf a^{\prime}\right)=R\left(\mathbf a^{\prime}\right)=1$ thus proving that the statement holds for $4m+13$. Define: $a^{\prime}_j=\left\{ \begin{array}{rl} a_{m+2}+2, &\mbox{ if } j=m+2\\ a_{j}+1, &\mbox{ if } j\in\{2m+2, 2m+3, 2m+4, 2m+5,2m+6\}\\ a_{\frac{j}2}+1, &\mbox{ if } j\in\{4m+4, 4m+6, 4m+8, 4m+10,4m+12\}\\ a_{m+2}+3, &\mbox{ if } j\in\{4m+3, 4m+5, 4m+7, 4m+9, 4m+11, 4m+13\}\\ a_j, &\mbox{ otherwise.} \end{array}\right.$
We now have $L\left(\mathbf a^{\prime}\right)=L(\mathbf a)-\frac1{2^{a_{m+2}}}- \sum_{j=2}^6 \frac1{2^{a_{2m+j}}}+\frac1{2^{a_{m+2}+2}}+\sum_{j=2}^6\frac1{2^{a_{2m+j}+1}}+\sum_{j=2}^6 \frac1{2^{a_{2m+j}+1}}+6\cdot \frac1{2^{a_{m+2}+3}}=1.$ It remains to verify that $R\left(\mathbf a^{\prime}\right)=R(\mathbf a)=1$. We write $R\left(\mathbf a^{\prime}\right)-R(\mathbf a)=R\left( \begin{array}{ccccccc}a^{\prime}_{m+2}, &a^{\prime}_{4m+3}, &a^{\prime}_{4m+5}, &a^{\prime}_{4m+7}, &a^{\prime}_{4m+9}, &a^{\prime}_{4m+11}, &a^{\prime}_{4m+13}\\ m+2, & 4m+3,& 4m+5,& 4m+7,& 4m+9,& 4m+11,& 4m+13\end{array}\right)-R\left(\begin{array}{c}a_{m+2}\\ m+2\end{array}\right)$ $+ \sum_{j=2}^6 \left( R\left(\begin{array}{cc}a^{\prime}_{2m+j}, &a^{\prime}_{4m+2j}\\ 2m+j,& 4m+j\end{array}\right)-R\left(\begin{array}{c}a_{2m+j}\\ 2m+j\end{array}\right)\right),$ where $R\left(\begin{array}{ccc} c_1, &\dots, &c_k\\ d_1, &\dots, &d_k\end{array}\right)=\frac{d_1}{3^{c_1}}+\cdots+\frac{d_k}{3^{c_k}}.$ For each $j\in\{2,3,4,5,6\}$ we have $R\left(\begin{array}{cc}a^{\prime}_{2m+j}, &a^{\prime}_{4m+2j}\\ 2m+j,& 4m+j\end{array}\right)-R\left(\begin{array}{c}a_{2m+j}\\ 2m+j\end{array}\right)= \frac{2m+j}{3^{a_{2m+j}+1}}+ \frac{4m+2j}{3^{a_{2m+j}+1}}-\frac{2m+j}{3^{a_{2m+j}}}=0.$ The first term in the expression for $R\left(\mathbf a^{\prime}\right)-R(\mathbf a)$ is also equal to $0$ because $R\left( \begin{array}{ccccccc}a^{\prime}_{m+2}, &a^{\prime}_{4m+3}, &a^{\prime}_{4m+5}, &a^{\prime}_{4m+7}, &a^{\prime}_{4m+9}, &a^{\prime}_{4m+11}, &a^{\prime}_{4m+13}\\ m+2, & 4m+3,& 4m+5,& 4m+7,& 4m+9,& 4m+11,& 4m+13\end{array}\right)-R\left(\begin{array}{c}a_{m+2}\\ m+2\end{array}\right)=$ $=\frac{m+2}{3^{a_{m+2}+2}}+\sum_{j=1}^6 \frac{4m+2j+1}{3^{a_{m+2}+3}}-\frac{m+2}{3^{a_{m+2}}}=0.$ Thus $R\left(\mathbf a^{\prime}\right)=0$ and the statement holds for $4m+13$. It remains to verify that there are sequences of lengths $1$, $5$, $9$, $13$, and $17$. One way to choose these sequences is: $(1), \;\;\; (2,1,3,4,4), \;\;\; (2,3,3,3,3,4,4,4,4),\;\;\; (2,3,3,4,4,4,5,4,4,5,4,5,5),$ $(3,2,2,4,4,5,5,6,5,6,6,6,6,6,6,6,5).$

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