0730 FYH

综合练习1

Problem 1

Suppose that $\{x_n\}$ is a positive real number sequence, prove that: there exist infinitely many positive integers $n$ such that $$x_1+\cdots +x_{n+1}\ge 3.99 x_n$$
Core idea
This is not a difficult problem for there is only a view turn that:
We will use prove reversely, suppose that as $n>n_0,$ $x_1+\cdots +x_{n+1}< 3.99 x_n$

• Consider Sum: $S_n=x_1+\cdots+x_n$
  It is easy to consider sum first, and if it is just = we gonna use the characteristic polynomial to determine the sum directly, but now it is useless.
• Consider this $$a_n=\frac{x_1+\cdots +x_{n-1}}{x_n}$$ We have that $$\frac{x_{n+1}}{x_n}= \left( \frac{x_{n+1}}{x_1+\cdots +x_n} \right)\left(\frac{x_1+\cdots +x_n}{x_n} \right)=\frac{a_n+1}{a_{n+1}}$$ So from the equality, we know that $(a_n+1)(1+\frac{1}{a_{n+1}})<3.99$. It is known that $(1+a)(1+\frac{1}{a})\ge 4$
  We multiple them all that $3.99^{n-n_0+1}>(1+a_{n_0})\cdots (1+1/a_{n+1})\ge 4^{n-n_0}(1+a_{n_0})(1+1/a_{n+1})$. And thus we know that here comes the contradiction: that right hand side after division can be vanished, but right hiand side still >4.
  To Summary: while it is not some really hard technique, just the “magic” algebra form that matters, and some basic inequality.

Problem 2

Suppose $m$ is a positive integer, strictly increasing positive integer sequence $\{a_n\}$, such that $a_{n+1}-a_n<m$ is followed by all n. Prove that: there exist pairwise different positive integers $p,q,r,s$, such that $a_pa_q=a_ra_s$.
Find the density.
Suppose $M= \max(a_1,m)$ for convenience,
then we have that $a_n\le nM$, so $a_pa_q\le pqM^2\le AM^2\Leftrightarrow pq\le A$. What we do is to count $\#(m,n):mn\le A$. For which, we fix $m=k<n$, then $n=k+1,\cdots,\lfloor\frac{A}{k}\rfloor$.
So the num is $$\sum_{k=1}^{\lfloor\sqrt{A}\rfloor}(\lfloor\frac{A}{k}\rfloor-k)\ge \sum_{k=1}^{\lfloor\sqrt{A}\rfloor}(\frac{A}{k}-k-1)\ge A(\ln\lfloor\sqrt A\rfloor) - \frac{\lfloor\sqrt A\rfloor^2+\lfloor\sqrt A\rfloor}{2}-\lfloor\sqrt A\rfloor >A$$ which is easy to prove as $A$ big enough.
To Summarize
Actually, dealing with the product, what we do is to find the density, is somewhere really interesting and seems quite useful. Not that hard to keep further, but hard on the inspire.

Problem 3

We call $S$ as an Aristotle Set such that, if for every $x<y<z\in S$ we all have $\frac{z-x}{y}\in S$, find all positive integer $n\ge 4$, such that there exist an $n$ elements Aristotle Set. (To be expanded)
Step 1
We can understand that $n$ won’t be many. So we will try to estimate a basic upper bound of $n$.
here is the key:
$$\frac{a_n-a_{n-2}}{a_{n-1}}<\frac{a_n-a_{n-3}}{a_{n-1}}<\cdots<\frac{a_n-a_1}{a_{n-1}}<\frac{a_{n-1}-a_1}{a_2}<\cdots<\frac{a_3-a_1}{a_2}$$ here are totally $n-2+n-3\le n\Rightarrow n\le 5$
Step 2: Talk about $n=5$:
We can construct one but
Step 3: Talk about n=4: which is possible.
0.51,0.7,1,1.21

Problem 4

Given an integer $n \ge 100$, find the smallest positive real number $r$ such that for any complex number $c$, there exist complex numbers $a_1, a_2, \dots, a_n$ satisfying all of the following conditions:
$|a_1| \ge |a_2| \ge \dots \ge |a_n|$
$|a_2| \le r$
$a_1 + a_2 + \dots + a_n = c$
$a_1 a_2 \dots a_n = 1$
Ideas:
We will first hold the feeling of number that if $c$ is really large, then it is easy to achieve just make $a_1$ close to c.
So we first let $c=0$.
$|a_1|\le |a_2|+\cdots +|a_n|\le (n-1)r$
So we know that $1\le (n-1)r^n\Rightarrow r\ge \frac{1}{\sqrt [n]{n-1}}$
So we guess that the answer is so.
Prove part:

1. stretch, r to 1
2. take the construction as $b_2=\cdots=b_k, b_{k+1}=\cdots =b_{n-1}=1/b_2$ (by even and odd talk)
3. We talk about the existence of $b_1$ and $b_n$

Problem 5

Prove that for any positive integer $N$, there exists positive integers a, b such that $b>a>N$, such that
$a-1\mid b-1$ and for all positive integer n: $a^n-1\not \mid b^n-1$.
Inspire from this problem:
prove that if $m,n$ are odd numbers, then $2^m-1\not\mid 3^n-1$.
The proof is by: if p is their factors, then we have that 2,3 are the quadratic residue of p, so we get that $p\equiv 1,-1\pmod{12}$(which is all prime factors). However, $2^m-1\equiv 7\pmod{12}$ as $m>1$.
Back to this question:
We want to show that n matters in factor 2, we construct that $v_2(a)=1$, and $3\mid b$. So easily, we take that $a=12k+2,b=3l^2$.
talk n odd and even

综合练习二

Problem 1

Some of the subsets in $\{1,2,\cdots,2026\}$ satisfied that any of its two subsets, if their intersection is not empty, then their intersection forms an arithmetic sequence (1 element is also allowed). Find the max of its subsets.

Problem 2

Given positive integer $n$, we call $f(x_1,\cdots,x_n)\in\mathbb{R}^n[x]$ is good, if it can be written in the form of $f=x_1^2f_1+\cdots +x_n^2f_n$,where $f_1,\cdots,f_n$ are all polynomials.Prove that; if $\deg f<\frac{d}{2}$, and $(x_1+\cdots+x_n)f$ is good, then $f$ itself is also good
In this problem, I will provide 2 solutions here and trying to find the connections between each.

METHOD 1

Definitely, we will start with the condition of $f$ is bad.
We can find that, if terms in $(x_1+\cdots+x_n)f$ have at least power 2 at one term, we can just ignore it, so we only care about terms like $x_1x_2\cdots x_k$ And we can see that as the product occurs, “bad” terms will be vanished. We can hypothesis that $f$ is an homogeneous polynomial. So we define like this:
$k:=\deg f$, and $S_r:=${choose $r$ terms in $x_1,\cdots,x_k$ and $(k-r+1)$ terms in $x_{k+1},\cdots,x_n$} then product them all.So $S_r$ are terms in good polynomials.
We know that coefficients of $S_r$ is $0$.
Similar definition to …

My BAD: I can’t find out the final part in this prove, if you have any ideas, feel free to contact me! (at any platform)

METHOD 2

Part 1:
This is where I learned from AI(LOL)
For Convenience, we still only consider when only one term $\deg g=k$, which $g=\prod_{i\in I}x_i$
We can make our conversation environment into the quotient ring: $\mathbb{R}[x_1,\cdots,x_n]/(x_1^2,\cdots,x_n^2)$ (this is called the square-free algebra.) which mean that every “Good” polynomial is 0 here.Then we define 2 types of operator, we can detect the reason of doing this later.
Lefschetz operator: $L(g)=(x_1+\cdots+x_n)$ (While it seems there is a little distinct with what stated on Google, I will stated here later)
Partial Derivative: $D=\frac{\partial}{\partial x_1}+\cdots+\frac{\partial}{\partial x_n}$
Part 2:
We compute $(DL-LD)g$ (why?)$$DLg-LDg=D(\sum_{j\not \in I}x_jg)-L(\sum_{k\in I}\frac{g}{x_k})=(n-k)g+kg=(n-2k)g$$
Actually, $(DL-LD)=[D,L]$ is the exterior product of D,L, so this equals (the way of adding(L))-(The way of deleting(D))=n-2k.

Part 3: Introduce a new inner product.
$$\langle x_I,x_J\rangle= \begin{cases} 1,&I=J,\\ 0,&I\neq J. \end{cases}$$
We want to look for $L*$ i.e. $\langle Lx_I,x_J\rangle=\langle x_I,Dx_J\rangle$, which is easy to check.

Part 4: prove the final part
$$Lf=0\Rightarrow -LDf=(n-2k)f\Rightarrow\langle -LDf,f\rangle=(n-2k)\langle f,f\rangle$$ so $$-||Df||_2 = (n-2d)||f||_2$$ but left is negative and right is positive, contradiction!

Problem 8

Find all function $f:\mathbb{R}\to\mathbb{R}$, so that $\forall x,y$, we all have $$f([f(x)+f(y)]^2)=(x+y)(f(x)+f(y))$$

0731 FYH

综合练习三

Problem 3

We have a map $f:\mathbb C\to\mathbb C$ such that:

1. There exists $c>0$ with $|f(z)-z|<|z|^{0.99}+c,\qquad \forall z\in\mathbb C$.
2. Whenever four points form a convex quadrilateral, their images also form a convex quadrilateral.
   Prove: $\boxed{\exists a\in\mathbb C,\quad f(z)=z+a}$

Mainly idea:
We first claim 3 points f(A), f(B), f(C) form a triangle then f(A), f(B) form 3 lines and cut the plane into 6 parts outwardly. And 3 parts can’t put f(D) (OR it will be a concave quadrilateral)

So what we want to prove is $$AB \parallel f(A)f(B)$$If not, around $\triangle f(A)f(B)f(C)$ one end of line AB will get into one of the “forbidden part” We claim D as really close to line AB. Because as $f(D)$ is inside a circle centered at D, radius $|D|^{0.99}<|D|$ But actually, the distance from D to the “forbidden part” border is linear increase, so as D far enough, it will must be totally controlled into the “forbidden part”

So $AB \parallel f(A)f(B)$. Then every 2 points will be projected into a parallel line. This must be a linear projection. (WHY???)

Problem 5

If ABCD is a regular tetrahedron, M N is 2 points in the space, prove that $$AM\cdot AN+BM\cdot BN+CM\cdot CN\ge DM\cdot DN$$
This can be proved by Ptolemy’s Theorem, but can also be completed by 3D Inversion(Which I am pretty interested in)

Problem 6

A test paper consists of $n$ true-or-false questions, where the answers are only True or False. The hardworking Xiaoming does not know how to solve any of the questions, so he decides to submit $m$ fully completed answer sheets that he prepared beforehand to the teacher at the same time. After grading every answer sheet, the teacher informs Xiaoming of the total score of each valid answer sheet (1 point for each correct question, 0 points for an incorrect question), but does not tell him which specific questions were answered correctly. Incomplete answer sheets (i.e., answer sheets with answers to fewer than $n$ questions) are considered invalid answer sheets and are not graded. Xiaoming can design the answers for each answer sheet in advance so that, regardless of the circumstances, he can deduce the correct answer to every question based on the score of each answer sheet.

Xiaoming can minimize $m$ through carefully designed answer sheets. Let this minimum $m$ be denoted as $f(n)$. Prove: There exist positive real numbers $c_1, c_2$ such that for any integer $n > 1$,
$$c_1 \frac{n}{\ln n} \le f(n) \le c_2 \frac{n}{\ln n}.$$

Ideas:
Lower Bound
The answer has $2^n$ types
Actually, we can so some basic setups here. That the feedback has 0,1,…, n, totally n+1 kinds. so we know that $(n+1)^m\ge2^n$ it means that Xiaoming need to make enough answer sheets to definitely cover all kinds of answer properties. To be more exact: left hand side means that hand in m papers, there are at most $(n+1)^m$ kinds of feedback, so if m papers are enough to guess the whole answer, so much situations must cover all kinds of answers. So We can derive that $$m\ln (n+1)\ge n\ln 2\Rightarrow m> \frac{\ln2}{2}\frac{n}{\ln n}$$ Upper Bound
We need to find an upper bound, which means we need to design a strategy for Xiaoming to help him finally guess the problems.

We’ll do some basic transformation of this problem to an algebra one.
The Check system can be turned into an inner product. So define $S=\{1,-1\}^n$ the answer $\vec \alpha\in S$, and every paper $\vec\lambda_i\in S$ So the feedback can be written as $$f(\lambda_i)=\frac{n+\lambda_i\cdot\alpha}{2}$$then, the question has transformed to this:

Find the least m where $\#L=m,L\subset S$ We can construct such $L$ so that $\forall \vec\alpha\in S$, $\exists \vec\lambda \in L$ such that $f(\lambda)=n$.

To write it in a more convenient way, we will describe in linear algebra like this: We construct a matrix $A_{m\times n}$, that contains each kind of Xiaomings trial paper work. And the feedback set will be $A\vec x$ so if $\forall \vec x\not=\vec y$ and $Ax\not = Ay$ then Xiaoming can really find out the answer here.

The logic is here: the equation is $Ax=b$, teacher gives xiaoming b, What xiaoming doing is to decode $x$. But Xiaoming can design a fixed A, and if we can make sure that $\ker A\cap \{1,0,-1\}^n-0=\emptyset$, then for any $\#\{x|Ax=b\}\le 1$, because $A(x-y)\not =0$ if $x\not = y$. Moreover, $x-y = \{1,0,-1\}^n$

This part quite stuck me for a long while, but after understanding the logic words, I can finally understand it.

So the problem turned to:

$M(n)=\min\left\{m:\exists A\in\{-1,1\}^{m\times n} \text{ such that } \ker A\cap\bigl(\{-1,0,1\}^n\setminus\{0\}\bigr)=\varnothing \right\}$
We want an upper bound for $M(n)\le c\frac{n}{\ln n}$

Fix a nonzero vector $d\in\{-1,0,1\}^n$, and let $k$ be the number of its nonzero coordinates.
What we want to do is to Compute the following probability, and if it <1, then we can say that there exists a solution of A. Then we use this as the condition to derive the upper bound of $M(n)$.

Because there are $\binom nk2^k$ such $x$ with a fixed $k$, the total probability is

For one random row $r$ of $A$.
$r\cdot d$ is the sum of $k$ independent random signs. Therefore (Which is we choose k numbers in {1,-1}, and a random row in A did so) To add: $k\le n$ because there may have 0 element in $x$.

To explain more: The whole sample space have $2^m$ in total, and we choose $k$ 1s, so there are $\binom{k}{k/2}$ possibilities to make the inner product 0.

Using the standard central-binomial estimate,

(By Using Stirling’s formula, $r!\sim \sqrt{2\pi r}\left(\frac re\right)^r$)

So

The $m$ rows are independent, so

Split $k$ into

• small supports $k\le n/(\ln n)^2$; (This is somewhere after we get everything and decide afterwards)
• large supports $k>n/(\ln n)^2$
  And the logic is like when if we set $m\le c\frac{n}{\ln n}$ we can always construct such A

For small $k$, we use the simpler bound

$\Pr(r\cdot d=0)\le\frac12\Rightarrow\Pr(Ad=0)\le 2^{-m}$
The number of such vectors is at most

How to Prove? I am so dumb!!!!! just $\binom nk\le n^k$

So then

For large $k$, $k>\frac n{(\ln n)^2}$,

How to Prove? just $\binom nk\le n^k$

Thus, for sufficiently large $n$,

Later we choose $m=C\frac{n}{\ln n}$, then

So as we let $C=2$, we can get $\Pr_{\text{k small}}\to 0$ as $n$ grow.

For large $k$, $k>\frac n{(\ln n)^2}$,

So

We choose $m=C\frac{n}{\ln n}$

Then

As n grow larger, we know that $\frac{\ln\ln n}{\ln n}\to 0$, so we just need to let $C=10000000000000$ lol, then $-\frac{C}2+\ln 3<0$ So $\Pr_{\text{k large}}\to 0$

So $\Pr<1$ Finished

Problem 8

Let $n$ be a positive integer. A circle $\Gamma$ has $n$ inscribed triangles whose vertices are all distinct. Prove: It is possible to select $2n$ vertices from these $3n$ vertices, and then have $n$ boys and $n$ girls stand at these chosen vertices, such that:

(1) At the three vertices of each triangle, there stands exactly one boy and one girl;

(2) Considering the relative positions of the children along the circumference of the circle, both neighbors of every child are children of the opposite sex.

综合练习四

Problem 1

Given a positive integer $n$, find the maximum integer $m$ satisfying the following condition:
There exists an irrational number $\alpha$, pairwise distinct rational numbers $a_1, a_2, \dots, a_m$, and a degree-$n$ polynomial $f(x)$ with rational coefficients such that $f(\alpha + a_i)$ is rational for all $i = 1, 2, \dots, m$.

Problem 3

Starting from a triplet of non-negative integers $(a, b, c)$, the following operation is allowed: select two numbers from the triplet, let them be $x$ and $y$, and change one of them to $x + y$ or $|x - y|$. For example: $(3, 5, 7) \rightarrow (3, 5, 4)$ is considered a single operation.

Prove: There exists a constant $r > 0$ such that for any positive integers $a, b, c, n$, if $a, b, c$ are all less than $2^n$, then one can perform no more than $rn$ operations on $(a, b, c)$ to make one of the numbers in the triplet equal to $0$.

Problem 5

Let integer $n \ge 2$. Let $S_1, S_2, \dots, S_{2^n}$ be $2^n$ subsets of the set $A = \{1, 2, 3, \dots, 2^{n+1}\}$ satisfying the following condition: there do not exist indices $a < b$ and three elements $x, y, z \in A$ such that $x < y < z$, where $x, z \in S_a$ and $y, z \in S_b$.

Prove: At least one of $S_1, S_2, \dots, S_{2^n}$ contains no more than $4n$ elements.

Problem 7

Let $n, d > 1$ be given positive integers. On a regular $n$-gon paperboard, there is an identical button at each vertex. Player A and Player B play the following game: Before the game begins, Player B may press any button any number of times, and then the game starts. In each round, Player A can look at the paperboard and choose to press certain buttons, pressing each chosen button any number of times. Then, Player A hands the paperboard to Player B. Player B may rotate the paperboard without being seen by Player A (but cannot flip it over). This process is called one round of operations. If after a certain round of operations, the total number of times each button has been pressed is a multiple of $d$, then Player A wins.

Prove: Player A can guarantee a win in a finite number of steps, independent of Player B’s operations, if and only if $n$ and $d$ are powers of the same prime number.

Problem 8

Let $S$ be a set consisting of 2025 distinct non-zero real numbers. For each odd-element subset of $S$, sum its elements and take the absolute value of the sum, then sum all these resulting absolute values together, denoted as $O$; for each even-element subset of $S$, sum its elements and take the absolute value of the sum, then sum all these resulting absolute values together, denoted as $E$. That is:

Prove:

0801 HLB

Idea 1: Plane sections of a tetrahedron

I found an interesting problem: can we compute the area or perimeter of the cross-section cut from a regular tetrahedron by an arbitrary plane?

Conversely, can the direction and position of the plane be determined from the area and perimeter of its cross-section?

Idea 2: Inversions in orthogonal circles

Two inversions whose defining circles are orthogonal commute. This can also be understood through a basic application of Miquel’s theorem.

While studying this idea, I encountered the following geometry problem:

In $\triangle ABC$, let $D\in AB$ and $E\in AC$. The circle $(ADE)$ is tangent to $BE$ at $F$ and to $CD$ at $G$. Let $H$ be the second intersection of circles $(BDG)$ and $(CDF)$. Is there a relation between $FG$ and $d(H,BC)$?

Idea 3: Great-circle arrangement (Problem 6)

A number of great circles are drawn on the surface of a sphere, dividing the spherical surface into a number of triangular regions and quadrilateral regions. It is known that no three great circles pass through the same point, and there is at least one quadrilateral region. Prove that there are exactly 8 triangular regions and 6 quadrilateral regions.

Let there be $m$ great circles, $T$ triangular regions, and $Q$ quadrilateral regions. Since each pair of circles meets at two antipodal points,

Every circle is divided into $2(m-1)$ arcs, so

Euler’s formula $V-E+F=2$ gives

Counting incidences between edges and regions gives

Solving these equations yields

Thus the number of triangular regions is always $8$. If $m=4$, then $Q=6$. As written, the problem needs an additional assumption forcing $m=4$ in order to conclude that there are exactly six quadrilateral regions.

Idea 4: A cubic congruence (Problem 12)

Let $p$ be a prime satisfying

Prove that there exists a positive integer $n$ such that

Because $9\mid p-1$, the cyclic group $\mathbb F_p^\times$ contains an element $\zeta$ of order $9$. Define

Using $(x+x^{-1})^3=x^3+x^{-3}+3(x+x^{-1})$, we obtain

Since $\zeta$ has order $9$, the element $\zeta^3$ is a primitive cube root of unity. Therefore

Also, $\zeta^{-3}=\zeta^6$, so

Consequently,

Finally, $n\ne0$: otherwise $\zeta^2=-1$, which is impossible for an element of order $9$. Choosing the representative $n\in\{1,2,\dots,p-1\}$ gives the required positive integer. Hence

0802

Problem 1

Let $n\ge 2$, and let $a_1,a_2,\dots,a_n$ be nonzero real numbers satisfying

Prove that

1. Introduce the given integers

Use cyclic notation

For every $i=1,2,\dots,n$, define

By assumption,

The defining equation can be rearranged as

or equivalently,

This is a fractional-linear, or Möbius, transformation.

2. Encode each step by a matrix

For every $i$, define

Because $b_i\in\mathbb Z$, every $M_i$ has integer entries. Moreover,

Therefore,

Now define the column vector

Then

Since

we obtain

Factor out $1/a_{i+1}$:

Thus

3. Go once around the cycle

Apply the preceding identity repeatedly:

and, continuing inductively,

Because the indices are cyclic,

and

Let

It follows that

Define the total matrix

Then

Since $v_1\ne 0$, this means that

4. Use the determinant

Since every $M_i$ has determinant $1$,

For a $2\times2$ matrix, the product of its two eigenvalues equals its determinant.

One eigenvalue of $M$ is $1/P$. If the other eigenvalue is denoted by $\lambda$, then

Therefore,

Hence the two eigenvalues of $M$ are

Their sum equals the trace of $M$, so

But $M$ is a product of integer matrices, so $M$ itself has integer entries. Consequently,

Therefore,

What Is Hidden Inside the Problem?

1. Möbius transformations

The equation

defines a Möbius transformation

It is represented by the matrix

Indeed, the general matrix

acts projectively by

For $M_i$, this gives

Thus the original cyclic system is an iteration of integer Möbius transformations.

2. Monodromy

After going once around the cycle, we obtain the composition

Because

the number $a_1$ is a fixed point of $T$:

The matrix

records the effect of going once around the cycle. It is called a monodromy matrix or a transfer matrix.

The product

appears as an eigenvalue of this monodromy matrix.

3. Why the expression $P+P^{-1}$ appears

Any matrix

has determinant $1$. Therefore, if one eigenvalue is $\lambda$, the other must be $\lambda^{-1}$.

Hence its characteristic polynomial has the form

The sum

is exactly the trace:

In this problem,

Therefore,

So the target expression is not an accidental choice: it is the natural trace invariant of a determinant-one matrix.

4. A stronger conclusion

Let

Since the eigenvalues of $M$ are $P$ and $P^{-1}$, the characteristic polynomial of $M$ is

Therefore $P$ satisfies

Thus we obtain the stronger statement

In particular, $P$ is a quadratic algebraic integer, unless it is already rational.

Its algebraic conjugate is

and its algebraic norm is

Therefore $P$ is an algebraic unit.

5. Classification when the $a_i$ are real

Because $P\in\mathbb R\setminus\{0\}$, the quantity

satisfies

Indeed, if $P>0$, then

while if $P<0$, then

Since $k\in\mathbb Z$, necessarily

Moreover,

Thus the product $P=a_1a_2\cdots a_n$ is highly restricted: it must be a real quadratic unit of norm $1$, or one of the special values $P=\pm1$.

Main structural principle

The proof can be summarized as

and then

The central hidden idea is therefore:

Whenever a cyclic recurrence can be represented by determinant-one integer matrices, products accumulated around the cycle often become eigenvalues, and expressions of the form

$$\lambda+\lambda^{-1}$$

become integer traces.

0803

1: 东南赛P3

 In $△𝐴𝐵𝐶$, $𝐴𝐵 < 𝐴𝐶$. Let 𝑀 be the midpoint of side 𝐵𝐶, and let 𝐷 be the intersection of the angle bisector of $∠𝐵𝐴𝐶$ with 𝐵𝐶. Let 𝑃 be the reflection of 𝐷 over 𝐴, and let 𝑄 be a point on the circumcircle of $△𝑃 𝐵𝐶$ such that $𝑀 𝑄 ∥ 𝐴𝐷$, with 𝑃 and 𝑄 on opposite sides of 𝐵𝐶. Let 𝑅 be the intersection of 𝑃𝑄 and 𝐴𝑀 . Prove that $∠𝑅𝐵𝑄 = ∠𝑅𝐶𝑄$.

This problem really make me stuck at the very first step: how to use that AD:bisector of $∠𝐵𝐴𝐶$.
While the problem construct a $F$, so that $ABFC$ are concyclic, and $FB=FC$

0806

Problems today

I did a piece of paper

While, in the exam system, how to give answers quickly and correctly. We have 1h20min, 8 short problems just ask for answer, and 3 problems that require process. So the tricks should be really steady and trained.

To summarize here, the score board is here

Problem 2

The number of integer solutions to the inequality $\log_6(1 + \sqrt{x}) > \log_{25} x$ is $\underline{\qquad}$.

$x=25^t$, then $1+5^t>6^t$, there is only $t<1$, so $x=1,\cdots 24$, totally 24 solutions.

Problem 4

In tetrahedron $ABCD$, one edge has length 3, and the remaining five edges all have length 2. The radius of its circumscribed sphere is $\underline{\qquad}$.

In such questions, don’t set coordinates directly, discover the geometry property first. We set triangle of (2,2,3) as base, then $A-BCD$, The center $O$ of plain triangle $\triangle BCD$ and $A$ are in the same vertical line.

Just compute this carefully. The answer is $\sqrt21/3$ .

Problem 5

A fair six-sided die, with faces labeled 1, 2, 3, 4, 5, 6, is thrown three times. The probability that the three numbers appearing on the top face can form the side lengths of a triangle whose perimeter is divisible by 3 is $\underline\qquad$.

Where I wrong here is I count less and lost the times that as (a,b,c), there are 6 kinds of order to get access to this set.

Then $(i,i,i): 6\times 1$, $(i,i,j):3\times 3$, $(i,j,k): 3\times 6$, and the result is $\frac{6+9+18}{6\times 6\times 6}=11/72$

Problem 7

Let point $P$ move on the right branch of the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ excluding the vertices, and let $E, F$ be its left and right foci, respectively. Point $A$ is the ex-center of $\triangle PEF$ within $\angle PEF$. The equation of the locus of point $A$ is $\underline\qquad$.

At first, I think it is P-excenter, so my wrong answer is $x=-4$ Carefully check the problem!!!!

It seems that even if I read it rightly, I still have no idea on it. I can find that $I_E(PF+4,r)$ buw what to do then?

Use the property of Appolonius Circle.
We suppose A is the ex-center, $B=PA\cap x$ axis. Then $$\frac{BA}{AP}=\frac{BF}{FP}=\frac{BE}{EP}=\frac{BE-BF}{EP-FP}=\frac{EF}{2a}=\frac{10}{8}=\frac{5}{4}$$But we need the coordinates of $B(x_B,0)$, $$\frac54=\frac{BF}{PF}=\frac{x_B-5}{ex_0-a}\Rightarrow\frac{x_0}{x_B}=\frac{16}{25}$$ Use the steady ratio at A, solves it