Category Archives: Elementary Math

Elementary Math

李善兰 恒等式

Education, Elementary Math, Uncategorized Leave a comment

这是一个非常经典的组合恒等式。李善兰恒等式（Li Shanlan’s Identity），也称为李善兰卷积，是中国清代数学家李善兰在其著作《则古昔斋算学》中提出并证明的一个重要恒等式。

它在现代组合数学中是一个标准结果，是超几何函数恒等式的一个特例，通常被称为 Chu-Vandermonde 恒等式 或 Vandermonde 卷积 的一种形式。李善兰的工作早于或独立于西方数学家，因此在中国数学史上具有重要意义。

  1. 恒等式的基本形式

李善兰恒等式可以表述为：

\displaystyle \sum_{k=0}^{m} \binom{r}{k} \binom{s}{m-k} = \binom{r+s}{m}

其中：

· r, s 是任意复数（通常为实数或整数）。
· m 是一个非负整数。
· \binom{x}{k} = \frac{x(x-1)\cdots(x-k+1)}{k!} 是广义二项式系数。

当 r 和 s 是非负整数时，这个恒等式有非常直观的组合解释。

  1. 组合解释（当 r, s \in \mathbb{N} 时）

假设你有两个集合：

· 集合 A 有 r 个元素。
· 集合 B 有 s 个元素，且 A 与 B 不相交。

问题： 从并集 A \cup B （共 r+s 个元素）中，选出 m 个元素，总共有多少种方法？

答案显然是： \binom{r+s}{m} 。

另一种数法： 我们可以按“从 A 中选多少个”来分类：

· 如果从 A 中选 k 个元素，那么从 B 中就必须选 m-k 个元素。
· 从 A 中选 k 个有 \binom{r}{k} 种方法，从 B 中选 m-k 个有 \binom{s}{m-k} 种方法。
· 对 k 从 0 到 m 求和，得到总方法数为 \displaystyle \sum_{k=0}^{m} \binom{r}{k} \binom{s}{m-k} 。

因为两种方法数的是同一个东西，所以等式成立。

  1. 与 Vandermonde 卷积的关系

标准的 Vandermonde 卷积 通常写作：

\displaystyle \sum_{k=0}^{m} \binom{r}{k} \binom{s}{m-k} = \binom{r+s}{m}

这与李善兰恒等式完全相同。因此，李善兰恒等式就是 Vandermonde 卷积。

· 历史背景：Alexandre-Théophile Vandermonde（1772年）和李善兰（19世纪中叶）都独立发现了这个公式。
· 更一般的形式：超几何函数形式为 \displaystyle {}_2F_1(-r, -s; -r-s; 1) = \frac{(r+s)!}{r!s!} ，经过适当变形可得到上述恒等式。

  1. 一个特例与应用

一个著名的特例是令 r = s = n ， m = n ：

\displaystyle ,{\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}}

这个结果在组合学中非常常见，它说明在 2n 个物品中选 n 个的方法数，等于从两组 n 个物品中各选 k 个（对所有 k 求和）的方法数平方和。

应用领域：

  1. 组合数学：计数问题、恒等式证明。
  2. 概率论：超几何分布的概率归一化证明。
  3. 特殊函数：超几何函数的特殊值。
  4. 数论：二项式系数的同余性质研究。
  1. 证明思路

代数证明（使用生成函数）：
考虑

\displaystyle (1+x)^r (1+x)^s = (1+x)^{r+s}。

展开左边：

\displaystyle \left( \sum_{i=0}^{r} \binom{r}{i} x^i \right) \left( \sum_{j=0}^{s} \binom{s}{j} x^j \right) = \sum_{m=0}^{r+s} \left( \sum_{k=0}^{m} \binom{r}{k} \binom{s}{m-k} \right) x^m

右边展开：

\displaystyle (1+x)^{r+s} = \sum_{m=0}^{r+s} \binom{r+s}{m} x^m

比较两边 x^m 的系数，即得恒等式。

组合证明：如上文第2点所述。

  1. 李善兰的贡献

李善兰（1811-1882）是中国近代数学的先驱。除了这个恒等式，他在数论（“李善兰素数定理”，即对伯特兰猜想的早期研究）、级数求和（“李善兰迭积术”，与斯特林数相关）以及微积分学翻译（与伟烈亚力合译《代微积拾级》）方面都有重要贡献。

李善兰恒等式是他独立研究成果的一个代表，展示了中国数学家在这一领域达到的深度。

Linear Derivative Eigen Value

Elementary Math, Modern Math Leave a comment
@donovinrussell

Linear transformations are one of the coolest things to study in math. Sure they’re somewhat restrictive but they have so many neat and generalizable properties. In this video I show off how you can use a matrix to represent taking a derivative of a second degree polynomial. #math #linearalgebra #matricies #lineartransformations #derivative

♬ Siberian Khatru (2025 Steven Wilson Edit) – Yes

胖瘦定理

Elementary Math Leave a comment

Make BP = CQ,PQ common=> BQ=CPAB=AC （等腰）Angles ∠B=∠CCongruence (SAS):ΔABQ Ξ ΔACP=> AP = AQ

八年级几何（难题）

Solution: x = 60°

👍 The key is 胖瘦定理。
Also “ai” guess:
None length given to find x°, then must be 3 cases of special angles :

  • 60° (regular△ )
  • 45° or 30° (right angle △ + 等腰).
    => Try to build a Regular△ CHD with CH 补助线

为何最后 他说 “第一步补助线 DG ” 是多余的？

https://vt.tiktok.com/ZSUusAeKc/

@corneliusgoh110

#中国数学

♬ original sound – Cornelius – Cornelius

Eigen-Value = 1

Chinese, Elementary Math Leave a comment

Many profs teach eigen-vector/-value by definition with
Av = λv ,
only 1 American Prof Paul Halmos revealed λ = 1 could be Very useful too.

例子：
孙悟空(v)被二郎神的天犬追咬，老孙变(A)成一间庙（w）， 但庙前的旗却在庙后(猴尾巴)。天犬被骗， 但二郎神猜出庙是🐒变！

Av(=w )= λv
λ = 1
w= 1v = v🐒

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数学老师分流

Education, Elementary Math Leave a comment

Trump is a “1st 流 Math teacher” 😄

Math Teachers 分4流：
4流：解题 但不解释
3流：文字解释清楚，不套公式
2流：“数学”导游，引发数学兴趣
1流： 学生motivated， 爱他的数学老师
“超”一流： 培养学生成为数学家。

[Note]
还有一个种 “不入流”的数学老师, 恰好是“反1流”：学生害怕凶巴巴的数学老师， 专门fail 人 by “Killer Math subject”.

中学/JC /大学 碰到许多这种数学老师/Prof， 就很倒霉。

几何5模型： 燕尾， 鸟头

Elementary Math Leave a comment
https://mp.weixin.qq.com/s?search_click_id=17933235565938826214-1753952940188-3957477798&__biz=MzUzMDg5NjkxNw==&mid=2247542887&idx=1&sn=ef43230951a5aa8a2516f798e63464ba&chksm=fb0f991d201f83c32e8ec0dd6b6e72f8024af5719a9db5f813ed73a9eb51450662adddf3a883&scene=7&subscene=10000&clicktime=1753952940&enterid=1753952940&sessionid=0&ascene=65&fasttmpl_type=0&fasttmpl_fullversion=7842931-en_US-zip&fasttmpl_flag=0&realreporttime=1753952940267#rd

Circle Inversion

Elementary Math, Modern Math Leave a comment

The term “反演变换” in the context of Chinese middle school (中考) mathematics is :

Inversion Transformation (or Geometric Inversion)

Key Details:

  1. Definition:
  • Inversion is a geometric transformation that maps a point ( P ) to a point ( P’ ) such that: OP. OP’ = r^2, where ( O ) is the center of inversion, ( r ) is the radius of inversion, and ( P’ ) lies on the line ( OP ).
  1. Properties:
  • Preserves angles (conformal map).
  • Maps circles/lines to circles/lines.
  • The center ( O ) has no inverse (maps to “infinity”).
  1. Common in Exams:
  • Problems often involve finding the inverse of a point, circle, or line.
  • Applications in solving geometry problems (e.g., tangent circles, orthogonal circles).

Example:

  • Given: A circle with center ( O ) and radius ( r ), and a point ( P ) inside the circle.
  • Inversion: The inverse point ( P’ ) lies outside the circle such that ( OP .OP’ = r^2 ). Note:
    In advanced math, it’s also called “Circle Inversion” or “Möbius Transformation” (when generalized to the complex plane).


Application in China Secondary School  (O-level) :

中考 O-level Geometry: 反演变换
Circle Inversion

https://vt.tiktok.com/ZSShW7eUE/
@corneliusgoh110

#circle inversion#反演变换

♬ original sound – Cornelius – Cornelius

Singapore Math: Internal Transfer (Total Unchanged)

Elementary Math Leave a comment

Note:

This 4th strategy of the 8 Singapore Math Strategies is called
“Internal Transfer with Total unchanged”.

(Total = 100, Peter and Mary internal transfer).

Tricky is by working backward.
Tough for PSLE students unless déjà-vu similar type problems / techniques before.

I tried algebra also, complex equations to solve:

P = M

¾P + 2/7(100-¾P ) = 5/7(100-¾P)

P = 40

Tricky Singapore Math: Assumption Method

Elementary Math Leave a comment

1st Conventional Method: By 1-variable simple Algebra

  1. This is PSLE Strategy #7: “Number x Value”
    Draw a table.
    Pokemon:  28 units

2nd Assumption Method: Assume all 50 units of  P

Grouping D, M as 1 category D&M

Assume all (50 units) are P at 30¢
Group D&M as one category at 90¢ (50+40)
Total cost of 50 xP
=50*30= 1500
Short of 1830-1500 = 330 

=>
330/ 30¢ = 11 units

[Note]
This 30 ¢ is the total price difference of
D-P: 50¢-30=20¢
M-P : 40¢-30=10¢
Total difference = 20+10=30¢ which affects the difference 330¢


Hence 330/30 = 11 units of “D&M” group
Since D and M have same number of  unit,
Hence D=11 units= M
P = 50-11-11 = 28 units

Tuition Center Method:

What’s wrong logic  by Tuition Center  in STEP 3) below:

1) Assume HALF of total 50 = 25 pairs are D&M. So it would cost a total of = 25*0.9 = $22.5
This is 22.5-18.3=$4.2 in excess of the actual expenses.

2) $4.2÷ $0.3= 14 (excess due to P)

3) Half *2 = 50 pairs (WRONG !!!)

=> 14 excess ×2 = 28 P stickers (coincidence = answer )

4)  D&M stickers are 50 – 28 =  22 stickers.

[Note]

Why TC  assumed only 25 D&M (but not 50 D&M?)
BCOS:
25 D&M = 25 D + 25 M = 50 = ALL pieces

3rd Method: Guess and Check – simple logic but lengthy

Analytical Geometry (DeepSeek)

Elementary Math, Uncategorized Leave a comment

Given the corrected labels:

  • A: (a, 0)
  • B: (0, b)
  • C: (0, 0)
  • E: (0, e) — A point on the y-axis (BC)
  • F: (f, 0) — A point on the x-axis (AC)
  • D: (\frac{f}{2}, \frac{e}{2}) — Midpoint of EF
  • P: (\frac{a}{2}, \frac{b}{2}) — Midpoint of AB

Given:

  • EF = 4 (distance between E(0,e) and F(f,0))
  • We need to find the minimum distance |PD|.

Step 1: Express |PD| Mathematically

The distance between P(\frac{a}{2}, \frac{b}{2}) and D(\frac{f}{2}, \frac{e}{2}) is:

|PD| = \sqrt{\left(\frac{a}{2} - \frac{f}{2}\right)^2 + \left(\frac{b}{2} - \frac{e}{2}\right)^2}

\boxed{|PD|= \frac{1}{2} \sqrt{(a - f)^2 + (b - e)^2}}

Step 2: Incorporate Constraint EF = 4

Since EF = 4, the distance between E(0,e) and F(f,0) is:

\sqrt{f^2 + e^2} = 4 \implies e^2 + f^2 = 16.

Step 3: Minimize |PD|

We need to minimize \sqrt{(a - f)^2 + (b - e)^2} under the constraint e^2 + f^2 = 16.

This is equivalent to finding the point (f, e) on the circle e^2 + f^2 = 16 that is closest to (a, b).

The minimal distance occurs when (f, e) lies along the line from (0,0) to (a,b), scaled to the circle:

\boxed{(f, e) = \left(\frac{4a}{\sqrt{a^2 + b^2}}, \frac{4b}{\sqrt{a^2 + b^2}}\right)}   ….[See Note 1]

Thus, the minimal distance is:

\sqrt{(a - f)^2 + (b - e)^2} = \sqrt{a^2 + b^2} - 4.

Therefore: (From Step 1 above)

\boxed{|PD|_{\text{min}} = \frac{1}{2} \left(\sqrt{a^2 + b^2} - 4\right)}

Step 4: Incorporate AB + AC = 12

Assuming AB + AC = 12 (as in the original problem):

  • AB = \sqrt{a^2 + b^2}
  • AC = a (distance from A(a,0) to C(0,0))

Thus:

\sqrt{a^2 + b^2} + a = 12.

Let \sqrt{a^2 + b^2} = 12 - a. Squaring both sides:

a^2 + b^2 = 144 - 24a + a^2 \implies b^2 = 144 - 24a \implies b = \sqrt{144 - 24a}.

Step 5: Substitute into |PD|

Now, \sqrt{a^2 + b^2} = 12 - a, so:

|PD|_{\text{min}} = \frac{1}{2} \left((12 - a) - 4\right) = \frac{1}{2} (8 - a) = 4 - \frac{a}{2}.

Step 6: Apply Constraint BC ≥ AC ≥ 4

From the original problem:

  • BC = \sqrt{0^2 + b^2} = b \geq AC = a \geq 4.

Given b = \sqrt{144 - 24a} \geq a, and a \geq 4.

Solve \sqrt{144 - 24a} \geq a:

Square both sides

(valid since a \geq 4 > 0 ):

144 - 24a \geq a^2 \implies a^2 + 24a - 144 \leq 0.

Solve the quadratic a^2 + 24a - 144 = 0:

a = \frac{-24 \pm \sqrt{576 + 576}}{2} = \frac{-24 \pm \sqrt{1152}}{2} = \frac{-24 \pm 24\sqrt{2}}{2} = -12 \pm 12\sqrt{2}.

The relevant root is a = -12 + 12\sqrt{2} \approx 4.97.

Thus, a \in [4, -12 + 12\sqrt{2}].

Step 7: Find Minimal |PD|

We have |PD|_{\text{min}} = 4 - \frac{a}{2}.

To minimize |PD|, maximize a (since |PD| decreases as a increases).

The maximum a is a = -12 + 12\sqrt{2}.

Thus:

|PD|_{\text{min}} = 4 - \frac{-12 + 12\sqrt{2}}{2} = 4 + 6 - 6\sqrt{2} = 10 - 6\sqrt{2}.

Final Answer

The minimum value of |PD| is:

\boxed{10 - 6\sqrt{2}}

[Note 1]

Parametric Derivation of (f, e) Coordinates

We want to find the point (f, e) on the circle e^2 + f^2 = 16 that minimizes the distance to (a, b).

1. Parametric Representation of the Circle

Any point on the circle can be expressed as:

f(t) = 4\cos t, \quad e(t) = 4\sin t \quad \text{for} \quad t \in [0, 2\pi)

2. Distance Squared Function

The squared distance between (f(t), e(t)) and (a, b) is:

D(t) = (a - 4\cos t)^2 + (b - 4\sin t)^2

Expanding:

D(t) = a^2 - 8a\cos t + 16\cos^2 t + b^2 - 8b\sin t + 16\sin^2 t

Simplifying using \cos^2 t + \sin^2 t = 1:

D(t) = (a^2 + b^2 + 16) - 8(a\cos t + b\sin t)

3. Minimizing the Distance

To minimize D(t), we maximize a\cos t + b\sin t.

The maximum value is:

\max(a\cos t + b\sin t) = \sqrt{a^2 + b^2}

achieved when:

\cos t = \frac{a}{\sqrt{a^2 + b^2}}, \quad \sin t = \frac{b}{\sqrt{a^2 + b^2}}

4. Optimal Coordinates (f, e)

Substituting the optimal t:

f = 4\cos t = \frac{4a}{\sqrt{a^2 + b^2}}, \quad e = 4\sin t = \frac{4b}{\sqrt{a^2 + b^2}}

5. Verification

Verify (f, e) lies on the circle:

f^2 + e^2 = \left(\frac{4a}{\sqrt{a^2 + b^2}}\right)^2 + \left(\frac{4b}{\sqrt{a^2 + b^2}}\right)^2 = \frac{16a^2 + 16b^2}{a^2 + b^2} = 16

6. Minimal Distance Calculation

The minimal distance is:

\boxed{\sqrt{(a - f)^2 + (b - e)^2} = \sqrt{a^2 + b^2} - 4} …[Note 2]

Thus:

|PD|_{\text{min}} = \frac{1}{2}\left(\sqrt{a^2 + b^2} - 4\right)

Final Result

The coordinates that minimize |PD| are:

\boxed{(f, e) = \left( \frac{4a}{\sqrt{a^2 + b^2}}, \frac{4b}{\sqrt{a^2 + b^2}} \right)}

This parametric derivation confirms the geometric solution, showing explicitly how the angle parameter t leads to the optimal point on the circle.

[Note 2]

Let’s verify the equation:

\boxed{\sqrt{(a - f)^2 + (b - e)^2} = \sqrt{a^2 + b^2} - 4}

Where we have:

f = \frac{4a}{\sqrt{a^2 + b^2}}, \quad e = \frac{4b}{\sqrt{a^2 + b^2}}

Step-by-Step Verification:

  1. Substitute f and e into the left side: \sqrt{\left(a - \frac{4a}{\sqrt{a^2 + b^2}}\right)^2 + \left(b - \frac{4b}{\sqrt{a^2 + b^2}}\right)^2}
  2. Factor out common terms:
    Substitute f and e into the left side:
    \sqrt{\left(a - \frac{4a}{\sqrt{a^2+b^2}}\right)^2 + \left(b - \frac{4b}{\sqrt{a^2+b^2}}\right)^2}
    Factor out common terms:
    = \sqrt{a^2\left(1 - \frac{4}{\sqrt{a^2+b^2}}\right)^2 + b^2\left(1 - \frac{4}{\sqrt{a^2+b^2}}\right)^2}
    Combine terms:
    = \sqrt{(a^2 + b^2)\left(1 - \frac{4}{\sqrt{a^2+b^2}}\right)^2}
  3. Combine terms: = \sqrt{(a^2 + b^2)\left(1 - \frac{4}{\sqrt{a^2 + b^2}}\right)^2}
  4. Simplify the square root: = \sqrt{a^2 + b^2} \cdot \left(1 - \frac{4}{\sqrt{a^2 + b^2}}\right)
  5. Distribute the multiplication: = \sqrt{a^2 + b^2} - 4

Conclusion:

The verification confirms that:

\sqrt{(a - f)^2 + (b - e)^2} = \sqrt{a^2 + b^2} - 4

is indeed correct.

This shows that the minimal distance between point (a,b) and the closest point (f,e) on the circle is exactly the distance from (a,b) to the origin minus the circle’s radius (4).

[Note 3]

Minimum PD Derivation \begin{aligned} &\textbf{1. Starting Point} \\ &\text{Point } P \text{ at } \left(\frac{a}{2}, \frac{b}{2}\right)\text{ (midpoint of AB)} \\ &\text{Point } D \text{ moves along path from EF's midpoint (EF=4)} \\[5pt] &\textbf{2. Circle Constraint} \\ &\text{Parameterize using angle } \theta: \\ &f = 4\cos\theta,\; e = 4\sin\theta \\ &\Rightarrow D = (2\cos\theta, 2\sin\theta) \\[5pt] &\textbf{3. Distance Function} \\ &PD^2 = \left(\frac{a}{2} - 2\cos\theta\right)^2 + \left(\frac{b}{2} - 2\sin\theta\right)^2 \\[5pt] &\textbf{4. Critical Point} \\ &\frac{d}{d\theta}PD^2 = 0 \Rightarrow \tan\theta = \frac{b}{a} \\ &\text{Optimal angle aligns with } (a,b) \text{ direction} \\[5pt] &\textbf{5. Optimal Coordinates} \\ &\cos\theta = \frac{a}{\sqrt{a^2+b^2}},\; \sin\theta = \frac{b}{\sqrt{a^2+b^2}} \\ &PD = \frac{1}{2}\sqrt{a^2+b^2} - 2 \\[5pt] &\textbf{6. AB + AC Constraint} \\ &\sqrt{a^2+b^2} + a = 12 \\ &\Rightarrow \sqrt{a^2+b^2} = 12 - a \\ &b = \sqrt{144 - 24a} \\[5pt] &\textbf{7. Optimization} \\ &PD = 4 - \frac{a}{2} \\ &\text{Minimize PD by maximizing } a \\ & \\ &a = -12 + 12\sqrt{2} \approx 4.97 \\[5pt] &\textbf{8. Final Calculation} \\ &PD_{\min} = 4 - \frac{-12 + 12\sqrt{2}}{2} \\ &= 10 - 6\sqrt{2} \\[5pt] &\textbf{Why This Works} \\ &\bullet \text{Finds closest } D \text{ to } P \text{ with EF=4 constraint} \\ &\bullet \text{Converts geometry to optimization problem} \\ &\bullet \text{Minimum when } D \text{ aligns with } OP \text{ direction} \\ &\bullet \text{Exact solution from system of equations} \\[5pt] &\textbf{Verification} \\ &a = -12 + 12\sqrt{2} \Rightarrow \\ &\sqrt{a^2+b^2} \approx 7.03 \\ &PD \approx 1.515 \text{ matches } 10 - 6\sqrt{2} \end{aligned}

Fermat Point

Elementary Math Leave a comment

We as a kid making kite using toilet-roll thin cover-paper and bamboo sticks from mom’s bamboo sweeper 竹扫帚, nobody had ever discovered this trick!!

https://vt.tiktok.com/ZShRTqQNQ/

https://vt.tiktok.com/ZShRopefX/

He used GeoGebra tool to draw the animation geometry.

@tim_brzezinski

Triangle insanity! #mathematics #math #maths #geometry #mathtok #mathchallenge

♬ Powerful dark futuristic science fiction film music(1536393) – Azure Glitch

Proof:

https://en.m.wikipedia.org/wiki/Fermat_point