A problem of finding the area – simple, but interesting


What is the area of the middle rectangle (coloured/shaded)? (Please note none of the rectangles in this figure is a square.) Areas Tri-Rectangle

Select the correct answer from the choices below:

(A) 13 (B) 15 (C) 22 (D) 31 (E)
(F) cannot be determined from the given condition.

It’s a simple question, eh? Which choice you’ve taken as the answer? Try not to guess randomly.

Let some hints be revealed.

If employing algebra, then which variables/unknowns shall we introduce? Obviously, the area sought shall be one unknown (let’s denote it as A), and the width that the three triangles share (denote it by h) can be another. So the length of the leftmost rectangle is (25/h), the length of the rightmost rectangle is (32/h). This leads to:

25 + A = 8h, and
A+32 = 9h

Eliminate h to find A will lead to the result.

Alternatively, we can use proportion to help solve this. Since all three rectangles share the same length, so the area ratio shall be equal to the ratio of the lengths:

(25+A) : (A+32) = 8 : 9

We only have one variable A, not two, as in the previous approach. This makes the solution even simpler.

You already got the answer, Eh? Hopefully, it’s neither (E) nor (A). It’s the one you should figure out from the system of equation(s).

If you found it already, it’s not necessary to look at others to ask “Am I correct?” Just bring the result to check if it fits those equations.

数学练习资源 Links for Practice

Practice Makes Perfect. 熟能生巧!

要学好数学,研习请教 数学方法和技巧 之外,最重要的就是 多练习了。为此,我们推荐如下的数学练习资源 (这些网站都是英文)。

本篇提到的数学练习资源面向初中(6-8 年级)学生;在高水平的老师指导下使用更佳。


首先介绍一本图绘数学词典 – 大人孩子都可以用。如果数学学习和做题时有些词看不懂,这些可以帮你忙。

它收录了 630 个数学常用词汇,有彩图,有例子。易懂,给力!

(Tools)A Maths Dictionary for Kids 2017

如果你经常使用这本数学词典,可以做个书签 (Bookmark);以这样的方式,也是向作者致敬喽!


接下来是数学练习资源。点击(或触摸)如下链接, 就可以看到相应内容。刷到本页后面,我们有简要介绍!

  1. 由英国剑桥大学牵头 关怀下一代数学教育的人士开发的:

Lower Secondary Page – Grades 6-8

Interactivities, Practices, Games, Articles, etc.     ENrich Maths,
Cambridge, England.


  1. 由美国的教师 教育工作者提供的:

AGMath.com — for Grade 8

(本链接指向资源以 8 年级为主: Primary for Grade 8)

Algebra and Geometry for the Academically Gifted


  1. 网上教育开发商

AdaptedMind Math — for Grade 6

(有些内容也可以供7 年级学生练习参考 Grade 7 can also use)

注意:需要加入他们的会员(Member)才可以使用;包含一个月的免费会员(One Month of Free Membership)




上面提到的数学练习资源在内容设计与风格安排上各有千秋。Lower Secondary Math 的形式活泼,内容更新较快,是本栏目推荐的主要原因;配合的是英式教学体系– 他山之石,可以攻玉;AGMath 主要面向 8 年级,收集了代数和几何的基础内容,取材经典,而且有讲义有习题,可配合起来,在老师指导下自学最好。其中包含了美国数学竞赛如 MathCount 与 AMC 8  一些模拟练习与答案(Mock Questions and Keys), 对于有志于挑战的学生,不妨感觉一下,作入门指导之用。

Adapted Mind Math – for Grade 6 在开发上下了些功夫,形式上吸引人,内容丰富;较紧密配合了美国基础教育的大纲。


Lower Secondary Math 有灵气,而AG Math 注重基础,较为系统。Adapted Math 呢,结合了开发学生智力与提升兴趣的理念。– 是本栏作者的意见。


网上数学阅览室 开通了!

我们的网上数学阅览室开通了!这个栏目是为今天读中小学的学生(5-11 年级)写的,兼及学生的家长。希望读者们喜欢。

为什么开这个栏目?学好数学,最重要的是用数学爱数学,在课堂教学外还要加油!这个栏目贴出的文章都不是在堆砌数学讲义,而从故事,从有趣的问题开始。数学同样需要广泛阅读,把理论与解题联系起来。我自己还有我的一些教育同行,我们对于数学方法的深刻体验 都在读书和解决实际问题中得到!


本阅览室有三个主栏目,把课内课外结合起来。所贴内容中英文都有。Beyond Class(开卷有益):此栏目的目的是扩展知识面。Enhance(数海逸致)的内容有挑战但是不艰深。为什么叫“逸致”呢?就是到你平时不常去的地方,好像周末的远足一样。想做点题目的孩子可以在Practice(熟能生巧)这个栏目中由相关链接找到你喜欢的分年级问题。打开首页,这些栏目都显示在菜单上,一目了然。又因涉及到的年级跨度大,为帮助读者,我们使用了 Tag,分为 G5/6/7 (5-7 年级),G8/9 (8 – 9 年级)以及 G10/11(10-11 年级),以帮助学生读者选择适合的内容。(顺便多讲一句:如果你需要翻译 拿来 Google Translator 就可以了。)

最后说明两点。1. 本栏目的不少帖子是以前写的;值当前新冠状肺炎肆虐之时,望学生们能够更生动地学习,停课不停学,故勉力整理出来。2. 本栏目的帖子暂未开通每帖的评论功能,主要因技术原因;未来我们将考虑加进交互反馈的功能。


Smart Solutions for Some Equations (Quadratic, Polynomial and Rational)


Let’s look at a few problems on solving equations of different forms.

Find the smaller root for the following.

Problem (a).  (2-x) (3-x) = 2 / 3

Problem (b).   x (x-1) (7-2x) = (3-x) (4-x) (4-2x)

Problem (c).   [x (x-1) ⁄ (3-x)] + [x (x-1) ⁄ (4-x)] = 4 – 2x

Relating to equations above, we plot several curves using Geogebra (an app which can produce the graph for given functions – of course, this is only one feature of the many).

What’s interesting is all the three equations as in (a)(b)(c) share a common solution, and it can be found through algebraic manipulation. Take a forward look (click) here – you will see amazing graphs; the solution (roots) shall be some intersection points.


Of these questions, the challenging levels are in the order of ascending: from average level like (a), to somewhat challenging like (b), to very challenging in (c). For better understanding, we suggest the reader to try solve questions (a) (b) (c) first; and then compare with the solution provided in this file (pdf),
where we solve algebraically only for (a), but applying a combination of algebraic and graphic solution to tackle problems (b) and (c).  It’s called smart solution — we are applying a variety of tools at hand, not just following a fixed procedure.


Combining graphing with algebra facilitate us to guess where the root(s) of the equation are, or (sometimes) make observation of the roots easier. It’s also fun!

Try work out these equations first, then you can check out the solutions (pdf) here.

Now you can take a second look at the graphic solution. Have you seen and understood more, about the question, and about the how algebraic manipulation plays out in function graphs?

Smart scrambling of digits onto pyramid to match eee


The Problem A tetrahedron (as shown below) has four sides and six edges. Numbers from 1 to 11, except number 10, are each assigned either to a vertex or to an edge. The number assigned to an edge must equal to the sum of the numbers assigned to the two endpoints. Each of the numbers must be used exactly once.

问题  一个四面体有 4 个顶点和 6 条边。从 1 – 11 的数 (没有10),一一地被分配到边和顶点。每条边上的数字是该边两个端点的数字之和。上述每个数字必须刚好使用一次。

Note: Another name for the tetrahedron (四面体) is triangular pyramid (三角锥).


Can you find a solution that satisfies ALL of the above conditions?Let’s work on this problem until we find a solution!


First step is always the most challenging, as we are usually not sure which path to take when facing a difficult problem. Don’t be afraid to try; when it does not lead to where we expect, reflect on why it does not work and have another try where we have a better chance.

(第一步)试了几次后,发现最小的两个数1 和 2应该被分配到顶点而不是边上。在连接1,2的边上的数字应是3。

(Stage 1) After a few trial-and-errors (or guess-and-checks), we come to focus on the two smallest numbers 1 and 2, and decide that they must be assigned to the vertices, not the edges.
Why? Because if one of them (say 1) is assigned to an edge, they must be the sum of two numbers assigned to its two endpoints, therefore the two numbers at the endpoints must be even smaller. However, this is impossible, as the number 1 and 2 are already the smallest.
Tetra -1-s1

Which vertices to assigned numbers ‘1’ & ‘2’? In a tetrahedron, any two vertices are adjacent, and the whole solid is highly symmetric – by rotating /reflecting, any two adjacent vertices can be transformed to the two designated ones. This is settled! We can assign numbers ‘1’ & ‘2’ to any two vertices, and the edge connecting those two vertices shall be assigned number ‘3’.

Well begun, half done! But there is other half, for which we start by some frantic trying.

(第二步) 最大的数字11 应该分配到一边上。 这条边的端点不是 1 或者2。

(Stage 2) This time we look at the largest number 11. Suppose we assign it to an edge that’s adjacent to ‘2’. For the edge, one endpoint is ‘2’, the other endpoint – it must be ‘9’. We are now in a situation that we cannot go on any further. ConTetra -1-s15sider any edge with endpoint ‘9’ (other than the edge we already assigned ’11’): which number shall we assign? It can neither be ‘1’ nor ‘2’ nor ‘3’ (since all these numbers have been used, we are not supposing to repeat numbers). Smallest available number is ‘4’, but 9 + 4 = 13, which is not in the given set of numbers.

So the edge to which number ’11’ is assigned cannot have any endpoint ‘1’ or ‘2’; and take another look at the figure, we decided that it must be the edge that is the furthest from the edge ‘3’. And we mark 11 there.

(第三步) 分解数字 11 成两数之和。11 = 4 + 7 = 5 + 6. (注意,数字 1,2,3 已被用过了。) 只有这几种可能性,试试就得了。

(Stage 3) Tetra -1-s3

Decompose 11 into the sum of two positive numbers (but without using number ‘1’ ‘2’ ‘3’), we have the following two: 11 = 4 + 7 = 5 + 6. A couple of trials will finally lead to the solution! We leave it to be worked out by the reader.

What knowledge we have used in this whole process?



  • Numbers and its operation – you have to understand addition and subtractions of integers.
  • Shapes – you are not asked to do any calculation like perimeters and areas, but you have to understand an edge is constituted with two endpoints, and which edges meet and which do not (we call them the furthest edges).
  • Logic deduction – see how we’ve used the fact that 1 and 2 are the SMALLEST to deduct that they MUST be assigned to vertices, not edges; and we’ve used the fact that 11 is the LARGEST to deduct that they must be assigned to an edge.
  • Also see how we have used some symmetry to reduce the situation we need to consider.

What is the lesson we learned? Try – Think – Try again where we have a better chance. Use thinking / logic reasoning as a guidance so the trial can efficiently leads to result.