# 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.) 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.

Practice Makes Perfect. 熟能生巧！

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:

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

1. 网上教育开发商

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

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。

# 不一样的数学探索 —有趣，直观又有用

3个，都不能把另一点包含在其中。第五点呢，比方说在正方形内部。还是没办法取出含另两点在内部的三角形。 教育中从简单开放的问题出发，易于参与。比方说 什么叫“最外面的点”？反驳的孩子的思路也好棒，他是举例说明有不行的情况。提醒注意，要鼓励的是探索，动脑筋思考，发表意见要重事实和逻辑，不嘲笑攻击别人。这才是安全探索的环境。和同行讨论时有共识，启发思考，东西方都有好的做法，有共通的东西。简单贴上东方,西方教育的标签，捧一个打一个，是我们一直不赞成的。

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# 竞赛问题：正整数的关联Murray数

(English version is being prepared and corrected – to be posted soon)

For each positive integer n, the Murray number of n is the smallest positive integer M, with M >n, for which there exists one or more distinct integers greater than n and less than or equal to M whose product times n is a perfect square.

（甲）6 Murray 数被发现是 12。说明为什么。

（乙）决定 8 Murray 数。

8 x 18 = 12 2

9 18 都大于8，且不超过18

8 x 10 x 12 x 15 = 1202

**********  ***********  **********  **********  **********

（丙）证明有无限多个这样的正整数n 满足：n 不是完全平方数，n Murray 数小于 2n.

（丙）其实不难，有兴趣的可钻研。我们给两个提示（窗户纸已捅破 – 再往前走一步便是证明了）。

A. 我已经找到了所有 n关联Murray数小于2n — 即原数2 倍的情形。一共有有限个。

B. 你找到所有这种情形的数了，并且有限个。那其中一定有最大的。能告诉我最大的n 吗？

A. 当然。n = xxxx.

B. 哦，4n Murray 数也一定会小于4n 2 ; 所以你把 4n 漏掉了。

A. 那就再加上 4n, 其中 n 是我刚讲的数。

B. 你一定还漏掉了4 (4n), 也就是 16n.

# 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?

# Online Math Reading Room is Calling !

Jonah’s math corner has created the online math reading room for you — if you are a math amateur, a student learning math, or just concerned parents! All aboard!

# Smart scrambling of digits onto pyramid to match eee

### 三角锥上的数字

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.

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. 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. Con sider 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。

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?

（解决本题时用到了什么？）用到了数和运算，用到了形状（点对和连接的线段），加上一些逻辑推演。所以这个问题是综合性的呀！

• 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.