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)

还有更多的练习资源,本栏以后会介绍;欢迎常回来看看!

 

我们简要介绍一下。

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。

 

不一样的数学探索 —有趣,直观又有用

可以吗?在平面上任给5个点 (where no three points are on the same line),可以选3个点用直线连成三角形,让另外2点都在这三角形的内部吗?有些情况下这能做到,但是总能作到吗?当这个仿佛天边飘过来的问题(既简单又似无从下手)提给 15左右的学生时,不少就表现出跃跃欲试的兴趣。有一个立即发表意见:"总是可以,只要找到最外面的三个点,余下的就在里面。"

Yes Interior -2同伴反驳了,他举了一个例子:取呈正方形的四个点,无论取哪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。
有学生补充说,如果五个点在一个圆上,都不能取出三角形把另两点含在内部喔。

Not-Interior - 1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。


******

数学家,数学教育家 Polya 讲过(大意),好老师善于提问题引导学生深入。通过上面讨论,文章开头的问题不是总能做得到。有时能,有时不能。那么什么时候能,什么时候不能呢?可以想一下,答案在文中说明能和不能的两幅插图中找。先注重观察特征,然后再想怎么表达清楚。有兴趣的切磋一下,问问你的孩子。

问题的意义也许不在于记住答案,在于启发思考和表达。在技术和社会领域,要能够跳出框框,这也是能力。对比传统几何教育的套定理证明,这样的探索更有意义,能启发心智(传统几何自有存在意义,此另当别论)。在从众的潮流中能思考堪称可贵。

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竞赛问题:正整数的关联Murray数

竞赛问题:正整数的关联Murray

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

你听到过正整数的关联Murray数吗?其实是一个蛮有趣的问题,基础在于完全平方数的简单推理–简单归简单,可是要严密哦!今年的CIMC(加拿大中级数学竞赛 – 910年级组)就出了这样的问题。

先来看一下定义:

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.

对于正整数n, 指定n 的(关联)Murray数是有如下性质的正整数中最小的那个:存在一或多个不同的大于 n 但不超过M的整数, 这些数的乘积 再乘上 n 得到一个完全平方数。

本文注:(关联)Murray数可就叫Murray 数。“关联”一词主要强调 Murray M和给定整数n 间是一个数学关系。完全平方数就是整数的平方,如1, 4, 9, 16, 25, .. 一直下去。

例子:3 × 6 × 8 = 122, 且找不到一或多个大于3 但是都比8更小的整数使其积再乘3是完全平方。

这个定义要多读两遍,确保理解。定义是写在题目开头的,所以无需事先知道,但是做题前一定要弄清意义。然后看下面的–就是CIMC 的问题 。

(甲)6 Murray 数被发现是 12。说明为什么。

(乙)决定 8 Murray 数。

如果稍有点数学底子,这似乎不是难题,尤其问题(甲),看来唾手可得。主要的挑战!其实是在严密上。(“说明为什么”就是证明的意思。)

比如这样的说明:因为 6 × 8 × 12 = 576 = 242 而且 6 < 8 < 12, 所以 12 合乎Murray数的定义。注意了!这个说明有点问题。因为按定义,任何数的关联Murray数只有一个,要尽量的小。所以验证 12 符合条件还不够,必得说明在6 12之间其他的数(7, 8, 9, 10, 11)都不行,从而排除。

现在说明 7 (或 11)不能成为 6 Murray数。他们不能在定义的完全平方式中出现。小于7的数字不能含素数因子7,所以7再次出现要等以后(至少等到14 吧)。11 是同样道理。9 是完全平方,假使在完全平方式中现身,那末去掉因子9以后还是完全平方数。8 10 也不行(理由从略)。

现在来看问题(乙)。答卷中有认为 8 Murray 数是18 的,理由是:

8 x 18 = 12 2
或 8 x 9 x 18 = 36 2

9 18 都大于8,且不超过18

这解不正确!有些细节可能被忽略了。第一是Murray 数只有一个,是满足如此如此性质中最小的那个。以上验证不足说明 8 Murray 数是 18,只是说明Murray数肯定小于 18 (依定义也知大于8)。此外在 完全平方的表达式中应有几个因子?定义里没说,唯一要求是这些因子相异。(不能因为例子中给出3 个因子,就把特例当一般。)看下面的式子:
8 x 10 x 12 x 15 = 1202
或者 8 x 9 x 10 x 12 x 15 = 3602

按定义,8 Murray 数必得不超过15. 且找不到更小M <15可用n, M 间的因子做积再乘上8是完全平方数。故 15 是解。

这问题真让人痒痒 – 答题时或者说明不严谨,或者理解有点偏差;于是与正确的解失之交臂。

至此文初提到的两个问题,已经完全解答完毕。如果有兴趣,不妨再往下读。

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

原题中还有另两个问题,我们举其中一个分析一下。

(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.

这题的证明很容易跑题。要求既不是说明 n Murray 数可以小于 2n,也不是要说明有无限多个正整数n。也非 n Murray 数一定小于2n。这样说吧(也许从反面理解更容易)假如有人声称他已经找到了所有关联Murray数小于2n — 即原数2 倍的情形;那你一定可以对他讲“且慢,你一定漏掉了一些”。

(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)。

假如有人声称他已经找到了所有关联Murray数小于2n — 即原数2 倍的情形; 那么一定可以告诉你最大的一组。现在你对他讲:我还有更大的一组,你一定漏掉了。这样开启了如下对话:(假定你是 B

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.

看出来了吧…… 这样下去一定没完没了。

有趣吗?弄明白了肯定有意思。知识点不难,但这题对于思维素质是有要求的。

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!

通过阅读,了解数学方法的原理,背景,数学在社会中的作用,从而爱上数学,增强学习动力,这与基础训练和系统学习同样重要。光抱怨今天的中小学生不喜欢数学无济于事,要反思大人们花了多少时间去引导兴趣。本栏目就为此而办。有许多文章较浅显,但是生动活泼,包含的数学原理是重要的,用到的推理过程有助提高思维;我们力求深入浅出,寓教于乐。有些学生的数学能力强,又喜欢挑战,数学阅览室也为他们安排了更深的内容;为适应网上阅读,我们采取一题一议,把要解的问题和中学数学方法有机结合起来,并且视角独到。

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。

请大家欣赏数学阅览室的帖子。祝大家开卷有益,学习进步。如果你喜欢,多多推荐给你的朋友喔!

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

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.

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).

Tetrahedron

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

第一步总是最难的。要大胆地试!

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?

(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!

还有,从图形的对称性,大大减少要考虑的情况。

  • 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.