Category Archives: Enhance for School Learning

Enhancement for what students have learned in school

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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?

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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.

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Mysterious Unbalanced Sheets on Loans

– story told by hunchbacked shopkeeper

Once there was a hunch-backed shopkeeper. And he told the following story:
(For readers at grade 6/7 level and up)

“Once I lent 100 dinars, 50 to a sheik from Medina and another 50 to a merchant from Cairo.

“The sheik paid the debt in four instalments, in the amount 20, 15, 10 and 5. .. .. Note that the total of his debt balance is 50 dinars.

1st Installment: Paid 20 & Still Owe 30
2nd installment: Paid 15 & Still Owe 15
3rd Installment: Paid 10 & Still Owe 5
4th Installment: Paid 5 & Owe 0
Total Paid 50 & Total Owe 50

“Meanwhile the merchants from Cairo also paid the debt of 50 dinars in four instalments, in the following amount: 20, 18, 3 and 9. And here is the balance sheet for his debt.

1st Installment: Paid 20 & Still Owe 30
2nd installment: Paid 18 & Still Owe 12
3rd Installment: Paid 3 & Still Owe 9
4th Installment: Paid 9 & Owe 0
Total Paid 50 & Total Owe 51

“But note his total owed is 51 dinars”, remarked by the hunchbacked shopkeeper; apparently this should not have occurred.

Using a bit math, can you help our shopkeeper to solve the mystery?

Think on your own first!

Then you might want to take a look at our explanation. Do you agree with us?

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Amazing Detection of Parallel-o-gram: Unusual But Fairly Reasonable

Talking about shapes, one of the most familiar is the parallelogram. And about it we have a problem as follows.

Given quadrilateral ABCD, E, F is the midpoint of AD, resp. DC. Lines BE and AF intersect at point G. (And the figure is as below.)

Suppose it happens that GE : GA : GF : GB = 1 : 2 : 3 : 4. Attempt to prove ABCD is a parallelogram.

How can the division ratio GE : GA : GF : GB as given, be used as the hints to deduct the parallelogram?

Well, to put the known condition in proper perspective is important. If first attention is directed to GA : GF = 2 : 3, you might wonder how that can be utilized. Suppose you notice that GE : GA = GA : GB = 1 : 2 : 3 : 4, and suppose AF and BE are perpendicular to each other, then we conclude that the two triangles EGA and AGB are similar right triangles .. But how that can be used to help recognize a parallelogram?

By observing AF = AG + GF, and BE = BG + GE, and we find that AF = BE (by applying properties of the proportion, AF : BE = (2 + 3) : (4 + 1) = 1 : 1). So we have two hanging line segments AF and BE that are congruent (equal in length), and both segments are drawn from one vertex (of ABCD) to midpoint of a non-adjacent side. The information in its current form seems much helpful.

Now passing through D let’s draw a parallel line of BE; suppose the line intersects BC at point J: DJ // BE. Suppose that DJ intersects AF at point Q (Q should be between G and F). We work out that DQ = 2 EG (notice E is the midpoint of AD), as well as GQ = AG. So if it held that DQ : QJ = 2 : 3, we would prove that DJ = BE.

(Why doing that? If BE = DJ were proved, then JDEB would be a parallelogram. Parallelogram JDEB implies that AD // BC. Can you see that? Just note ED // BJ, and  segment ED is on line AD, resp. BJ is on BC; so we have, finally, AD // BC). So we’ve proved that AD and BC, as one pair of the opposite sides of a quadrilateral, is parallel. We can analogously attempt for AB // CD. Now by definition, ABCD is a parallelogram.

However to work out DQ : QJ = 2 : 3 some effort is needed, and the good news is we are on the track.

A rather interesting way (of course challenging but a pleasant experience), isn’t it? For anyone who wants to do more systematically, or automatically, is there another way?

Yes, there is. That way is to use coordinate system.

Suppose the two hanging segments AF and BE are perpendicular to each other, and let their intersection (G) be the origin: so G(0,0). With the given GE : GA = GA : GB = 1 : 2 : 3 : 4 (we’ll call it1-2-3-4 condition), we draw the following diagram that involves G, E, A, B, F to help us understand, and to figure out coordinates to be E(0,1), A (2,0), F (-3,0) and B (0,-4).

Coordinates of the other two vertices can be worked out from the conditions that E is the midpoint of AD, as well as F is the midpoint of DC. (Noting the positive direction of x-axis points from G to A.) We worked out D (-2,2) and C (-4,-2). If it’s indeed that ABCD is one parallelogram, then the two diagonals AC and BD bisect each other, which means that the midpoint of the two diagonals are the same, as verified straightforwardly:

Midpoint of AC: ((2-4)/2, (0-2)/2) ) = (-1, -1)

Midpoint of BD: ((0-2)/2, (-4+2)/2 ) = (-1,-1)

Since AC and BD bisect each other, ABCD is a parallelogram. But what if the AB and BE are not perpendicular? Well ABCD is still a parallelogram in case – but explanation is needed – for which we can pursue later. It deserves to point out, by changing the x- and y- axis from being orthogonal to arbitrary, whether the shape is parallelogram or not will have the same answer.

Let me ask if you would like this problem. If like, then share it with your friends!

Continuing from the problem discussed above, suppose that apart from AE, BF, we draw CJ and DK, where J is midpoint of AB and K is the midpoint of BC. Here is your turn to take a ramp-up challenge:
(If you truly understand what’s going on in the above discussion, then it’s just a piece of cake!)

(i) Can you show that there are exactly four intersections among the four lines AE, BF, CJ and DK; and the four intersection points connect to make a parallelogram?

(ii) What type of parallelogram is the one mentioned in question (i)? Is it a rectangle or rhombus?

*(iii) Is it possible for the original parallelogram to be a rectangle ABCD? In that case, if GE : GA : GF : GB = 1 : 2 : 3 : 4, what would be the aspect ratio AB : BC (aspect ratio is the ratio of width-to-length in a rectangle shape).

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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数

(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 – 9，10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 数 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 的 Murray 数被发现是 12。说明为什么。

（乙）决定 8 的 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 和12之间其他的数(有 7, 8, 9, 10, 11)都不行，从而排除。

现在说明 7 （或 11）不能成为 6 的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 和10 也不行（理由从略）。

现在来看问题（乙）。答卷中有认为 8 的Murray 数是18 的，理由是：

8 x 18 = 12 2
或 8 x 9 x 18 = 36 2

而 9 或 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 的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 的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 的Murray 数小于 2n.

这题的证明很容易跑题。要求既不是说明 n 的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 的 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 的Murray 数也一定会小于4n 的2 倍; 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。

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化三角为正方的问题： 一波三折，耐人寻味

亲爱的读者，这是一个蛮有趣的问题：给你一个正三角形（即等边三角形），你能把它切割成几块，然后拼成一个正方形吗？

也许你完全摸不着头脑：其实这个问题是一个经典老问题。有意思的是，曾广为流传的一个解法其实是有毛病的，但却瞒过了许多人。后来才有人火眼金睛地发现了问题，经过一波三折，终于有了漂亮而完美的新解法！

下面这张图就是曾经广为流传的一个解。（图是新画的，画的人也知道图中有误 — 但是作为数学教育，我们必须学会审视错误和偏差。）

Triag-CutGlue-Square看明白了吗？参照上图：沿着ABC 的方向是一个正三角形，即AB=BC=CA （三边长相等）。把它切成四块（图中分别涂了黄绿蓝粉色），然后让绿色块以E为中心旋转180 度，粉色块以D为中心旋转 180 度，再把蓝色块移到最上端，诺，看到新成的正方形了吗？从MD，ME 分别延长出去的是两个相邻边，MA再延长到最上面的顶点是对角线。看起来活脱一个标准正方形呵，只是旋转了 45 度，对吗？从图形的对称性，新成四边形的四边都相等，应没有问题。四个角呢，那恰是环绕M的四个直角，经过翻转移动，最后变成了新四边形的四个直角。四条边相等，四个直角，那不就是正方形了吗？

没有真地拿剪子裁开，拿胶水粘上去试一把，可能有些想当然了。试一试，还真能粘上。几乎大功告成之际，有人发现了破绽，看看四个角，好像不是直角哎？如果不是直角，但是四边都相等，那就只是菱形，不是正方形了。

再细看看。D，E是AB，AC 两边的中点（就是说DA=DB，EA=EC），F，G是BC的两个四分点（BF 和 GC各是BC的四分之一，FG 占了BC的一半）。（略去推导）DEGF应该是一个长方形。DG和EF呢，是这长方形的两条对角线。长方形的两对角线交成直角吗？一般不是，除非长宽相等（不信就试试）。那DE = DF 吗？DE 是正三角形边长的一半，DF呢，是三角形高的一半。高线总比边长要短些的，对不对？所以DF < DE; 于是两条对角线的交线不是直角，即环绕M的四个角都不是直角。经过翻转成为新四边形的四个角，也非直角也。

结论：新成的四边形真的不是正方形，只是菱形（也有人把它叫做钻石形–Diamond Shape）。我们计算了一下，角 FMG 大约成 98.2 度的角度。

假使没注意到交角上的破绽（不是直角），还有一个办法看出问题：既然新四边形是由原三角形裁剪拼成的，那么他们的面积应该相等咯。面积的计算稍复杂些，不再详细讨论：只提一下如果正三角形的边长是 2，那么它的面积是 3 的平方根（约为 1.73）；而新四边形按照构造方法，边长应为2的平方根，所以如果是正方形，那么面积是 2，而2 不等于 3 的平方根，所以新四边形一定不是正方形！

还有操作性特强的办法发现问题：从E出发，作 DG的垂线，延长到和BC 相交。传统几何中非常讲究作图法，而且只用直尺和圆规。如果不熟悉，还是可以借助量角器完成作图，对吗？一画图就发现问题了，垂线不通过M，延长后与BC 的交点也不在F，有偏差。

能不能改正这个偏差，裁剪拼出真正的正方形呢？有的学生脑子快，马上想到一个新招。菱形是特殊的平行四边形，而平行四边形总是可以裁剪的.通过裁剪形成了四个直角。于是这位学生说，基于菱形，因为四边已经相等，所以裁剪整出四个直角后，那就是正方形了。有人反对，要裁剪两次，太麻烦了吧？而老师的意见是，麻烦不麻烦其实是第二位的，我们首先要较真一下，按这个过程得到的果真是正方形吗？

确是个好的尝试，可是有有点问题。细想一下裁剪平行四边形的过程，就发现裁剪再粘上的结果，有一边长度未变，而另一边却成了原平行四边形的高线，因而长度比原边缩短。既然裁剪前长度一样，作了这个过程后就不再相等了。当然在这问题中差别蛮小，所以得到的是长宽很接近的一个长方形。

培根有句名言：“数学使人精细。”正确求解一个可能途径是，在BC 上求一点X，使得EX刚好等于新正方形的边长。F 和 X 其实就差那么一点点。我们要做的：从原三角形的面积等于新正方形的面积，求出新正方形的边长 s，然后求F在BC上的位置, 使得 EF = s. 这是求解的第一步。正解已经呼之欲出。

说明：
1. 前辈蒋声先生最先指出开明书店旧版的 《数学万花筒》 中的一幅插图是错的，那图所指的就是化三角形为正方形的问题。真正的火眼金睛！在他后来80 年代的著作《几何变换》 中亲自改正了原图。

2. 插图取自公众号 数学教学研究，向邵勇老师致意。（他同时给出完全使用直尺圆规作图的解法。）

本文目的不在于建立严格解法，而讨论如何以审视的目光发现问题和解法的破绽。

看到和改正错误是数学的基本功之一。