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.

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).

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。

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。





(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。


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。


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。


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


(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。


17 年蝉 — 素数生命周期蝉猜想

自然界中有一种蝉,拉丁名叫 Magicicada Septendecim,是昆虫里生命周期最长的:十七年。他们的生命从地下开始,幼虫从树根中耐心地吮吸汁液。过17年后,成虫从地里冒头,成群结队,一时间泛滥了整块地界。几周之后,他们交配产卵,后悄然死去。产下的卵也要待17年再冒头。(另有一种叫 Magicicada Tredecim,生命周期是13 年)。

一个数学爱好者客串回答,他注意到13 和 17 都是素数(即除了1和自身以外无其他约数),于是就提出了下面有趣的假说:

设有一种寄生于蝉的虫子。蝉冒头时一定要避开这种寄生虫的大年(即最繁盛的一年)。因为碰到寄生虫不是好事,会影响到这一代蝉的群体健康和生活质量。如果该虫的生命周期是 2 年,那么为了蝉自身的好处,蝉的生命周期是单数才好 (两代蝉中只有一代会遇到寄生虫的大年,比每代都撞上好。)同样地,为了蝉自己,生命周期最好也不是3的倍数,5 的倍数,…… 于是,比较合理的就是一个不太小的素数了。

对于17年生命周期的蝉,如果寄生虫的生命周期是 1 年,那么假设这一代蝉不幸遭遇寄生虫的话,过17代才会再次遭劫。如果寄生虫的生命周期是 2 年,那么也是再过17 代 (年头当然更长,是34 年)。对虫子来讲,17 代中只一代遇到十七年蝉的这个结果是不变的,只要寄生虫的生命周期是整数年(合理的假设)。而对蝉来讲,当寄生虫的周期从1年-16年 变化时,蝉已经越来越难遇到寄生虫的大年。如果寄生虫的生命周期是17年,糟糕了:可能每次都撞车:蝉被虫子克了。


通过这个有趣的例子,让我们对于素数也有了更多认识!你作分数加法时,假定两个分数的分母是不同的素数,知道怎末做的人可能会说,My God!因为通分以后的公分母一定很大,刚好是两个分母的乘积。比方你如果做 (2/7) + (1/13) = ? (七分之二 加上 十三分之一 等于什么)公分母是 7 乘 13 得 91。蝉和寄生虫的公共大年,也就在找最小公倍数,和通分的概念是一样的。