# 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.

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.

# Mysterious Unbalanced Sheets on Loans

### – story told by hunchbacked shopkeeper

Once there was a hunch-backed shopkeeper. And he told the following story:

“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?

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

# 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

(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.

# 化三角为正方的问题： 一波三折，耐人寻味

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MDME 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？

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。）