Category: High School Mathematics

GeoGebra Tutorial 7 – Sliders and Rotation

High School Geometry, High School Mathematics, Software Tutorials

This is the seventh tutorial of the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series.

In the Graphs and Sliders posts (click here and here ), we have discussed how to use number sliders.  In this tutorial, we use the Angle slider to rotate a triangle in order to show that its angle sum is 180 degrees. This is the same GeoGebra worksheet shown in my Parallel Lines and Transversals post, but we will change some of the labels. Although this tutorial is the seventh of the GeoGebra Tutorial Series.

Figure 1 – Rotated triangles using sliders.

Construction Overview

The construction will start by drawing line AB and constructing triangle ABC using the Polygon tool. Afterwards, we reveal the interior angle measures of the triangle and create two angle sliders namely \alpha  and \beta. Next, we rotate the triangle 180 clockwise about the midpoint of BC producing triangle A’B’C’ (see Figure 1-B). We then repeat the process, and rotate triangle A’B’C’ 180 degrees clockwise about the midpoint of A’C’ to produce A’’B’’C’’.

Part I – Constructing Triangle ABC

1.) Open GeoGebra and select Geometry from the Perspectives menu.
2.) Click the Line through Two Points tool, and click two distinct locations on the Graphics view to construct line AB.
3.) If the labels of the points are not displayed, click the Move tool, right click each point and click Show label from the context menu.
4.) Click the New Point tool and construct a point C not on line AB.
5.) Display the name of the third point. GeoGebra would automatically name it C, otherwise right click and rename it C.
6.) Click the Polygon tool and click the points in the following order: point A, point B, point C, and click again on point A to close the polygon. Your drawing should look like Figure 1.

Figure 2 – Triangle ABC on line AB.
7.) Move the vertices of the polygon. What do you observe?
8.) Now we construct two angle sliders \alpha and \beta. To do this, click the Slider tool, and click on the Graphics view.
9.) In the Slider dialog box (see Figure 3), choose the Angle radio button, and then leave the name angle as \alpha.  In the Interval tab, choose 0° as minimum, 180° as maximum and 1°, and then click the Apply button when finished.

Figure 3 – The Slider dialog box
10.) Using steps 8-9, create another slider with the same specifications shown in Figure 3 and name it \beta. You can find the Greek letters by pressing the \alpha button located at the right of the text box.
11. ) We reveal the angle measures of the interior angles of the triangle, the change the colors of the angle symbols (green sectors). To do this, click the Angle tool and then click the interior of triangle ABC.
12.)  We now hide the measures of the angles. To do this, right click each angle symbol and uncheck Show label from the context menu.
13.  We set angle colors: angle A red, angle B blue and angle C green. To change the color of the angle symbol of angle A, right click the angle symbol (not point A) and click Object Properties from the context menu to display the Preferences window.
14.  In the Preferences window, click the Color tab and choose the color you want from the color palette then click the Close button.
15.  Change the color of angle B to blue and leave angle C as is.  Your drawing should look line Figure 1-A after step 15.

Part II – Rotating the Triangle

We already have the sliders ready. The next thing that we will do is to rotate the triangle. The idea is to create a rotation point. Our choice would be the midpoint of BC. That is because if we rotate ABC by 180 degrees producing A’B’C’, angle A’C’B’ will be adjacent to angle ABC (see Figure 1-B). This is also the idea when we rotate A’B’C’ producing A’’B’’C’’.

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1.) To construct D, the midpoint of BC, click the Midpoint or Center tool, and click side BC (the segment, not the points).
2.) Note that we want ABC to rotate around D \alpha degrees clockwise. To do this, choose Rotate around a Point by Angle tool, click the interior of the triangle and click point D to reveal the Rotate Object dialog box.
3.) In the Rotate Object around Point by Angle dialog box, change the measure of the angle to \alpha, choose the clockwise radio button, and then click the OK button.

Figure 4 – The Rotate Dialog Box
4.) Now move slider \alpha. What do you observe?
5.) Adjust slider \alpha to 90 degrees, and show the labels of the vertices of the rotated triangle. (Refer to Part I – Step 3).
6.) While the triangle is still rotated 90 degrees, click the Angle tool and click the interior of triangle A’B’C’. Hide the labels of the angles symbols.
7.) Change the colors of the angle measures. Refer to Part I – Steps 13 through 15. Be sure that angle A and A’ have the same color, B and B’ have the same color, and C and C’ have the same color. Your drawing should look like the drawing in Figure 6.

Figure 5 – The Rotated Triangle

Part III – Creating the Third Triangle

The idea of creating the third triangle is basically the same as that of creating the second triangle, so I will just enumerate the steps and left the construction as an exercise.

  1. Get the midpoint of A’C’. (Refer to Part II Step 1)
  2. Rotate triangle A’B’C’ \beta degrees clockwise around the midpoint of A’C’.  (Refer to Part II – Steps 2 – 3).
  3. Reveal the labels of the vertices of the third triangle which is A’’B’’C’’.  (Refer to Part II – Step 5 and Part I – Step 3).
  4. Reveal the angle symbols of triangle A’’B’’C’’.  (Refer to Part I – Step 11)
  5. Hide the labels of the angle symbols, and change the colors of the angle symbols of triangle A’’B’’C’’. (Refer to Part I – Step 13-15)

The explanation of the theory behind this construction is in my Parallel Lines and Transversal post.

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How to use the summation symbol

High School Algebra, High School Mathematics

If we want to add the expression x_1,x_2 all the way up to x_{10}, it is quite cumbersome to write x_1 +x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 +x_9 +x_{10}. Mathematical notations permit us to shorten such addition using the \cdots symbol to denote “all the way up to” or “all the way down to”. Using the this symbol, the expression above can be written as x_1 + x_2 + \cdots + x_{10}.

There is, however, a more compact way of writing sums. We can use the Greek letter \Sigma as shown below.

Figure 1 – The Sigma Notation

In the figure above, a is the first index, and letter b is the last index.   The variable(s) are the letters or the numbers that appear constantly in all terms. In the expression

x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}

1 is the first index, 10 is the last index and x is the variable. We use the letter i as our index variable, or the variable that will hold the changing quantities.  Hence, if we are going to use the sigma or the summation notation for the expression x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}, we have

\displaystyle\sum_{i=1}^{10} x_i

Some of the examples are shown below.  Observe the colors of the indices and the variables, to familiarize yourself how the summation symbol works.

Figure 2 – Examples of Summation Notation

In using the summation symbol, take note of the following:

  • An index variable is just a “dummy” variable. It means that you can use a different index variable without changing the value of the sum. The sum \displaystyle\sum_{i=1}^{10} a_i is the same as \displaystyle\sum_{j=1}^{10} a_j and is the same as \displaystyle\sum_{k=1}^{10} a_k.
  • The indices are the natural numbers 1, 2, 3, \cdots and so on.
  • The last index is always greater than the first index.
  • A variable without an index most of the time represent an infinite sum or a sum from 1 through n

More Examples

1. (a - 1) + (a^2 - 2) + (a^3 - 3) + (a^4 -4) \displaystyle\sum_{i=1}^{4} (a^i - i)
2. 3p_5 + 3p_6 + 3p_7 + 3p_8 \displaystyle\sum_{j=5}^{8} 3p_j
3. 5 + 5 + 5 + 5 + 5 + 5 + 5 \displaystyle\sum_{k=1}^{7} 5
4. 1 + 2 + 3 + \cdots + 99 + 100 \displaystyle\sum_{m=1}^{100} m
5. (a_3 + b_3) + (a_4 + b_4) +(a_5 + b_5) \displaystyle\sum_{n=3}^{5} (a_n + b_n)

Properties of the Summation Symbol

1.) The expression (x_1 + x_2 + x_3 + x_4) + (x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}) equals (x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}) which means that \displaystyle\sum_{i=1}^{4} x_i + \displaystyle\sum_{j=5}^{10} x_j = \displaystyle\sum_{i=1}^{10} x_i . In general, \displaystyle\sum_{i=1}^{m} x_i + \displaystyle\sum_{j=m+1}^{n} x_j = \displaystyle\sum_{i=1}^{n} x_i.

2.)The expression (x_1 + x_2 + x_3 + x_4) + (y_1 + y_2 + y_3 + y_4) = \displaystyle\sum_{i=1}^{4} x_i + \displaystyle\sum_{j=1}^{4} y_j. Regrouping the expression, we have  (x_1 + y_1) + (x_2 + y_2) + (x_3 + y_3) + (x_4 + y_4) = \displaystyle\sum_{i=1}^{4} (x_i + y_i). This means that \displaystyle\sum_{i=1}^{4} x_i + \displaystyle\sum_{i=j}^{4} y_j = \displaystyle\sum_{i=1}^{4} (x_i + y_i) Generalizing, we have \displaystyle\sum_{i=1}^{n} x_i \pm \displaystyle\sum_{j=1}^{n} y_j = \displaystyle\sum_{i=1}^{n} (x_i \pm y_i).

3.) The expression c + c + c + \cdots + c (k of them) = \displaystyle\sum_{i=1}^{k} c. But c + c + c + \cdots + c = c( 1 + 1 + 1 + \cdots + 1) (k of them) = kc. Therefore, \displaystyle\sum_{i=1}^{k} c = kc.

4.) The expression 2x_1 + 2x_2 + 2x_3 + 2x_4 = \displaystyle\sum_{i=1}^{4} 2x_i. But 2x_1 + 2x_2 + 2x_3 + 2x_4 = 2(x_1 + x_2 + x_3 + x_4) = 2 \displaystyle\sum_{i=1}^{4} x_i.  In general, \displaystyle\sum_{i=1}^{k} cx_i = c \displaystyle\sum_{i=1}^{k} x_i.

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Introduction to Combinations

College Mathematics, Combinatorics, High School Algebra

Introduction to Combinations

In my Introduction to Permutations post, we have learned that the number of permutations (or arrangements) of n objects taken at n at a time written as P(n,n) is equal to n! = n(n-1)(n-2) \cdots (3)(2)(1), and we have also learned that the number of permutations of n objects taken k at a time written as P(n,k) is equal to \displaystyle\frac{n!}{(n - k)!}.

In Figure 1, shown are the permutations of 4 letters, A, B, C and D taken 4 at a time.  From the figure, we can see that there are indeed 4!= (4)(3)(2)(1) = 24 of such arrangement. In Figure 2, shown are the permutations of 4 letters taken 3 at a time, and we have shown that the number of permutations is equal to \displaystyle\frac{4!}{ (4 - 3)!} = \frac{4!}{1!} = (4)(3)(2)(1) = 24.  In Figure 3, we have again listed the permutations of 4 letters taken 2 at a time, and have shown that the number of permutations is equal to \displaystyle\frac{4!}{(4-2)!} =12.

Figure 1 – Permutations of ABCD, taken 4 at a time.

Figure 2 – Permutations of ABCD, taken 3 at a time.

Figure 3 – Permutations of ABCD, taken 2 at a time.

If we talk about combinations, however, the arrangement of objects does not matter. For example, if we want to buy a milk shake and we are allowed to choose to combine any 3 flavors from Apple, Banana, Cherry and Durian*, then the combination of Apple, Banana and Cherry is the same as the combination Cherry, Apple, Banana.

Try to list all the possible combinations of 3 flavors taken from 4 before proceeding.

If we choose to shorten the name the fruits by selecting the first letter of their names, we only have 4 possible combinations for question above: ABC, ABD, ACD, and BCD. Notice that these are the only possible combinations. Also, observe that if we list the permutations of ABC, we have ACB, BAC,BCA, CAB and CBA.  This means that in permutations, we have counted each combination of 3 flavors from 4 flavors 6 times (or 3! times instead of one.

In other words, a combination is just like a subgroup of a group. For instance, if we want to find the number of subgroups containing 3 objects taken from 4 objects (or the combination of 4 objects taken 3 at a time), it is the same as asking  “how many possible groups of 3 objects can be taken from 4 objects?” In Figure 4, all the possible subgroups of 3 letters taken from 4 letters are displayed by the orange border. You also would have realized that the number of permutations is an overcounting of the number of combinations.

Figure 4 – The combinations of 4 objects taken 3 at a time is the same as the number of subgroups of 3 objects taken from 4 objects.

In Figure 2, ABC, ACB, BAC, BCA, CAB and CBA are permutations of Apples, Banana and Cherry. For each subgroup of 3, we realized that we counted 3! = 6 times. So, to get the number of combinations, we divide our number of permutations P(4,3) by the number of permutations of our subgroupP(3,3) = 3!. Therefore, we can say that the number of combinations of 4 objects taken 3 at a time is equal to

\displaystyle\frac{P(4,3)}{P(3,3)} = \frac{\frac{4!}{(4-3)!}}{3!} = \frac {4!}{(4-3)! 3!}

In general, to get the number of combinations of n objects taken k at a time, we have to divide the number of permutations of P(n,k) by the number of permutations of the subgroup P(k,k).

\displaystyle\frac{P(n,k)}{P(k,k)} = \frac{\frac{n!}{(n-k)!}}{n!} = \frac {n!}{(n-k)! k!}

The combinations of n objects taken k is usually denoted by C(n,k) or \displaystyle n \choose k

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*Durian is a fruit which can be found in the Philippines. It looks like a jackfruit.

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