# How to prove that the two angles below the apex of an isosceles triangle are equivalent

*You will need a pair of compasses, a ruler, pen and pencil to formulate this proof.

How would you go about proving that an isosceles triangle has two angles (below its apex) equal to one another?

Well, first of all – let’s start off by drawing a circle…

Now… We can tell that the circle we’ve just drawn has a centre (point at the centre). Next, what we have to do is add a couple of points to the edge of this circle. Like this…

Let’s name all these points A, B and C…

Now, let’s connect these points together with a few lines – to create an isosceles triangle ABC…

Ok… So far, so good… What you will need to do now is – place the needle of your compass on the point C and your pencil on the point B, like this…

Now spin your compass – and create an arc…

Next, get the needle of your compass and place it on the point B and put your pencil on the point C…

Draw another arc, like this…

Where the two arcs you’ve just drawn intersect, create a point… Call this point D…

Now, draw a line going through the points A and D. Call this line L. Line L will be perpendicular to the line BC…

Where the line L intersects the line BC, create a point E…

Now, it turns out, within the isosceles triangle ABC, we’ve created two right angles… This is because the line L is perpendicular to the line BC. Remember that the line L cuts the isosceles triangle down its centre. Let’s name these right angles big R…

If you look at the diagram above carefully, what you will notice is that the radius of the circle is equal in length to the line AB and also the line AC. Let’s name the lines AB and AC… We’ll call them r.

Let’s also name the line BC… We’ll call it x. This means that the line BE is equal to half of x, and because of this, the line CE must also be equal to half of x…

Finally (I know you must be tired of drawing), let’s call the angle ABC alpha and the angle ACB beta…

With our diagram complete, we can now prove that alpha and beta are equivalent to each other.

*You will need to know a bit of trigonometry to pass this point. SOH CAH TOA rules to be precise.

It turns out that:

$\cos { \left( \alpha \right) } =\frac { A }{ H } =\frac { \frac { x }{ 2 } }{ r } =\frac { x }{ 2r }$

And also:

$\cos { \left( \beta \right) } =\frac { A }{ H } =\frac { \frac { x }{ 2 } }{ r } =\frac { x }{ 2r }$

This means that:

$\cos { \left( \alpha \right) } =\cos { \left( \beta \right) } \\ \\ \therefore \quad \alpha =\beta$

Hence, we’ve proven that an isosceles triangle has two angles (below its apex) equal to one another.

# Proof: Opposite angles formed when two lines intersect, are equal to one another

How can we prove that opposite angles (when two lines intersect) are in fact equal to one another?

Well, first of all – let’s draw a circle…

We know that in a full circle, there are 360 degrees. This is an indisputable fact. Now, what happens if we split this circle in two with a straight line (going through its centre)?

Well, each half of the circle (top and bottom) – will now contain 180 degrees. We know this because:

$\frac { 360 }{ 2 } =180$

Ok, so far so good… Now, let’s draw another line through the circle (going through its centre) which intersects the first line we have drawn…

As we can see, because we have done this, we now have 4 different angles. Let’s name the two angles which are situated in the top half of the circle α and β

Earlier in this demonstration, we remarked that the top half of the circle (when it was split in two) contained 180 degrees. Mathematically and logically speaking, as this is the case, we must say that:

$\alpha +\beta =180$

Great, now let’s name the angles in the bottom half of the circle x and y

It follows, because the angles in the top half of the circle add up to 180 degrees, we must deduce that:

$x+y=180$

So, it turns out we now have two useful equations:

$\alpha +\beta =180$

$x+y=180$

Do we have enough to form our proof though? Unfortunately, not quite… We have to look at our most recent figure again, but this time from a different perspective…

You see, there are different top halves and bottom halves…

• There exists top halves α+β and also… β+y
• There exists bottom halves x+y but also α+x

You may ask, why is this important? Well, here’s what’s crucial:

$\beta +y=180$

$\alpha +x=180$

And now we have 4 different equations, 3 of which – will help us finally complete our proof.

$\alpha +\beta =180$

$x+y=180$

$\beta +y=180$

$\alpha +x=180$

Here’s why we need these equations…

α+β and α+x are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\alpha +x\\ \\ \therefore \quad \beta =x$

*Subtract α from both sides of the equation.

α+β and β+y are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\beta +y\\ \\ \therefore \quad \alpha =y$

*Subtract β from both sides of the equation.

Hence, we’ve proven that: Opposite angles (when two lines intersect) are equal to one another.

β=x and α=y:

# How to quickly double the area of a square (simple geometry lesson)

In this post, I’ll be demonstrating how you can quickly double the area of a square using a simple geometrical trick.

Let’s say you have an ordinary square, like the one below…

Firstly, what you have to do is name the area of this square “A”…

Then, what you do next is divide this square (diagonally) into 4 equal parts…

After you have done this, you then name each part of this square “1/4 x A”…

Notice now, that to double the area of this square, all you have to do, is double the number of the 1/4 x A right angled triangles which currently exist – then configure them – like this…

As you can see, you’ve now got eight of these 1/4 x A right angled triangles neatly configured…

Not only are you left with a new square, double the size of your original square (follow the lines on the outside of the shape), but a handy equation, which proves that you doubled the area of the square you started off with…

$8\times \frac { 1 }{ 4 } A=2A$

# How to derive the formula for a circle from scratch

If you’d like to derive the formula for a circle from absolute scratch, then your best option would be to draw a diagram such as the one below:

If you look at this diagram carefully, what you will notice is:

• A circle exists and each point on this circle has the coordinate (x, y).
• The centre of the circle can be found at (a, b).
• The circle has a radius ‘r’.
• The right angled triangles in the diagram each have an adjacent length, opposite length and hypotenuse (r).

Once you’ve prepared a similar diagram, your next aim should be to turn your attention towards the right angled triangles which exist within the circle. You should also think about the many different right angled triangles which could fit within the circle provided they emanate from the centre point (a, b).

The reason I’ve mentioned these right angled triangles is because according to Pythagoras’ theorem, when you have a right angled triangle – its adjacent length squared plus its opposite length squared is equal to the length of its hypotenuse squared:

Now, in this case – the adjacent lengths of the right angled triangles which can fit within the circle on the diagram can be described using the expression:

$\left( x-a \right)$ or $\left| x-a \right|$

The opposite lengths can be described using the expression:

$\left( y-b \right)$ or $\left| y-b \right|$

Also, very interestingly:

• Each of the right angled triangles you can think of has a hypotenuse ‘r’.
• ${ \left( x-a \right) }^{ 2 }={ \left| x-a \right| }^{ 2 }$
• ${ \left( y-b \right) }^{ 2 }={ \left| y-b \right| }^{ 2 }$

When you combine all the information above, what you get is a neat formula which looks like this:

${ \left( x-a \right) }^{ 2 }+{ \left( y-b \right) }^{ 2 }={ r }^{ 2 }$

And it turns out… This is the formula for a circle on the x, y plane, whereby, (a, b) is the centre of the circle and ‘r’ is the length of its radius. How spectacular is that? 🙂

# How to prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically

In this post I’ll be demonstrating how one can prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically…

First of all, let me show you this diagram…

sin(A-B)=sin(A)cos(B)-cos(A)sin(B) proof

*If you click on the diagram, you will be able to see its full size version.

Now, to begin with, I will have to write about some of the properties related to the diagram…

Property 1:

Angle B + (A – B) = B + A – B = A

Therefore, angle POR = A.

Property 2:

Angle OPS = 90 degrees

Property 3:

Length OS = 1

Also note:

All angles within a triangle on a flat plane should add up to 180 degrees. If you understand this rule, you will be able to discover why the angles shown on the diagram are correct. Angles which are 90 degrees are shown on the diagram too.

PROVING THAT SIN(A-B)=SIN(A)COS(B)-COS(A)SIN(B)

Since I’ve noted down some of the important properties related to the diagram, I can now focus on demonstrating why the formula above is true. I will demonstrate why the formula above is true using mathematics and the SOH CAH TOA rule…

$\sin { \left( A-B \right) } =\frac { O }{ H } =\frac { ST }{ 1 } =ST$

But it turns out that…

$ST=PR-PQ$

Because:

$QR=ST$

Now, what is PR and what is PQ?

$\sin { \left( B \right) } =\frac { O }{ H } =\frac { PS }{ 1 } =PS\\ \\ \cos { \left( B \right) } =\frac { A }{ H } =\frac { OP }{ 1 } =OP\\ \\ \sin { \left( A \right) } =\frac { O }{ H } =\frac { PR }{ \cos { \left( B \right) } } \quad \\ \\ \therefore \quad \sin { \left( A \right) } \cos { \left( B \right) } =PR\\ \\ \cos { \left( A \right) } =\frac { A }{ H } =\frac { PQ }{ \sin { \left( B \right) } } \\ \\ \therefore \quad \cos { \left( A \right) } \sin { \left( B \right) } =PQ$

And finally, to sum it all up:

$ST=PR-PQ\\ \\ \therefore \quad \sin { \left( A-B \right) =\sin { \left( A \right) } \cos { \left( B \right) } -\cos { \left( A \right) } \sin { \left( B \right) } }$

Need a better explanation? Watch this video…

Related Videos:

Related posts:

Simple But Elegant Way To Prove That sin(A+B)=sinAcosB+cosAsinB (Edexcel Proof Simplified)

# Properties of C squared, Pythagorean Theorem

In this post, I’ll be writing about some peculiar properties of C squared in Pythagoras’ theorem.

Look at this diagram very carefully…

*What are the weird properties of C^2..? It turns out that A1=A2 and A3=A4. A2 + A4 = C^2.

It turns out out that area A1 is equal to area A2, and that area A3 is equal to area A4:

A1 = A2

A3 = A4

This can be proven because:

1. ${ A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$
2. ${ x }^{ 2 }+{ D }^{ 2 }={ B }^{ 2 }$
3. ${ \left( C-x \right) }^{ 2 }+{ D }^{ 2 }={ A }^{ 2 }$

Now, due to the above:

${ D }^{ 2 }={ B }^{ 2 }-{ x }^{ 2 }\\ \\ { D }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ \therefore \quad { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-\left\{ { C }^{ 2 }-2Cx+{ x }^{ 2 } \right\} \\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx-{ x }^{ 2 }\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-\left\{ { A }^{ 2 }+{ B }^{ 2 } \right\} +2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ A }^{ 2 }-{ B }^{ 2 }+2Cx\\ \\ { B }^{ 2 }=-{ B }^{ 2 }+2Cx\\ \\ 2{ B }^{ 2 }=2Cx\\ \\ \therefore \quad { B }^{ 2 }=Cx\\ \\$

But… B^2 is actually the area A1 and Cx is the area A2, which means that A1=A2.

Now, if B^2=Cx, this means that:

${ A }^{ 2 }+Cx={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }={ C }^{ 2 }-Cx\\ \\ { A }^{ 2 }=C\left( C-x \right) \\ \\$

However, A^2 is equal to the area A3, and C(C-x) is equal to the area A4 – which means that A3=A4. Hence, we’ve proven that:

A1=A2

A3=A4

Related:

2 ways to derive Pythagoras’ equation from scratch

# 2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements…

*Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is:

${ \left( A+B \right) }^{ 2 }$

It is also:

$4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }$

Which means that:

${ \left( A+B \right) }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that:

$4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right) }^{ 2 }={ C }^{ 2 }$

And since:

${ \left( B-A \right) }^{ 2 }\\ \\ ={ \left( A+B \right) }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }$

I was able to say that:

$4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??

Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation

# How to add up all the even numbers from 0 onwards quickly

In this post, I’ll be demonstrating how you can add up all the even numbers from 0 onwards.

Adding up all the even numbers from 0 to 2:

In this diagram, we are going to say that n=2. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 2 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 4:

In this diagram, we are going to say that n=4. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 4 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 6:

In this diagram, we are going to say that n=6. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 6 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 8:

In this diagram, we are going to say that n=8. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 8 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

What we’ve discovered:

We’ve discovered that a simple formula can be used to add up all the even numbers from 0 to “n”, whereby “n” is an even number. This formula is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Alternative method:

There is also an alternative formula you can use to add up even numbers, from 0 onwards. That is:

# How to add up odd numbers from 0 upwards

In this post, I’ll be demonstrating how to add up all the odd numbers from 0 to any specific odd number. To create a robust demonstration, I’ll be taking the footsteps below:

• I’ll first be showing you how to add up all the odd numbers from 0 to 1, using a diagram and formula.
• I’ll then be showing you how to add up all the odd numbers from 0 to 3, using a diagram and formula.
• I’ll also be showing you how to add up all the odd numbers from 0 to 5, using a diagram and also the same formula which was used to count up all the odd numbers from 0 to 1 and 0 to 3.
• And finally, I’ll be using similar diagrams and formulas used to count odd numbers from 0 to 1, 0 to 3 and 0 to 5 to count odd numbers from 0 to 7 and 0 to 9.

What you will find, after I complete the tasks above – is that a pattern emerges. You will notice that the formula I use to count odd numbers from 0 to n (n which is an odd number) is very robust and will allow you to count all the odd numbers from 0 to n very easily.

COUNTING ALL THE ODD NUMBERS FROM 0 to 1:

If you count all the odd numbers from 0 to 1, what you will get is obviously 1. Furthermore, what you will also get as a formula (if n=1, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 1 into n, you will get 1. 1 is the value of all the odd numbers added up from 0 to 1.

COUNTING ALL THE ODD NUMBERS FROM 0 to 3:

If you count all the odd numbers from 0 to 3, what you will get is 4. Furthermore, what you will also get as a formula (if n=3, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 3 into n, you will get 4. 4 is the value of all the odd numbers added up from 0 to 3.

COUNTING ALL THE ODD NUMBERS FROM 0 to 5:

If you count all the odd numbers from 0 to 5, what you will get is 9. Furthermore, what you will also get as a formula (if n=5, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 5 into n, you will get 9. 9 is the value of all the odd numbers added up from 0 to 5.

COUNTING ALL THE ODD NUMBERS FROM 0 to 7:

If you count all the odd numbers from 0 to 7, what you will get is 16. Furthermore, what you will also get as a formula (if n=7, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 7 into n, you will get 16. 16 is the value of all the odd numbers added up from 0 to 7.

COUNTING ALL THE ODD NUMBERS FROM 0 to 9:

If you count all the odd numbers from 0 to 9, what you will get is 25. Furthermore, what you will also get as a formula (if n=9, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 9 into n, you will get 25. 25 is the value of all the odd numbers added up from 0 to 9.

THE FORMULA WHICH CAN BE USED TO ADD UP ALL THE ODD NUMBERS FROM 0 TO n, WHEREBY n IS AN ODD NUMBER:

If you look at each and every diagram and formula above, what you will notice is that the formula

$Formula=\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

will allow you to add up all the odd numbers from 0 to n, whereby n is an odd number. The diagrams above have demonstrated why this formula is robust and completely logical. If you need to add up all the odd numbers from 0 to n (n is an odd number), the formula above is one you can trust.

ALTERNATIVE METHOD:

Using the table below, we can come up with an alternative method of calculating every odd number from 0 to n (n is an odd number):

n: Sum Total Total (Exponential form)
1 1 1 1^2
3 1+3 4 2^2
5 1+3+5 9 3^2
7 1+3+5+7 16 4^2
9 1+3+5+7+9 25 5^2

It turns out that:

*Note that 2x+1 can be used to denote an odd number.

# Solving The Student Handshake Problem

The other day, a question came up on a site called Brainly.com.

It went like this…

In a cafeteria, all students shook hands with one another. There were 66 handshakes in total. How many students were in the cafeteria?

As this question is quite interesting, I’m going to explain how you can answer it, and in the process – I’ll also be revealing its answer.

Now, to answer such a question we first have to perform a few experiments and ask ourselves mini questions. The data from these experiments and mini questions will have to be recorded, so that we can spot potential patterns which may ultimately help us create a formula to solve the main problem.

EXPERIMENTS + MINI QUESTIONS

Will a pattern emerge??

1. Firstly, let’s think about how many handshakes there’d be with only one student in this cafeteria. Well, we can say 0. Why would someone shake their own hand?

2. Secondly, how many handshakes would there be if there are 2 students in this cafeteria? Well, the answer to this question is 1. These two students would be able to shake hands with one another.

3. Thirdly, how many handshakes would there be if there are 3 students in the cafeteria? Haha, now things get a little more complicated… To answer this mini question, let’s attach the variables A, B and C to these students {A, B, C}.

It turns out that:

• A can shake hands with B (A,B).
• A can shake hands with C (A,C).
• B can shake hands with C (B, C).

*Possible combinations: (A, B), (A, C) and (B, C).

So the answer to this mini question has to be 3.

4. Fourthly, how many handshakes would there be if there are 4 students in this cafeteria? To answer this question we can use the same strategy we used to answer the third question. Let’s attach the variables A, B, C and D to these students {A, B, C, D}.

It turns out that:

• A can shake hands with B (A,B).
• A can shake hands with C (A, C).
• A can shake hands with D. (A, D).
• B can shake hands with C (B, C).
• B can shake hands with D (B, D).
• C can shake hands with D (C,D).

* Possible combinations: (A, B), (A, C), (A, D), (B, C), (B, D) and (C, D).

So, the answer to this mini question would have to be 6.

5. Fifthly, how many handshakes would there be if there are 5 students in this cafeteria? Using the same strategy we used to answer mini questions 3 and 4 – we will answer this question too. Let’s attach the variables A, B, C, D and E to these students {A, B, C, D, E}.

It turns out that:

• A can shake hands with B (A, B).
• A can shake hands with C (A, C).
• A can shake hands with D (A, D).
• A can shake hands with E (A, E).
• B can shake hands with C (B, C).
• B can shake hands with D (B, D).
• B can shake hands with E (B, E).
• C can shake hands with D (C, D).
• C can shake hands with E (C, E).
• D can shake hands with E (D, E).

So, the answer to this mini question would have to be 10.

Possible combinations: (A,B), (A, C), (A, D), (A, E), (B, C), (B, D), (B, E), (C, D), (C, E) and (D, E).

CAN WE SOLVE THE MAIN PROBLEM WITH A FORMULA? WHAT PATTERN WILL DEFINE THE FORMULA?

Alright… Now that we’ve performed a few experiments and have answered a few mini questions – let’s see if we can spot a pattern in our data. If we can spot a pattern in our data, we may be able to solve the problem relating to 66 handshakes. We need to find a pattern so that we don’t have to answer the main question using brute force and hundreds, if not, thousands of calculations. Remember, solving mathematical problems is all about spotting patterns.

To spot patterns, the best tool we can use is a table. Let’s create a table which contains the information we’ve just produced, related to the mini questions…

Student(s) Handshakes Pattern (Related to handshakes)
1 0 0
2 1 1
3 3 1+2
4 6 1+2+3
5 10 1+2+3+4

Ok… Let’s look at this table carefully. It turns out that a pattern has emerged… As a pattern, we get tidy little sums. The kind of sums that Carl Friedrich Gauss was able to add up, thanks to diagrams such as the one below…

Diagram Explanation:

To add up the sum 1+2+3+4, you simply have to multiply 5 (which is the variable ‘s’ in this case) by (5-1) which is 4, then divide their product ( 5 x (5-1) ) by 2.

(5×4)/2 = 10 = 1+2+3+4.

Notice that:

• When there was one student in the cafeteria, there were 0 handshakes. (0) is 1 less than the number 1.
• When there were two students in the cafeteria, there was 1 handshake. (1) is 1 less than 2.
• When there were 3 students in the cafeteria, there were 3 handshakes. 3 =1+(2). 2 is 1 less than 3.
• When there were 4 students in the cafeteria, there were 6 handshakes. 6=1+2+(3). 3 is 1 less than 4.
• When there were 5 students in the cafeteria, there were 10 handshakes. 10=1+2+3+(4). 4 is 1 less than 5.

Also notice that:

*To understand the pattern below and how it was intuitively discovered, see the diagram which helped Carl Friedrich Gauss neatly add up sums such as 1+2+3+4.

• [ 1 x (1-1) ] / 2 = 0 which is the same as : [ 1 x 0 ] / 2 = 0
• [ 2 x (2-1) ] / 2 = 1 which is the same as : [ 2 x 1 ] / 2 = 1
• [ 3 x (3-1) ] / 2 = 3 which is the same as : [ 3 x 2 ] / 2 = 3
• [ 4 x (4-1) ] / 2 = 6 which is the same as : [ 4 x 3 ] / 2 = 6
• [ 5 x (5-1) ] / 2 = 10  which is the same as : [ 5 x 4 ] / 2 = 10

With this information, we can conclude that:

s = number of students

h = handshakes

[ s x (s-1) ] / 2 = h

And this is the formula we can use to solve all student handshake problems such as the one mentioned at the top of this post. If we plug the value 66 into this formula, we will discover how many students there were in the cafeteria whereby 66 handshakes took place. At the beginning of this post, I said that I would reveal the answer to the main question. To reveal it though, I will have to solve a quadratic equation by completing the square. I will also have to turn the variable ‘h’ into 66. Let’s do this…

$\frac { s\left( s-1 \right) }{ 2 } =66\\ \\ s\left( s-1 \right) =132\\ \\ { s }^{ 2 }-s=132\\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=132\\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=132+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }$

${ \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=132+\frac { 1 }{ 4 } \\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 4\left( 100+30+2 \right) }{ 4 } +\frac { 1 }{ 4 } \\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 400+120+8 }{ 4 } +\frac { 1 }{ 4 } \\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 529 }{ 4 } \\ \\ s-\frac { 1 }{ 2 } =\sqrt { \frac { 529 }{ 4 } } \\ \\ s-\frac { 1 }{ 2 } =\frac { 23 }{ 2 } \\ \\ s=\frac { 23 }{ 2 } +\frac { 1 }{ 2 } \\ \\ s=\frac { 24 }{ 2 } \\ \\ \therefore \quad s=12$

We now know that there were 12 students in the cafeteria. Obviously, this problem could have been solved when we knew that s x (s-1) = 132, because 12 x 11 = 132. However, if you get a larger problem, you will need to produce a quadratic formula and complete the square to get an answer.

I hope that this post has shed light on how to solve handshake / people problems. If you have any questions or feedback, please leave a comment below. 🙂