At summer schools we often asked to state **Pythagoras’ theorem**. The most popular answer was short: \(a^2+b^2=c^2\). Initially we were rather shocked that many students haven’t probably seen any *properly stated* mathematical theorems.

Now we are more experienced and will give more hints. The first *light hint*: could you write Pythagoras’ theorem in more details, e.g. by using words? Here is the next popular attempt (literally!): “a squared plus b squared equals c squared”.

Then we ask: what do you mean by \(a^2+b^2=c^2\)? Is it an equation you are going to solve or what? Let us give a parallel example from history.

**Question**: who is the Queen of England?

**Answer**: the second!

The numerical answer is correct, but can you expect a full mark? Similarly in mathematics, the equation \(a^2+b^2=c^2\) says very little about Pythagoras’ theorem.

At this moment most students draw a triangle with sides a,b,c. With some extra help the triangle is marked as *right-angled*, not “right triangle”, which is another common confusion. So the statement now says: “A right-angled triangle with sides a,b,c satisfies \(a^2+b^2=c^2\)“. Almost correct: Pythagoras’ theorem is about *any* right-angled triangle, not just “a triangle”.

There is a slightly longer, but more *logical* statement: *if *a triangle with sides \(a\leq b<c\) is right-angled, *then* \(a^2+b^2=c^2\). Please notice that we have clearly separated the *geometric condition* (a triangle is right-angled) and the *algebraic conclusion* (\(a^2+b^2=c^2\) holds).

We continue: great, this is the correct first half of Pythagoras’ theorem, could you state the second half? Even good students who have easily written the first half are often stuck and admit that they don’t know the second half. Another hint: is the triangle with sides 3,4,5 right-angled? Everyone says “yes”, so we ask why. The popular answer: by Pythagoras’ theorem.

Then we repeat: do you really claim that our statement above implies that the triangle with sides 3,4,5 is right-angled? Most students are confused, but sometimes we receive this answer: “it’s just the opposite”. Opposite sides of most coins are different and so are opposite statements in mathematics.

It is easier to reverse the longer logical statement: “if \(a,b,c>0\) satisfy \(a^2+b^2=c^2\), then the triangle with the sides a,b,c is right-angled.” Notice how we swapped the two key parts of the statement: if the *algebraic condition* is given, then the *geometric conclusion* holds.

Now we ask to write two statements together as full Pythagoras’ theorem. A popular attempt is to write one phrase after another: “Any right-angled triangle with sides a,b,c satisfies \(a^2+b^2=c^2\). If \(a,b,c>0\) satisfy \(a^2+b^2=c^2\), then the triangle with the sides a,b,c is right-angled.” We clarify: can you write a shorter single phrase by making a logical connection between the algebraic and geometric parts?

After talking about “*if… then*“, “*only if*” or “*only when*“, we arrive at the following statement of full Pythagoras’ theorem: “Numbers \(a,b,c>0\) satisfy \(a^2+b^2=c^2\) *if and only if* the triangle with the sides a,b,c is right-angled.” The most important part is the keywords “*if and only if*” saying that the theorem works in both directions.

So any mathematical statement consists of 3 *key parts*:

- a
*given condition*(that is given and can be used later in a proof) - a
*logical connection*(between the condition and the conclusion) - a
*resulting conclusion*(that should be deduced from the condition).

Many statements are *one-sided*: if A then B (if a condition A is given, then a conclusion B holds). These statements play the role of a one-side bridge between villages A and B. Namely, villagers can go from A to B, but not from B to A.

All great theorems in mathematics are *two-sided*: A if and only if B, which establishes a two-sided bridge between A and B, so villagers can freely travel between A and B. For instance, Pythagoras’ theorem is a two-sided bridge between geometry and algebra.

Similar and more general statements led to the development of analytic geometry where we rephrase a geometric problem in algebraic terms, solve by using algebraic methods and then translate our solution back to geometry.

You have hopefully appreciated the fascinating fact that *triples of positive numbers* satisfying \(a^2+b^2=c^2\) are **equivalent** to *geometric shapes* of right-angled triangles. Simple Pythagoras’ theorem stated above can be generalised to many (even infinite) dimensions, also works for functions instead of segments and makes sense in very general so-called Hilbert spaces.

In all these cases Pythagoras’ theorem illustrates the **powerful principle**: *2-sided bridges are much better than 1-sided ones*.

**Riddle 5**: prove converse Pythagoras’ theorem (from algebra to geometry).**How to submit**: to write your full answer, simply submit a comment.**Hint**: create a new account, complete free tutorial on Pythagoras’ theorem.**Warning**: write few lines using simpler tools, not more general results.**Restriction**: only the first fully correct public answer will be rewarded.**Prize**: free 1-year access to one of our interactive web tutorials.

If you wish to receive automatic e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

viewing the inspirational videos i am actually surprised. thank you quite much for the blog. You have accomplished an excellent job

This website was… how do you say it? Relevant!! Finally I’ve found something which helped me. Thanks a lot!

Let us take some triangle (not necessarily right-angled) with sides of length a, b and c such that c^2 = a^2 + b^2.

The angle between a and b is opposite c, which we will label as C.

Using the cosine formula, c^2 = a^2 + b^2 – 2abCos(C). However, since c^2=a^2+b^2, 2abCos(C) = 0. Since a and b must be positive, non-zero numbers, 2abCos(C) = 0 when Cos(C)=0, i.e. when C is some multiple of 90. However, 0 degrees would create an impossible triangle, as would 180, thus C must be 90 degrees, and the triangle is right-angled.

Dear Henry, thank you for your attempt at the proof of Pythagoras’ theorem. First, you might have read our warning at the end of the post to use simpler tools, not more general results. Of course, Pythagoras’ theorem is a particular case of the more general cosine theorem. However, new results in mathematics are proved in the opposite direction: from simpler to more general. Second, you claimed that cos(C)=0 when C is a multiple of 90 (degrees). This claim fails for C=0 or C=180 (or infinitely many other angles). Moreover, the values C=0 and C=180 should be excluded not because of impossible triangles (3 points on the same lines are perfectly possible), but because cos(0)=1 and cos(180)=-1. We would advise you to follow our hint and try our free web tutorial on Pythagoras’ theorem at http://mastermathtest.com/tutorials/login/index.php. Then you can learn a 2-line proof of converse Pythagoras’ theorem, whose main idea is accessible to primary school students who never had about cosines. As a one-off exception, we could give you one more chance, so you could submit a better attempt and still win a prize.