I'm back with another mathematical fallacy which I found to be dangerously devious when I saw it for the first time. To be honest, I couldn't spot the mistake in this one until I'd drawn an accurate picture. Before we dive into this fallacious proof, a word of advice: never assume anything without proof!

We have a circle with center \(A\) and radius \(r\). \(C\) is a point inside the circle. \(B\) is a point such that \(AB\cdot AC=r^2\). We're going to draw the perpendicular bisector of \(BC\) and let that bisector intersect the circle at \(E\) and \(D\). \(F\) is the midpoint of \(BC\).

Now on to the proof!

\[r^2=AB\cdot AC\]

\[\Rightarrow r^2=(AF+BF)(AF-CF)=(AF+CF)(AF-CF)=AF^2-CF^2 \cdots(1)\]

Now we're going to use the Pythagorean theorem.

\(AF^2=AE^2-EF^2=r^2-EF^2\) and

\(CF^2=CE^2-EF^2\).

Let's make these substitutions in \((1)\).

We have,

\[r^2=r^2-EF^2-CE^2+EF^2\].

Now, if you do the calculations, you have :

\[CE=0\]

Take a moment to understand what this statement is saying. Take a look at the picture if you have to.

I know it's hard to believe but this means that \(C\) is actually *on the circle*. But that is insane. By definition, \(C\) is a point inside the circle. What went wrong?

Before you read the rest of the article, think about this for a while. Review the steps. And most importantly, don't assume anything without proof.

Now how do we solve this apparent contradiction? As I said earlier, if you try to draw an accurate picture, you'll see what's happening here. But I find the picture here a bit *uninteresting*. So, draw an accurate picture if you want to. I'm not going to do that here. Instead, we're going solve this with pure logic. Let's begin!

Okay, since \(r>AC\), we have \((r-AC)^2>0\).

Rewrite this as \(r^2+AC^2>2\cdot r \cdot AC\).

Multiply both the sides by \(\frac{1}{2 \times AC}\) to get:

\[\frac{1}{2}(\frac{r^2}{AC} +AC)>r\]

Remember that \(\frac{r^2}{AC}=AB\). So, we have:

\[\frac{1}{2}(AB+AC)>r\]

\[\Rightarrow \frac{1}{2} (AC+BC+AC)>r\]

\[\Rightarrow AC+\frac{1}{2}\times BC>r\]

\[\Rightarrow AC+CF>r\]

\[\Rightarrow AF>r\]

This actually solves the whole thing! Because it tells us that the point \(F\) is actually *outside* the circle and the points \(D\) and \(E\) don't even exist! We only *assumed* they existed but we never proved it. If these points don't exist, all the arguments mentioned above are meaningless.

As we'll see again and again, implicit [& incorrect] assumptions will come back to haunt us.

## Comments

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TopNewestI'm back with another post. Feel free to post your feedback in the comments section.

I see that Cosines Group is "For Level 2-3 members, an immersion in concepts beyond the standard curriculum". So, also let me know if this falls outside the Cosines Group range. – Mursalin Habib · 3 years ago

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– Snehal Shekatkar · 3 years ago

Hi Mursalin, can you suggest me good way to start studying Euclidean geometry?Log in to reply

For starters, I'd recommend the 'Geometry for Americans' section in Paul Zeitz's book, "The Art & Craft of Problem Solving" (2nd edition). For solving Olympiad level problems, THE book I can recommend you is 'Geometry Revisited' by Coxeter and Greitzer.

The most important thing while studying geometry is making sure that you understand formal proof structures and checking if you can apply the theorems you've learnt. For example, when you have an angle bisector and anything with ratios, you have to start thinking about the angle bisector theorem. And there's no better way to do that than solving problems, lots of them.

I hope this helps! – Mursalin Habib · 3 years ago

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– Snehal Shekatkar · 3 years ago

Thanks Mursalin.. but I can see that these books are quite expensive.. how did you get them?Log in to reply

If you don't want to buy the books, no problem! All you need is the index or the 'contents' page. And you can view that on Amazon. Then you can learn about those topics from free sources like Wikipedia, Mathworld or Aops. Yes, that way you can't learn the topics from the author's own words, but you can learn. There are free resources all over the internet. – Mursalin Habib · 3 years ago

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I got a little suspicious when I saw that \(F\) was assumed to be inside the circle. Since as \(C\) approaches \(A\), then \(B\) approaches infinity; surely \(F\) cannot go left of the intersection of \(AB\) and the circle...

Turns out I was right :) – Daniel Liu · 3 years ago

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– Mursalin Habib · 3 years ago

That is actually a great way to think about it. The more time you spend with fallacies like these, the better you get spotting the fallacious steps. Good job!Log in to reply

– Tanya Gupta · 2 years, 10 months ago

I didnt think of it that way...nice!!Log in to reply

Nice one :) Got me trick for a while too :) – Happy Melodies · 3 years ago

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So we can't argue with no proof before. Really nice post! – Muh. Amin Widyatama · 3 years ago

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so coool!! – Jordi Bosch · 2 years, 12 months ago

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