Pythagorean theorem by contradiction

Question

This proof cannot be found in cut-the-knot.org nor in Loomis' collection.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Let $D$, $E$ and $F$ be the contact points of the incircle with $BC$, $AC$ and $AB$, respectively. Also, let $AE=AF=x$; $BD=BF=y$; $CD=CE=r$, where $r$ is the inradius of $\triangle{ABC}$.

Assume to the contrary that $a^2+b^2>c^2$. Then,

$$(r+y)^2+(x+r)^2>(x+y)^2.$$

Expanding, collecting like terms and simplifying we get

$$ry+rx+r^2>xy.$$

Notice that $ry+rx+r^2=r(y+x+r)=rs=\Delta$, where $\Delta$ denotes the area of $\triangle{ABC}$ and $s$ its semiperimeter. Moreover, it is well-known that $xy=\Delta$. So $ry+rx+r^2>xy$ is equivalent to write $\Delta>\Delta$, which is a contradiction. A similar situation arise if you assume $a^2+b^2<c^2$.

I got this proof rejected by an editor because, accoding to him, this is the same proof as his, just more complicated since I presented it as a contradiction.

This is the editor's proof:

For any triangle, $\Delta=rs$. (Standard argument by areas)

So we have $\frac{1}{2}ab=(s-c)s=\frac{1}{4}(a+b-c)(a+b+c)$ and this, after a couple of lines of simple algebra, yields $c^2=a^2+b^2$.

Is this really the same argument? Is always proof by contradiction more complicated than direct proofs?

Answer 1

Here's a direct proof using your symbols and formulas.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Let $D$, $E$ and $F$ be the contact points of the incircle with $BC$, $AC$ and $AB$, respectively. Also, let $AE=AF=x$; $BD=BF=y$; $CD=CE=r$, where $r$ is the inradius of $\triangle{ABC}$. Let $\Delta$ denote the area of $\triangle{ABC}$ and $s$ its semiperimeter.

Then $ry+rx+r^2=r(y+x+r)=rs=\Delta$ and it is well-known that $\Delta = xy,$ so $$ ry+rx+r^2 = xy.$$ By algebraic manipulation (on each side multiply by $2,$ then add $x^2 + y^2,$ then factor) we obtain $$(r+y)^2+(x+r)^2 = (x+y)^2,$$ that is, $a^2 + b^2 = c^2.$

---

Sometimes direct proof is difficult and proof by contradiction is simpler. In this case it is not.

You might have had more luck presenting a direct proof. I would argue that your proof is different because you require knowing not only that $\Delta = rs$ but also $\Delta = xy.$ This is not necessarily a point in favor of your proof; but my impression of the sites that have large collections of proofs is that they look for variation in methods, not strictly for optimality.

Answer 2

"Is always proof by contradiction more complicated than direct proofs?" Always? No. Here? Yes.

Your assumption that $a^2+b^2>c^2$ has no bearing on how the proof unfolds. You seem to assume an inequality just to have two sides of an in/equation to manipulate in tandem, which is unnecessary.

Without making any assumption, you could walk through your proof replacing "$>$" with, say, "$\bigcirc$" to represent an unknown, and unassumed, comparator ($>$, $<$, or $=$). So, you start with $a^2+b^2\bigcirc c^2$, and end with $\Delta\bigcirc\Delta$. This reveals that "$\bigcirc$" must have been "$=$" all along, and you're done. No contradiction required.

Of course, the "$\bigcirc$" formulation is a little unusual; all it really affords us is that two-sided in/equation structure, but we can do without that. A better approach is to combine those sides, and to consider the nature of the expression $a^2+b^2-c^2$. Unpacking, we get

$$a^2+b^2-c^2=(r+x)^2+(y+r)^2-(x+y)^2=2\left(\;r(y+x+r)-xy\;\right)=2\left(\;\Delta-\Delta\;\right)=0$$

This gives the result, again without contradiction.

Can you see how your contradiction is somewhat "artificial"?

---

For further opinions about the relative (de)merits of proofs-by-contradiction, search the site. You'll find questions like these

• Why some people don't like proofs by contradiction

• Are the “proofs by contradiction” weaker than other proofs?