## miércoles, 20 de mayo de 2020

### Another Simple Proof of Johnson's Theorem

Other proofs can be found in cut-the-knot.org. See also Johnson's Three Circles Theorem Revisited and Johnson's theorem proof.

Johnson's theorem: Let three equal circles with centers $J_a$, $J_b$, and $J_c$ intersect in a single point $H$ and intersect pairwise in the points $A$, $B$, and $C$. Then the circumcircle of the triangle $\triangle{ABC}$ is congruent to the original three.

Proof. Since $\odot{ABH}$ and $\odot{BCH}$ are congruent and as $\angle{BAH}$ and $\angle{BCH}$ are angles subtended by the same arc, it follows that $\angle{BAH}=\angle{BCH}$. Analogously, $\angle{CAH}=\angle{CBH}$. Let $O$, $J_a$ be the centers of $\odot{ABC}$ and $\odot{BCH}$, respectively, then

$$\angle{COB}=2\angle{CAB}=2\angle{CAH}+2\angle{BAH}=2\angle{CBH}+2\angle{BCH}=\angle{BJ_aC}.$$

It follows that $\triangle{BCO}$ and $\triangle{BCJ_a}$ are similar isosceles triangles sharing a common side, $BC$, so by $ASA$ we deduce that $\triangle{BCO}\cong{\triangle{BCJ_a}}$. Consequently, $\odot{ABC}$ is congruent to the original three.

$\square$

Note: The point $H$ may cross the side lines of the triangle $\triangle{ABC}$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.

## sábado, 9 de mayo de 2020

### The Pythagorean Theorem by Reductio ad Absurdum

Other proofs by contradiction can be found in cut-the-knot.org (see proof #122) and in Loomis' collection (see proofs # 16/2 and # 32).

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 $a^2+b^2<c^2$.

There is a simple direct proof starting from $\Delta=rs$ and $\Delta=xy$, can you find it? My sincerest thanks to Andrius Navas and José Hernández for pointing me out this.

## jueves, 7 de mayo de 2020

### Yet Another Proof of the Pythagorean Theorem - Simpler version

In cut-the-knot.org you can find more than a hundred proofs of the Pythagorean theorem. Here I give another proof which I hope to be new.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Below I will be using standard notations: $a$, $b$, $c$, for the side length, $s$ for the semiperimeter, $\Delta$ for the area and $r$ for the inradius.

$$s=c+r$$
$$s-\frac{c}{2}=\frac{c}{2}+r$$
$$\frac{a}{2}+\frac{b}{2}=\frac{c}{2}+r$$
$$a+b=c+2r$$

Squaring both sides,

$$a^2+2ab+b^2=c^2+4cr+4r^2$$

It is well-known $\Delta=rs$, which implies $2ab=4r(c+r)$. So the equation $a^2+2ab+b^2=c^2+4cr+4r^2$ can be rewritten like this $a^2+b^2=c^2$, which is the Pythagorean theorem.

Below is the comment by John Molokach, whom I consider an expert on this topic (he has proved the theorem in more than ten different ways).

## miércoles, 6 de mayo de 2020

### Yet Another Proof of the Pythagorean Theorem

In cut-the-knot.org you can find more than a hundred proofs of the Pythagorean theorem. Here we give another proof which we hope to be new.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Denote $D$, $E$ the contact points of the incircle and sides $BC$ and $AC$, respectively. Below I shall be using standard notations: $a$, $b$, $c$, for the side length, $p$ for the semiperimeter and $r$ for the inradius.

$$p=c+r$$
$$p-\frac{c}{2}=\frac{c}{2}+r$$
$$\frac{a}{2}+\frac{b}{2}=\frac{c}{2}+r$$
$$a+b=c+2r$$

Squaring both sides,

$$a^2+2ab+b^2=c^2+4cr+4r^2$$

Construct a parallel line, $l$, to $AB$ passing through $I$, the incenter of $\triangle{ABC}$. Call $H$ and $G$ the orthogonal projections of $B$ and $A$ onto $l$, respectively. Then, $[ABC]=[CDIE]+[ABHG]$. Indeed, let $K$, $J$ be the intersections of $l$ with $AC$, $BC$, respectively. Notice that $[ABC]=[ABJK]+[CDIE]+[EIK]+[DJI]$. Clearly, $\triangle{AGK}\cong{\triangle{EIK}}$ and $\triangle{BHJ}\cong{\triangle{DJI}}$, hence $[ABC]=[CDIE]+[ABGH]=r^2+cr$. As a consequence, $2ab=4r^2+4cr$. Now, the equation $a^2+2ab+b^2=c^2+4cr+4r^2$ can be rewritten like this $a^2+b^2=c^2$, which is the Pythagorean theorem.

$\square$

I realized the construction was not necessary, so I have published a simpler version of this proof here.

## lunes, 4 de mayo de 2020

### Galileo's Genius: The Experiment That Never Was.

Stolen from Clueless Fundatma.

Many of us have this mental picture of Galileo standing on top of the Leaning Tower of Pisa and dropping a feather and stone to watch them fall on the ground simultaneously, thus upsetting the Aristotelian assertion that "heavier objects fall faster".

Of course, we know Galileo couldn't possibly have done that experiment, because if he actually had, the presence of air would have reaffirmed the Aristotelian world-view, and the feather would have fallen later.

So what exactly did Galileo do?

I haven't read his original work, but my PhD advisor - Ron Larson - told us this charming story during a class I took with him, and its stuck with me since.

Apparently Galileo did a thought experiment.

He used the idea of "reductio ad absurdum", which begins by assuming something that we wish to prove as false, and hitting a contradiction.

The proof is beautiful, and here is a rough outline.

Consider two objects, with the heavier object of mass M, and the lighter object of mass m. Let us assume that "heavier objects fall faster" - which is opposite of what we wish to prove.

Further consider a really tall tower. If you can think of an infinitely tall tower, that is even better.

Now we've assumed that mass M falls faster than mass m. Here's the beautiful construction: Consider tying the two masses with a piece of string - to create a new object of mass M + m, as shown below.

Let us drop this composite object from the top of our tall tower, and think about what may happen. Because of what we've assumed, the mass M is going to fall faster than m, and in a short while the composite object (that is still falling, remember) will have adopted a configuration that looks more like the following.

Let us now consider the tension inside the string (red line in my diagram). The big mass tugs on the small mass, while the small mass tugs back.

Thus, the small mass slows the big mass down.

Consequently, the composite object of mass (M+m) falls slower than the unfettered object of mass M. But the composite object is heavier, and should fall faster.

### Reductio ad absurdum or the contrapositive?

Response by
Although the other answers correctly explain the basic logical equivalence of the two proof methods, I believe an important point has been missed:

With good reason, we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal)

"What is the reason? The reason is the fecundity of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r1, r2, before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r1 and r2 and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds."

Similarly, when we prove the contrapositive (¬q implies ¬p) directly, we assume ¬q, make intermediary conclusions r1, r2, and then finally conclude ¬p. Thus, we have also established not only that ¬q implies ¬p, but also, that it implies r1 and r2 and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (¬r1 implies q), we learn about many different hypotheses that all imply q.

These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q).

With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r1, r2, and so on, before arriving at a contradiction. The statements r1 and r2 are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistent, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.

I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof.

Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p1, ..., pn are all the primes. It follows that since none of them divide the product-plus-one p1...pn+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p1, ..., pn are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p1, ..., pn were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p1, ..., pn are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p1...pn+1 has merely a prime divisor not on the original list.