Mathematical Biscuit I

 Can you find two irrational numbers $a,b$ such that $a^b$ is rational? Surprisingly, yes and the argument is very easy. 

If $\sqrt{2}^{\sqrt{2}}$ is rational then we are done.

If $\sqrt{2}^{\sqrt{2}}$ is irrational then $$(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}=\sqrt{2}^{\sqrt{2} \sqrt{2}}=\sqrt{2}^2=2$$ and we are done :)

PS: It turns out that $\sqrt{2}^{\sqrt{2}}$ is indeed irrational (infact transcendental). This is a consequence of an advanced result called Gelfond-Schneider Theorem. However this was irrelevant to our argument.

Ceva's Theorem

This is a classical result which is very often quite useful. Consider $\triangle ABC$ and three points $P,Q$ and $R$ on $BC, AC$ and $AB$ respectively (segments of this form are called cevians). Then Ceva's theorem states that:

The three cevians $AP, BQ$ and $CR$ are concurrent (meet at a point) if and only if $$\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}=1$$

The quantity $\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}$ is called Ceva's Ratio and it is determined upto reciprocal.

Convince yourself by playing with the following GeoGebra applet (use $P, Q$ and $R$, vertices won't change the ratio)

 

Let us first assume the given cevians are concurrent (as shown below):

We will use the fact that ratio of areas of triangles with equal altitudes is equal to ratio of their bases. Area of $\triangle XYZ$ will be denoted as $XYZ$.

$$\frac{ARC}{RBC}=\frac{AR}{RB}=\frac{ARO}{RBO}$$

Using properties of ratio:

$$\frac{AR}{RB}=\frac{ARC-ARO}{RBC-RBO}=\frac{ACO}{BCO}$$

Similarly $\frac{BP}{PC}=\frac{BAO}{CAO}$ and $\frac{CQ}{QA}=\frac{CBO}{ABO}$. Thus,

$$\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}=\frac{ACO}{BCO}\frac{BAO}{CAO}\frac{CBO}{ABO}=1$$

Conversely, assume $\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}=1$. Suppose $AP$ and $BQ$ meet at $O$. Suppose $CO$ intersect line $AB$ at $R'$ ($R'$ lies on segment $AB$, for more details see below). Then from above, $\frac{AR'}{R'B}\frac{BP}{PC}\frac{CQ}{QA}=1$. Using the given hypothesis,  $\frac{AR'}{R'B}\frac{BP}{PC}\frac{CQ}{QA}=\frac{AR}{RB}\frac{BP}{PC}\frac{CQ}{QA}$ ie. $\frac{AR'}{R'B}=\frac{AR}{RB}$. Since $R$ and $R'$ lie between $A$ and $B$, by uniqueness of internal ratio we have $R=R'$.

PS: Some of the assertions like $R'$ lying on segment $AB$, $O$ being in interior of $\triangle ABC$ or the uniqueness assertion which gives us $R=R'$ need rigorous proof which can be found in George E Martin's book given in the reference. It is not an easy read, so beware.

References:

• The foundations of geometry and the non-Euclidean plane by George Edward Martin
• A Sequel to the First Six Books of Euclid by John Casey
• David Joyce's web version of Euclid's Elements