# When Is Triangle Isosceles: Miguel Ochoa Sanchez's Criterion

### Problem

In $\Delta ABC,\;$ point $P\;$ is on the angle bisector $AD.\;$ $BP\;$ meets $AC\;$ in $E;\;$ $CP\;$ meets $AB\;$ in $F.$

Prove that if

$\displaystyle \frac{1}{AB^3}+\frac{1}{AE^3}=\frac{1}{AC^3}+\frac{1}{AF^3},$

the triangle is isosceles.

### Lemma 1

Under the conditions of the above statement, if $AC\gt AB,\;$ then $[\Delta BCE]\gt [\Delta BCF],\;$ where $[X]\;$ denotes the area of shape $X.$

The lemma is equivalent to the assertion that point $E\;$ is "above" point $F.$

The lemma has an important consequence.

If $AC\gt AB,\;$ then

$AB\cdot AE\lt AC\cdot AF,$

Indeed, the lemma implies that

$\displaystyle\frac{1}{2}\sin A\cdot AB\cdot AE=[ABE]\lt [ACF]=\frac{1}{2}\sin A\cdot AC\cdot AF.$ Thus we have $AB\cdot AE\lt AC\cdot AF.$

### Lemma 2

Under the conditions of the above statement,

$\displaystyle\frac{1}{AB}+\frac{1}{AE}=\frac{1}{AC}+\frac{1}{AF}.$

Indeed, by Ceva's theorem, $\displaystyle\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1.\;$ In addition, using the fact that $AD\;$ is an angle bisector, $\displaystyle\frac{AB}{AC}=\frac{BD}{DC}.\;$ Combining the two we get,

$\displaystyle\frac{AB-AF}{AB\cdot AF}=\frac{AC-AE}{AC\cdot AE}.$

This reduces to the required identity.

The two lemmas combined give the following:

If $AC\gt AB,\;$ then

$AB+AE\lt AC+AF,$

Indeed, by Lemma 2, $\displaystyle\frac{AB+AE}{AB\cdot AE}=\frac{AC+AF}{AC\cdot AF}.\;$ So, since $AB\cdot AE\lt AC\cdot AF,\;$ so too $AB+AE\lt AC+AF.$

### A Simplified Version

First we prove the following

In $\Delta ABC,\;$ point $P\;$ is on the angle bisector $AD.\;$ $BP\;$ meets $AC\;$ in $E;\;$ $CP\;$ meets $AB\;$ in $F.$

Prove that if

$\displaystyle \frac{1}{AB^2}+\frac{1}{AE^2}=\frac{1}{AC^2}+\frac{1}{AF^2},$

the triangle is isosceles.

The proof is by contradiction. WLOG, assume $AC\gt AB.\;$ Then, by Lemma 2

$\displaystyle\left(\frac{1}{AB}+\frac{1}{AE}\right)^2=\left(\frac{1}{AC}+\frac{1}{AF}\right)^2,$

implying, under our assumption, $\displaystyle\frac{2}{AB\cdot AE}=\frac{2}{AC\cdot AF}.$ But this contradicts Lemma 1.

### Proof 1

We shall assume, as before, $AC\gt AB\;$ and now also $\displaystyle \frac{1}{AB^3}+\frac{1}{AE^3}=\frac{1}{AC^3}+\frac{1}{AF^3}.\;$ By Lemma 2,

$\displaystyle\left(\frac{1}{AB}+\frac{1}{AE}\right)^3=\left(\frac{1}{AC}+\frac{1}{AF}\right)^3,$

and subsequently,

$\displaystyle \frac{3}{AB\cdot AE\cdot (AB+AE)}=\frac{3}{AC\cdot AF\cdot (AC+AF)}.$

This is false by the corollaries to the two lemmas.

### Proof 2

This is the original proof by Miguel Ochoa Sanchez.

### Proof 3

Denote $\displaystyle\frac{1}{AB}+\frac{1}{AE}=s,\;$ $\displaystyle\frac{1}{AB}\cdot \frac{1}{AE}=p,\;$ $\displaystyle\frac{1}{AC}+\frac{1}{AF}=q,\;$ $\displaystyle\frac{1}{AC}\cdot\frac{1}{AF}=r.\;$

In $\Delta ABE,\;$ $\displaystyle AP=\frac{2\cos\phi}{s}\;$ and in $\Delta ACF,\;$ $\displaystyle AP=\frac{2\cos\phi}{q},\;$ implying $q=s.\;$ The given condittion is equivalent to $s^3-3sp=s^3-3sr\;$ such that $r=p.\;$ From here

$\displaystyle\left\{\frac{1}{AB},\frac{1}{AE}\right\}=\left\{\frac{1}{AC},\frac{1}{AF}\right\}.$

There are two possibilities: $AB=AF,\;AC=AE\;$ or $AB=AC,\;AE=AF.\;$ The former may only hold when $P=D,\;$ in which case the condtion of the problem becomes an inconsequential tautology and needs to be excluded. The latter possibility confirms the conclusion of the problem.

### Extra Challenge

In $\Delta ABC,\;$ point $P\;$ is on the angle bisector $AD.\;$ $BP\;$ meets $AC\;$ in $E;\;$ $CP\;$ meets $AB\;$ in $F.$

Prove that if

$\displaystyle \frac{1}{AB^4}+\frac{1}{AE^4}=\frac{1}{AC^4}+\frac{1}{AF^4},$

the triangle is isosceles.

Think of other exponents.

### Answer to the Extra Challenge

As in Proof 3, $q=s.\;$ From the given condition, $s^4-4s^2p+2p^2=s^4-4s^2r+2r^2,\;$ implying $4s^2p-2p^2=4s^2r-2r^2.\;$ But $\displaystyle 0\lt p,r\lt\frac{s^2}{4}\;$ and the quadratic $f(x)=x^2-2s^2x\;$ is injectivebon $[0,s^2]\;$ such that $r=p.\;$ The ending as in Proof 3.

### Acknowledgment

The problem above has been posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook page. Proof 2 is by Miguel; Proof 3 and the resoponse to the Extra Challenge is by Leo Giugiuc.