# Alexandru Lupas' Property of the Incenter

### Problem

### Solution

Let $BD=ka$ and $CD=(1-k)a,$ $0\lt k\lt 1.$ From the Intersecting Chords theorem,

$\displaystyle AD\cdot DX=k(1-k)a^2\,\Rightarrow\,DX=\frac{k(1-k)a^2}{AD}\,\Rightarrow\,\frac{AD}{DX}=\frac{AD^2}{k(1-k)a^2}.$

From Stewart's theorem,

$\displaystyle AD^2+k(1-k)a^2=kb^2+(1-k)C^2\,\Rightarrow\,\frac{AD}{DX}=\frac{1}{a^2}\left(\frac{b^2}{1-k}+\frac{c^2}{k}\right)-1.$

Consider the function $f:\,(0,1)\to\mathbb{R}$ defined y $\displaystyle f(k)=\frac{b^2}{1-k}+\frac{c^2}{k}.$ We have $\displaystyle f'(k)=\left(\frac{c}{k}\right)^2-\left(\frac{b}{1-k}\right)^2,$ so that the only critical point of $f$ is $\displaystyle k_0=\frac{c}{b+c}.$ Further, $f(k)\ge f(k_0)=(b+c)^2.$ It follows that $\displaystyle \min_{D\in(BC)}\frac{AD}{DX}=\left(\frac{b+c}{a}\right)^2-1$ and the minimum is attained only if $D$ is the foot of the angle bisector from $A.$

Similarly, $\displaystyle \frac{BE}{EY}$ attains its minimum when $E$ is the foot of the angle bisector $BE$ and $\displaystyle \frac{CF}{FZ}$ attains its minimum when $F$ is the foot of the angle bisector $CF.$ Since the three angle bisectors intersect at the incenter $I$ of $\Delta ABC,$ it follows that $F(P)$ attains its minimum when $P=I.$

The minimum is equal to $\displaystyle \left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2+\left(\frac{a+b}{c}\right)^2-3.$

### Acknowledgment

Leo Giugiuc has kindly communicated to me his solution of a problem posted at the facebook by Marian Dinca in 2017 in memory of Alexandru Lupas, on the 10th anniversary of the latter passing. The above is a slightly modified version of that problem, with Leo's solution perfectly fitting in.

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