Constrained Area in Triangle

Leo Giugiuc has kindly reported a problem from the forum, followed by with solution by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc.

Given $\Delta ABC$ and points $M,$ $N,$ $P$ on the sides $BC,$ $AC,$ and $AB,$ respectively.

Constrained Area in Triangle

Assume points $M,$ $N,$ $P$ have the property

$AP + BM + CN = PB + MC + NA.$

Then $[\Delta MNP]\le \frac{1}{4}[\Delta ABC].$


Denote the side lengths of $\Delta ABC$ as $a,b,c.$ As usual, $[F]$ will denote the area of shape $F.$ Let $BM=ma,$ $CM=(1-m)a,$ $CN=nb,$ $AN=(1-n)b,$ $AP=pc$ and $BP=(1-p)c,$ where $m,n,p\in (0,1).$ By hypothesis we get $(2m-1)a+(2n-1)b+(2p-1)c=0.$

On the other hand, $2\cdot [MCN]=(1-m)a\cdot nb\cdot\sin C=2(1-m)n\cdot [\Delta ABC]$ and, similarly, $2\cdot [\Delta NAP]=2(1-n)p\cdot [\Delta ABC],$ $2\cdot [\Delta PBM]=2(1-p)m\cdot [\Delta ABC].$ Since $[\Delta MNP]=[\Delta ABC]-([\Delta MCN]+[\Delta NAP]+[\Delta PBM]),$ the problem is equivalent to proving $\displaystyle m+n+p-(mn+np+pm)\ge \frac{3}{4}.$

Let denote $2m-1=\alpha ,$ $2n-1=\beta$ and $2p-1=\gamma .$ Then $\alpha ,\beta ,\gamma\in(-1,1),$ $a\alpha+b\beta+c\gamma=0,$ $\displaystyle m=\frac{\alpha +1}{2},$ $\displaystyle m=\frac{\beta +1}{2},$ $\displaystyle p=\frac{\gamma +1}{2}$ and we need to prove that $\alpha\beta +\beta\gamma +\gamma\alpha\le 0.$ If $0\in \{\alpha ,\beta ,\gamma\},$ let WLOG $\alpha=0.$ Then $b\beta+c\gamma=0,$ implying $\beta\gamma=0,$ and the required inequality follows.

Assume now that none of $\alpha,$ $\beta,$ $\gamma$ is $0.$ Since $a\alpha +b\beta +c\gamma=0,$ then two of $\alpha,$ $\beta,$ $\gamma$ have the same sign and the third has the opposite sign. WLOG $\alpha, \beta \gt 0$ and $\gamma\lt 0.$ let $-\gamma=k\gt 0.$ We know that $a\alpha +b\beta=kc$ and we need to show that $\alpha\beta\le k\alpha+k\beta.$ But by the triangle inequality, $kc\lt k(a+b),$ such that $a\alpha+b\beta \lt k(a+b).$ In other words, $(\alpha -k)a+(\beta -k)b\lt 0.$ This says that either $\alpha\lt k$ (and $\alpha\beta\lt k\beta)$ or $\beta\lt k$ (and $\alpha\beta\lt k\alpha)$ and in either case $\alpha\beta\le k\alpha+k\beta,$ as required.

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