217. A nice and interesting Murray S. Klamkin's problem.

by Virgil Nicula, Jan 28, 2011, 8:07 AM

Murray S. Klamkin's problem. Let $ABC$ be an acute non-equilateral triangle. Prove that there is an interior

point $P$ other than the circumcenter such that $a^2\cdot (PB-PC)+ b^2\cdot (PC-PA) + c^2\cdot (PA-PB) = 0$ .

Proof. $\blacktriangleright$ Existence of the point[/u] $P$ . For an interior point $P$ of $\triangle ABC$ define the function $a^2\cdot (PB-PC)+ b^2\cdot (PC-PA) + c^2\cdot (PA-PB) = 0$ .

Thus, $\left\|\begin{array}{c} f(A) = (b-c)(b^2+c^2 +bc - a^2) = bc(b-c)(1 + 2\cos A)\\\\ f(B) = (c-a)(c^2+a^2 +ca - b^2) = ca(c-a)(1 + 2\cos B)\\\\ f(C) = (a-b)(a^2+b^2 +ab - c^2) = ab(a-b)(1 + 2\cos C)\end{array}\right\|$ . Observe that $1 + 2\cos A>0$ , $1 + 2\cos B>0$ , $1 + 2\cos C > 0$ and

$\triangle ABC$ not equilateral and $(b-c) + (c-a) + (a-b) = 0$ $\implies$ at least one of $b-c$ , $c-a$ , $a-b$ is positive, and at least one from them is negative. It means

at least one of $f(A), f(B), f(C)$ is positive, and at least one of $f(A), f(B), f(C)$ is negative. Is clearly that clearly $f(O) = 0$ for $O$ the circumcenter of $\triangle ABC$ .

Assume w.l.o.g. $f(A) < 0<f(B)$ . Since $f$ is clearly continuous function, on any arc $\gamma$ connecting $A$ and $B$ there will lie (at least) a point $P_{\gamma}$ such that $f(P_{\gamma}) = 0$ .

All we need is take arcs contained in $\triangle ABC$ which avoid $O$ ; we find infinitely many such points $P$ .

$\blacktriangleright$ Find at least a point[/u] $P$ . Choose $P:=F$ - the first Fermat's point of $\triangle ABC$ , i.e. $m(\angle BFC)=m(\angle CFA)=m(\angle AFB)=120^{\circ}$ .

By cosine law for $\triangle AFC$ and $\triangle AFB$ we obtain $\left\|\begin{array}{c} b^2=FA^2+FC^2-2 \cdot FA \cdot FC \cdot \cos 120^{\circ}=FA^2+FC^2+FA \cdot FC\\\\ c^2=FA^2+FB^2-2 \cdot FA \cdot FB \cdot \cos 120^{\circ}=FA^2+FB^2+FA \cdot FB\end{array}\right\|$ $\implies$

$\ (b^2-c^2) \cdot FA=FA \cdot FC^2+FC \cdot FA^2-FA \cdot FB^2-FB \cdot FA^2 \ \ (1)$ . By permutating $a$ , $b$ , $c$ and $FA$ , $FB$ , $FC$ cyclically we get

the identities $\left\|\begin{array}{c} (c^2-a^2) \cdot FB=FA \cdot FB^2+FB \cdot FA^2-FB \cdot FC^2-FC \cdot FB^2 \ \ (2)\\\\ (a^2-b^2) \cdot FC=FC \cdot FB^2+FB \cdot FC^2-FC \cdot FA^2-FA \cdot FC^2 \ \ (3)\end{array}\right\|$ . Summing up the identities $(1)$ , $(2)$ and $(3)$

yields $a^2 \cdot (FB-FC)+b^2 \cdot (FC-FA)+c^2 \cdot (FA-FB)=0$ .
This post has been edited 7 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:46 PM

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