Geometric inequality with Fermat point

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Geometric inequality with Fermat point

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Source: Balkan MO Shortlist 2024 G2

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#1 h Apr 27, 2025, 10:21 PM

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Let $ABC$ be an acute triangle and let $P$ be an interior point for it such that $\angle APB = \angle BPC = \angle CPA$ . Prove that
$\frac{PA^2 + PB^2 + PC^2}{2S} + \frac{4}{\sqrt{3}} \leq \frac{1}{\sin \alpha} + \frac{1}{\sin \beta} + \frac{1}{\sin \gamma}.$ When does equality hold?

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#2 h Apr 27, 2025, 11:15 PM • 3 Y

Y by sami1618, ItsBesi, solidgreen

Here's my solution:
The inequality can be rewritten as $2S(\frac{1}{sin \alpha}+\frac{1}{sin \beta}+\frac{1}{sin \gamma}-\frac{4}{\sqrt{3}})\geq PA^2+PB^2+PC^2$ .

Note that $AB\cdot AC\cdot\sin\alpha=AB\cdot BC\cdot\sin\beta=AC\cdot BC\cdot\sin\gamma=2S$ and

$2S=PA\cdot PB\cdot\sin 120^{\circ}+PB\cdot PC\cdot\sin 120^{\circ}+PC\cdot PA\cdot\sin 120^{\circ}=(PA\cdot PB+PB\cdot PC +PC\cdot PA)\sin 120^{\circ}=(PA\cdot PB+PB\cdot PC +PC\cdot PA)\frac{\sqrt{3}}{2}$ .

The inequality can be further rewritten as $AB\cdot AC + AC\cdot BC+BC\cdot AB - 2(PA\cdot PB+PB\cdot PC+ PC\cdot PA)\geq PA^2+PB^2+PC^2$ that is equivalen to $AB\cdot AC + AC\cdot BC+BC\cdot AB\geq (PA+PB+PC)^2$

Consider the points $X$ and $Y$ such that $\bigtriangleup XAB$ and $\bigtriangleup YAC$ are equilateral ( $X$ and $C$ lie on different halfplanes with respect to $AB$ , similarly $Y$ and $B$ with respect to $AC$ ). $\angle AXB+\angle APB=180^{\circ}=\angle AYC+\angle APC\Rightarrow XAPB$ and $YAPC$ are cyclic quadrilaterals.

Also note that $\angle BPA+\angle APY=120^{\circ}+\angle ACY=120^{\circ}+60^{\circ}=180^{\circ}$ hence $B$ , $P$ and $Y$ are collinear. Similarly, points $C$ , $P$ and $X$ are collinear.

From Ptolemy's Theorem in cyclic quadrilateral $XAPB$ we have that $PA\cdot XB+ PB\cdot XA=PX\cdot AB$ , but since $\bigtriangleup XAB$ is equilateral, then $XA=XB=AB$ and we have that $PA+PB=PX$ . From here, $CX=PX+PC=PA+PB+PC$ . Similarly, $BY=PA+PB+PC$ .

Now, we apply Ptolemy's Inequality in quadrilateral $XBCY$ and get that $XB\cdot YC+XY\cdot BC\geq CX\cdot BY$ . From Triangle Inequality we have that $AX+AY\geq XY$ so $XB\cdot YC+(AX+AY)BC\geq CX\cdot BY$ . Rewriting the inequality based on the above relations we have that $AB\cdot AC+AB\cdot BC +AC\cdot BC\geq (PA+PB+PC)^2$ .

The equality occurs if and only if both equality cases of Ptolemy's Inequality and Triangle's Inequality occur. The equality case of Triangle's Inequality occurs when $X$ , $A$ and $Y$ are collinear $\iff 180^{\circ}=\angle XAB+\angle BAC+\angle CAY=60^{\circ}+\angle BAC+60^{\circ}\Rightarrow \angle BAC=60^{\circ}$ . The Ptolemy's Inequality equality case occurs if and only if $XBCY$ is cyclic $\iff\ 180^{\circ}=\angle BXY+\angle BCY= 60^{\circ}+\angle BCA+\angle ACY=60^{\circ}+\angle BCA+60^{\circ}\Rightarrow \angle BCA=60^{\circ}$ . So the equality case happens if and only if $\bigtriangleup ABC$ is equilateral.

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Quantum-Phantom

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Quantum-Phantom

#3 h Apr 28, 2025, 4:07 AM

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We use standard notation. Point $P$ is the Fermat point, so by the Napoleon's theorem configuration,
$\begin{align*} (PA+PB+PC)^2&=a^2+b^2-2ab\cos\left(C+\frac\pi3\right)\\ &=a^2+b^2-2ab\left(\frac12\cdot\frac{a^2+b^2-c^2}{2ab}-\frac{\sqrt3}2\cdot\frac c{2R}\right)\\ &=\frac{\left(\sum\limits a\right)^2}2-\sum_{\rm cyc}ab+\frac{\sqrt3abc}{2R}.\tag{1} \end{align*}$ Using the identities $ab+bc+ca=s^2+4Rr+r^2$ , $abc=4sRr$ , we have
$(1)=\frac{(2s)^2}2-s^2-4Rr-r^2+\frac{4\sqrt3sRr}{2R}=s^2-4Rr-r^2+2\sqrt3sr.\tag{2}$ By the law of cosines,
$\left(\sum_{\rm cyc}a\right)^2-2\sum_{\rm cyc}ab=\sum_{\rm cyc}a^2=\sum_{\rm cyc}\left(PA^2+PB^2+PA\cdot PB\right),$ so
$2\sum_{\rm cyc}PA^2+\sum_{\rm cyc}PA\cdot PB=(2s)^2-2\left(s^2+4Rr+r^2\right)=2s^2-8Rr-2r^2.\tag{3}$ Therefore,
$\sum_{\rm cyc}PA^2=\frac{2\times(3)-(2)}3=s^2-4Rr-r^2-\frac2{\sqrt3}sr.$ Now we need to show that
$\frac{s^2-4Rr-r^2-\frac2{\sqrt3}sr}{2sr}+\frac4{\sqrt3}\le\sum_{\rm cyc}\frac1{\sin A}=\frac{s^2+4Rr+r^2}{2rs},$ which reduces to $\sqrt3s\le4R+r$ . Indeed, by Gerretsen's inequality and Euler's inequality,
$\sqrt3s\le\sqrt{3\left(4R^2+4Rr+3r^2\right)}\le4R+r.$

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mathuz

#4 h Apr 28, 2025, 6:41 AM

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We have $\sum PA\cdot PB = \frac{4}{\sqrt{3}}S$ . Therefore, it suffices to show that $\left( PA+PB+PC\right)^2 \le ab+bc+ca.$ By Cauchy-Schwarz, we have $4b^2c^2 = (PA^2+PC^2+(PA+PC)^2)(PA^2+PB^2+(PA+PB)^2)$ $\ge (PA^2+PB\cdot PC+(PA+PC)(PA+PB))^2.$ So $bc \ge PA^2 + PB\cdot PC +\frac{PA\cdot PB + PA\cdot PC}{2}.$ Summing up, we obtain our desired inequality.

Added: Equality holds precisely when equality occurs in the Cauchy–Schwarz applied above. So, it should be when $PA=PB=PC$ , i.e. $AB=BC=CA$ .

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#6 h Apr 28, 2025, 11:22 AM

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This problem was proposed by Circumcircle Dren Neziri, Albania.

This post has been edited 1 time. Last edited by ItsBesi, Apr 28, 2025, 1:08 PM

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sqing

#7 h Apr 30, 2025, 2:48 PM

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https://drive.google.com/file/d/15S5byn0y4RLfceYbc-8XA9BItE3rz49H/view

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arqady

#8 h Apr 30, 2025, 6:09 PM

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Assassino9931 wrote:

Let $ABC$ be an acute triangle and let $P$ be an interior point for it such that $\angle APB = \angle BPC = \angle CPA$ . Prove that
$\frac{PA^2 + PB^2 + PC^2}{2S} + \frac{4}{\sqrt{3}} \leq \frac{1}{\sin \alpha} + \frac{1}{\sin \beta} + \frac{1}{\sin \gamma}.$ When does equality hold?

It's just Hadwiger–Finsler

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