Geometry Triangle inequality

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Source: Irish MO

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#1 h May 10, 2012, 6:54 PM • 2 Y

Y by Adventure10, Mango247

Suppose P is a point in the interior of a triangle ABC, that x; y; z are
the distances from P to A; B; C, respectively, and that p; q; r are the per-
pendicular distances from P to the sides BC; CA; AB, respectively. Prove
that
$xyz \geq 8pqr$ ;
with equality implying that the triangle ABC is equilateral.

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AleSP

#2 h May 10, 2012, 10:25 PM • 2 Y

Y by Adventure10, Mango247

Let $\angle PAC=\alpha_1$ , $\angle PAB=\alpha_2$ , $\angle PBA = \beta_1$ , $\angle PBC = \beta_2$ , $\angle PCB=\gamma_1$ , $\angle PCA =\gamma_2$ .
We need to prove that $\sin(\alpha_1).\sin(\beta_1).\sin(\gamma_1) = \sin(\alpha_2).\sin(\beta_2).\sin(\gamma_2) \leq \frac {1}{8}$ $\Leftrightarrow \sqrt{\sin(\alpha_1).\sin(\beta_1).\sin(\gamma_1).\sin(\alpha_2).\sin(\beta_2).\sin(\gamma_2)} \leq \frac{1}{8}$ .
But $\sin\left(\frac{\alpha_1+\alpha_2}{2}\right)\geq \sqrt{\sin(\alpha_1).\sin(\alpha_2)}\Leftrightarrow$
$\sin\left(\frac{\alpha_1}{2}\right).\cos\left(\frac{\alpha_2}{2}\right)+\cos\left(\frac{\alpha_1}{2}\right).\sin\left(\frac{\alpha_2}{2}\right)\geq$ $\sqrt{4.\sin\left(\frac{\alpha_1}{2}\right).\cos\left(\frac{\alpha_1}{2}\right).\sin\left(\frac{\alpha_2}{2}\right).\cos\left(\frac{\alpha_2}{2}\right)}\Leftrightarrow$ $\sin^2\left(\frac{\alpha_1}{2}\right).\cos^2\left(\frac{\alpha_2}{2}\right)+\sin^2\left(\frac{\alpha_2}{2}\right).\cos^2\left(\frac{\alpha_1}{2}\right) \geq$ $2.\cos\left(\frac{\alpha_1}{2}\right).\cos\left(\frac{\alpha_2}{2}\right).\sin\left(\frac{\alpha_1}{2}\right).\sin\left(\frac{\alpha_2}{2}\right)$ , which is true by the AM-GM inequality.
So we just need to prove that $\sin\left(\frac{\alpha_1+\alpha_2}{2}\right).\sin\left(\frac{\beta_1+\beta_2}{2}\right).\sin\left(\frac{\gamma_1+\gamma_2}{2}\right)\leq \frac{1}^{8}$.
But $\frac{\alpha_1+\alpha_2}{2}$ , $\frac{\beta_1+\beta_2}{2}$ , $\frac{\gamma_1+\gamma_2}{2}$ $\in (0,\pi/2)$ , and $f(x)=\sin(x)$ is concave down in this interval, so we can apply Jensen's inequality and AM-GM inequality to get $\sin\left(\frac{\alpha_1+\alpha_2}{2}\right).\sin\left(\frac{\beta_1+\beta_2}{2}\right).\sin\left(\frac{\gamma_1+\gamma_2}{2}\right)\leq$ $\left(\frac{\sin\left(\frac{\alpha_1+\alpha_2}{2}\right)+\sin\left(\frac{\beta_1+\beta_2}{2}\right)+\sin\left(\frac{\gamma_1+\gamma_2}{2}\right)}{3}\right)^3\leq$ $\left(\sin\left(\frac{\alpha_1+\alpha_2+\beta_1+\beta_2+\gamma_1+\gamma_2}{6}\right)\right)^3$ $=\sin^3\left(\frac{\pi}{6}\right)=\frac{1}{8}.$ Equality occurs $\Leftrightarrow$ the angles are all equal, that is, the triangle is equilateral and $P$ is the incenter. $\blacksquare$

This post has been edited 1 time. Last edited by AleSP, Oct 1, 2012, 6:44 PM

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#3 h May 11, 2012, 6:37 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Proposition. Let $P$ be a point inside $\triangle ABC$ . The line through $PA,PB,PC$ intersects $BC,CA,AB$ respectively at $A',B',C'$ respectively. Let $\left\|\begin{array}{ccc}PA=x\\PB=y\\PC=z\end{array}\right\|$ and $\left\|\begin{array}{ccc}PA'=p\\PB'=q\\PC'=r\end{array}\right\|$ . Then $\frac{x}{p}\cdot\frac{y}{q}\cdot\frac{z}{r}\ge 8$ .

Proof. Let $\left\|\begin{array}{ccc}A'B=a_b\\A'C=a_c\end{array}\right\|$ and $\left\|\begin{array}{ccc} B'A=b_a\\B'C=b_c\end{array}\right\|$ and $\left\|\begin{array}{ccc} C'B=c_b\\C'A=c_a\end{array}\right\|$ .

Van Aubel's theorem states that $\frac{x}{p}=\frac{b_a}{b_c}+\frac{c_a}{c_b}$ and $\frac{y}{q}=\frac{c_b}{c_a}+\frac{a_b}{a_c}$ and $\frac{z}{r}=\frac{a_c}{a_b}+\frac{b_c}{b_a}$ .

Then by AM-GM within each pair of brackets, $\frac{x}{p}\cdot\frac{y}{q}\cdot\frac{z}{r}= \left(\frac{b_a}{b_c}+\frac{c_a}{c_b}\right)\cdot\left(\frac{c_b}{c_a}+\frac{a_b}{a_c}\right)\cdot\left(\frac{a_c}{a_b}+\frac{b_c}{b_a}\right)\ge 8$ The equality case is the result of some trivial 2-line algebra which I leave to the diligent reader. $\blacksquare$

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jjax

#4 h May 11, 2012, 12:37 PM • 2 Y

Y by Adventure10, Mango247

Like BigSams, I will prove that $\frac{AP}{PA'} \cdot \frac{BP}{PB'} \cdot \frac{CP}{PC'} \geq 8$ , where $A'$ is the intersection of $AP$ with $BC$ , and $B',C'$ likewise defined.

Let $[PBC]=d,[PCA]=e,[PAB]=f$ , where $[S]$ denotes the area of $S$ .
Then $\frac{AP}{PA'} = \frac{e+f}{d} \geq 2 \sqrt{ \frac{ef}{d^2} }$ .
Obtaining similar bounds for $\frac{BP}{PB'}, \frac{CP}{PC'}$ and multiplying them gives the result.
Equality occurs when $d=e=f$ , or $P$ is the centroid of the triangle.

$PA' \cdot PB' \cdot PC' \geq$ product of perpendicular distances.
If equality holds for this, then $P$ is the orthocenter. As the centroid and orthocenter coincide, it follows that $ABC$ is equilateral.

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#5 h Dec 12, 2022, 7:38 PM

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Suppose $P$ is a point in the interior of a triangle $ABC$ , that $x, y, z$ are the distances from $P$ to $A, B, C$ , respectively, and that $p,q,r$ are the perpendicular distances from $P$ to the sides $BC, CA,AB$ , respectively. Prove that $xyz \geq 8pqr$ with equality implying that the triangle $ABC$ is equilateral.

This post has been edited 2 times. Last edited by parmenides51, Dec 21, 2022, 8:04 PM

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soryn

#6 h Dec 13, 2022, 5:31 AM

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Romania NMO 1986,10-th grade

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mihaig

#7 h Dec 14, 2022, 11:00 AM

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Indeed, Sorin

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